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1l6hz0rdm | maths | 3d-geometry | plane-in-space | <p>A vector $$\vec{a}$$ is parallel to the line of intersection of the plane determined by the vectors $$\hat{i}, \hat{i}+\hat{j}$$ and the plane determined by the vectors $$\hat{i}-\hat{j}, \hat{i}+\hat{k}$$. The obtuse angle between $$\vec{a}$$ and the vector $$\vec{b}=\hat{i}-2 \hat{j}+2 \hat{k}$$ is :</p> | [{"identifier": "A", "content": "$$\\frac{3 \\pi}{4}$$"}, {"identifier": "B", "content": "$$\\frac{2 \\pi}{3}$$"}, {"identifier": "C", "content": "$$\\frac{4 \\pi}{5}$$"}, {"identifier": "D", "content": "$$\\frac{5 \\pi}{6}$$"}] | ["A"] | null | <p>If $${\overrightarrow n _1}$$ is a vector normal to the plane determined by $$\widehat i$$ and $$\widehat i + \widehat j$$ then</p>
<p>$${\overrightarrow n _1} = \left| {\matrix{
{\widehat i} & {\widehat j} & {\widehat k} \cr
1 & 0 & 0 \cr
1 & 1 & 0 \cr
} } \right| = \widehat k$$</p>
<p>If $${\overr... | mcq | jee-main-2022-online-26th-july-evening-shift |
1l6jd01uv | maths | 3d-geometry | plane-in-space | <p>If the plane $$P$$ passes through the intersection of two mutually perpendicular planes $$2 x+k y-5 z=1$$ and $$3 k x-k y+z=5, k<3$$ and intercepts a unit length on positive $$x$$-axis, then the intercept made by the plane $$P$$ on the $$y$$-axis is :</p> | [{"identifier": "A", "content": "$$\\frac{1}{11}$$"}, {"identifier": "B", "content": "$$\\frac{5}{11}$$"}, {"identifier": "C", "content": "6"}, {"identifier": "D", "content": "7"}] | ["D"] | null | <p>$${P_1}:2x + ky - 5z = 1$$</p>
<p>$${P_2}:3kx - ky + z = 5$$</p>
<p>$$\because$$ $${P_1}\, \bot \,{P_2} \Rightarrow 6k - {k^2} + 5 = 0$$</p>
<p>$$ \Rightarrow k = 1,5$$</p>
<p>$$\because$$ $$k < 3$$</p>
<p>$$\therefore$$ $$k = 1$$</p>
<p>$${P_1}:2x + y - 5z = 1$$</p>
<p>$${P_2}:3x - y + z = 5$$</p>
<p>$$P:(2x + y - ... | mcq | jee-main-2022-online-27th-july-morning-shift |
1l6m62lhe | maths | 3d-geometry | plane-in-space | <p>The foot of the perpendicular from a point on the circle $$x^{2}+y^{2}=1, z=0$$ to the plane $$2 x+3 y+z=6$$ lies on which one of the following curves?</p> | [{"identifier": "A", "content": "$$(6 x+5 y-12)^{2}+4(3 x+7 y-8)^{2}=1, z=6-2 x-3 y$$"}, {"identifier": "B", "content": "$$(5 x+6 y-12)^{2}+4(3 x+5 y-9)^{2}=1, z=6-2 x-3 y$$"}, {"identifier": "C", "content": "$$(6 x+5 y-14)^{2}+9(3 x+5 y-7)^{2}=1, z=6-2 x-3 y$$"}, {"identifier": "D", "content": "$$(5 x+6 y-14)^{2}+9(3 ... | ["B"] | null | <p>Any point on $${x^2} + {y^2} = 1$$, $$z = 0$$ is $$p(\cos \theta ,\,\sin \theta ,\,0)$$</p>
<p>If foot of perpendicular of p on the plane $$2x + 3y + z = 6$$ is $$(h,k,l)$$ then</p>
<p>$${{h - \cos \theta } \over 2} = {{k - \sin \theta } \over 3} = {{l - 0} \over 1}$$</p>
<p>$$ = - \left( {{{2\cos \theta + 3\sin \... | mcq | jee-main-2022-online-28th-july-morning-shift |
1l6nnoffv | maths | 3d-geometry | plane-in-space | <p>A plane P is parallel to two lines whose direction ratios are $$-2,1,-3$$ and $$-1,2,-2$$ and it contains the point $$(2,2,-2)$$. Let P intersect the co-ordinate axes at the points $$\mathrm{A}, \mathrm{B}, \mathrm{C}$$ making the intercepts $$\alpha, \beta, \gamma$$. If $$\mathrm{V}$$ is the volume of the tetrahedr... | [{"identifier": "A", "content": "$$(48,-13)$$"}, {"identifier": "B", "content": "$$(24,-13)$$"}, {"identifier": "C", "content": "$$(48,11)$$"}, {"identifier": "D", "content": "$$(24,-5)$$"}] | ["B"] | null | <p>Let $${\overrightarrow a _1} = ( - 2,1, - 3)$$ and $${\overrightarrow a _2} = ( - 1,2, - 2)$$</p>
<p>Vector normal to plane $$\overline n = {\overrightarrow a _1} \times {\overrightarrow a _2}$$</p>
<p>$$\overline n = (4, - 1, - 3)$$</p>
<p>Plane through $$(2,2, - 2)$$ and normal to $$\overline n $$</p>
<p>$$(x - ... | mcq | jee-main-2022-online-28th-july-evening-shift |
1l6reqbxr | maths | 3d-geometry | plane-in-space | <p>Let $$Q$$ be the foot of perpendicular drawn from the point $$P(1,2,3)$$ to the plane $$x+2 y+z=14$$. If $$R$$ is a point on the plane such that $$\angle P R Q=60^{\circ}$$, then the area of $$\triangle P Q R$$ is equal to :</p> | [{"identifier": "A", "content": "$$\\frac{\\sqrt{3}}{2}$$"}, {"identifier": "B", "content": "$$ \\sqrt{3}$$"}, {"identifier": "C", "content": "$$2 \\sqrt{3}$$"}, {"identifier": "D", "content": "3"}] | ["B"] | null | <img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l7xrxbtg/d0e01ee5-c717-48d4-b80a-6a1b9ae27182/345bdd40-320e-11ed-bab2-5faeb0367b9d/file-1l7xrxbth.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l7xrxbtg/d0e01ee5-c717-48d4-b80a-6a1b9ae27182/345bdd40-320e-11ed-bab2-5faeb0367b9d/fi... | mcq | jee-main-2022-online-29th-july-evening-shift |
1l6resdnr | maths | 3d-geometry | plane-in-space | <p>If $$(2,3,9),(5,2,1),(1, \lambda, 8)$$ and $$(\lambda, 2,3)$$ are coplanar, then the product of all possible values of $$\lambda$$ is:</p> | [{"identifier": "A", "content": "$$\\frac{21}{2}$$"}, {"identifier": "B", "content": "$$\\frac{59}{8}$$"}, {"identifier": "C", "content": "$$\\frac{57}{8}$$"}, {"identifier": "D", "content": "$$\\frac{95}{8}$$"}] | ["D"] | null | $\because A(2,3,9), B(5,2,1), C(1, \lambda, 8)$ and D$(\lambda, 2,3)$ are coplanar.
