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1ldo4vr0v | maths | application-of-derivatives | maxima-and-minima | <p>The sum of the absolute maximum and minimum values of the function $$f(x)=\left|x^{2}-5 x+6\right|-3 x+2$$ in the interval $$[-1,3]$$ is equal to :</p> | [{"identifier": "A", "content": "13"}, {"identifier": "B", "content": "24"}, {"identifier": "C", "content": "10"}, {"identifier": "D", "content": "12"}] | ["C"] | null | The sum of the absolute maximum and minimum values of the function $$f(x) = \left|x^2 - 5x + 6\right| - 3x + 2$$ in the interval $$[-1, 3]$$ can be found by finding the maximum and minimum values of $$f(x)$$ in this interval and then adding them.
<br/><br/>
First, let's find the critical points of $$f(x)$$. To do this,... | mcq | jee-main-2023-online-1st-february-evening-shift | 4,703 |
1ldpsc9p1 | maths | application-of-derivatives | maxima-and-minima | <p>A wire of length $$20 \mathrm{~m}$$ is to be cut into two pieces. A piece of length $$l_{1}$$ is bent to make a square of area $$A_{1}$$ and the other piece of length $$l_{2}$$ is made into a circle of area $$A_{2}$$. If $$2 A_{1}+3 A_{2}$$ is minimum then $$\left(\pi l_{1}\right): l_{2}$$ is equal to :</p> | [{"identifier": "A", "content": "6 : 1"}, {"identifier": "B", "content": "1 : 6"}, {"identifier": "C", "content": "4 : 1"}, {"identifier": "D", "content": "3 : 1"}] | ["A"] | null | $ \ell_{1}+\ell_{2}=20 \Rightarrow \frac{\mathrm{d} \ell_{2}}{\mathrm{~d} \ell_{1}}=-1$
<br/><br/>$\mathrm{A}_{1}=\left(\frac{\ell_{1}}{4}\right)^{2}$ and $\mathrm{A}_{2}=\pi\left(\frac{\ell_{2}}{2 \pi}\right)^{2}$
<br/><br/>Let $\mathrm{S}=2 \mathrm{~A}_{1}+3 \mathrm{~A}_{2}=\frac{\ell_{1}^{2}}{8}+\frac{3 \ell_{2}^{... | mcq | jee-main-2023-online-31st-january-morning-shift | 4,704 |
ldqwo36h | maths | application-of-derivatives | maxima-and-minima | If the functions $f(x)=\frac{x^3}{3}+2 b x+\frac{a x^2}{2}$
<br/><br/>and $g(x)=\frac{x^3}{3}+a x+b x^2, a \neq 2 b$ <br/><br/>have a common extreme point, then $a+2 b+7$ is equal to : | [{"identifier": "A", "content": "6"}, {"identifier": "B", "content": "$\\frac{3}{2}$"}, {"identifier": "C", "content": "3"}, {"identifier": "D", "content": "4"}] | ["A"] | null | <p>$$f'(x)=x^2+2b+ax$$</p>
<p>$$g'(x)=x^2+a+2bx$$</p>
<p>$$\Rightarrow x=1$$ is common root</p>
<p>$$a+2b+1=0$$</p> | mcq | jee-main-2023-online-30th-january-evening-shift | 4,705 |
lgnz80z6 | maths | application-of-derivatives | maxima-and-minima | Consider the triangles with vertices $A(2,1), B(0,0)$ and $C(t, 4), t \in[0,4]$.
<br/><br/>If the maximum and the minimum perimeters of such triangles are obtained at <br/><br/>$t=\alpha$ and $t=\beta$ respectively, then $6 \alpha+21 \beta$ is equal to ___________. | [] | null | 48 | We have a triangle with vertices $A(2,1)$, $B(0,0)$, and $C(t, 4)$, where $t$ belongs to the interval $[0,4]$.
<br><br>We want to find the maximum and minimum perimeters of such triangles, which occur at $t=\alpha$ and $t=\beta$, respectively.
<br><br>To find the minimum perimeter, we use a geometric approach. Reflec... | integer | jee-main-2023-online-15th-april-morning-shift | 4,708 |
1lgq0kk6g | maths | application-of-derivatives | maxima-and-minima | <p>$$\max _\limits{0 \leq x \leq \pi}\left\{x-2 \sin x \cos x+\frac{1}{3} \sin 3 x\right\}=$$</p> | [{"identifier": "A", "content": "$$\\frac{5 \\pi+2+3 \\sqrt{3}}{6}$$"}, {"identifier": "B", "content": "0"}, {"identifier": "C", "content": "$$\\frac{\\pi+2-3 \\sqrt{3}}{6}$$"}, {"identifier": "D", "content": "$$\\pi$$"}] | ["A"] | null | <p>Given the function:</p>
<p>$$
f(x) = x - 2\sin{x}\cos{x} + \frac{1}{3}\sin{3x}
$$</p>
<p>We want to find the maximum value of this expression for $$0 \leq x \leq \pi$$.</p>
<p>Step 1: Rewrite the expression
<br/><br/>Notice that we can rewrite the expression as:
<br/><br/>$$
f(x) = x - \sin{2x} + \frac{1}{3}\sin{3x}... | mcq | jee-main-2023-online-13th-april-morning-shift | 4,709 |
1lgrg9l7x | maths | application-of-derivatives | maxima-and-minima | <p>If the local maximum value of the function $$f(x)=\left(\frac{\sqrt{3 e}}{2 \sin x}\right)^{\sin ^{2} x}, x \in\left(0, \frac{\pi}{2}\right)$$ , is $$\frac{k}{e}$$, then $$\left(\frac{k}{e}\right)^{8}+\frac{k^{8}}{e^{5}}+k^{8}$$ is equal to</p> | [{"identifier": "A", "content": "$$e^{3}+e^{6}+e^{10}$$"}, {"identifier": "B", "content": "$$e^{3}+e^{5}+e^{11}$$"}, {"identifier": "C", "content": "$$e^{3}+e^{6}+e^{11}$$"}, {"identifier": "D", "content": "$$e^{5}+e^{6}+e^{11}$$"}] | ["C"] | null | $$
\text { Let } y=\left(\frac{\sqrt{3 e}}{2 \sin x}\right)^{\sin ^2 x}
$$
<br/><br/>$$
\ln \mathrm{y}=\sin ^2 \mathrm{x} \cdot \ln \left(\frac{\sqrt{3 \mathrm{e}}}{2 \sin \mathrm{x}}\right)
$$
<br/><br/>$$
\frac{1}{y} y^{\prime}=\ln \left(\frac{\sqrt{3 e}}{2 \sin x}\right) 2 \sin x \cos x+\sin ^2 x \frac{2 \sin x}{\sq... | mcq | jee-main-2023-online-12th-april-morning-shift | 4,710 |
1lh00jkrz | maths | application-of-derivatives | maxima-and-minima | <p>If $$a_{\alpha}$$ is the greatest term in the sequence $$\alpha_{n}=\frac{n^{3}}{n^{4}+147}, n=1,2,3, \ldots$$, then $$\alpha$$ is equal to _____________.</p> | [] | null | 5 | $$
\begin{aligned}
& \text { Let } y=\frac{x^3}{x^4+147} \\\\
& \Rightarrow \frac{d y}{d x}=\frac{\left(x^4+147\right) \times 3 x^2-x^3\left(4 x^3\right)}{\left(x^4+147\right)^2} \\\\
& =\frac{3 x^6+441 x^2-4 x^6}{\left(x^4+147\right)^2}=\frac{441 x^2-x^6}{\left(x^4+147\right)^2}
\end{aligned}
$$
<br><br>Fo... | integer | jee-main-2023-online-8th-april-morning-shift | 4,712 |
jaoe38c1lsf0sbol | maths | application-of-derivatives | maxima-and-minima | <p>Let $$f(x)=2^x-x^2, x \in \mathbb{R}$$. If $$m$$ and $$n$$ are respectively the number of points at which the curves $$y=f(x)$$ and $$y=f^{\prime}(x)$$ intersect the $$x$$-axis, then the value of $$\mathrm{m}+\mathrm{n}$$ is ___________.</p> | [] | null | 5 | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lt307bt9/dd10aeba-887c-4185-a5fe-218e2ee9b2bf/09af1ae0-d4af-11ee-8384-811001421c41/file-1lt307bta.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lt307bt9/dd10aeba-887c-4185-a5fe-218e2ee9b2bf/09af1ae0-d4af-11ee-8384-811001421c41... | integer | jee-main-2024-online-29th-january-morning-shift | 4,714 |
1lsg4fvwc | maths | application-of-derivatives | maxima-and-minima | <p>Let $$f(x)=(x+3)^2(x-2)^3, x \in[-4,4]$$. If $$M$$ and $$m$$ are the maximum and minimum values of $$f$$, respectively in $$[-4,4]$$, then the value of $$M-m$$ is</p> | [{"identifier": "A", "content": "108"}, {"identifier": "B", "content": "392"}, {"identifier": "C", "content": "608"}, {"identifier": "D", "content": "600"}] | ["C"] | null | <p>$$\begin{aligned}
& \mathrm{f}^{\prime}(\mathrm{x})=(\mathrm{x}+3)^2 \cdot 3(\mathrm{x}-2)^2+(\mathrm{x}-2)^3 2(\mathrm{x}+3) \\
& =5(\mathrm{x}+3)(\mathrm{x}-2)^2(\mathrm{x}+1) \\
& \mathrm{f}^{\prime}(\mathrm{x})=0, \mathrm{x}=-3,-1,2
\end{aligned}$$</p>
<p><img src="https://app-content.cdn.examgoal.ne... | mcq | jee-main-2024-online-30th-january-evening-shift | 4,716 |
luy9clmh | maths | application-of-derivatives | maxima-and-minima | <p>Let the set of all positive values of $$\lambda$$, for which the point of local minimum of the function $$(1+x(\lambda^2-x^2))$$ satisfies $$\frac{x^2+x+2}{x^2+5 x+6}<0$$, be $$(\alpha, \beta)$$. Then $$\alpha^2+\beta^2$$ is equal to _________.</p> | [] | null | 39 | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lw3i47n9/78e3808f-b6b1-4774-ad57-06e016fc0e17/bd7fae50-1059-11ef-abcd-c333ada72a30/file-1lw3i47na.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lw3i47n9/78e3808f-b6b1-4774-ad57-06e016fc0e17/bd7fae50-1059-11ef-abcd-c333ada72a30... | integer | jee-main-2024-online-9th-april-morning-shift | 4,718 |
lv0vxd5u | maths | application-of-derivatives | maxima-and-minima | <p>Let the sum of the maximum and the minimum values of the function $$f(x)=\frac{2 x^2-3 x+8}{2 x^2+3 x+8}$$ be $$\frac{m}{n}$$, where $$\operatorname{gcd}(\mathrm{m}, \mathrm{n})=1$$. Then $$\mathrm{m}+\mathrm{n}$$ is equal to :</p> | [{"identifier": "A", "content": "217"}, {"identifier": "B", "content": "182"}, {"identifier": "C", "content": "201"}, {"identifier": "D", "content": "195"}] | ["C"] | null | <p>$$\begin{aligned}
& f(x)=\frac{2 x^2-3 x+8}{2 x^2+3 x+8}=y, 2 x^2+3 x+8>0 \quad \forall x \in \mathbb{R} \\
& \Rightarrow \quad x^2(2 y-2)+x(3 y+3)+8 y-8=0
\end{aligned}$$</p>
<p>Since $$x \in \mathbb{R}$$, the equation has real roots</p>
<p>$$\begin{aligned}
& \Rightarrow \quad D \geq 0 \\
& \Rightarrow(3 y+3)^2-4(... | mcq | jee-main-2024-online-4th-april-morning-shift | 4,719 |
lv2er3i1 | maths | application-of-derivatives | maxima-and-minima | <p>Let $$f(x)=3 \sqrt{x-2}+\sqrt{4-x}$$ be a real valued function. If $$\alpha$$ and $$\beta$$ are respectively the minimum and the maximum values of $$f$$, then $$\alpha^2+2 \beta^2$$ is equal to</p> | [{"identifier": "A", "content": "42"}, {"identifier": "B", "content": "38"}, {"identifier": "C", "content": "24"}, {"identifier": "D", "content": "44"}] | ["A"] | null | <p>$$\begin{aligned}
& f(x)=3 \sqrt{x-2}+\sqrt{4-x} \\
& \text { Let } x=2 \sin ^2 \theta+4 \cos ^2 \theta \\
& =3 \sqrt{2 \sin ^2 \theta+4 \cos ^2 \theta-2}+\sqrt{4-2 \sin ^2 \theta-4 \cos ^2 \theta} \\
& =3 \sqrt{2 \cos ^2 \theta}+\sqrt{2 \sin ^2 \theta} \\
& =3 \sqrt{2}|\cos \theta|+\sqrt{2}|\sin \theta|
\end{aligne... | mcq | jee-main-2024-online-4th-april-evening-shift | 4,720 |
lv3ve65y | maths | application-of-derivatives | maxima-and-minima | <p>If the function $$f(x)=2 x^3-9 \mathrm{ax}^2+12 \mathrm{a}^2 x+1, \mathrm{a}> 0$$ has a local maximum at $$x=\alpha$$ and a local minimum at $$x=\alpha^2$$, then $$\alpha$$ and $$\alpha^2$$ are the roots of the equation :</p> | [{"identifier": "A", "content": "$$x^2-6 x+8=0$$\n"}, {"identifier": "B", "content": "$$8 x^2-6 x+1=0$$\n"}, {"identifier": "C", "content": "$$8 x^2+6 x-1=0$$\n"}, {"identifier": "D", "content": "$$x^2+6 x+8=0$$"}] | ["A"] | null | <p>$$\begin{aligned}
& f(x)=6 x^2-18 a x+12 a^2 \\
& =6\left(x^2-3 a+2 a^2\right) \\
& =6(x-a)(x-2 a)=0 \\
& x=a, 2 a
\end{aligned}$$</p>
<p>$$a=\alpha, \quad 2 a=\alpha^2 \quad \Rightarrow \alpha=0,2$$</p>
<p>$$\begin{array}{lll}
a>0 & \therefore & \alpha=2 \\
& & \alpha^2=4
\end{array}$$</p>
<p>$$\therefore x^2-6 x+8... | mcq | jee-main-2024-online-8th-april-evening-shift | 4,721 |
lv3vef38 | maths | application-of-derivatives | maxima-and-minima | <p>Let $$\mathrm{A}$$ be the region enclosed by the parabola $$y^2=2 x$$ and the line $$x=24$$. Then the maximum area of the rectangle inscribed in the region $$\mathrm{A}$$ is ________.</p> | [] | null | 128 | <p>$$\begin{aligned}