<br/><br/>$$ \therefore $$ $[\overrightarrow{\mathrm{AB}} \,\,\,\,\overrightarrow{\mathrm{AC}} \,\,\,\, \overrightarrow{\mathrm{AD}}]=0$
<br/><br/>$\left|\begin{array}{ccc}3 & -1 & -8 \\ -1 & \lambda-3 & -1 \\ \lambda-2 & -1 & -6\end{... | mcq | jee-main-2022-online-29th-july-evening-shift |
ldo9tm7i | maths | 3d-geometry | plane-in-space | If a point $\mathrm{P}(\alpha, \beta, \gamma)$ satisfying
<br/><br/>$$\left( {\matrix{
\alpha & \beta & \gamma \cr
} } \right)\left( {\matrix{
2 & {10} & 8 \cr
9 & 3 & 8 \cr
8 & 4 & 8 \cr
} } \right) = \left( {\matrix{
0 & 0 & 0 \cr
} } \right)$$... | [{"identifier": "A", "content": "$\\frac{11}{5}$"}, {"identifier": "B", "content": "11"}, {"identifier": "C", "content": "$-1$"}, {"identifier": "D", "content": "$\\frac{5}{4}$"}] | ["B"] | null | Point $\mathrm{P}(\alpha, \beta, \gamma)$ lies on the plane $2 x+4 y+3 z=5$,
<br/><br/>$$ \therefore $$ $2 \alpha+4 \beta+3 \gamma=5$ ........(1)
<br/><br/>Given, $$\left( {\matrix{
\alpha & \beta & \gamma \cr
} } \right)\left( {\matrix{
2 & {10} & 8 \cr
9 & 3 & 8 \cr
8 & 4 & 8 \cr
} } \right... | mcq | jee-main-2023-online-31st-january-evening-shift |
1ldonegez | maths | 3d-geometry | plane-in-space | <p>Let the image of the point $$P(2,-1,3)$$ in the plane $$x+2 y-z=0$$ be $$Q$$. <br/><br/>Then the distance of the plane $$3 x+2 y+z+29=0$$ from the point $$Q$$ is :</p> | [{"identifier": "A", "content": "$$2\\sqrt{14}$$"}, {"identifier": "B", "content": "$$\\frac{22\\sqrt2}{7}$$"}, {"identifier": "C", "content": "$$\\frac{24\\sqrt2}{7}$$"}, {"identifier": "D", "content": "$$3\\sqrt{14}$$"}] | ["D"] | null | <img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lea9v38f/9887ff8f-08e7-4f18-9711-257b854f4854/0e8003e0-afb7-11ed-a5f0-99851f9df37c/file-1lea9v38g.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lea9v38f/9887ff8f-08e7-4f18-9711-257b854f4854/0e8003e0-afb7-11ed-a5f0-99851f9df37c/fi... | mcq | jee-main-2023-online-1st-february-morning-shift |
1ldr83jqr | maths | 3d-geometry | plane-in-space | <p>If $$\lambda_{1} < \lambda_{2}$$ are two values of $$\lambda$$ such that the angle between the planes $$P_{1}: \vec{r}(3 \hat{i}-5 \hat{j}+\hat{k})=7$$ and
$$P_{2}: \vec{r} \cdot(\lambda \hat{i}+\hat{j}-3 \hat{k})=9$$ is $$\sin ^{-1}\left(\frac{2 \sqrt{6}}{5}\right)$$, then the square of the length of perpendicu... | [] | null | 315 | <p>$${P_1}:\overrightarrow r \,.\,(3\widehat i - 5\widehat j + \widehat k) = 7$$</p>
<p>$${P_2}:\overrightarrow r \,.\,(\lambda \widehat i + \widehat j - 3\widehat k) = 9$$</p>
<p>Let angle between P<sub>1</sub> and P<sub>2</sub> is $$\theta$$</p>
<p>Then $$\cos \theta = {{3\lambda - 5 - 3} \over {\sqrt {35} \sqrt {{... | integer | jee-main-2023-online-30th-january-morning-shift |
1ldyb64xl | maths | 3d-geometry | plane-in-space | <p>The distance of the point (7, $$-$$3, $$-$$4) from the plane passing through the points (2, $$-$$3, 1), ($$-$$1, 1, $$-$$2) and (3, $$-$$4, 2) is :</p> | [{"identifier": "A", "content": "$$4\\sqrt2$$"}, {"identifier": "B", "content": "4"}, {"identifier": "C", "content": "5"}, {"identifier": "D", "content": "$$5\\sqrt2$$"}] | ["D"] | null | $A(2,-3,1), B(-1,1,-2), C(3,-4,2)$
<br/><br/>
$$
\begin{aligned}
& \overrightarrow{A B}=-3 \hat{i}+4 \hat{j}-3 \hat{k} \quad \overrightarrow{A C}=\hat{i}-\hat{j}+\hat{k} \\\\
& \vec{n}=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\\\
-3 & 4 & -3 \\\\
1 & -1 & 1
\end{array}\right|=\hat{i}-\hat{k}
\end{aligned}
... | mcq | jee-main-2023-online-24th-january-morning-shift |
lgnwzi53 | maths | 3d-geometry | plane-in-space | Let the foot of perpendicular of the point $P(3,-2,-9)$ on the plane passing through the points $(-1,-2,-3),(9,3,4),(9,-2,1)$ be $Q(\alpha, \beta, \gamma)$. Then the distance of $Q$ from the origin is : | [{"identifier": "A", "content": "$\\sqrt{38}$"}, {"identifier": "B", "content": "$\\sqrt{29}$"}, {"identifier": "C", "content": "$\\sqrt{42}$"}, {"identifier": "D", "content": "$\\sqrt{35}$"}] | ["C"] | null | <p>The equation of the plane passing through points $A(-1, -2, -3)$, $B(9, 3, 4)$, and $C(9, -2, 1)$ can be written using the determinant :</p>
<p>$$
\left|\begin{array}{ccc}
x+1 & y+2 & z+3 \\
10 & 5 & 7 \\
10 & 0 & 4
\end{array}\right|=0
$$</p>
<p>Expanding the determinant, we get :</p>
<p>$$
2x + 3y - 5z - 7 = 0
$$<... | mcq | jee-main-2023-online-15th-april-morning-shift |
lgny51ak | maths | 3d-geometry | plane-in-space | Let the system of linear equations
<br/><br/>$-x+2 y-9 z=7$
<br/><br/>$-x+3 y+7 z=9$
<br/><br/>$-2 x+y+5 z=8$
<br/><br/>$-3 x+y+13 z=\lambda$
<br/><br/>has a unique solution $x=\alpha, y=\beta, z=\gamma$. Then the distance of the point
<br/><br/>$(\alpha, \beta, \gamma)$ from the plane $2 x-2 y+z=\lambda$ is : | [{"identifier": "A", "content": "11"}, {"identifier": "B", "content": "7"}, {"identifier": "C", "content": "13"}, {"identifier": "D", "content": "9"}] | ["B"] | null | $$
\begin{aligned}
& -x+2 y-9 z=7-(1) \\\\
& -x+3 y-7 z=9-(2) \\\\
& -2 x+y+5 z=8-(3) \\\\
& (2)-(1) \\\\
& y+16 z=2 \quad(4) \\\\
& (3)-2 \times(1) \\\\
& -3 y+23 z=-6-(5) \\\\
& 3 \times(4)+(5) \\\\
& 71 z=0 \Rightarrow z=0 \\\\
& \quad y=2 \\\\
& (-3,2,0) \rightarrow(\alpha, \beta, \gamma) \\\\
& \text { Put in }-3 ... | mcq | jee-main-2023-online-15th-april-morning-shift |
1lgoxhquj | maths | 3d-geometry | plane-in-space | <p>Let $$\mathrm{N}$$ be the foot of perpendicular from the point $$\mathrm{P}(1,-2,3)$$ on the line passing through the points $$(4,5,8)$$ and $$(1,-7,5)$$. Then the distance of $$N$$ from the plane $$2 x-2 y+z+5=0$$ is :</p> | [{"identifier": "A", "content": "7"}, {"identifier": "B", "content": "6"}, {"identifier": "C", "content": "9"}, {"identifier": "D", "content": "8"}] | ["A"] | null | <img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lh2hc8pe/d4ba8260-fd3b-48a4-aed6-cb4a8b6ea85d/ac53ff20-e6d2-11ed-b683-39a79c7e644a/file-1lh2hc8pf.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lh2hc8pe/d4ba8260-fd3b-48a4-aed6-cb4a8b6ea85d/ac53ff20-e6d2-11ed-b683-39a79c7e644a/fi... | mcq | jee-main-2023-online-13th-april-evening-shift |
1lgpxn2hg | maths | 3d-geometry | plane-in-space | <p>Let the equation of plane passing through the line of intersection of the planes $$x+2 y+a z=2$$ and $$x-y+z=3$$ be $$5 x-11 y+b z=6 a-1$$. For $$c \in \mathbb{Z}$$, if the distance of this plane from the point $$(a,-c, c)$$ is $$\frac{2}{\sqrt{a}}$$, then $$\frac{a+b}{c}$$ is equal to :</p> | [{"identifier": "A", "content": "$$-$$2"}, {"identifier": "B", "content": "4"}, {"identifier": "C", "content": "2"}, {"identifier": "D", "content": "$$-$$4"}] | ["D"] | null | Given the equation of the plane passing through the intersection of the two given planes:
<br/><br/>$$P: (x + 2y + az - 2) + \lambda(x - y + z - 3) = 0$$
<br/><br/>$$\Rightarrow x(\lambda+1)+y(2-\lambda)+z(a+\lambda)-2-3 \lambda=0$$
<br/><br/>This is the same as the given equation $$5x - 11y + bz = 6a - 1$$.