& y^2=2 x \\
& a=\left(\frac{1}{2}\right)
\end{aligned}$$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lw4uhftn/eaae4a9d-2d8a-4b86-93c7-b8b3dfc45e4c/e388a2b0-1116-11ef-b9cb-b5e0fe4ba33b/file-1lw4uhfto.png?format=png" data-orsrc="https://app-content.cdn.examgoal.ne... | integer | jee-main-2024-online-8th-april-evening-shift | 4,722 |
lv5grwiz | maths | application-of-derivatives | maxima-and-minima | <p>The number of critical points of the function $$f(x)=(x-2)^{2 / 3}(2 x+1)$$ is</p> | [{"identifier": "A", "content": "2"}, {"identifier": "B", "content": "1"}, {"identifier": "C", "content": "0"}, {"identifier": "D", "content": "3"}] | ["A"] | null | <p>To find the number of critical points of the function $$f(x)=(x-2)^{2 / 3}(2 x+1)$$, we need to determine where its derivative $$f'(x)$$ is equal to zero or undefined. Critical points occur where the derivative is zero or does not exist.</p>
<p>First, let's find the derivative of the function:</p>
<p>$$f(x)=(x-2)^... | mcq | jee-main-2024-online-8th-april-morning-shift | 4,724 |
lv9s20kb | maths | application-of-derivatives | maxima-and-minima | <p>Let the maximum and minimum values of $$\left(\sqrt{8 x-x^2-12}-4\right)^2+(x-7)^2, x \in \mathbf{R}$$ be $$\mathrm{M}$$ and $$\mathrm{m}$$, respectively. Then $$\mathrm{M}^2-\mathrm{m}^2$$ is equal to _________.</p> | [] | null | 1600 | <p>$$\begin{aligned}
& \text { Let } y=\sqrt{8 x-x^2-12} \Rightarrow(x-4)^2+y^2=2^2 \\
& \Rightarrow d=(y-4)^2+(x-7)^2
\end{aligned}$$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lweqs4gr/cb92f82c-cbcf-4ca2-8ad5-d05029ed7bb2/3ea356b0-1688-11ef-9ee8-13752e98d8d6/file-1lweqs4gs.png?for... | integer | jee-main-2024-online-5th-april-evening-shift | 4,726 |
WKbUFfnFg5soehXh | maths | application-of-derivatives | mean-value-theorem | If $$2a+3b+6c=0,$$ $$\left( {a,b,c \in R} \right)$$ then the quadratic equation $$a{x^2} + bx + c = 0$$ has | [{"identifier": "A", "content": "at least one root in $$\\left[ {0,1} \\right]$$"}, {"identifier": "B", "content": "at least one root in $$\\left[ {2,3} \\right]$$"}, {"identifier": "C", "content": "at least one root in $$\\left[ {4,5} \\right]$$"}, {"identifier": "D", "content": "none of these "}] | ["A"] | null | Let $$f\left( x \right) = {{a{x^3}} \over 3} + {{b{x^2}} \over 2} + cx \Rightarrow f\left( 0 \right) = 0$$ and $$f(1)$$
<br><br>$$ = {a \over 3} + {b \over 2} + c = {{2a + 3b + 6c} \over 6} = 0$$
<br><br>Also $$f(x)$$ is continuous and differentiable in $$\left[ {0,1} \right]$$ and
<br><br>$$\left[ {0,1\left[ {.\,\,} ... | mcq | aieee-2002 | 4,727 |
s1Rt0Q9slH5xcy55 | maths | application-of-derivatives | mean-value-theorem | If $$2a+3b+6c=0$$, then at least one root of the equation
<br/>$$a{x^2} + bx + c = 0$$ lies in the interval | [{"identifier": "A", "content": "$$(1, 3)$$ "}, {"identifier": "B", "content": "$$(1, 2)$$ "}, {"identifier": "C", "content": "$$(2, 3)$$ "}, {"identifier": "D", "content": "$$(0, 1)$$ "}] | ["D"] | null | Let us define a function
<br><br>$$f\left( x \right) = {{ax{}^3} \over 3} + {{b{x^2}} \over 2} + cx$$
<br><br>Being polynomial, it is continuous and differentiable, also,
<br><br>$$f\left( 0 \right) = 0\,$$ and $$\,\,f\left( 1 \right) = {a \over 3} + {b \over 2} + c$$
<br><br>$$ \Rightarrow f\left( 1 \right) = {{2a + ... | mcq | aieee-2004 | 4,728 |
FvFBxKtfz58VB2r3 | maths | application-of-derivatives | mean-value-theorem | If the equation $${a_n}{x^n} + {a_{n - 1}}{x^{n - 1}} + ........... + {a_1}x = 0$$
<br/>$${a_1} \ne 0,n \ge 2,$$ has a positive root $$x = \alpha $$, then the equation
<br/>$$n{a_n}{x^{n - 1}} + \left( {n - 1} \right){a_{n - 1}}{x^{n - 2}} + ........... + {a_1} = 0$$ has a positive root, which is | [{"identifier": "A", "content": "greater than $$\\alpha $$ "}, {"identifier": "B", "content": "smaller than $$\\alpha $$ "}, {"identifier": "C", "content": "greater than or equal to smaller than $$\\alpha $$ "}, {"identifier": "D", "content": "equal to smaller than $$\\alpha $$ "}] | ["B"] | null | Let $$f\left( x \right) = {a_n}{x^n} + {a_{n - 1}}{x^{n - 1}} + ........... + {a_1}x = 0$$
<br><br>The other given equation,
<br><br>$$na{}_n{x^{n - 1}} + \left( {n - 1} \right){a_{n - 1}}{x^{n - 2}} + .... + {a_1} = 0 = f'\left( x \right)$$
<br><br>Given $${a_1} \ne 0 \Rightarrow f\left( 0 \right) = 0$$
<br><br>Again... | mcq | aieee-2005 | 4,729 |
UJUItPkT64O9QtSG | maths | application-of-derivatives | mean-value-theorem | Let f be differentiable for all x. If f(1) = -2 and f'(x) $$ \ge $$ 2 for
<br/>x $$ \in \left[ {1,6} \right]$$, then | [{"identifier": "A", "content": "f(6) $$ \\ge $$ 8"}, {"identifier": "B", "content": "f(6) < 8"}, {"identifier": "C", "content": "f(6) < 5"}, {"identifier": "D", "content": "f(6) = 5"}] | ["A"] | null | As $$\,\,f\left( 1 \right) = - 2\,\,\& \;\,f'\left( x \right) \ge 2\,\forall x \in \left[ {1,6} \right]$$
<br><br>Applying Lagrange's mean value theorem
<br><br>$${{f\left( 6 \right) - f\left( 1 \right)} \over 5} = f'\left( c \right) \ge 2$$
<br><br>$$ \Rightarrow f\left( 6 \right) \ge 10 + f\left( 1 \right)$$
<br... | mcq | aieee-2005 | 4,730 |
yVtBIed6SEB53zdc | maths | application-of-derivatives | mean-value-theorem | A value of $$c$$ for which conclusion of Mean Value Theorem holds for the function $$f\left( x \right) = {\log _e}x$$ on the interval $$\left[ {1,3} \right]$$ is | [{"identifier": "A", "content": "$${\\log _3}e$$ "}, {"identifier": "B", "content": "$${\\log _e}3$$"}, {"identifier": "C", "content": "$$2\\,\\,{\\log _3}e$$ "}, {"identifier": "D", "content": "$${1 \\over 2}{\\log _3}e$$ "}] | ["C"] | null | Using Lagrange's Mean Value Theorem
<br><br>Let $$f(x)$$ be a function defined on $$\left[ {a,b} \right]$$
<br><br>then, $$f'\left( c \right) = {{f\left( b \right) - f\left( a \right)} \over {b - a}}\,\,\,\,\,\,\,\,\,\,\,\,....\left( i \right)$$
<br><br>$$c\,\, \in \left[ {a,b} \right]$$
<br><br>$$\therefore$$ Given $$... | mcq | aieee-2007 | 4,731 |
9i0za95SuBx6OsEG | maths | application-of-derivatives | mean-value-theorem | If $$f$$ and $$g$$ are differentiable functions in $$\left[ {0,1} \right]$$ satisfying
<br/>$$f\left( 0 \right) = 2 = g\left( 1 \right),g\left( 0 \right) = 0$$ and $$f\left( 1 \right) = 6,$$ then for some $$c \in \left] {0,1} \right[$$ | [{"identifier": "A", "content": "$$f'\\left( c \\right) = g'\\left( c \\right)$$ "}, {"identifier": "B", "content": "$$f'\\left( c \\right) = 2g'\\left( c \\right)$$"}, {"identifier": "C", "content": "$$2f'\\left( c \\right) = g'\\left( c \\right)$$"}, {"identifier": "D", "content": "$$2f'\\left( c \\right) = 3g'\\left... | ["B"] | null | Since, $$f$$ and $$g$$ both are continuous function on $$\left[ {0,1} \right]$$
<br><br>and differentiable on $$\left( {0,1} \right)$$ then $$\exists c \in \left( {0,1} \right)$$ such that
<br><br>$$f'\left( c \right) = {{f\left( 1 \right) - f\left( 0 \right)} \over 1} = {{6 - 2} \over 1} = 4$$
<br><br>and $$g'\left( c... | mcq | jee-main-2014-offline | 4,732 |
RiuUixc2KvzB1Mn3kM7k9k2k5e32k5u | maths | application-of-derivatives | mean-value-theorem | Let the function, Ζ:[-7, 0]$$ \to $$R be continuous on [-7,0] and differentiable on (-7, 0). If Ζ(-7) = -
3 and Ζ'(x) $$ \le $$ 2, for all x $$ \in $$ (-7,0), then for all such functions Ζ, Ζ(-1) + Ζ(0) lies in the interval:
| [{"identifier": "A", "content": "$$\\left[ { - 6,20} \\right]$$"}, {"identifier": "B", "content": "$$\\left( { - \\infty ,\\left. {20} \\right]} \\right.$$"}, {"identifier": "C", "content": "$$\\left[ { - 3,11} \\right]$$"}, {"identifier": "D", "content": "$$\\left( { - \\infty ,\\left. {11} \\right]} \\right.$$"}] | ["B"] | null | Using Lagrangeβs Mean Value Theorem in [β7, β1]
<br><br>$${{f\left( { - 1} \right) - f\left( { - 7} \right)} \over { - 1 - \left( { - 7} \right)}}$$ = f'(c<sub>1</sub>)
<br><br>As Ζ'(x) $$ \le $$ 2 then f'(c<sub>1</sub>) $$ \le $$ 2
<br><br>$$ \therefore $$ $${{f\left( { - 1} \right) - f\left( { - 7} \right)} \over { -... | mcq | jee-main-2020-online-7th-january-morning-slot | 4,733 |
SY1QQZrwAcg8toamm57k9k2k5fnb4sd | maths | application-of-derivatives | mean-value-theorem | The value of c in the Lagrange's mean value theorem for the function <br/>Ζ(x) = x<sup>3</sup>
- 4x<sup>2</sup>
+ 8x + 11,
when x $$ \in $$ [0, 1] is:
| [{"identifier": "A", "content": "$${2 \\over 3}$$"}, {"identifier": "B", "content": "$${{\\sqrt 7 - 2} \\over 3}$$"}, {"identifier": "C", "content": "$${{4 - \\sqrt 5 } \\over 3}$$"}, {"identifier": "D", "content": "$${{4 - \\sqrt 7 } \\over 3}$$"}] | ["D"] | null | Ζ(x) = x<sup>3</sup>
- 4x<sup>2</sup>
+ 8x + 11
<br><br>f(0) = 11
<br><br>f(1) = 16
<br><br>Using LMVT
<br><br>f'(c) = $${{f\left( 1 \right) - f\left( 0 \right)} \over {1 - 0}}$$
<br><br>$$ \Rightarrow $$ 3c<sup>2</sup>
β 8c + 8 = $${{16 - 11} \over {1 - 0}}$$
<br><br>$$ \Rightarrow $$ 3c<sup>2</sup>
β 8c + 3 = 0
<... | mcq | jee-main-2020-online-7th-january-evening-slot | 4,734 |
7HHI3mgBSQbnJgC9CJ7k9k2k5gyu0fm | maths | application-of-derivatives | mean-value-theorem | If c is a point at which Rolle's theorem holds
for the function,
<br/>f(x) = $${\log _e}\left( {{{{x^2} + \alpha } \over {7x}}} \right)$$ in the
interval [3, 4], where a $$ \in $$ R, then Ζ''(c) is equal
to | [{"identifier": "A", "content": "$${1 \\over {12}}$$"}, {"identifier": "B", "content": "$${{\\sqrt 3 } \\over 7}$$"}, {"identifier": "C", "content": "$$-{1 \\over {12}}$$"}, {"identifier": "D", "content": "$$-{1 \\over {24}}$$"}] | ["A"] | null | For Rolleβs theorem to be applicable in [3, 4]
<br><br>Ζ(3) = Ζ(4)
<br><br>$$ \Rightarrow $$ $${\log _e}\left( {{{9 + \alpha } \over {21}}} \right) = {\log _e}\left( {{{16 + \alpha } \over {28}}} \right)$$
<br><br>$$ \Rightarrow $$ $$\left( {{{9 + \alpha } \over {21}}} \right) = \left( {{{16 + \alpha } \over {28}}} \r... | mcq | jee-main-2020-online-8th-january-morning-slot | 4,735 |
hRKCTLC497lJlskzXd1kls4t2ij | maths | application-of-derivatives | mean-value-theorem | If Rolle's theorem holds for the function $$f(x) = {x^3} - a{x^2} + bx - 4$$, $$x \in [1,2]$$ with $$f'\left( {{4 \over 3}} \right) = 0$$, then ordered pair (a, b) is equal to : | [{"identifier": "A", "content": "($$-$$5, $$-$$8)"}, {"identifier": "B", "content": "(5, $$-$$8)"}, {"identifier": "C", "content": "($$-$$5, 8)"}, {"identifier": "D", "content": "(5, 8)"}] | ["D"] | null | $$f(1) = f(2)$$<br><br>$$ \Rightarrow 1 - a + b - 4 = 8 - 4a + 2b - 4$$<br><br>$$3a - b = 7$$ ..... (1)<br><br>$$f'(x) = 3{x^2} - 2ax + b$$<br><br>$$ \Rightarrow f'\left( {{4 \over 3}} \right) = 0 \Rightarrow 3 \times {{16} \over 9} - {8 \over 3}a + b = 0$$<br><br>$$ \Rightarrow - 8a + 3b = - 16$$ ..... (2)<br><br>$$... | mcq | jee-main-2021-online-25th-february-morning-slot | 4,736 |
IvgThkO2TD7WaLN3jF1klui2sw9 | maths | application-of-derivatives | mean-value-theorem | Let f be any function defined on R and let it satisfy the condition : $$|f(x) - f(y)|\, \le \,|{(x - y)^2}|,\forall (x,y) \in R$$<br/><br/>If f(0) = 1, then : | [{"identifier": "A", "content": "f(x) can take any value in R"}, {"identifier": "B", "content": "$$f(x) < 0,\\forall x \\in R$$"}, {"identifier": "C", "content": "$$f(x) > 0,\\forall x \\in R$$"}, {"identifier": "D", "content": "$$f(x) = 0,\\forall x \\in R$$"}] | ["C"] | null | $$|f(x) - f(y)|\, \le \,|{(x - y)^2}|$$<br><br>$$ \Rightarrow \left| {{{f(x) - f(y)} \over {x - y}}} \right|\, \le \,|x - y|$$<br><br>$$ \Rightarrow \left| {\mathop {\lim }\limits_{x \to y} {{f(x) - f(y)} \over {x - y}}} \right|\, \le \,|\mathop {\lim }\limits_{x \to y} (x - y)|$$<br><br>$$ \Rightarrow |f'(x)|\, \le 0$... | mcq | jee-main-2021-online-26th-february-morning-slot | 4,737 |
lsblgir9 | maths | application-of-derivatives | mean-value-theorem | Let for a differentiable function $f:(0, \infty) \rightarrow \mathbf{R}, f(x)-f(y) \geqslant \log _{\mathrm{e}}\left(\frac{x}{y}\right)+x-y, \forall x, y \in(0, \infty)$. Then $\sum\limits_{n=1}^{20} f^{\prime}\left(\frac{1}{n^2}\right)$ is equal to ____________. | [] | null | 2890 | <p>$$\begin{aligned}
& f(x)-f(y) \geq \ln x-\ln y+x-y \\
& \frac{f(x)-f(y)}{x-y} \geq \frac{\ln x-\ln y}{x-y}+1
\end{aligned}$$</p>
<p>Let $$x>y$$</p>
<p>$$\lim _\limits{y \rightarrow x} f^{\prime}\left(x^{-}\right) \geq \frac{1}{x}+1\quad\text{.... (1)}$$</p>
<p>Let $$x< y$$</p>
<p>$$\lim _\limits{y \rightarrow x} f^{... | integer | jee-main-2024-online-27th-january-morning-shift | 4,739 |
jaXMH0IezPGTzEA1 | maths | application-of-derivatives | monotonicity | The function $$f\left( x \right) = {\tan ^{ - 1}}\left( {\sin x + \cos x} \right)$$ is an incresing function in | [{"identifier": "A", "content": "$$\\left( {0,{\\pi \\over 2}} \\right)$$ "}, {"identifier": "B", "content": "$$\\left( { - {\\pi \\over 2},{\\pi \\over 2}} \\right)$$ "}, {"identifier": "C", "content": "$$\\left( { {\\pi \\over 4},{\\pi \\over 2}} \\right)$$"}, {"identifier": "D", "content": "$$\\left( { - {\\pi... | ["D"] | null | Given $$f\left( x \right) = {\tan ^{ - 1}}\left( {\sin x + \cos x} \right)$$
<br><br>$$f'\left( x \right) = {1 \over {1 + {{\left( {\sin x + \cos x} \right)}^2}}}.\left( {\cos x - \sin x} \right)$$
<br><br>$$ = {{\sqrt 2 .\left( {{1 \over {\sqrt 2 }}\cos x - {1 \over {\sqrt 2 }}\sin x} \right)} \over {1 + {{\left( {\si... | mcq | aieee-2007 | 4,741 |
1hCumHByo0ZaOkXiKDLcf | maths | application-of-derivatives | monotonicity | Let f(x) = sin<sup>4</sup>x + cos<sup>4</sup> x. Then <i>f</i> is an increasing function in the interval : | [{"identifier": "A", "content": "$$] 0, \\frac{\\pi}{4}[$$"}, {"identifier": "B", "content": "$$] \\frac{\\pi}{4}, \\frac{\\pi}{2}[$$"}, {"identifier": "C", "content": "$$] \\frac{\\pi}{2}, \\frac{5 \\pi}{8}[$$"}, {"identifier": "D", "content": "$$] \\frac{5 \\pi}{8}, \\frac{3 \\pi}{4}[$$"}] | ["B"] | null | f(x) = sin<sup>4</sup>x + cos<sup>4</sup>x
<br><br>$$ \therefore $$ f'(x) = 4sin<sup>3</sup>x cosx + 4cos<sup>3</sup>x ($$-$$ sinx)
<br><br>= 4sinx cosx (sin<sup>2</sup>x $$-$$ cos<sup>2</sup>x)
<br><br>= $$-$$ 2sin2x cos2x
<br><br>= $$-$$ sin4x
<br><br>As, f(x) is increasing function when f'(x) > 0... | mcq | jee-main-2016-online-10th-april-morning-slot | 4,743 |
QxDDYKNpsGnt8PZt0K0rq | maths | application-of-derivatives | monotonicity | The function f defined by
<br/><br/>f(x) = x<sup>3</sup> $$-$$ 3x<sup>2</sup> + 5x + 7 , is : | [{"identifier": "A", "content": "increasing in <b>R</b>."}, {"identifier": "B", "content": "decreasing in <b>R</b>."}, {"identifier": "C", "content": "decreasing in (0, $$\\infty $$) and increasing in ($$-$$ $$\\infty $$, 0)"}, {"identifier": "D", "content": "increasing in (0, $$\\infty $$) and decreasing in ($$-$$ $... | ["A"] | null | <p>The given function is</p>
<p>$$f(x) = {x^2} - 3{x^2} + 5x + 7$$</p>
<p>$$f'(x) = 3{x^2} - 6x + 5$$</p>
<p>The discriminant of the above quadratic equation is</p>
<p>$$\Delta = 36 - 4(3)(5) = 36 - 60 < 0$$</p>
<p>Therefore, $$f'(x) > 0\,\forall x \in {R^ + }$$</p>
<p>Also, $$f'(x) > 0\,\forall x \in {R^ - }$$</p>
<p... | mcq | jee-main-2017-online-9th-april-morning-slot | 4,744 |
XB56wgM0Ps8wqSh3VSc4L | maths | application-of-derivatives | monotonicity | Let f(x) = $${x \over {\sqrt {{a^2} + {x^2}} }} - {{d - x} \over {\sqrt {{b^2} + {{\left( {d - x} \right)}^2}} }},\,\,$$ x $$\, \in $$ R, where a, b and d are non-zero real constants. Then : | [{"identifier": "A", "content": "f is an increasing function of x"}, {"identifier": "B", "content": "f is neither increasing nor decreasing function of x"}, {"identifier": "C", "content": "f ' is not a continuous function of x"}, {"identifier": "D", "content": "f is a decreasing function of x"}] | ["A"] | null | $$f\left( x \right) = {x \over {\sqrt {{a^2} + {x^2}} }} - {{d - x} \over {\sqrt {{b^2} + {{\left( {d - x} \right)}^2}} }}$$
<br><br>$$f'\left( x \right) = {{{a^2}} \over {{{\left( {{a^2} + {x^2}} \right)}^{3/2}}}} + {{{b^2}} \over {{{\left( {{b^2} + {{\left( {d - x} \right)}^2}} \right)}^{3/2}}}} > 0\forall x \in R... | mcq | jee-main-2019-online-11th-january-evening-slot | 4,745 |
kHdxiJODeiP3oavxynlS5 | maths | application-of-derivatives | monotonicity | Let Ζ : [0, 2] $$ \to $$ R be a twice differentiable
function such that Ζ''(x) > 0, for all x $$ \in $$ (0, 2).
If $$\phi $$(x) = Ζ(x) + Ζ(2 β x), then $$\phi $$ is : | [{"identifier": "A", "content": "decreasing on (0, 2)"}, {"identifier": "B", "content": "decreasing on (0, 1) and increasing on (1, 2)"}, {"identifier": "C", "content": "increasing on (0, 2)"}, {"identifier": "D", "content": "increasing on (0, 1) and decreasing on (1, 2)"}] | ["B"] | null | $$\phi $$(x) = Ζ(x) + Ζ(2 β x)
<br><br>$$ \Rightarrow $$ $$\phi $$'(x) = Ζ'(x) - Ζ'(2 β x)
<br><br>Since Ζ''(x) > 0 for all x $$ \in $$ (0, 2)
<br><br>$$ \Rightarrow $$ Ζ'(x) is an increasing function for all x $$ \in $$ (0, 2).
<br><br><b>Case 1 : When $$\phi $$(x) is increasing function</b>
<br><br>So $$\phi $$'(x... | mcq | jee-main-2019-online-8th-april-morning-slot | 4,747 |
5qPzYxvBb1sYWL6hEVjgy2xukf0vzp5w | maths | application-of-derivatives | monotonicity | The function, f(x) = (3x β 7)x<sup>2/3</sup>, x $$ \in $$ R, is
increasing for all x lying in : | [{"identifier": "A", "content": "$$\\left( { - \\infty ,0} \\right) \\cup \\left( {{3 \\over 7},\\infty } \\right)$$"}, {"identifier": "B", "content": "$$\\left( { - \\infty ,0} \\right) \\cup \\left( {{{14} \\over {15}},\\infty } \\right)$$"}, {"identifier": "C", "content": "$$\\left( { - \\infty ,{{14} \\over {15}}} ... | ["B"] | null | f(x) = (3x β 7)x<sup>2/3</sup>
<br><br>fβ(x) = $$\left( {3x - 7} \right){2 \over {3{x^{1/3}}}} + {x^{{2 \over 3}}}.3$$
<br><br>= $${{6x - 14 - 9x} \over {3{x^{1/3}}}}$$
<br><br>= $${{15x - 14} \over {3{x^{1/3}}}}$$
<br><br>As f(x) increasing so f'(x) > 0
<br><br>$$ \therefore $$ $${{15x - 14} \over {3{x^{1/3}}}}$$ &... | mcq | jee-main-2020-online-3rd-september-morning-slot | 4,749 |
FBiTnaSpw8Lymvky7rjgy2xukf8z47kh | maths | application-of-derivatives | monotonicity | Let f be a twice differentiable function on (1, 6). If f(2) = 8, fβ(2) = 5, fβ(x) $$ \ge $$ 1 and f''(x) $$ \ge $$ 4, for all x $$ \in $$ (1, 6), then :
| [{"identifier": "A", "content": "f(5) $$ \\le $$ 10"}, {"identifier": "B", "content": "f(5) + f'(5) $$ \\ge $$ 28"}, {"identifier": "C", "content": "f(5) + f'(5) $$ \\le $$ 26"}, {"identifier": "D", "content": "f'(5) + f''(5) $$ \\le $$ 20"}] | ["B"] | null | Given, $$f'(x) \ge 1$$<br><br>$$ \therefore $$ $$\int_2^5 {f'(x)} dx\, \ge \,\int_2^5 {dx} $$<br><br>$$ \Rightarrow f(5) - f(2) \ge 3$$<br><br>$$ \Rightarrow f(5) - 8 \ge 3$$<br><br>$$ \Rightarrow f(5) \ge 11$$ ...(1)<br><br>Also, $$f''(x) \ge 4$$<br><br>$$ \therefore $$ $$\int_2^5 {f''(x)} dx\, \ge \,\int_2^5 {4dx} $$... | mcq | jee-main-2020-online-4th-september-morning-slot | 4,750 |
Zpp13XvDvvswz7dWtl7k9k2k5gzwf3e | maths | application-of-derivatives | monotonicity | Let Ζ(x) = xcos<sup>β1</sup>(βsin|x|), $$x \in \left[ { - {\pi \over 2},{\pi \over 2}} \right]$$, then
which of the following is true? | [{"identifier": "A", "content": "\u0192' is decreasing in $$\\left( { - {\\pi \\over 2},0} \\right)$$ and increasing\nin $$\\left( {0,{\\pi \\over 2}} \\right)$$"}, {"identifier": "B", "content": "\u0192 '(0) = $${ - {\\pi \\over 2}}$$"}, {"identifier": "C", "content": "\u0192 is not differentiable at x = 0"}, {"ide... | ["A"] | null | We know, cos<sup>-1</sup>(-x) = $$\pi $$ - cos<sup>-1</sup>x
<br><br>$$ \therefore $$ Ζ(x) = x($$\pi $$ - cos<sup>β1</sup>(sin|x|))
<br><br>= x($$\pi $$ - $${\pi \over 2}$$ + sin<sup>β1</sup>(sin|x|))
<br><br>= x($$\pi $$ - $${\pi \over 2}$$ + sin<sup>β1</sup>(sin|x|))
<br><br>= x$${\pi \over 2}$$ + x|x|
<br><br>$$ ... | mcq | jee-main-2020-online-8th-january-morning-slot | 4,752 |
blzO06JKs65FBoRsOx1klrlnsa6 | maths | application-of-derivatives | monotonicity | Let $$f:R \to R$$ be defined as<br/><br/>$$f(x) = \left\{ {\matrix{
{ - 55x,} & {if\,x < - 5} \cr
{2{x^3} - 3{x^2} - 120x,} & {if\, - 5 \le x \le 4} \cr
{2{x^3} - 3{x^2} - 36x - 336,} & {if\,x > 4,} \cr
} } \right.$$<br/><br/>Let A = {x $$ \in $$ R : f is increasing}. Then A is equal... | [{"identifier": "A", "content": "$$( - 5,\\infty )$$"}, {"identifier": "B", "content": "$$( - \\infty , - 5) \\cup (4,\\infty )$$"}, {"identifier": "C", "content": "$$( - 5, - 4) \\cup (4,\\infty )$$"}, {"identifier": "D", "content": "$$( - \\infty , - 5) \\cup ( - 4,\\infty )$$"}] | ["C"] | null | $$f(x) = \left\{ {\matrix{
{ - 55x,} & {if\,x < - 5} \cr
{2{x^3} - 3{x^2} - 120x,} & {if\, - 5 \le x \le 4} \cr
{2{x^3} - 3{x^2} - 36x - 336,} & {if\,x > 4,} \cr
} } \right.$$
<br><br>Now, $$f'(x) = \left\{ {\matrix{
{ - 55} & ; & {x < - 5} \cr
{6({x^2} - x - 20)}... | mcq | jee-main-2021-online-24th-february-evening-slot | 4,754 |