<br/><b... | mcq | jee-main-2023-online-13th-april-morning-shift |
1lgq12wu6 | maths | 3d-geometry | plane-in-space | <p>Let the image of the point $$\left(\frac{5}{3}, \frac{5}{3}, \frac{8}{3}\right)$$ in the plane $$x-2 y+z-2=0$$ be P. If the distance of the point $$Q(6,-2, \alpha), \alpha > 0$$, from $$\mathrm{P}$$ is 13 , then $$\alpha$$ is equal to ___________.</p> | [] | null | 15 | Image of point $\left(\frac{5}{3}, \frac{5}{3}, \frac{8}{3}\right)$
<br/><br/>$$
\begin{aligned}
& \frac{x-\frac{5}{3}}{1}=\frac{y-\frac{5}{3}}{-2}=\frac{\mathrm{z}-\frac{8}{3}}{1}=\frac{-2\left(1 \times \frac{5}{3}+(-2) \times \frac{8}{3}+1 \times \frac{8}{3}-2\right)}{1^2+2^2+1^2} =\frac{1}{3} \\\\
& \therefore x=2, ... | integer | jee-main-2023-online-13th-april-morning-shift |
1lgrgo0q2 | maths | 3d-geometry | plane-in-space | <p>Let the plane $$x+3 y-2 z+6=0$$ meet the co-ordinate axes at the points A, B, C. If the orthocenter of the triangle $$\mathrm{ABC}$$ is $$\left(\alpha, \beta, \frac{6}{7}\right)$$, then $$98(\alpha+\beta)^{2}$$ is equal to ___________.</p> | [] | null | 288 | $$
\begin{aligned}
& \mathrm{A}(-6,0,0) \quad \mathrm{B}(0,-2,0) \mathrm{C}=(0,0,3) \\\\
& \overrightarrow{\mathrm{AB}}=6 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}, \quad \overrightarrow{\mathrm{BC}}=2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}, \\\\
& \overrightarrow{\mathrm{AC}}=6 \hat{\mathrm{i}}+3 \hat{\mathrm{k}}
\... | integer | jee-main-2023-online-12th-april-morning-shift |
1lgsuokhr | maths | 3d-geometry | plane-in-space | <p>Let the line passing through the points $$\mathrm{P}(2,-1,2)$$ and $$\mathrm{Q}(5,3,4)$$ meet the plane $$x-y+z=4$$ at the point $$\mathrm{R}$$. Then the distance of the point $$\mathrm{R}$$ from the plane $$x+2 y+3 z+2=0$$ measured parallel to the line $$\frac{x-7}{2}=\frac{y+3}{2}=\frac{z-2}{1}$$ is equal to :</p> | [{"identifier": "A", "content": "$$\\sqrt{31}$$"}, {"identifier": "B", "content": "$$\\sqrt{189}$$"}, {"identifier": "C", "content": "$$\\sqrt{61}$$"}, {"identifier": "D", "content": "3"}] | ["D"] | null | Equation of line $P Q$ :
<br><br>$$
\frac{x-2}{3}=\frac{y+1}{4}=\frac{z-2}{2}=\lambda
$$
<br><br>Let $R$ be $(3 \lambda+2,4 \lambda-1,2 \lambda+2)$
<br><br>$\mathrm{R}$ lies on plane $x-y+z=4$
<br><br>$$
\begin{aligned}
& \therefore \quad 3 \lambda+2-4 \lambda+1+2 \lambda+2=4 \\\\
& \Rightarrow \quad \lambda=-1... | mcq | jee-main-2023-online-11th-april-evening-shift |
1lgsvumqy | maths | 3d-geometry | plane-in-space | <p>Let P be the plane passing through the points $$(5,3,0),(13,3,-2)$$ and $$(1,6,2)$$.