jnHf1y8tK5XeR4tPda1kluz1e2i | maths | application-of-derivatives | monotonicity | Let a be an integer such that all the real roots of the polynomial <br/>2x<sup>5</sup> + 5x<sup>4</sup> + 10x<sup>3</sup> + 10x<sup>2</sup> + 10x + 10 lie in the interval (a, a + 1). Then, |a| is equal to ___________. | [] | null | 2 | Let, $$f(x) = 2{x^5} + 5{x^4} + 10{x^3} + 10{x^2} + 10x + 10$$<br><br>$$ \Rightarrow f'(x) = 10({x^4} + 2{x^3} + 3{x^2} + 2x + 1)$$<br><br>$$ = 10\left( {{x^2} + {1 \over {{x^2}}} + 2\left( {x + {1 \over x}} \right) + 3} \right)$$<br><br>$$ = 10\left( {{{\left( {x + {1 \over x}} \right)}^2} + 2\left( {x + {1 \over x}} ... | integer | jee-main-2021-online-26th-february-evening-slot | 4,755 |
mQLintGwdQfFf2qnw61kmiw0lr1 | maths | application-of-derivatives | monotonicity | Let f be a real valued function, defined on R $$-$$ {$$-$$1, 1} and given by <br/><br/>f(x) = 3 log<sub>e</sub> $$\left| {{{x - 1} \over {x + 1}}} \right| - {2 \over {x - 1}}$$.<br/><br/>Then in which of the following intervals, function f(x) is increasing? | [{"identifier": "A", "content": "($$-$$$$\\infty $$, $$-$$1) $$\\cup$$ $$\\left( {[{1 \\over 2},\\infty ) - \\{ 1\\} } \\right)$$"}, {"identifier": "B", "content": "($$-$$$$\\infty $$, $$\\infty $$) $$-$$ {$$-$$1, 1)"}, {"identifier": "C", "content": "($$-$$$$\\infty $$, $${{1 \\over 2}}$$] $$-$$ {$$-$$1}"}, {"identifi... | ["A"] | null | f(x) = 3 log<sub>e</sub> $$\left| {{{x - 1} \over {x + 1}}} \right| - {2 \over {x - 1}}$$
<br><br>$$f'(x) = {{3(x + 1)} \over {x - 1}} \times {{(x + 1) - (x - 1)} \over {{{(x + 1)}^2}}} + {2 \over {{{(x - 1)}^2}}} > 0$$<br><br>$$ = {6 \over {{x^2} - 1}} + {2 \over {{{(x - 1)}^2}}} > 0$$<br><br>$$ = {{2(3(x - 1) +... | mcq | jee-main-2021-online-16th-march-evening-shift | 4,756 |
4M0JODSlLdB22Grotz1kmkn7lpe | maths | application-of-derivatives | monotonicity | Consider the function f : R $$ \to $$ R defined by
<br/><br/>$$f(x) = \left\{ \matrix{
\left( {2 - \sin \left( {{1 \over x}} \right)} \right)|x|,x \ne 0 \hfill \cr
0,\,\,x = 0 \hfill \cr} \right.$$. Then f is : | [{"identifier": "A", "content": "not monotonic on ($$-$$$$\\infty $$, 0) and (0, $$\\infty $$)"}, {"identifier": "B", "content": "monotonic on (0, $$\\infty $$) only"}, {"identifier": "C", "content": "monotonic on ($$-$$$$\\infty $$, 0) only"}, {"identifier": "D", "content": "monotonic on ($$-$$$$\\infty $$, 0) $$\\cup... | ["A"] | null | $$f(x) = \left\{ {\matrix{
{ - \left( {2 - \sin {1 \over x}} \right)x} & , & {x < 0} \cr
0 & , & {x = 0} \cr
{\left( {2 - \sin {1 \over x}} \right)x} & , & {x > 0} \cr
} } \right.$$<br><br>$$f'(x) = \left\{ \matrix{
- x\left( { - \cos {1 \over x}} \right)\left( { - {1 \... | mcq | jee-main-2021-online-17th-march-evening-shift | 4,757 |
1krvrw9s4 | maths | application-of-derivatives | monotonicity | Let $$f(x) = 3{\sin ^4}x + 10{\sin ^3}x + 6{\sin ^2}x - 3$$, $$x \in \left[ { - {\pi \over 6},{\pi \over 2}} \right]$$. Then, f is : | [{"identifier": "A", "content": "increasing in $$\\left( { - {\\pi \\over 6},{\\pi \\over 2}} \\right)$$"}, {"identifier": "B", "content": "decreasing in $$\\left( {0,{\\pi \\over 2}} \\right)$$"}, {"identifier": "C", "content": "increasing in $$\\left( { - {\\pi \\over 6},0} \\right)$$"}, {"identifier": "D", "cont... | ["D"] | null | $$f(x) = 3{\sin ^4}x + 10{\sin ^3}x + 6{\sin ^2}x - 3,x \in \left[ { - {\pi \over 6},{\pi \over 2}} \right]$$<br><br>$$f'(x) = 12{\sin ^3}x\cos x + 30{\sin ^2}x\cos x + 12\sin x\cos x$$<br><br>$$ = 6\sin x\cos x(2{\sin ^2}x + 5\sin x + 2)$$<br><br>$$ = 6\sin x\cos x(2\sin x + 1)(\sin + 2)$$<br><br><img src="https://... | mcq | jee-main-2021-online-25th-july-morning-shift | 4,759 |
1kto3gifb | maths | application-of-derivatives | monotonicity | The function $$f(x) = {x^3} - 6{x^2} + ax + b$$ is such that $$f(2) = f(4) = 0$$. Consider two statements :<br/><br/>Statement 1 : there exists x<sub>1</sub>, x<sub>2</sub> $$\in$$(2, 4), x<sub>1</sub> < x<sub>2</sub>, such that f'(x<sub>1</sub>) = $$-$$1 and f'(x<sub>2</sub>) = 0.<br/><br/>Statement 2 : there exist... | [{"identifier": "A", "content": "both Statement 1 and Statement 2 are true"}, {"identifier": "B", "content": "Statement 1 is false and Statement 2 is true"}, {"identifier": "C", "content": "both Statement 1 and Statement 2 are false"}, {"identifier": "D", "content": "Statement 1 is true and Statement 2 is false"}] | ["A"] | null | $$f(x) = {x^3} - 6{x^2} + ax + b$$<br><br>$$f(2) = 8 - 24 + 2a + b = 0$$<br><br>$$2a + b = 16$$ .... (1)<br><br>$$f(4) = 64 - 96 + 4a + b = 0$$<br><br>$$4a + b = 32$$ .... (2)<br><br>Solving (1) and (2)<br><br>a = 8, b = 0<br><br>$$f(x) = {x^3} - 6{x^2} + 8x$$<br><br>$$f'(x) = 3{x^2} - 12x + 8$$<br><br>$$f''(x) = 6x - ... | mcq | jee-main-2021-online-1st-september-evening-shift | 4,761 |
1l566qk0u | maths | application-of-derivatives | monotonicity | <p>The number of real solutions of <br/><br/>$${x^7} + 5{x^3} + 3x + 1 = 0$$ is equal to ____________.</p> | [{"identifier": "A", "content": "0"}, {"identifier": "B", "content": "1"}, {"identifier": "C", "content": "3"}, {"identifier": "D", "content": "5"}] | ["B"] | null | <img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lc8ekncq/3bb86ec8-2e21-4aa5-b7aa-771daae2cbcc/79bb29a0-8717-11ed-b3ec-0bde88094e1e/file-1lc8ekncr.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lc8ekncq/3bb86ec8-2e21-4aa5-b7aa-771daae2cbcc/79bb29a0-8717-11ed-b3ec-0bde88094e1e/fi... | mcq | jee-main-2022-online-28th-june-morning-shift | 4,762 |
1l58a7w47 | maths | application-of-derivatives | monotonicity | <p>Let $$f(x) = 2{\cos ^{ - 1}}x + 4{\cot ^{ - 1}}x - 3{x^2} - 2x + 10$$, $$x \in [ - 1,1]$$. If [a, b] is the range of the function f, then 4a $$-$$ b is equal to :</p> | [{"identifier": "A", "content": "11"}, {"identifier": "B", "content": "11 $$-$$ $$\\pi$$"}, {"identifier": "C", "content": "11 + $$\\pi$$"}, {"identifier": "D", "content": "15 $$-$$ $$\\pi$$"}] | ["B"] | null | <p>$$f(x) = 2{\cos ^{ - 1}}x + 4{\cot ^{ - 1}}x - 3{x^2} - 2x + 10\,\forall x \in [ - 1,1]$$</p>
<p>$$ \Rightarrow f'(x) = - {2 \over {\sqrt {1 - {x^2}} }} - {4 \over {1 + {x^2}}} - 6x - 2 < 0\,\forall x \in [ - 1,1]$$</p>
<p>So f(x) is decreasing function and range of f(x) is [f(1), f($$-$$1)], which is [$$\pi$$ + 5,... | mcq | jee-main-2022-online-26th-june-morning-shift | 4,763 |
1l5c19dx8 | maths | application-of-derivatives | monotonicity | <p>For the function <br/><br/>$$f(x) = 4{\log _e}(x - 1) - 2{x^2} + 4x + 5,\,x > 1$$, which one of the following is NOT correct?</p> | [{"identifier": "A", "content": "f is increasing in (1, 2) and decreasing in (2, $$\\infty$$)"}, {"identifier": "B", "content": "f(x) = $$-$$1 has exactly two solutions"}, {"identifier": "C", "content": "$$f'(e) - f''(2) < 0$$"}, {"identifier": "D", "content": "f(x) = 0 has a root in the interval (e, e + 1)"}] | ["C"] | null | Lets draw the curve $y=f(x)=4 \log _e(x-1)-2 x^2$ $+4 x+5, x>1$
<br><br><img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lkavtohl/5b2f07ad-6815-4447-8b5a-bda6bf0cb5b3/3c6a7290-26d6-11ee-b52b-3728f15f4ced/file-6y3zli1lkavtohm.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image... | mcq | jee-main-2022-online-24th-june-morning-shift | 4,765 |
1l6gjnyuy | maths | application-of-derivatives | monotonicity | <p>Let the function $$f(x)=2 x^{2}-\log _{\mathrm{e}} x, x>0$$, be decreasing in $$(0, \mathrm{a})$$ and increasing in $$(\mathrm{a}, 4)$$. A tangent to the parabola $$y^{2}=4 a x$$ at a point $$\mathrm{P}$$ on it passes through the point $$(8 \mathrm{a}, 8 \mathrm{a}-1)$$ but does not pass through the point $$\left... | [] | null | 45 | <p>$$\delta '(x) = {{4{x^2} - 1} \over x}$$ so f(x) is decreasing in $$\left( {0,{1 \over 2}} \right)$$ and increasing in $$\left( {{1 \over 2},\infty } \right) \Rightarrow a = {1 \over 2}$$</p>
<p>Tangent at $${y^2} = 2x \Rightarrow y = ,x + {1 \over {2m}}$$</p>
<p>It is passing through $$(4,3)$$</p>
<p>$$3 = 4m + {1 ... | integer | jee-main-2022-online-26th-july-morning-shift | 4,766 |
1l6nm1791 | maths | application-of-derivatives | monotonicity | <p>The function $$f(x)=x \mathrm{e}^{x(1-x)}, x \in \mathbb{R}$$, is :</p> | [{"identifier": "A", "content": "increasing in $$\\left(-\\frac{1}{2}, 1\\right)$$"}, {"identifier": "B", "content": "decreasing in $$\\left(\\frac{1}{2}, 2\\right)$$"}, {"identifier": "C", "content": "increasing in $$\\left(-1,-\\frac{1}{2}\\right)$$"}, {"identifier": "D", "content": "decreasing in $$\\left(-\\frac{1}... | ["A"] | null | <p>$$f(x) = x{e^{x(1 - x)}},\,x \in R$$</p>
<p>$$f'(x) = x{e^{x(1 - x)}}\,.\,(1 - 2x) + {e^{x(1 - x)}}$$</p>
<p>$$ = {e^{x(1 - x)}}[x - 2{x^2} + 1]$$</p>
<p>$$ = - {e^{x(1 - x)}}[2{x^2} - x - 1]$$</p>
<p>$$ = - {e^{x(1 - x)}}(2x + 1)(x - 1)$$</p>
<p>$$\therefore$$ $$f(x)$$ is increasing in $$\left( { - {1 \over 2},1}... | mcq | jee-main-2022-online-28th-july-evening-shift | 4,767 |
1ldv2lvfb | maths | application-of-derivatives | monotonicity | <p>Let $$f:(0,1)\to\mathbb{R}$$ be a function defined $$f(x) = {1 \over {1 - {e^{ - x}}}}$$, and $$g(x) = \left( {f( - x) - f(x)} \right)$$. Consider two statements</p>
<p>(I) g is an increasing function in (0, 1)</p>
<p>(II) g is one-one in (0, 1)</p>
<p>Then,</p> | [{"identifier": "A", "content": "Both (I) and (II) are true"}, {"identifier": "B", "content": "Neither (I) nor (II) is true"}, {"identifier": "C", "content": "Only (II) is true"}, {"identifier": "D", "content": "Only (I) is true"}] | ["A"] | null | $g(x)=f(-x)-f(x)$
<br/><br/>
$$
\begin{aligned}
& =\frac{1}{1-e^{x}}-\frac{1}{1-e^{-x}} \\\\
& =\frac{1}{1-e^{x}}-\frac{e^{x}}{e^{x}-1} \\\\
& =\frac{1+e^{x}}{1-e^{x}} \\\\
g^{\prime}(x) & =\frac{\left(1-e^{x}\right) e^{x}-\left(1+e^{x}\right)\left(-e^{x}\right)}{\left(1-e^{x}\right)^{2}} \\\\
& =\frac{e^{x}-2 e^{x}+e^... | mcq | jee-main-2023-online-25th-january-morning-shift | 4,768 |
1lgvpk0py | maths | application-of-derivatives | monotonicity | <p>Let $$\mathrm{g}(x)=f(x)+f(1-x)$$ and $$f^{\prime \prime}(x) > 0, x \in(0,1)$$. If $$\mathrm{g}$$ is decreasing in the interval $$(0, a)$$ and increasing in the interval $$(\alpha, 1)$$, then $$\tan ^{-1}(2 \alpha)+\tan ^{-1}\left(\frac{1}{\alpha}\right)+\tan ^{-1}\left(\frac{\alpha+1}{\alpha}\right)$$ is equal t... | [{"identifier": "A", "content": "$$\\frac{3 \\pi}{4}$$\n"}, {"identifier": "B", "content": "$$\\pi$$"}, {"identifier": "C", "content": "$$\\frac{5 \\pi}{4}$$"}, {"identifier": "D", "content": "$$\\frac{3 \\pi}{2}$$"}] | ["B"] | null | We have, $g(x)=f(x)+f(1-x)$
<br/><br/>Differentiating both side, we get
<br/><br/>$g^{\prime}(x)=f^{\prime}(x)-f^{\prime}(1-x)$
<br/><br/>As $f^{\prime \prime}(x)>0, f^{\prime}(x)$ is an increasing function.
<br/><br/>Also, $g(x)=f(x)+f(2 a-x)$ is always symmetric about $x=a$
<br/><br/>So, $g(x)=f(x)+f(1-x)$ is also... | mcq | jee-main-2023-online-10th-april-evening-shift | 4,769 |
lsaps6wb | maths | application-of-derivatives | monotonicity | If $5 f(x)+4 f\left(\frac{1}{x}\right)=x^2-2, \forall x \neq 0$ and $y=9 x^2 f(x)$, then $y$ is strictly increasing in : | [{"identifier": "A", "content": "$\\left(0, \\frac{1}{\\sqrt{5}}\\right) \\cup\\left(\\frac{1}{\\sqrt{5}}, \\infty\\right)$"}, {"identifier": "B", "content": "$\\left(-\\frac{1}{\\sqrt{5}}, 0\\right) \\cup\\left(\\frac{1}{\\sqrt{5}}, \\infty\\right)$"}, {"identifier": "C", "content": "$\\left(-\\frac{1}{\\sqrt{5}}, 0\\... | ["B"] | null | $$
5 f(x)+4 f(1 / x)=x^2-2
$$ ........(1)
<br><br>Replace $x$ by $1 / x$
<br><br>$$
5 f(1 / x)+4 f(x)=\frac{1}{x^2}-2
$$ ..........(2)
<br><br>Multiply equation (1) by 5 and multiply equation (2) by 4 and then subtract equation (2) from (1)
<br><br>$\begin{aligned} & 25 f(x)-16 f(x)=5 x^2-10-\frac{4}{x^2}+8 \\\\ ... | mcq | jee-main-2024-online-1st-february-morning-shift | 4,770 |
jaoe38c1lsd4jja0 | maths | application-of-derivatives | monotonicity | <p>Let $$f: \rightarrow \mathbb{R} \rightarrow(0, \infty)$$ be strictly increasing function such that $$\lim _\limits{x \rightarrow \infty} \frac{f(7 x)}{f(x)}=1$$. Then, the value of $$\lim _\limits{x \rightarrow \infty}\left[\frac{f(5 x)}{f(x)}-1\right]$$ is equal to</p> | [{"identifier": "A", "content": "0"}, {"identifier": "B", "content": "4"}, {"identifier": "C", "content": "1"}, {"identifier": "D", "content": "7/5"}] | ["A"] | null | <p>$$\begin{aligned}
& f: R \rightarrow(0, \infty) \\
& \lim _{x \rightarrow \infty} \frac{f(7 x)}{f(x)}=1
\end{aligned}$$</p>
<p>$$\because \mathrm{f}$$ is increasing</p>
<p>$$\begin{aligned}
& \therefore \mathrm{f}(\mathrm{x})<\mathrm{f}(5 \mathrm{x})<\mathrm{f}(7 \mathrm{x}) \\
& \because \frac{\mathrm{f}(\mathrm{x}... | mcq | jee-main-2024-online-31st-january-evening-shift | 4,771 |
jaoe38c1lsf073t9 | maths | application-of-derivatives | monotonicity | <p>Consider the function $$f:\left[\frac{1}{2}, 1\right] \rightarrow \mathbb{R}$$ defined by $$f(x)=4 \sqrt{2} x^3-3 \sqrt{2} x-1$$. Consider the statements</p>
<p>(I) The curve $$y=f(x)$$ intersects the $$x$$-axis exactly at one point.</p>
<p>(II) The curve $$y=f(x)$$ intersects the $$x$$-axis at $$x=\cos \frac{\pi}{1... | [{"identifier": "A", "content": "Both (I) and (II) are correct.\n"}, {"identifier": "B", "content": "Only (I) is correct.\n"}, {"identifier": "C", "content": "Both (I) and (II) are incorrect.\n"}, {"identifier": "D", "content": "Only (II) is correct."}] | ["A"] | null | <p>$$\begin{aligned}
& \mathrm{f}^{\prime}(\mathrm{x})=12 \sqrt{2} \mathrm{x}^2-3 \sqrt{2} \geq 0 \text { for }\left[\frac{1}{2}, 1\right] \\
& \mathrm{f}\left(\frac{1}{2}\right)<0
\end{aligned}$$</p>
<p>$$\mathrm{f}(1)>0 \Rightarrow(\mathrm{A})$$ is correct.</p>
<p>$$f(x)=\sqrt{2}\left(4 x^3-3 x\right)-1=0$$</p>
<p>Le... | mcq | jee-main-2024-online-29th-january-morning-shift | 4,773 |
jaoe38c1lsfkgqg0 | maths | application-of-derivatives | monotonicity | <p>The function $$f(x)=\frac{x}{x^2-6 x-16}, x \in \mathbb{R}-\{-2,8\}$$</p> | [{"identifier": "A", "content": "decreases in $$(-\\infty,-2) \\cup(-2,8) \\cup(8, \\infty)$$\n"}, {"identifier": "B", "content": "increases in $$(-\\infty,-2) \\cup(-2,8) \\cup(8, \\infty)$$\n"}, {"identifier": "C", "content": "decreases in $$(-2,8)$$ and increases in $$(-\\infty,-2) \\cup(8, \\infty)$$\n"}, {"identif... | ["A"] | null | <p>$$f(x)=\frac{x}{x^2-6 x-16}$$</p>
<p>Now,</p>
<p>$$\begin{aligned}
& \mathrm{f}^{\prime}(\mathrm{x})=\frac{-\left(\mathrm{x}^2+16\right)}{\left(\mathrm{x}^2-6 \mathrm{x}-16\right)^2} \\
& \mathrm{f}^{\prime}(\mathrm{x})<0
\end{aligned}$$</p>
<p>Thus $$f(x)$$ is decreasing in</p>
<p>$$(-\infty,-2) \cup(-2,8) \cup(8, ... | mcq | jee-main-2024-online-29th-january-evening-shift | 4,774 |
luxwdep3 | maths | application-of-derivatives | monotonicity | <p>Let the set of all values of $$p$$, for which $$f(x)=\left(p^2-6 p+8\right)\left(\sin ^2 2 x-\cos ^2 2 x\right)+2(2-p) x+7$$ does not have any critical point, be the interval $$(a, b)$$. Then $$16 a b$$ is equal to _________.</p> | [] | null | 252 | <p>$$\begin{aligned}
& f(x)=\left(p^2-6 p+8\right)\left(\sin ^2 2 x-\cos ^2 2 x\right) +2(2-p) x+7 \\
& f(x)=-\cos 4 x\left(p^2-6 p+8\right)+2(2-p) x+7 \\
& f^{\prime}(x)=4 \sin 4 x\left(p^2-6 p+8\right)+2(2-p) \neq 0 \\
& 2(2-p)+\left[-4\left(p^2-6 p+8\right), 4\left(p^2-6 p+8\right)\right] \\
& \Rightarrow\left[-4 p^... | integer | jee-main-2024-online-9th-april-evening-shift | 4,775 |
lv7v3jxe | maths | application-of-derivatives | monotonicity | <p>For the function</p>
<p>$$f(x)=\sin x+3 x-\frac{2}{\pi}\left(x^2+x\right), \text { where } x \in\left[0, \frac{\pi}{2}\right],$$</p>
<p>consider the following two statements :</p>
<p>(I) $$f$$ is increasing in $$\left(0, \frac{\pi}{2}\right)$$.</p>
<p>(II) $$f^{\prime}$$ is decreasing in $$\left(0, \frac{\pi}{2}\rig... | [{"identifier": "A", "content": "only (I) is true.\n"}, {"identifier": "B", "content": "both (I) and (II) are true.\n"}, {"identifier": "C", "content": "only (II) is true.\n"}, {"identifier": "D", "content": "neither (I) nor (II) is true."}] | ["B"] | null | <p>$$\begin{aligned}
& f(x)=\sin x+3 x-\frac{2}{\pi}\left(x^2+x\right), \text { where } x \in\left[0, \frac{\pi}{2}\right] \\
& f^{\prime}(x)=\cos x+3-\frac{2}{\pi}(2 x+1) \\
& =\cos x-\frac{4 x}{\pi}-\frac{2}{\pi}+3 \\
& \text { as } x \in\left[0, \frac{\pi}{2}\right] \\
& \frac{4 x}{\pi} \in[0,2]
\end{aligned}$$</p>
... | mcq | jee-main-2024-online-5th-april-morning-shift | 4,777 |
lvc57uea | maths | application-of-derivatives | monotonicity | <p>The interval in which the function $$f(x)=x^x, x>0$$, is strictly increasing is</p> | [{"identifier": "A", "content": "$$(0, \\infty)$$\n"}, {"identifier": "B", "content": "$$\\left(0, \\frac{1}{e}\\right]$$\n"}, {"identifier": "C", "content": "$$\\left[\\frac{1}{e^2}, 1\\right)$$\n"}, {"identifier": "D", "content": "$$\\left[\\frac{1}{e}, \\infty\\right)$$"}] | ["D"] | null | <p>$$\begin{aligned}
& f(x)=x^x \\
& f(x)=x^x(\log x+1) \\
& f(x) \geq 0 \\
& \Rightarrow 1+\log x \geq 0 \\
& \Rightarrow \log x \geq-1 \\
& \Rightarrow x \geq e^{-1} \\
& \therefore x \in\left[\frac{1}{e^{\prime}}, \infty\right)
\end{aligned}$$</p> | mcq | jee-main-2024-online-6th-april-morning-shift | 4,778 |
5hg8moK4Iyz89JC7 | maths | application-of-derivatives | rate-of-change-of-quantity | A point on the parabola $${y^2} = 18x$$ at which the ordinate increases at twice the rate of the abscissa is | [{"identifier": "A", "content": "$$\\left( {{9 \\over 8},{9 \\over 2}} \\right)$$ "}, {"identifier": "B", "content": "$$(2, -4)$$ "}, {"identifier": "C", "content": "$$\\left( {{-9 \\over 8},{9 \\over 2}} \\right)$$"}, {"identifier": "D", "content": "$$(2, 4)$$ "}] | ["A"] | null | $${y^2} = 18x \Rightarrow 2y{{dy} \over {dx}} = 18 \Rightarrow {{dy} \over {dx}} = {9 \over y}$$
<br><br>Given $${{dy} \over {dx}} = 2 \Rightarrow {9 \over 2} = 2 \Rightarrow y = {9 \over 2}$$
<br><br>Puting in $${y^2} = 18x \Rightarrow x = {9 \over 8}$$
<br><br>$$\therefore$$ Required point is $$\left( {{9 \over 8},{... | mcq | aieee-2004 | 4,779 |
eSua2dp9jsk4cv3A | maths | application-of-derivatives | rate-of-change-of-quantity | A spherical iron ball $$10$$ cm in radius is coated with a layer of ice of uniform thickness that melts at a rate of $$50$$ cm$$^3$$ /min. When the thickness of ice is $$5$$ cm, then the rate at which the thickness of ice decreases is
| [{"identifier": "A", "content": "$${1 \\over {36\\pi }}$$ cm/min"}, {"identifier": "B", "content": "$${1 \\over {18\\pi }}$$ cm/min"}, {"identifier": "C", "content": "$${1 \\over {54\\pi }}$$ cm/min"}, {"identifier": "D", "content": "$${5 \\over {6\\pi }}$$ cm/min"}] | ["B"] | null | Given that
<br><br>$${{dv} \over {dt}} = 50\,c{m^3}/\min $$
<br><br>$$ \Rightarrow {d \over {dt}}\left( {{4 \over 3}\pi {r^3}} \right) = 50$$
<br><br>$$ \Rightarrow 4\pi {r^2}{{dr} \over {dt}} = 50$$
<br><br>$$ \Rightarrow {{dr} \over {dt}} = {{50} \over {4\pi {{\left( {15} \right)}^2}}} = {1 \over {18\pi }}\,\,cm/\min... | mcq | aieee-2005 | 4,780 |
ljle24hk | maths | application-of-derivatives | rate-of-change-of-quantity | A lizard, at an initial distance of 21 cm behind an insect moves from rest with an acceleration of $2 \mathrm{~cm} / \mathrm{s}^2$ and pursues the insect which is crawling uniformly along a straight line at a speed of $20 \mathrm{~cm} / \mathrm{s}$. Then the lizard will catch the insect after : | [{"identifier": "A", "content": "20 s"}, {"identifier": "B", "content": "1 s"}, {"identifier": "C", "content": "21 s"}, {"identifier": "D", "content": "24 s"}] | ["C"] | null | <p>The motion of the lizard, which starts from rest and accelerates at a rate of $a = 2 \, \text{cm/s}^2$, can be described by the equation of motion :</p>
<p>$D_l = \frac{1}{2} a t^2$</p>
<p>where $D_l$ is the distance the lizard travels, $a$ is its acceleration, and $t$ is the time.</p>
<p>The insect, moving at a con... | mcq | aieee-2005 | 4,781 |
ZlHAOfk1vBZtgBRm | maths | application-of-derivatives | rate-of-change-of-quantity | A spherical balloon is filled with $$4500\pi $$ cubic meters of helium gas. If a leak in the balloon causes the gas to escape at the rate of $$72\pi $$ cubic meters per minute, then the rate (in meters per minute) at which the radius of the balloon decreases $$49$$ minutes after the leakage began is : | [{"identifier": "A", "content": "$${{9 \\over 7}}$$"}, {"identifier": "B", "content": "$${{7 \\over 9}}$$"}, {"identifier": "C", "content": "$${{2 \\over 9}}$$"}, {"identifier": "D", "content": "$${{9 \\over 2}}$$"}] | ["C"] | null | Volume of spherical balloon $$ = V = {4 \over 3}\pi {r^3}$$
<br><br>$$ \Rightarrow 4500\pi = {{4\pi {r^3}} \over 3}$$
<br><br>( as Given, volume $$ = 4500\pi {m^3}$$ )
<br><br>Differentiating both the sides, $$w.r.t't'$$ we get,
<br><br>$${{dV} \over {dt}} = 4\pi {r^2}\left( {{{dr} \over {dt}}} \right)$$
<br><br>Now... | mcq | aieee-2012 | 4,782 |
j6dKprjfuC9iNtm3Qv18hoxe66ijvwuw2ce | maths | application-of-derivatives | rate-of-change-of-quantity | A water tank has the shape of an inverted right
circular cone, whose semi-vertical angle is
$${\tan ^{ - 1}}\left( {{1 \over 2}} \right)$$. Water is poured into it at a constant
rate of 5 cubic meter per minute. The the rate
(in m/min.), at which the level of water is rising
at the instant when the depth of water in th... | [{"identifier": "A", "content": "$${1 \\over {15\\pi }}$$"}, {"identifier": "B", "content": "$${1 \\over {5\\pi }}$$"}, {"identifier": "C", "content": "$${1 \\over {10\\pi }}$$"}, {"identifier": "D", "content": "$${2 \\over \\pi }$$"}] | ["B"] | null | <picture><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265192/exam_images/spujcpaesz26bmcciwz1.webp"><source media="(max-width: 500px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264494/exam_images/cmldsgcfwufq2oz2bpff.webp"><source media="(max-wid... | mcq | jee-main-2019-online-9th-april-evening-slot | 4,783 |
kaw0SkAcvfOydLXkka3rsa0w2w9jx1zj510 | maths | application-of-derivatives | rate-of-change-of-quantity | A spherical iron ball of radius 10 cm is coated with a layer of ice of uniform thickness that melts at a rate of
50 cm<sup>3</sup>
/min. When the thickness of the ice is 5 cm, then the rate at which the thickness (in cm/min) of the ice
decreases, is : | [{"identifier": "A", "content": "$${5 \\over {6\\pi }}$$"}, {"identifier": "B", "content": "$${1 \\over {9\\pi }}$$"}, {"identifier": "C", "content": "$${1 \\over {36\\pi }}$$"}, {"identifier": "D", "content": "$${1 \\over {18\\pi }}$$"}] | ["D"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263695/exam_images/a99ifvlxnsbj5wrmjdqh.webp" style="max-width: 100%; height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2019 (Online) 10th April Evening Slot Mathematics - Application of Derivatives Question 140 English Explanatio... | mcq | jee-main-2019-online-10th-april-evening-slot | 4,784 |
D5MTcIGTKq9EGcY3D23rsa0w2w9jx65rjtw | maths | application-of-derivatives | rate-of-change-of-quantity | A 2 m ladder leans against a vertical wall. If the top of the ladder begins to slide down the wall at the rate
25 cm/sec, then the rate (in cm/sec.) at which the bottom of the ladder slides away from the wall on the
horizontal ground when the top of the ladder is 1 m above the ground is : | [{"identifier": "A", "content": "$${{25} \\over 3}$$"}, {"identifier": "B", "content": "25"}, {"identifier": "C", "content": "25$$\\sqrt 3 $$"}, {"identifier": "D", "content": "$${{25} \\over {\\sqrt 3 }}$$"}] | ["D"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266309/exam_images/axa7ph2cq7yfizyo5afc.webp" style="max-width: 100%; height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2019 (Online) 12th April Morning Slot Mathematics - Application of Derivatives Question 137 English Explanatio... | mcq | jee-main-2019-online-12th-april-morning-slot | 4,785 |
WdFaC9JydJYDDwFq5b7k9k2k5ioxa2s | maths | application-of-derivatives | rate-of-change-of-quantity | A spherical iron ball of 10 cm radius is
coated with a layer of ice of uniform
thickness the melts at a rate of 50 cm<sup>3</sup>/min.