For $$\alpha \in \mathbb{N}$$, if the distances of the points $$\mathrm{A}(3,4, \alpha)$$ and $$\mathrm{B}(2, \alpha, a)$$ from the plane P are 2 and 3 respectively, then the positive value of a is :</p> | [{"identifier": "A", "content": "6"}, {"identifier": "B", "content": "4"}, {"identifier": "C", "content": "5"}, {"identifier": "D", "content": "3"}] | ["B"] | null | <img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lig7g1t0/e607e161-c3d1-45fc-96f7-d8420574551c/2b69fe40-022b-11ee-8eea-09e354b91938/file-1lig7g1t1.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lig7g1t0/e607e161-c3d1-45fc-96f7-d8420574551c/2b69fe40-022b-11ee-8eea-09e354b91938/fi... | mcq | jee-main-2023-online-11th-april-evening-shift |
1lguu3law | maths | 3d-geometry | plane-in-space | <p>Let $$(\alpha, \beta, \gamma)$$ be the image of the point $$\mathrm{P}(2,3,5)$$ in the plane $$2 x+y-3 z=6$$. Then $$\alpha+\beta+\gamma$$ is equal to :</p> | [{"identifier": "A", "content": "10"}, {"identifier": "B", "content": "9"}, {"identifier": "C", "content": "5"}, {"identifier": "D", "content": "12"}] | ["A"] | null | <img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1ln74ubxe/c878026c-5267-42b0-b10f-7150edf4c2a2/04392c20-602b-11ee-84d4-f1c1aec911f3/file-6y3zli1ln74ubxf.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/6y3zli1ln74ubxe/c878026c-5267-42b0-b10f-7150edf4c2a2/04392c20-602b-11ee-84... | mcq | jee-main-2023-online-11th-april-morning-shift |
1lguuhrui | maths | 3d-geometry | plane-in-space | <p>If equation of the plane that contains the point $$(-2,3,5)$$ and is perpendicular to each of the planes $$2 x+4 y+5 z=8$$ and $$3 x-2 y+3 z=5$$ is $$\alpha x+\beta y+\gamma z+97=0$$ then $$\alpha+\beta+\gamma=$$</p> | [{"identifier": "A", "content": "15"}, {"identifier": "B", "content": "16"}, {"identifier": "C", "content": "17"}, {"identifier": "D", "content": "18"}] | ["A"] | null | The equation of plane that passes through the point $(-2,3,5)$ is
<br/><br/>$$
a(x+2)+b(y-3)+c(z-5)=0
$$ ..........(i)
<br/><br/>The plane is perpendicular to
<br/><br/>$$
\begin{array}{ll}
2 x+4 y+5 z =8 \text { and } 3 x-2 y+3 z=5 \\\\
\end{array}
$$
<br/><br/>$$
\begin{array}{ll}
\therefore 2 a+4 b+5 c=0 ......... | mcq | jee-main-2023-online-11th-april-morning-shift |
1lgvpmh4j | maths | 3d-geometry | plane-in-space | <p>Let the image of the point $$\mathrm{P}(1,2,6)$$ in the plane passing through the points $$\mathrm{A}(1,2,0), \mathrm{B}(1,4,1)$$ and $$\mathrm{C}(0,5,1)$$ be $$\mathrm{Q}(\alpha, \beta, \gamma)$$. Then $$\left(\alpha^{2}+\beta^{2}+\gamma^{2}\right)$$ is equal to :</p> | [{"identifier": "A", "content": "76"}, {"identifier": "B", "content": "62"}, {"identifier": "C", "content": "70"}, {"identifier": "D", "content": "65"}] | ["D"] | null | Equation of plane passing through the points $A(1,2$, $0), B(1,4,1)$ and $C(0,5,1)$ is
<br/><br/>$$
\begin{aligned}
& \left|\begin{array}{ccc}
x-1 & y-2 & z-0 \\
0 & 2 & 1 \\
-1 & 3 & 1
\end{array}\right|=0 \\\\
& \Rightarrow x+y-2 z=3
\end{aligned}
$$
<br/><br/>Now $Q(\alpha, \beta, \gamma)$ is the image of the point ... | mcq | jee-main-2023-online-10th-april-evening-shift |
1lgvpsiky | maths | 3d-geometry | plane-in-space | <p>Let the line $$\frac{x}{1}=\frac{6-y}{2}=\frac{z+8}{5}$$ intersect the lines $$\frac{x-5}{4}=\frac{y-7}{3}=\frac{z+2}{1}$$ and $$\frac{x+3}{6}=\frac{3-y}{3}=\frac{z-6}{1}$$ at the points $$\mathrm{A}$$ and $$\mathrm{B}$$ respectively. Then the distance of the mid-point of the line segment $$\mathrm{AB}$$ from the pl... | [{"identifier": "A", "content": "3"}, {"identifier": "B", "content": "$$\\frac{10}{3}$$"}, {"identifier": "C", "content": "4"}, {"identifier": "D", "content": "$$\\frac{11}{3}$$"}] | ["C"] | null | We have, $\frac{x}{1}=\frac{6-y}{2}=\frac{z+8}{5}$ intersect the line
<br/><br/>$\frac{x-5}{4}=\frac{y-7}{3}=\frac{z+2}{1}$ and $\frac{x+3}{6}=\frac{3-y}{3}=\frac{z-6}{1}$
<br/><br/>$$
\begin{array}{ll}
& \text { Now, } \frac{x}{1}=\frac{6-y}{2}=\frac{z+8}{5}=\lambda ...........(i)\\\\
& \Rightarrow x=\lambda, y=6-2... | mcq | jee-main-2023-online-10th-april-evening-shift |
1lgyop0gk | maths | 3d-geometry | plane-in-space | <p>Let $$\mathrm{P}_{1}$$ be the plane $$3 x-y-7 z=11$$ and $$\mathrm{P}_{2}$$ be the plane passing through the points $$(2,-1,0),(2,0,-1)$$, and $$(5,1,1)$$. If the foot of the perpendicular drawn from the point $$(7,4,-1)$$ on the line of intersection of the planes $$P_{1}$$ and $$P_{2}$$ is $$(\alpha, \beta, \gamma)... | [] | null | 11 | Equation of plane $\mathrm{P}_2$ passing through $(2,-1,0),(2,0$, $-1)$ and $(5,1,1)$ is
<br/><br/>$$
\begin{aligned}
& \left|\begin{array}{ccc}
x-5 & y-1 & z-1 \\
3 & 2 & 1 \\
3 & 1 & 2
\end{array}\right|=0 \\\\
& \Rightarrow(x-5)(4-1)-(y-1)(6-3)+(z-1)(3-6)=0 \\\\
& \Rightarrow 3 x-15-3 y+3-3 z+3=0 \\\\
& \Rightarrow ... | integer | jee-main-2023-online-8th-april-evening-shift |
1lgzzukx0 | maths | 3d-geometry | plane-in-space | <p>If the equation of the plane containing the line <br/><br/>$$x+2 y+3 z-4=0=2 x+y-z+5$$ and perpendicular to the plane <br/><br/>$\vec{r}=(\hat{i}-\hat{j})+\lambda(\hat{i}+\hat{j}+\hat{k})+\mu(\hat{i}-2 \hat{j}+3 \hat{k})$<br/><br/> is $a x+b y+c z=4$, then $$(a-b+c)$$ is equal to :</p> | [{"identifier": "A", "content": "18"}, {"identifier": "B", "content": "22"}, {"identifier": "C", "content": "20"}, {"identifier": "D", "content": "24"}] | ["B"] | null | Equation of plane $\mathrm{P}$ containing the given lines is
<br/><br/>$$
\begin{aligned}
& (x+2 y+3 z-4)+\lambda(2 x+y-z+5)=0 \\\\
& \Rightarrow(1+2 \lambda) x+(2+\lambda) y+(3-\lambda) z+(-4+5 \lambda)=0
\end{aligned}
$$
<br/><br/>Now, plane $\mathrm{P}$ is perpendicular to plane $\mathrm{P}^{\prime}$
<br/><br/>$$
\v... | mcq | jee-main-2023-online-8th-april-morning-shift |