When the thickness of ice is 5 cm, then the rate
(in cm/min.) at which of the thickness of ice
decreases, is : | [{"identifier": "A", "content": "$${1 \\over {18\\pi }}$$"}, {"identifier": "B", "content": "$${1 \\over {36\\pi }}$$"}, {"identifier": "C", "content": "$${1 \\over {54\\pi }}$$"}, {"identifier": "D", "content": "$${5 \\over {6\\pi }}$$"}] | ["A"] | null | Let the thickness = h cm
<br><br>Volume of ice = v = $${{4\pi } \over 3}\left( {{{\left( {10 + h} \right)}^3} - {{10}^3}} \right)$$
<br><br>$$ \Rightarrow $$ $${{dv} \over {dt}} = {{4\pi } \over 3}\left( {3{{\left( {10 + h} \right)}^2}} \right).{{dh} \over {dt}}$$
<br><br>Given $${{dv} \over {dt}} = $$ 50 cm<sup>3</sup... | mcq | jee-main-2020-online-9th-january-morning-slot | 4,786 |
R9XIGlr0aLF6KGHJEFjgy2xukf46a92x | maths | application-of-derivatives | rate-of-change-of-quantity | If the surface area of a cube is increasing at a
rate of 3.6 cm<sup>2</sup>/sec, retaining its shape; then
the rate of change of its volume (in cm<sup>3</sup>/sec),
when the length of a side of the cube is
10 cm, is : | [{"identifier": "A", "content": "9"}, {"identifier": "B", "content": "10"}, {"identifier": "C", "content": "18"}, {"identifier": "D", "content": "20"}] | ["A"] | null | For cube of side 'a'<br><br>A = 6a<sup>2</sup> and V = a<sup>3</sup><br><br>Given $${{dA} \over {dt}} = 3.6$$<br><br>$$ \Rightarrow $$$$ 12a{{da} \over {dt}}$$ = 3.6<br><br>$${{dV} \over {dt}} = 3{a^2}.{{da} \over {dt}} = 3{a^2}\left( {{{3.6} \over {12a}}} \right)$$<br><br>at a = 10<br><br>$${{dV} \over {dt}} = 9$$ | mcq | jee-main-2020-online-3rd-september-evening-slot | 4,787 |
1l59ke5cr | maths | application-of-derivatives | rate-of-change-of-quantity | <p>Water is being filled at the rate of 1 cm<sup>3</sup> / sec in a right circular conical vessel (vertex downwards) of height 35 cm and diameter 14 cm. When the height of the water level is 10 cm, the rate (in cm<sup>2</sup> / sec) at which the wet conical surface area of the vessel increases is</p> | [{"identifier": "A", "content": "5"}, {"identifier": "B", "content": "$${{\\sqrt {21} } \\over 5}$$"}, {"identifier": "C", "content": "$${{\\sqrt {26} } \\over 5}$$"}, {"identifier": "D", "content": "$${{\\sqrt {26} } \\over {10}}$$"}] | ["C"] | null | <p>$$\because$$ $$V = {1 \over 3}\pi {r^2}h$$ and $${r \over h} = {7 \over {35}} = {1 \over 5}$$</p>
<p>$$ \Rightarrow V = {1 \over {75}}\pi {h^3}$$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l5th5qtk/a97fb8bf-1c8b-4d95-accc-6e78d7642980/88a53880-0818-11ed-98aa-f9038709a939/file-1l5th5qtl.p... | mcq | jee-main-2022-online-25th-june-evening-shift | 4,788 |
1l5c0t4qw | maths | application-of-derivatives | rate-of-change-of-quantity | <p>The surface area of a balloon of spherical shape being inflated, increases at a constant rate. If initially, the radius of balloon is 3 units and after 5 seconds, it becomes 7 units, then its radius after 9 seconds is :</p> | [{"identifier": "A", "content": "9"}, {"identifier": "B", "content": "10"}, {"identifier": "C", "content": "11"}, {"identifier": "D", "content": "12"}] | ["A"] | null | <p>We know,</p>
<p>Surface area of balloon (s) = 4$$\pi$$r<sup>2</sup></p>
<p>$$\therefore$$ $${{ds} \over {dt}} = {d \over {dt}}(4\pi {r^2})$$</p>
<p>$$ \Rightarrow {{ds} \over {dt}} = 4\pi (2r) \times {{dr} \over {dt}}$$</p>
<p>$$ \Rightarrow {{ds} \over {dt}} = 8\pi r \times {{dr} \over {dt}}$$</p>
<p>Given that, su... | mcq | jee-main-2022-online-24th-june-morning-shift | 4,789 |
1l5w1f9pv | maths | application-of-derivatives | rate-of-change-of-quantity | <p>A hostel has 100 students. On a certain day (consider it day zero) it was found that two students are infected with some virus. Assume that the rate at which the virus spreads is directly proportional to the product of the number of infected students and the number of non-infected students. If the number of infected... | [] | null | 90 | <p>Total students = 100</p>
<p>At t = 0 (zero day), infected student = 2</p>
<p>Let at t = t day infected student = x</p>
<p>$$\therefore$$ At t = t day non infected student = (100 $$-$$ x)</p>
<p>Rate of infection $$ = {{dx} \over {dt}}$$</p>
<p>Given, $${{dx} \over {dt}} \propto x(100 - x)$$</p>
<p>$$ \Rightarrow \i... | integer | jee-main-2022-online-30th-june-morning-shift | 4,790 |
1l6kljqrp | maths | application-of-derivatives | rate-of-change-of-quantity | <p>A water tank has the shape of a right circular cone with axis vertical and vertex downwards. Its semi-vertical angle is $$\tan ^{-1} \frac{3}{4}$$. Water is poured in it at a constant rate of 6 cubic meter per hour. The rate (in square meter per hour), at which the wet curved surface area of the tank is increasing, ... | [] | null | 5 | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l7qaiwsv/317e788b-9fe5-4c30-acd5-40a1d96c242d/abb0b900-2df0-11ed-a744-1fb8f3709cfa/file-1l7qaiwsw.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l7qaiwsv/317e788b-9fe5-4c30-acd5-40a1d96c242d/abb0b900-2df0-11ed-a744-1fb8f3709cfa... | integer | jee-main-2022-online-27th-july-evening-shift | 4,791 |
1lh2zk9lq | maths | application-of-derivatives | rate-of-change-of-quantity | <p> The number of points, where the curve $$y=x^{5}-20 x^{3}+50 x+2$$ crosses the $$\mathrm{x}$$-axis, is ____________.</p> | [] | null | 5 | Given equation of curve
<br><br>$$
\begin{aligned}
& y=x^5-20 x^3+50 x+2 \\\\
& \Rightarrow \frac{d y}{d x}=5 x^4-60 x^2+50
\end{aligned}
$$
<br><br>On putting $\frac{d y}{d x}=0$
<br><br>$$
\begin{array}{ll}
\Rightarrow & 5\left(x^4-12 x^2+10\right)=0 \\\\
\Rightarrow & x^2=\frac{12 \pm \sqrt{144-40}}{... | integer | jee-main-2023-online-6th-april-evening-shift | 4,792 |
lv7v4o2e | maths | application-of-derivatives | rate-of-change-of-quantity | <p>Let $$f(x)=x^5+2 x^3+3 x+1, x \in \mathbf{R}$$, and $$g(x)$$ be a function such that $$g(f(x))=x$$ for all $$x \in \mathbf{R}$$. Then $$\frac{g(7)}{g^{\prime}(7)}$$ is equal to :</p> | [{"identifier": "A", "content": "42"}, {"identifier": "B", "content": "7"}, {"identifier": "C", "content": "1"}, {"identifier": "D", "content": "14"}] | ["D"] | null | <p>$$\begin{aligned}
& f(x)=x^5+2 x^3+3 x+1 \\
& g(f(x))=x . \quad \Rightarrow g^{\prime}(f(x)) f^{\prime}(x)=1 \\
\end{aligned}$$</p>
<p>$$\begin{aligned}
&\begin{aligned}
& \text { Now } \frac{g(7)}{g^{\prime}(7)} \\
& g(7) \Rightarrow f(x)=7 \\
& x^5+2 x^3+3 x+1=7 \\
& \Rightarrow x\left(x^4+2 x^2+3\right)=0 \\
& \R... | mcq | jee-main-2024-online-5th-april-morning-shift | 4,794 |
WzB2UMLEAvI0j8Sz | maths | application-of-derivatives | tangent-and-normal | A function $$y=f(x)$$ has a second order derivative $$f''\left( x \right) = 6\left( {x - 1} \right).$$ If its graph passes through the point $$(2, 1)$$ and at that point the tangent to the graph is $$y = 3x - 5$$, then the function is : | [{"identifier": "A", "content": "$${\\left( {x + 1} \\right)^2}$$ "}, {"identifier": "B", "content": "$${\\left( {x - 1} \\right)^3}$$"}, {"identifier": "C", "content": "$${\\left( {x + 1} \\right)^3}$$"}, {"identifier": "D", "content": "$${\\left( {x - 1} \\right)^2}$$"}] | ["B"] | null | $$f''\left( x \right) = 6\left( {x - 1} \right).$$ Inegrating,
<br><br>we get $$f'\left( x \right) = 3{x^2} - 6x + c$$
<br><br>Slope at $$\left( {2,1} \right) = f'\left( 2 \right) = c = 3$$
<br><br>$$\left[ {\,\,} \right.$$ As slope of tangent at $$(2, 1)$$ is $$3$$ $$\left. {\,\,} \right]$$
<br><br>$$\therefore$$ $$... | mcq | aieee-2004 | 4,795 |
7xmtfPK7UFAAvV0N | maths | application-of-derivatives | tangent-and-normal | The normal to the curve x = a(1 + cos $$\theta $$), $$y = a\sin \theta $$ at $$'\theta '$$ always passes through the fixed point | [{"identifier": "A", "content": "$$(a, a)$$ "}, {"identifier": "B", "content": "$$(0, a)$$ "}, {"identifier": "C", "content": "$$(0, 0)$$ "}, {"identifier": "D", "content": "$$(a, 0)$$ "}] | ["D"] | null | $${{dx} \over {d\theta }} = - a\sin \theta $$ and $${{dy} \over {d\theta }} = a\cos \theta $$
<br><br>$$\therefore$$ $${{dy} \over {dx}} = - \cot \theta .$$
<br><br>$$\therefore$$ The slope of the normal at $$\theta $$ = $$ - {1 \over { - \cot \theta }}$$$$= \tan \theta $$
<br><br>$$\therefore$$ The equation of the ... | mcq | aieee-2004 | 4,796 |
xDpHAzVDO3wJz2Tq | maths | application-of-derivatives | tangent-and-normal | The normal to the curve
<br/>$$x = a\left( {\cos \theta + \theta \sin \theta } \right),y = a\left( {\sin \theta - \theta \cos \theta } \right)$$ at any point
<br/>$$\theta\, '$$ is such that
| [{"identifier": "A", "content": "it passes through the origin "}, {"identifier": "B", "content": "it makes an angle $${\\pi \\over 2} + \\theta $$ with the $$x$$-axis "}, {"identifier": "C", "content": "it passes through $$\\left( {a{\\pi \\over 2}, - a} \\right)$$ "}, {"identifier": "D", "content": "it is at a const... | ["D"] | null | $$x = a\left( {\cos \theta + \theta \sin \theta } \right)$$
<br><br>$$ \Rightarrow {{dx} \over {d\theta }} = a\left( { - \sin \theta + \sin \theta + \theta \cos \theta } \right)$$
<br><br>$$ \Rightarrow {{dx} \over {d\theta }} = a\theta \cos \theta \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 1 \right)$$
<br><br>$$y = ... | mcq | aieee-2005 | 4,797 |
DLqakneawRUrbWd3 | maths | application-of-derivatives | tangent-and-normal | Angle between the tangents to the curve $$y = {x^2} - 5x + 6$$ at the points $$(2,0)$$ and $$(3,0)$$ is | [{"identifier": "A", "content": "$$\\pi $$ "}, {"identifier": "B", "content": "$${\\pi \\over 2}$$ "}, {"identifier": "C", "content": "$${\\pi \\over 6}$$"}, {"identifier": "D", "content": "$${\\pi \\over 4}$$"}] | ["B"] | null | $${{dy} \over {dx}} = 2x - 5$$
<br><br>$$\therefore$$ $${m_1} = {\left( {2x - 5} \right)_{\left( {2,0} \right)}} = - 1,$$
<br><br> $${m_2} = {\left( {2x - 5} \right)_{\left( {3,0} \right)}} = 1 \Rightarrow {m_1}{m_2} = - 1$$
<br><br>i.e. the tangents are perpendicular to each other. | mcq | aieee-2006 | 4,798 |
TN45biXlr0c9O695 | maths | application-of-derivatives | tangent-and-normal | The equation of the tangent to the curve $$y = x + {4 \over {{x^2}}}$$, that
<br/>is parallel to the $$x$$-axis, is | [{"identifier": "A", "content": "$$y=1$$ "}, {"identifier": "B", "content": "$$y=2$$ "}, {"identifier": "C", "content": "$$y=3$$ "}, {"identifier": "D", "content": "$$y=0$$ "}] | ["C"] | null | Since tangent is parallel to $$x$$-axis,
<br><br>$$\therefore$$ $${{dy} \over {dx}} = 0 \Rightarrow 1 - {8 \over {{x^3}}} = 0 \Rightarrow x = 2 \Rightarrow y = 3$$
<br><br>Equation of tangent is $$y - 3 = 0\left( {x - 2} \right) \Rightarrow y = 3$$ | mcq | aieee-2010 | 4,799 |
da6RXvhPYMZWxEIr | maths | application-of-derivatives | tangent-and-normal | The shortest distance between line $$y-x=1$$ and curve $$x = {y^2}$$ is | [{"identifier": "A", "content": "$${{3\\sqrt 2 } \\over 8}$$ "}, {"identifier": "B", "content": "$${8 \\over {3\\sqrt 2 }}$$ "}, {"identifier": "C", "content": "$${4 \\over {\\sqrt 3 }}$$ "}, {"identifier": "D", "content": "$${{\\sqrt 3 } \\over 4}$$ "}] | ["A"] | null | Shortest distance between two curve occurred along -
<br><br>the common normal
<br><br>Slope of normal to $${y^2} = x$$ at point
<br><br>$$P\left( {{t^2},t} \right)$$ is $$-2t$$ and
<br><br>slope of line $$y - x = 1$$ is $$1.$$
<br><br>As they are perpendicular to each other
<br><br>$$\therefore$$ $$\left( { - 2t} ... | mcq | aieee-2011 | 4,800 |
i0kX6KoVoB3amGZz | maths | application-of-derivatives | tangent-and-normal | The intercepts on $$x$$-axis made by tangents to the curve,
<br/>$$y = \int\limits_0^x {\left| t \right|dt,x \in R,} $$ which are parallel to the line $$y=2x$$, are equal to : | [{"identifier": "A", "content": "$$ \\pm 1$$ "}, {"identifier": "B", "content": "$$ \\pm 2$$"}, {"identifier": "C", "content": "$$ \\pm 3$$"}, {"identifier": "D", "content": "$$ \\pm 4$$"}] | ["A"] | null | Since, $$y = \int\limits_0^x {\left| t \right|} dt,x \in R$$
<br><br>therefore $${{dy} \over {dx}} = \left| x \right|$$
<br><br>But from $$y = 2x,{{dy} \over {dx}} = 2$$
<br><br>$$ \Rightarrow \left| x \right| = 2 \Rightarrow x = \pm 2$$
<br><br>Points $$y = \int\limits_0^{ \pm 2} {\left| t \right|dt} = \pm 2$$
<br>... | mcq | jee-main-2013-offline | 4,801 |
nbnpebQ2nYcrhYcK | maths | application-of-derivatives | tangent-and-normal | The normal to the curve, $${x^2} + 2xy - 3{y^2} = 0$$, at $$(1,1)$$ | [{"identifier": "A", "content": "meets the curve again in the third quadrant. "}, {"identifier": "B", "content": "meets the curve again in the fourth quadrant. "}, {"identifier": "C", "content": "does not meet the curve again."}, {"identifier": "D", "content": "meets the curve again in the second quadrant."}] | ["B"] | null | Given curve is
<br><br>$${x^2} + 2xy - 3{y^2} = 0$$
<br><br>Difference $$w.r.t.x,$$
<br><br>$$2x + 2x{{dy} \over {dx}} + 2y - 6y{{dy} \over {dx}} = 0$$
<br><br>$${\left( {{{dy} \over {dx}}} \right)_{\left( {1,1} \right)}} = 1$$
<br><br>Equation of normal at $$(1,1)$$ is
<br><br>$$y=2-x$$
<br><br>Solving eq. $$(1)$... | mcq | jee-main-2015-offline | 4,802 |
k0fZVE4svRgWgQQCzYBRu | maths | application-of-derivatives | tangent-and-normal | If the tangent at a point P, with parameter t, on the curve x = 4t<sup>2</sup> + 3, y = 8t<sup>3</sup>β1, <i>t</i> $$ \in $$ <b>R</b>, meets the curve again at a point Q, then the coordinates of Q are : | [{"identifier": "A", "content": "(t<sup>2</sup> + 3, \u2212 t<sup>3</sup> \u22121)\n"}, {"identifier": "B", "content": "(4t<sup>2</sup> + 3, \u2212 8t<sup>3</sup> \u22121)\n"}, {"identifier": "C", "content": "(t<sup>2</sup> + 3, t<sup>3</sup> \u22121)\n"}, {"identifier": "D", "content": "(16t<sup>2</sup> + 3, \u2212 64... | ["A"] | null | Given, x = 4t<sup>2</sup> + 3 and y = 8t<sup>3</sup> $$-$$ 1
<br><br>$$ \therefore $$ P $$ \equiv $$ (4t<sup>2</sup> + 3, 8t<sup>3</sup> $$-$$ 1)
<br><br>$${{dx} \over {dt}} = 8t$$ and $${{dy} \over {dt}}$$ $$=$$ 24t<sup>2</sup>
<br><br>Slope of tangent at
<br><br>P $$=$$ $${{dy} \over {dx}}$$ $$=$$ $... | mcq | jee-main-2016-online-9th-april-morning-slot | 4,803 |
idGZjiTMQyZ5SsmMHXupk | maths | application-of-derivatives | tangent-and-normal | Let C be a curve given by y(x) = 1 + $$\sqrt {4x - 3} ,x > {3 \over 4}.$$ If P is a point
on C, such that the tangent at P has slope $${2 \over 3}$$, then a point through which the normal at P passes, is : | [{"identifier": "A", "content": "(2, 3)"}, {"identifier": "B", "content": "(4, $$-$$3)"}, {"identifier": "C", "content": "(1, 7) "}, {"identifier": "D", "content": "(3, $$-$$ 4), "}] | ["C"] | null | Given,
<br><br>y = 1 + $$\sqrt {4x - 3} $$
<br><br>$$ \therefore $$ $${{dy} \over {dx}}$$ = $${1 \over {2\sqrt {4x - 3} }} \times 4 = {2 \over 3}$$
<br><br>$$ \Rightarrow $$ 4x $$-$$ 3 = 9
<br><br>$$ \Rightarrow $$ x = 3
<br><br>$$ \therefore $$ y = 1 ... | mcq | jee-main-2016-online-10th-april-morning-slot | 4,804 |
v0O67CCyo8mfgAHJ | maths | application-of-derivatives | tangent-and-normal | Consider :
<br/>f $$\left( x \right) = {\tan ^{ - 1}}\left( {\sqrt {{{1 + \sin x} \over {1 - \sin x}}} } \right),x \in \left( {0,{\pi \over 2}} \right).$$
<p>A normal to $$y = $$ f$$\left( x \right)$$ at $$x = {\pi \over 6}$$ also passes through the point:</p> | [{"identifier": "A", "content": "$$\\left( {{\\pi \\over 6},0} \\right)$$ "}, {"identifier": "B", "content": "$$\\left( {{\\pi \\over 4},0} \\right)$$"}, {"identifier": "C", "content": "$$(0,0)$$ "}, {"identifier": "D", "content": "$$\\left( {0,{{2\\pi } \\over 3}} \\right)$$ "}] | ["D"] | null | $$f\left( x \right) = {\tan ^{ - 1}}\left( {\sqrt {{{1 + \sin \,x} \over {1 - \sin x}}} } \right)$$
<br><br>$$ = {\tan ^{ - 1}}\left( {\sqrt {{{{{\left( {\sin {x \over 2} + \cos {x \over 2}} \right)}^2}} \over {{{\left( {\sin {x \over x} - \cos {x \over 2}} \right)}^2}}}} } \right)$$
<br><br>$$ = {\tan ^{ - 1}}\left( ... | mcq | jee-main-2016-offline | 4,805 |
2Cp4wkEasxlGmlzJJmg0m | maths | application-of-derivatives | tangent-and-normal | A tangent to the curve, y = f(x) at P(x, y) meets x-axis at A and y-axis at B. If AP : BP = 1 : 3 and f(1) = 1, then the curve also passes through the point : | [{"identifier": "A", "content": "$$\\left( {{1 \\over 3},24} \\right)$$ "}, {"identifier": "B", "content": "$$\\left( {{1 \\over 2},4} \\right)$$ "}, {"identifier": "C", "content": "$$\\left( {2,{1 \\over 8}} \\right)$$ "}, {"identifier": "D", "content": "$$\\left( {3,{1 \\over 28}} \\right)$$"}] | ["C"] | null | <p>We have</p>
<p>$${{(y - {y_2})} \over {(x - {x_1})}} = f'({x_1})$$</p>
<p>$$ \Rightarrow y - {y_1} = f'({x_1})(x - {x_1})$$</p>
<p>$$\bullet$$ When y = 0: $${{ - {y_1}} \over {f'({x_1})}} = x - {x_1}$$</p>
<p>$$ \Rightarrow x = {x_1} - {{{y_1}} \over {f'({x_1})}}$$</p>
<p>Therefore, point A is $$A\left( {{x_1} - {{{... | mcq | jee-main-2017-online-9th-april-morning-slot | 4,808 |
boajRKD8w1QSY5sgpYNZA | maths | application-of-derivatives | tangent-and-normal | If $$\beta $$ is one of the angles between the normals to the ellipse, x<sup>2</sup> + 3y<sup>2</sup> = 9 at the points (3 cos $$\theta $$, $$\sqrt 3 \sin \theta $$) and ($$-$$ 3 sin $$\theta $$, $$\sqrt 3 \,\cos \theta $$); $$\theta \in \left( {0,{\pi \over 2}} \right);$$ then $${{2\,\cot \beta } \over {\sin 2\thet... | [{"identifier": "A", "content": "$${2 \\over {\\sqrt 3 }}$$"}, {"identifier": "B", "content": "$${1 \\over {\\sqrt 3 }}$$"}, {"identifier": "C", "content": "$$\\sqrt 2 $$"}, {"identifier": "D", "content": "$${{\\sqrt 3 } \\over 4}$$ "}] | ["A"] | null | Since, x<sup>2</sup> + 3y<sup>2</sup> = 9
<br><br>$$ \Rightarrow $$ 2x + 6y $${{dy} \over {dx}}$$ = 0
<br><br>$$ \Rightarrow $$ $${{dy} \over {dx}}$$ = $${{ - x} \over {3y}}$$
<br><br>Slope of normal is $$-$$ $${{dx} \over {dy}}$$ = $${{3y} \over x}$$
<br><br>$$ \Rightarrow $$ $${\left( { - {{dx} \over {dy}}} \right)_... | mcq | jee-main-2018-online-15th-april-morning-slot | 4,810 |
8ffCZxoD0TRKsnKUmshBK | maths | application-of-derivatives | tangent-and-normal | The tangent to the curve y = x<sup>2</sup> β 5x + 5, parallel to the line 2y = 4x + 1, also passes through the point : | [{"identifier": "A", "content": "$$\\left\\{ {{1 \\over 4},{7 \\over 2}} \\right\\}$$"}, {"identifier": "B", "content": "$$\\left( { - {1 \\over 8},7} \\right)$$"}, {"identifier": "C", "content": "$$\\left( {{7 \\over 2},{1 \\over 4}} \\right)$$"}, {"identifier": "D", "content": "$$\\left( {{1 \\over 8}, - 7} \\right)$... | ["D"] | null | y = x<sup>2</sup> $$-$$ 5x + 5
<br><br>$${{dy} \over {dx}} = 2x - 5 = 2 \Rightarrow x = {7 \over 2}$$
<br><br>at x = $${7 \over 2}$$, y = $${{ - 1} \over 4}$$
<br><br>Equation of tangent at
<br><br>$$\left( {{7 \over 2},{{ - 1} \over 4}} \right)$$ is 2x $$-$$ y $$-$$ $${{29} \over 4}$$ = 0
<br><br>Now ... | mcq | jee-main-2019-online-12th-january-evening-slot | 4,811 |
o4KkVYuBGGgiJsxAqi3rsa0w2w9jx2evlbl | maths | application-of-derivatives | tangent-and-normal | If the tangent to the curve $$y = {x \over {{x^2} - 3}}$$
, $$x \in \rho ,\left( {x \ne \pm \sqrt 3 } \right)$$, at a point ($$\alpha $$, $$\beta $$) $$ \ne $$ (0, 0) on it is parallel to the line
2x + 6y β 11 = 0, then : | [{"identifier": "A", "content": "| 6$$\\alpha $$ + 2$$\\beta $$ | = 9"}, {"identifier": "B", "content": "| 2$$\\alpha $$ + 6$$\\beta $$ | = 11"}, {"identifier": "C", "content": "| 2$$\\alpha $$ + 6$$\\beta $$ | = 19"}, {"identifier": "D", "content": "| 6$$\\alpha $$ + 2$$\\beta $$ | = 19"}] | ["D"] | null | $${{dy} \over {dx}}{|_{(\alpha ,\beta )}} = {{ - {\alpha ^2} - 3} \over {{{\left( {{\alpha ^2} - 3} \right)}^2}}}$$<br><br>
Given that $${{ - {\alpha ^2} - 3} \over {{{\left( {{\alpha ^2} - 3} \right)}^2}}} = - {1 \over 3}$$<br><br>
$$ \Rightarrow $$ $$\alpha $$ = 0, $$ \pm $$ 3 ($$\alpha \ne $$ 0) | mcq | jee-main-2019-online-10th-april-evening-slot | 4,812 |
FFjlUVpcYahbjG67Tj18hoxe66ijvwp8sn0 | maths | application-of-derivatives | tangent-and-normal | If the tangent to the curve, y = x<sup>3</sup> + ax β b at
the point (1, β5) is perpendicular to the line,
βx + y + 4 = 0, then which one of the following
points lies on the curve ? | [{"identifier": "A", "content": "(2, \u20132)"}, {"identifier": "B", "content": "(2, \u20131)"}, {"identifier": "C", "content": "(\u20132, 2)"}, {"identifier": "D", "content": "(\u20132, 1)"}] | ["A"] | null | Slope of the tangent to the curve y = x<sup>3</sup> + ax β b at point (1, β5)
<br><br>m<sub>1</sub> = $${\left. {{{dy} \over {dx}}} \right|_{\left( {1, - 5} \right)}}$$ = 3x<sup>2</sup> + a = 3 + a
<br><br>Slope of the line βx + y + 4 = 0,
<br><br>m<sub>2</sub> = 1
<br><br>As line and tangent to the curve are perpendic... | mcq | jee-main-2019-online-9th-april-morning-slot | 4,813 |
H0eqMDxBBYA939FXRvxgg | maths | application-of-derivatives | tangent-and-normal | Let S be the set of all values of x for which the
tangent to the curve
<br/>y = Ζ(x) = x<sup>3</sup> β x<sup>2</sup> β 2x at
(x, y) is parallel to the line segment joining the
points (1, Ζ(1)) and (β1, Ζ(β1)), then S is equal
to : | [{"identifier": "A", "content": "$$\\left\\{ { {1 \\over 3}, - 1} \\right\\}$$"}, {"identifier": "B", "content": "$$\\left\\{ { - {1 \\over 3}, 1} \\right\\}$$"}, {"identifier": "C", "content": "$$\\left\\{ { - {1 \\over 3}, - 1} \\right\\}$$"}, {"identifier": "D", "content": "$$\\left\\{ { {1 \\over 3}, 1} \\right... | ["B"] | null | Given Ζ(x) = x<sup>3</sup> β x<sup>2</sup> β 2x