1lh244z4l | maths | 3d-geometry | plane-in-space | <p>Let the image of the point $$\mathrm{P}(1,2,3)$$ in the plane $$2 x-y+z=9$$ be $$\mathrm{Q}$$. If the coordinates of the point $$\mathrm{R}$$ are $$(6,10,7)$$, then the square of the area of the triangle $$\mathrm{PQR}$$ is _____________.</p> | [] | null | 594 | Let $Q(x, y, z)$ be the image of $P(1,2,3)$ in the plane
<br><br>$$
\begin{aligned}
& 2 x-y+z=9 \\\\
& \therefore \frac{x-1}{2}=\frac{y-2}{-1}=\frac{z-3}{1}
=\frac{-2(2 \times 1+(-1)(2)+(1)(3)(-9)}{(2)^2+(-1)^2+(1)^2}
\end{aligned}
$$
<br><br>$$
\begin{aligned}
& \Rightarrow \frac{x-1}{2}=\frac{y-2}{-1}=\... | integer | jee-main-2023-online-6th-april-morning-shift |
1lh2xv9fi | maths | 3d-geometry | plane-in-space | <p>A plane P contains the line of intersection of the plane $$\vec{r} \cdot(\hat{i}+\hat{j}+\hat{k})=6$$ and $$\vec{r} \cdot(2 \hat{i}+3 \hat{j}+4 \hat{k})=-5$$. If $$\mathrm{P}$$ passes through the point $$(0,2,-2)$$, then the square of distance of the point $$(12,12,18)$$ from the plane $$\mathrm{P}$$ is :</p> | [{"identifier": "A", "content": "310"}, {"identifier": "B", "content": "620"}, {"identifier": "C", "content": "1240"}, {"identifier": "D", "content": "155"}] | ["B"] | null | Given plane $\vec{r} \cdot(\hat{i}+\hat{j}+\hat{k})=6$ and
<br/><br/>$$
\vec{r} \cdot(2 \hat{i}+3 \hat{j}+4 \hat{k})=-5
$$
<br/><br/>Equation of plane passing through both plane
<br/><br/>$$
\begin{aligned}
& \mathrm{P}_1 \rightarrow(x \hat{i}+y \hat{j}+2 \hat{k})(\hat{i}+\hat{j}+\hat{k})=6 \\\\
& \mathrm{P}_1=x+y+z=6... | mcq | jee-main-2023-online-6th-april-evening-shift |
i0c95WVQI0WQ06dx | maths | application-of-derivatives | maxima-and-minima | The maximum distance from origin of a point on the curve
<br/>$$x = a\sin t - b\sin \left( {{{at} \over b}} \right)$$
<br/>$$y = a\cos t - b\cos \left( {{{at} \over b}} \right),$$ both $$a,b > 0$$ is | [{"identifier": "A", "content": "$$a-b$$ "}, {"identifier": "B", "content": "$$a+b$$ "}, {"identifier": "C", "content": "$$\\sqrt {{a^2} + {b^2}} $$ "}, {"identifier": "D", "content": "$$\\sqrt {{a^2} - {b^2}} $$ "}] | ["B"] | null | Distance of origin from $$\left( {x,y} \right) = \sqrt {{x^2} + {y^2}} $$
<br><br>$$ = \sqrt {{a^2} + {b^2} - 2ab\cos \left( {t - {{at} \over b}} \right)} ;$$
<br><br>$$ \le \sqrt {{a^2} + {b^2} + 2ab} $$ $$\left[ {{{\left\{ {\cos \left( {t - {{at} \over b}} \right)} \right\}}_{\min }} = - 1} \right]$$
<br><br>$$=a+b... | mcq | aieee-2002 |
mM353u3qL1EToDRE | maths | application-of-derivatives | maxima-and-minima | The real number $$x$$ when added to its inverse gives the minimum sum at $$x$$ equal : | [{"identifier": "A", "content": "-2 "}, {"identifier": "B", "content": "2 "}, {"identifier": "C", "content": "1 "}, {"identifier": "D", "content": "-1"}] | ["C"] | null | $$f(x) = y = x + {1 \over x}$$ or $${{dy} \over {dx}} = 1 - {1 \over {{x^2}}}$$
<br><br>For max. or min, $$1 - {1 \over {{x^2}}} = 0 \Rightarrow x = \pm 1$$
<br><br>$${{{d^2}y} \over {d{x^2}}} = {2 \over {{x^3}}} $$
<br/><br/>$$ {\left( {{{{d^2}y} \over {d{x^2}}}} \right)_{x = 1}} = 2$$ ($$+ve$$)
<br><br>$$\therefor... | mcq | aieee-2003 |
xIrzYEUCC29YS3QC | maths | application-of-derivatives | maxima-and-minima | If the function $$f\left( x \right) = 2{x^3} - 9a{x^2} + 12{a^2}x + 1,$$ where $$a>0,$$ attains its maximum and minimum at $$p$$ and $$q$$ respectively such that $${p^2} = q$$ , then $$a$$ equals | [{"identifier": "A", "content": "$${1 \\over 2}$$ "}, {"identifier": "B", "content": "$$3$$"}, {"identifier": "C", "content": "$$1$$ "}, {"identifier": "D", "content": "$$2$$ "}] | ["D"] | null | $$f\left( x \right) = 2{x^3} - 9a{x^2} + 12{a^2}x + 1$$
<br><br>$$f'\left( x \right) = 6{x^2} - 18ax + 12{a^2};$$
<br><br>$$\,\,\,\,\,\,\,\,\,\,f''\left( x \right) = 12x - 18a$$
<br><br>For max. or min.
<br><br>$$6{x^2} - 18ax + 12{a^2} = 0$$
<br><br>$$ \Rightarrow {x^2} - 3ax + 2{a^2} = 0$$
<br><br>$$ \Rightarrow x = ... | mcq | aieee-2003 |
2tGr74lgODUL42wb | maths | application-of-derivatives | maxima-and-minima | Area of the greatest rectangle that can be inscribed in the
<br/>ellipse $${{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}} = 1$$ | [{"identifier": "A", "content": "$$2ab$$ "}, {"identifier": "B", "content": "$$ab$$ "}, {"identifier": "C", "content": "$$\\sqrt {ab} $$ "}, {"identifier": "D", "content": "$${a \\over b}$$ "}] | ["A"] | null | Area of rectangle $$ABCD$$ $$ = 2a\,\cos \,\theta $$
<br><br>$$\left( {2b\,\sin \,\theta } \right) = 2ab\,\sin \,2\theta $$
<br><br><img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263702/exam_images/e7w2mcfgqs0qwtwa1jkf.webp" loading="lazy" alt="AIEEE 2005 Mathematics - Applicati... | mcq | aieee-2005 |
BPl54gj84e83x2jD | maths | application-of-derivatives | maxima-and-minima | The function $$f\left( x \right) = {x \over 2} + {2 \over x}$$ has a local minimum at | [{"identifier": "A", "content": "$$x=2$$ "}, {"identifier": "B", "content": "$$x=-2$$ "}, {"identifier": "C", "content": "$$x=0$$ "}, {"identifier": "D", "content": "$$x=1$$ "}] | ["A"] | null | $$f\left( x \right) = {x \over 2} + {2 \over x} \Rightarrow f'\left( x \right) = {1 \over 2} - {2 \over {{x^2}}} = 0$$
<br><br>$$ \Rightarrow {x^2} = 4$$ or $$x=2,-2;$$ $$\,\,\,\,\,f''\left( x \right) = {4 \over {{x^3}}}$$
<br><br>$$f''{\left. {\left( x \right)} \right]_{x = 2}} = + ve \Rightarrow f\left( x \right)$$... | mcq | aieee-2006 |
AEdU5WTln3rLzwom | maths | application-of-derivatives | maxima-and-minima | A triangular park is enclosed on two sides by a fence and on the third side by a straight river bank. The two sides having fence are of same length $$x$$. The maximum area enclosed by the park is | [{"identifier": "A", "content": "$${3 \\over 2}{x^2}$$ "}, {"identifier": "B", "content": "$$\\sqrt {{{{x^3}} \\over 8}} $$ "}, {"identifier": "C", "content": "$${1 \\over 2}{x^2}$$ "}, {"identifier": "D", "content": "$$\\pi {x^2}$$ "}] | ["C"] | null | Area $$ = {1 \over 2}{x^2}\,\sin \,\theta $$
<br><br><img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265305/exam_images/jl4uz1aheqoemq1myxxa.webp" loading="lazy" alt="AIEEE 2006 Mathematics - Application of Derivatives Question 192 English Explanation">
<br><br>Maximum value of $... | mcq | aieee-2006 |
SOofWOrrmqiZbPr1 | maths | application-of-derivatives | maxima-and-minima | If $$p$$ and $$q$$ are positive real numbers such that $${p^2} + {q^2} = 1$$, then the maximum value of $$(p+q)$$ is | [{"identifier": "A", "content": "$${1 \\over 2}$$ "}, {"identifier": "B", "content": "$${1 \\over {\\sqrt 2 }}$$ "}, {"identifier": "C", "content": "$${\\sqrt 2 }$$ "}, {"identifier": "D", "content": "$$2$$"}] | ["C"] | null | Given that $${p^2} + {q^2} = 1$$
<br><br>$$\therefore$$ $$p = \cos \theta $$ and $$q = \sin \theta $$
<br><br>Then $$p+q$$ $$ = \cos \theta + \sin \theta $$
<br><br>We know that
<br><br>$$ - \sqrt {{a^2} + {b^2}} \le a\cos \theta + b\sin \theta \le \sqrt {{a^2} + {b^2}} $$
<br><br>$$\therefore$$ $$ - \sqrt 2 \le ... | mcq | aieee-2007 |
U4dcftyIinHNo7sg | maths | application-of-derivatives | maxima-and-minima | Suppose the cubic $${x^3} - px + q$$ has three distinct real roots
<br/>where $$p>0$$ and $$q>0$$. Then which one of the following holds? | [{"identifier": "A", "content": "The cubic has minima at $$\\sqrt {{p \\over 3}} $$ and maxima at $$-\\sqrt {{p \\over 3}} $$"}, {"identifier": "B", "content": "The cubic has minima at $$-\\sqrt {{p \\over 3}} $$ and maxima at $$\\sqrt {{p \\over 3}} $$"}, {"identifier": "C", "content": "The cubic has minima at both $$... | ["A"] | null | Let $$y = {x^3} - px + q$$
<br><br>$$ \Rightarrow {{dy} \over {dx}} = 3{x^2} - p$$
<br><br>For $${{dy} \over {dx}} = 0 \Rightarrow 3{x^2} - p = 0$$
<br><br>$$ \Rightarrow x = \pm \sqrt {{p \over 3}} $$
<br><br>$${{{d^2}y} \over {d{x^2}}} = 6x$$
<br><br>$${\left. {{{{d^2}y} \over {d{x^2}}}} \right|_{x = \sqrt {{p \over... | mcq | aieee-2008 |
h0h4CCP2OjJaHXK7 | maths | application-of-derivatives | maxima-and-minima | Given $$P\left( x \right) = {x^4} + a{x^3} + b{x^2} + cx + d$$ such that $$x=0$$ is the only
<br/>real root of $$P'\,\left( x \right) = 0.$$ If $$P\left( { - 1} \right) < P\left( 1 \right),$$ then in the interval $$\left[ { - 1,1} \right]:$$ | [{"identifier": "A", "content": "$$P(-1)$$ is not minimum but $$P(1)$$ is the maximum of $$P$$"}, {"identifier": "B", "content": "$$P(-1)$$ is the minimum but $$P(1)$$ is not the maximum of $$P$$"}, {"identifier": "C", "content": "Neither $$P(-1)$$ is the minimum nor $$P(1)$$ is the maximum of $$P$$"}, {"identifier": "... | ["A"] | null | We have $$P\left( x \right) = {x^4} + a{x^3} + b{x^2} + cx + d$$
<br><br>$$ \Rightarrow P'\left( x \right) = 4\,{x^3} + 3a{x^2} + 2bx + c$$
<br><br>But $$P'\left( 0 \right) = 0 \Rightarrow c = 0$$
<br><br>$$\therefore$$ $$P\left( x \right) = {x^4} + a{x^3} + b{x^2} + d$$
<br><br>As given that $$P\left( { - 1} \right) &... | mcq | aieee-2009 |
20SUnhwmW0otvw8z | maths | application-of-derivatives | maxima-and-minima | Let $$f:R \to R$$ be defined by
$$$f\left( x \right) = \left\{ {\matrix{
{k - 2x,\,\,if} & {x \le - 1} \cr
{2x + 3,\,\,if} & {x > - 1} \cr
} } \right.$$$
<p>If $$f$$has a local minimum at $$x=-1$$, then a possible value of $$k$$ is </p> | [{"identifier": "A", "content": "$$0$$ "}, {"identifier": "B", "content": "$$ - {1 \\over 2}$$ "}, {"identifier": "C", "content": "$$-1$$ "}, {"identifier": "D", "content": "$$1$$"}] | ["C"] | null | $$f\left( x \right) = \left\{ {\matrix{
{k - 2x,\,\,\,\,if\,\,\,\,x \le - 1} \cr
{2x + 3,\,\,\,\,if\,\,\,\,x > - 1} \cr
} } \right.$$
<br><br><img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264221/exam_images/up9xtippomu7kefhskum.webp" loading="lazy" alt="AIEEE 2... | mcq | aieee-2010 |
WAccCVwQ4VUOLsfn | maths | application-of-derivatives | maxima-and-minima | Let $$f:R \to R$$ be a continuous function defined by
$$$f\left( x \right) = {1 \over {{e^x} + 2{e^{ - x}}}}$$$
<p><b>Statement - 1 :</b> $$f\left( c \right) = {1 \over 3},$$ for some $$c \in R$$.</p>
<p><b>Statement - 2 :</b> $$0 < f\left( x \right) \le {1 \over {2\sqrt 2 }},$$ for all $$x \in R$$</p> | [{"identifier": "A", "content": "Statement - 1 is true, Statement -2 is true; Statement - 2 is <b>not</b> a correct explanation for Statement - 1."}, {"identifier": "B", "content": "Statement - 1 is true, Statement - 2 is false."}, {"identifier": "C", "content": "Statement - 1 is false, Statement - 2 is true."}, {"iden... | ["D"] | null | $$f\left( x \right) = {1 \over {{e^x} + 2{e^{ - x}}}} = {{{e^x}} \over {{e^{2x}} + 2}}$$
<br><br>$$f'\left( x \right) = {{\left( {{e^{2x}} + 2} \right)e{}^x - 2{e^{2x}}.{e^x}} \over {{{\left( {{e^{2x}} + 2} \right)}^2}}}$$
<br><br>$$f'\left( x \right) = 0 \Rightarrow {e^{2x}} + 2 = 2{e^{2x}}$$
<br><br>$${e^{2x}} = 2 \... | mcq | aieee-2010 |
Lcf95duImInmLqTv | maths | application-of-derivatives | maxima-and-minima | For $$x \in \left( {0,{{5\pi } \over 2}} \right),$$ define $$f\left( x \right) = \int\limits_0^x {\sqrt t \sin t\,dt.} $$ Then $$f$$ has | [{"identifier": "A", "content": "local minimum at $$\\pi $$ and $$2\\pi $$"}, {"identifier": "B", "content": "local minimum at $$\\pi $$ and local maximum at $$2\\pi $$"}, {"identifier": "C", "content": "local maximum at $$\\pi $$ and local minimum at $$2\\pi $$"}, {"identifier": "D", "content": "local maximum at $$\\p... | ["C"] | null | $$f'\left( x \right) = \sqrt x \sin x$$
<br><br>At local maxima or minima, $$f'\left( x \right) = 0$$
<br><br>$$ \Rightarrow x = 0$$ or $$sin$$ $$x=0$$
<br><br>$$ \Rightarrow x = 2\pi ,\,\,\pi \in \left( {0,{{5\pi } \over 2}} \right)$$
<br><br>$$f''\left( x \right) = \sqrt x \cos \,x + {1 \over {2\sqrt x }}\sin \,x$$... | mcq | aieee-2011 |
iXvsDjZDvJ3Y3hNO | maths | application-of-derivatives | maxima-and-minima | Let $$a,b \in R$$ be such that the function $$f$$ given by $$f\left( x \right) = In\left| x \right| + b{x^2} + ax,\,x \ne 0$$ has extreme values at $$x=-1$$ and $$x=2$$
<p><b>Statement-1 :</b> $$f$$ has local maximum at $$x=-1$$ and at $$x=2$$.</p>