<br><br>$$ \therefore $$ Ζ(1) = 1 β 1 β 2 = - 2
<br><br>and Ζ(-1) = -1 β 1 + 2 = 0
<br><br>So point A(1, Ζ(1)) = (1, -2)
<br><br>and point B(β1, Ζ(β1)) = (-1, 0)
<br><br>Slope of tangent at point (x, y) to the curve
<br><br> y = Ζ(x) = x<sup>3</sup> β x<sup>2</sup> β 2x ... | mcq | jee-main-2019-online-9th-april-morning-slot | 4,814 |
mV9YspGfsF6PJhUiFQYou | maths | application-of-derivatives | tangent-and-normal | Given that the slope of the tangent to a curve y
= y(x) at any point (x, y) is
$$2y \over x^2$$. If the curve passes through the centre of the circle x<sup>2</sup> + y<sup>2</sup> β 2x β 2y = 0, then its equation is : | [{"identifier": "A", "content": "x log<sub>e</sub>|y| = 2(x \u2013 1)"}, {"identifier": "B", "content": "x<sup>2</sup> log<sub>e</sub>|y| = \u20132(x \u2013 1)"}, {"identifier": "C", "content": "x log<sub>e</sub>|y| = x \u2013 1"}, {"identifier": "D", "content": "x log<sub>e</sub>|y| = \u20132(x \u2013 1)"}] | ["A"] | null | Slope, $${{dy} \over {dx}}$$ = $$2y \over x^2$$
<br><br>$$ \Rightarrow $$ $$\int {{{dy} \over y}} = \int {2{{dx} \over {{x^2}}}} $$
<br><br>$$ \Rightarrow $$ $${\log _e}|y| = - {2 \over x} + C$$ ....... (1)
<br><br>Center of the circle x<sup>2</sup> + y<sup>2</sup> β 2x β 2y = 0 is (1, 1)
<br><br>Equation (1) passes ... | mcq | jee-main-2019-online-8th-april-evening-slot | 4,815 |
KpnzSmbznFMg3YBKF4Sio | maths | application-of-derivatives | tangent-and-normal | The tangent to the curve, y = xe<sup>x<sup>2</sup></sup> passing through the point (1, e) also passes through the point | [{"identifier": "A", "content": "$$\\left( {{4 \\over 3},2e} \\right)$$"}, {"identifier": "B", "content": "(3, 6e)"}, {"identifier": "C", "content": "(2, 3e)"}, {"identifier": "D", "content": "$$\\left( {{5 \\over 3},2e} \\right)$$"}] | ["A"] | null | y = xe<sup>x<sup>2</sup></sup>
<br><br>$${\left. {{{dy} \over {dx}}} \right|_{(1,e)}}{\left. { = \left( {e.e{x^2}.2x + {e^{{x^2}}}} \right)} \right|_{(1,e)}}$$
<br><br>$$ = 2 \cdot e + e = 3e$$
<br><br>T : y $$-$$ e = 3e (x $$-$$ 1)
<br><br>y = 3ex $$-$$ 3e + e
<br><br>y = $$\left( {3e} \right)x - 2e$$
<br><br>$$\left(... | mcq | jee-main-2019-online-10th-january-evening-slot | 4,816 |
LUrxBmMKm8jszpS1sgjgy2xukg38q7pr | maths | application-of-derivatives | tangent-and-normal | If the tangent to the curve, y = f (x) = xlog<sub>e</sub> x,
<br/>(x > 0) at a point (c, f(c)) is parallel to the
line-segment<br/> joining the points (1, 0) and
(e, e), then c is equal to : | [{"identifier": "A", "content": "$${{e - 1} \\over e}$$"}, {"identifier": "B", "content": "$${e^{\\left( {{1 \\over {1 - e}}} \\right)}}$$"}, {"identifier": "C", "content": "$${e^{\\left( {{1 \\over {e - 1}}} \\right)}}$$"}, {"identifier": "D", "content": "$${1 \\over {e - 1}}$$"}] | ["C"] | null | y = f (x) = xlog<sub>e</sub> x
<br><br>$$ \Rightarrow $$ $${{dy} \over {dx}} = $$ 1 + log<sub>e</sub> x
<br><br>$$ \Rightarrow $$ $${\left. {{{dy} \over {dx}}} \right|_{\left( {c,f\left( c \right)} \right)}}$$ = 1 + log<sub>e</sub> e = m<sub>1</sub>
<br><br>This tangent parallel to the
line-segment<br> joining the poin... | mcq | jee-main-2020-online-6th-september-evening-slot | 4,817 |
EkDkOZ7NbLhXskYcSejgy2xukfqfuuc9 | maths | application-of-derivatives | tangent-and-normal | If the lines x + y = a and x β y = b touch the
<br/>curve y = x<sup>2</sup>
β 3x + 2 at the points where the
curve intersects the x-axis, then $${a \over b}$$ is equal
to _______. | [] | null | 0.50 | y = x<sup>2</sup>
β 3x + 2
<br><br>$$ \Rightarrow $$ y = (x β 1)(x β 2)
<br><br>At x-axis y = 0
<br>$$ \Rightarrow $$ x = 1, 2
<br><br>So this curve intersects the x-axis
at A(1, 0) and B(2, 0).
<br><br>$${{dy} \over {dx}} = 2x - 3$$
<br><br>$${\left( {{{dy} \over {dx}}} \right)_{x = 1}} = - 1$$ and $${\left( {{{dy}... | integer | jee-main-2020-online-5th-september-evening-slot | 4,818 |
pxlhjCCkd0Ohos8TCwjgy2xukfqb1rdv | maths | application-of-derivatives | tangent-and-normal | Which of the following points lies on the
tangent to the curve
<br/><br>x<sup>4</sup>e<sup>y</sup> + 2$$\sqrt {y + 1} $$ = 3 at the
point (1, 0)?</br> | [{"identifier": "A", "content": "(2, 2)"}, {"identifier": "B", "content": "(\u20132, 4)"}, {"identifier": "C", "content": "(2, 6)"}, {"identifier": "D", "content": "(\u20132, 6)"}] | ["D"] | null | x<sup>4</sup>e<sup>y</sup> + 2$$\sqrt {y + 1} $$ = 3
<br><br>Differentiating w.r.t. x, we get
<br><br>x<sup>4</sup>e<sup>y</sup>y' + e<sup>y</sup>4x<sup>3</sup> + $${{2y'} \over {2\sqrt {y + 1} }}$$ = 0
<br><br>At P(1, 0)
<br><br>$${y{'_P}}$$ + 4 + $${y{'_P}}$$ = 0
<br><br>$$ \Rightarrow $$ $${y{'_P}}$$ = -2
<br><br>Ta... | mcq | jee-main-2020-online-5th-september-evening-slot | 4,819 |
wcSiqoBtSVPf2VGnBMjgy2xukewra473 | maths | application-of-derivatives | tangent-and-normal | Let P(h, k) be a point on the curve
<br/>y = x<sup>2</sup>
+ 7x + 2, nearest to the line, y = 3x β 3.
<br/>Then the equation of the normal to the curve at
P is : | [{"identifier": "A", "content": "x \u2013 3y \u2013 11 = 0"}, {"identifier": "B", "content": "x \u2013 3y + 22 = 0"}, {"identifier": "C", "content": "x + 3y \u2013 62 = 0"}, {"identifier": "D", "content": "x + 3y + 26 = 0"}] | ["D"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266899/exam_images/knvyxnlyn6etgfelas4i.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2020 (Online) 2nd September Morning Slot Mathematics - Application of Derivatives Question 126 English Explanati... | mcq | jee-main-2020-online-2nd-september-morning-slot | 4,821 |
lwjOzJrH6jcRF5DUMS7k9k2k5hii08h | maths | application-of-derivatives | tangent-and-normal | The length of the perpendicular from the origin,
on the normal to the curve,<br/> x<sup>2</sup> + 2xy β 3y<sup>2</sup> = 0
at the point (2,2) is | [{"identifier": "A", "content": "$$\\sqrt 2 $$"}, {"identifier": "B", "content": "$$4\\sqrt 2 $$"}, {"identifier": "C", "content": "2"}, {"identifier": "D", "content": "$$2\\sqrt 2 $$"}] | ["D"] | null | x<sup>2</sup> + 2xy β 3y<sup>2</sup> = 0
<br><br>Differentiate the curve
<br><br> 2x + 2y + 2xy' β 6yy' = 0
<br><br>$$ \Rightarrow $$ $$ \Rightarrow $$ x + y + xy' β 3yy' = 0
<br><br>$$ \Rightarrow $$ y'(x β 3y) = β (x + y)
<br><br>$$ \Rightarrow $$ y' = $${{x + y} \over {3y - x}}$$
<br><br>Slope of normal = $$ - {{dx}... | mcq | jee-main-2020-online-8th-january-evening-slot | 4,822 |
NLUumf2kXLiqyI6O3a7k9k2k5h0679u | maths | application-of-derivatives | tangent-and-normal | Let the normal at a point P on the curve
<br/>y<sup>2</sup> β 3x<sup>2</sup> + y + 10 = 0 intersect the y-axis at $$\left( {0,{3 \over 2}} \right)$$
. <br/>If m is the slope of the tangent at P to
the curve, then |m| is equal to | [] | null | 4 | Given curve : y<sup>2</sup>
β 3x<sup>2</sup>
+ y + 10 = 0
<br><br>$$ \Rightarrow $$ 2y$${{dy} \over {dx}}$$ - 6x + $${{dy} \over {dx}}$$ = 0
<br><br>$$ \Rightarrow $$ $${{dy} \over {dx}}$$ = $${{6x} \over {2y + 1}}$$
<br><br>Let P be (x<sub>1</sub>, y<sub>1</sub>)
<br><br>Slope of tangent at P = $${{6{x_1}} \over {2{y... | integer | jee-main-2020-online-8th-january-morning-slot | 4,823 |
dDfWIivwafjQLW2Uenjgy2xukezarnpo | maths | application-of-derivatives | tangent-and-normal | The equation of the normal to the curve
<br/>y = (1+x)<sup>2y</sup> + cos
<sup>2</sup>(sin<sup>β1</sup>x) at x = 0 is : | [{"identifier": "A", "content": "y = 4x + 2"}, {"identifier": "B", "content": "x + 4y = 8"}, {"identifier": "C", "content": "y + 4x = 2"}, {"identifier": "D", "content": "2y + x = 4"}] | ["B"] | null | Given equation of curve
<br><br>y = (1+x)<sup>2y</sup> + cos
<sup>2</sup>(sin<sup>β1</sup>x)
<br><br>at x = 0
<br><br>$$ \Rightarrow $$ y = (1 + 0)<sup>2y</sup> + cos<sup>2</sup>(sin<sup>β1</sup>0)
<br><br>$$ \Rightarrow $$ y = 1 + 1
<br><br>$$ \Rightarrow $$ y = 2
<br><br>So we have to find the normal at (0, 2)
<br><b... | mcq | jee-main-2020-online-2nd-september-evening-slot | 4,824 |
CkjU7HfpXnSIHE1AlR1klrgv24w | maths | application-of-derivatives | tangent-and-normal | If the tangent to the curve y = x<sup>3</sup> at the point P(t, t<sup>3</sup>) meets the curve again at Q, then the
ordinate of the point which divides PQ internally in the ratio 1 : 2 is : | [{"identifier": "A", "content": "0"}, {"identifier": "B", "content": "2t<sup>3</sup>"}, {"identifier": "C", "content": "-2t<sup>3</sup>"}, {"identifier": "D", "content": "-t<sup>3</sup>"}] | ["C"] | null | Given $$P(t,{t^3})$$<br><br>Let $$Q = ({t_1},t_1^3)$$<br><br>Slope of tangent at point p,<br><br>$${{dy} \over {dx}} = 3{x^2}$$<br><br>$$ \Rightarrow {\left. {{{dy} \over {dx}}} \right|_{(t,{t^3})}} = 3{t^2}$$<br><br>This slope is same as slope of line PQ.<br><br>Slope of $$PQ = {{t_1^3 - {t_3}} \over {{t_1} - t}}$$<br... | mcq | jee-main-2021-online-24th-february-morning-slot | 4,825 |
WIZrag7XacI8iKDxjV1klrk8szp | maths | application-of-derivatives | tangent-and-normal | For which of the following curves, the line $$x + \sqrt 3 y = 2\sqrt 3 $$ is the tangent at the point $$\left( {{{3\sqrt 3 } \over 2},{1 \over 2}} \right)$$? | [{"identifier": "A", "content": "$$2{x^2} - 18{y^2} = 9$$"}, {"identifier": "B", "content": "$${y^2} = {1 \\over {6\\sqrt 3 }}x$$"}, {"identifier": "C", "content": "$${x^2} + 9{y^2} = 9$$"}, {"identifier": "D", "content": "$${x^2} + {y^2} = 7$$"}] | ["C"] | null | Tangent to $${x^2} + 9{y^2} = 9$$ at
<br><br>point $$\left( {{{3\sqrt 3 } \over 2},{1 \over 2}} \right)$$ is $$x\left( {{{3\sqrt 3 } \over 2}} \right) + 9y\left( {{1 \over 2}} \right) = 9$$<br><br>$$3\sqrt 3 x + 9y = 18 \Rightarrow x + \sqrt 3 y = 2\sqrt 3 $$<br><br>$$ \Rightarrow $$ option (1) is true. | mcq | jee-main-2021-online-24th-february-evening-slot | 4,826 |
F7Rio2iCpsDwUvu6La1klrmblum | maths | application-of-derivatives | tangent-and-normal | If the curve y = ax<sup>2</sup> + bx + c, x$$ \in $$R, passes through the point (1, 2) and the tangent line to this curve at origin is y = x, then the possible values of a, b, c are : | [{"identifier": "A", "content": "a = $$-$$ 1, b = 1, c = 1"}, {"identifier": "B", "content": "a = 1, b = 1, c = 0"}, {"identifier": "C", "content": "a = $${1 \\over 2}$$, b = $${1 \\over 2}$$, c = 1"}, {"identifier": "D", "content": "a = 1, b = 0, c = 1"}] | ["B"] | null | Given curve y = ax<sup>2</sup> + bx + c, x$$ \in $$R
<br><br>This curve passes through the point (1, 2)
<br><br>$$ \therefore $$ $$2 = a + b + c$$ ..... (i)
<br><br>Given, slope of tangent at origin is 1
<br><br>$$ \therefore $$ $${{dy} \over {dx}} = 2ax + b \Rightarrow {\left. {{{dy} \over {dx}}} \right|_{(0,0)}} = 1$... | mcq | jee-main-2021-online-24th-february-evening-slot | 4,827 |
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