<p><b>Statement-2 :</b> $$a = {1 \over 2}$$ and $$b = {-1 \over 4}$$</... | [{"identifier": "A", "content": "Statement - 1 is false, Statement - 2 is true."}, {"identifier": "B", "content": "Statement - 1 is true , Statement - 2 is true; Statement - 2 is a correct explanation for Statement - 1."}, {"identifier": "C", "content": "Statement - 1 is true, Statement - 2 is true; Statement - 2 is <b... | ["B"] | null | Given, $$f\left( x \right) = \ln \left| x \right| + b{x^2} + ax$$
<br><br>$$\therefore$$ $$f'\left( x \right) = {1 \over x} + 2bx + a$$
<br><br>At $$x=-1,$$ $$f'\left( x \right) = - 1 - 2b + a = 0$$
<br><br>$$ \Rightarrow a - 2b = 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( i \right)$$
<br><br>At $$x=2,$$ $$... | mcq | aieee-2012 |
wyiDjCvI5NPRr2DZ | maths | application-of-derivatives | maxima-and-minima | A line is drawn through the point $$(1, 2)$$ to meet the coordinate axes at $$P$$ and $$Q$$ such that it forms a triangle $$OPQ,$$ where $$O$$ is the origin. If the area of the triangle $$OPQ$$ is least, then the slope of the line $$PQ$$ is : | [{"identifier": "A", "content": "$$-{1 \\over 4}$$"}, {"identifier": "B", "content": "$$-4$$ "}, {"identifier": "C", "content": "$$-2$$ "}, {"identifier": "D", "content": "$$-{1 \\over 2}$$"}] | ["C"] | null | Equation of a line passing through $$\left( {{x_1},{y_1}} \right)$$ having
<br><br>slope $$m$$ is given by $$y - {y_1} = m\left( {x - {x_1}} \right)$$
<br><br>Since the line $$PQ$$ is passing through $$(1,2)$$ therefore its
<br><br>equation is
<br><br>$$\left( {y - 2} \right) = m\left( {x - 1} \right)$$
<br><br>where... | mcq | aieee-2012 |
uhpBt8tZl7jVNmvk | maths | application-of-derivatives | maxima-and-minima | If $$x=-1$$ and $$x=2$$ are extreme points of $$f\left( x \right) = \alpha \,\log \left| x \right|+\beta {x^2} + x$$ then | [{"identifier": "A", "content": "$$\\alpha = 2,\\beta = - {1 \\over 2}$$ "}, {"identifier": "B", "content": "$$\\alpha = 2,\\beta = {1 \\over 2}$$ "}, {"identifier": "C", "content": "$$\\alpha = - 6,\\beta = {1 \\over 2}$$ "}, {"identifier": "D", "content": "$$\\alpha = - 6,\\beta = -{1 \\over 2}$$ "}] | ["A"] | null | Let $$f\left( x \right) = \alpha \log \left| x \right| + \beta {x^2} + x$$
<br><br>Differentiating both sides,
<br><br>$$f'\left( x \right) = {\alpha \over x} + 2\beta x + 1$$
<br><br>Since $$x=-1$$ and $$x=2$$ are extreme points therefore
<br><br>$$f'\left( x \right) = 0$$ at these points.
<br><br>Put $$x = - 1$$... | mcq | jee-main-2014-offline |
JAEcF3gxW0HGEZRy | maths | application-of-derivatives | maxima-and-minima | Let $$f(x)$$ be a polynomial of degree four having extreme values
<br/>at $$x=1$$ and $$x=2$$. If $$\mathop {\lim }\limits_{x \to 0} \left[ {1 + {{f\left( x \right)} \over {{x^2}}}} \right] = 3$$, then f$$(2)$$ is equal to : | [{"identifier": "A", "content": "$$0$$ "}, {"identifier": "B", "content": "$$4$$ "}, {"identifier": "C", "content": "$$-8$$ "}, {"identifier": "D", "content": "$$-4$$ "}] | ["A"] | null | $$\mathop {\lim }\limits_{x \to 0} \left[ {1 + {{f\left( x \right)} \over {{x^2}}}} \right] = 3 \Rightarrow \mathop {Lim}\limits_{x \to 0} {{f\left( x \right)} \over {{x^2}}} = 2$$
<br><br>So, $$f(x)$$ contains terms in $$x{}^2,{x^3}$$ and $${x^4}$$
<br><br>Let $$f\left( x \right) = {a_1}{x^2} + {a_2}{x^3} + {a_3}{x^4}... | mcq | jee-main-2015-offline |
jrSbPCcMERZCP1ui | maths | application-of-derivatives | maxima-and-minima | A wire of length $$2$$ units is cut into two parts which are bent respectively to form a square of side $$=x$$ units and a circle of radius $$=r$$ units. If the sum of the areas of the square and the circle so formed is minimum, then: | [{"identifier": "A", "content": "$$x=2r$$ "}, {"identifier": "B", "content": "$$2x=r$$ "}, {"identifier": "C", "content": "$$2x = \\left( {\\pi + 4} \\right)r$$ "}, {"identifier": "D", "content": "$$\\left( {4 - \\pi } \\right)x = \\pi \\,\\, r$$ "}] | ["A"] | null | $$4x + 2\pi r = 2$$ $$\,\,\,$$ $$ \Rightarrow 2x + \pi r = 1$$
<br><br>$$S = {x^2} + \pi {r^2}$$
<br><br>$$S = {\left( {{{1 - \pi r} \over 2}} \right)^2} + \pi {r^2}$$
<br><br>$${{dS} \over {dr}} = 2\left( {{{1 - \pi r} \over 2}} \right)\left( {{{ - \pi } \over 2}} \right) + 2\pi r$$
<br><br>$$ \Rightarrow {{ - \pi } \... | mcq | jee-main-2016-offline |
byueGoFmiFA0tNnAffoxb | maths | application-of-derivatives | maxima-and-minima | The minimum distance of a point on the curve y = x<sup>2</sup>−4 from the origin is : | [{"identifier": "A", "content": "$${{\\sqrt {19} } \\over 2}$$"}, {"identifier": "B", "content": "$$\\sqrt {{{15} \\over 2}} $$"}, {"identifier": "C", "content": "$${{\\sqrt {15} } \\over 2}$$"}, {"identifier": "D", "content": "$$\\sqrt {{{19} \\over 2}} $$"}] | ["C"] | null | Let point on the curve
<br><br>y = x<sup>2</sup> $$-$$ 4 is ($$\alpha $$<sup>2</sup>, $$\alpha $$<sup>2</sup> $$-$$ 4)
<br><br>$$ \therefore $$ Distance of the point ($$\alpha $$<sup>2</sup>, $$\alpha $$<sup>2</sup> $$-$$ 4) from origin,
<br><br>D = $$\sqrt {{\alpha ^2} + {{\left( {{\alpha ^2} - ... | mcq | jee-main-2016-online-9th-april-morning-slot |
JRHASG5PZuMAMaza | maths | application-of-derivatives | maxima-and-minima | Twenty meters of wire is available for fencing off a flower-bed in the form of a circular sector. Then the
maximum area (in sq. m) of the flower-bed, is : | [{"identifier": "A", "content": "10"}, {"identifier": "B", "content": "25"}, {"identifier": "C", "content": "30"}, {"identifier": "D", "content": "12.5"}] | ["B"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265411/exam_images/o27yks7ncu1vj6sgzmlx.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2017 (Offline) Mathematics - Application of Derivatives Question 167 English Explanation">
We have
<br><br>Total... | mcq | jee-main-2017-offline |
YesRI3v7YkrCJy2r | maths | application-of-derivatives | maxima-and-minima | Let $$f\left( x \right) = {x^2} + {1 \over {{x^2}}}$$ and $$g\left( x \right) = x - {1 \over x}$$,
<br/>$$x \in R - \left\{ { - 1,0,1} \right\}$$.
<br/>If $$h\left( x \right) = {{f\left( x \right)} \over {g\left( x \right)}}$$, then the local minimum value of h(x) is | [{"identifier": "A", "content": "$$2\\sqrt 2 $$"}, {"identifier": "B", "content": "3"}, {"identifier": "C", "content": "-3"}, {"identifier": "D", "content": "$$-2\\sqrt 2 $$"}] | ["A"] | null | Given $$f\left( x \right) = {x^2} + {1 \over {{x^2}}}$$ and $$g\left( x \right) = x - {1 \over x}$$
<br><br>As $$h\left( x \right) = {{f\left( x \right)} \over {g\left( x \right)}}$$
<br><br>= $${{{x^2} + {1 \over {{x^2}}}} \over {x - {1 \over x}}}$$
<br><br>= $${{{{\left( {x - {1 \over x}} \right)}^2} + 2.x.{1 \over x... | mcq | jee-main-2018-offline |
45nNp849vryhuiO8gPGhJ | maths | application-of-derivatives | maxima-and-minima | If a right circular cone, having maximum volume, is inscribed in a sphere of radius 3 cm, then the curved surface area (in cm<sup>2</sup>) of this cone is : | [{"identifier": "A", "content": "$$6\\sqrt 2 \\pi $$"}, {"identifier": "B", "content": "$$6\\sqrt 3 \\pi $$"}, {"identifier": "C", "content": "$$8\\sqrt 2 \\pi $$"}, {"identifier": "D", "content": "$$8\\sqrt 3 \\pi $$"}] | ["D"] | null | Sphere of radius r = 3 cm
<br><br>Let b, h be base radius and height of cone respectively.
<br><br>So, volume of cone = $${1 \over 2}$$ $$\pi $$b<sup>2</sup>h
<br><br><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265946/exam_images/x9rxyugujvfepiew1pgy.webp" style="max-width: 100%; height: auto;disp... | mcq | jee-main-2018-online-15th-april-morning-slot |
rkBBF51Mq0IdrNDlgz0jp | maths | application-of-derivatives | maxima-and-minima | Let M and m be respectively the absolute maximum and the absolute minimum values of the function, f(x) = 2x<sup>3</sup> $$-$$ 9x<sup>2</sup> + 12x + 5 in the interval [0, 3]. Then M $$-$$m is equal to : | [{"identifier": "A", "content": "5"}, {"identifier": "B", "content": "9"}, {"identifier": "C", "content": "4"}, {"identifier": "D", "content": "1"}] | ["B"] | null | <p>To determine the absolute maximum (M) and absolute minimum (m) of the function $ f(x) = 2x^3 - 9x^2 + 12x + 5 $ over the interval $[0, 3]$, we need to examine its critical points and endpoints.</p>
<p>First, we find the derivative of the function, $ f'(x) $, to locate the critical points:</p>
<p>
<p>$$ f'(x) = \f... | mcq | jee-main-2018-online-16th-april-morning-slot |
R9TaH8Cmtts4XSPllBy4F | maths | application-of-derivatives | maxima-and-minima | A helicopter is flying along the curve given by y – x<sup>3/2</sup> = 7, (x $$ \ge $$ 0). A soldier positioned at the point $$\left( {{1 \over 2},7} \right)$$ wants to shoot down the helicopter when it is nearest to him. Then this nearest distance is - | [{"identifier": "A", "content": "$${1 \\over 6}\\sqrt {{7 \\over 3}} $$"}, {"identifier": "B", "content": "$${{\\sqrt 5 } \\over 6}$$"}, {"identifier": "C", "content": "$${1 \\over 2}$$"}, {"identifier": "D", "content": "$${1 \\over 3}$$$$\\sqrt {{7 \\over 3}} $$"}] | ["A"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264139/exam_images/woynacqz8rh7gnukj3c5.webp" style="max-width: 100%; height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2019 (Online) 10th January Evening Slot Mathematics - Application of Derivatives Question 153 English Explanat... | mcq | jee-main-2019-online-10th-january-evening-slot |
H7ek3BudnKTxOUvEFo54m | maths | application-of-derivatives | maxima-and-minima | If ƒ(x) is a non-zero polynomial of degree four,
having local extreme points at x = –1, 0, 1; then
the set
<br/>S = {x $$ \in $$ R : ƒ(x) = ƒ(0)}<br/>
Contains exactly : | [{"identifier": "A", "content": "four rational numbers."}, {"identifier": "B", "content": "four irrational numbers."}, {"identifier": "C", "content": "two irrational and one rational number."}, {"identifier": "D", "content": "two irrational and two rational numbes."}] | ["C"] | null | Local extreme points of f(x) is at x = –1, 0, 1.
<br><br>$$ \therefore $$ f'(x) = 0 has three solutions x = –1, 0, 1.
<br><br>$$ \therefore $$ f'(x) = k(x + 1)x(x - 1)
<br><br>$$\int {f'(x)dx} = \int {k(x + 1)x(x - 1)dx} $$
<br><br>f(x) = $$k\left[ {{{{x^4}} \over 4} - {{{x^2}} \over 2}} \right] + C$$
<br><br>Also giv... | mcq | jee-main-2019-online-9th-april-morning-slot |
2wavbu9DmtWZrZAiqfr4E | maths | application-of-derivatives | maxima-and-minima | The height of a right circular cylinder of maximum
volume inscribed in a sphere of radius 3 is | [{"identifier": "A", "content": "$$\\sqrt 3 $$"}, {"identifier": "B", "content": "$$2\\sqrt 3 $$"}, {"identifier": "C", "content": "$$\\sqrt 6 $$"}, {"identifier": "D", "content": "$${2 \\over 3} {\\sqrt 3} $$"}] | ["B"] | null | <picture><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265888/exam_images/flpfpj2fdznelp4hlns0.webp"><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265702/exam_images/kgucy7y7y3d2pmdomcey.webp" style="max-width: 100%;height: auto;display: block;margi... | mcq | jee-main-2019-online-8th-april-evening-slot |
bsO11rG2h8VURkIcQKDH2 | maths | application-of-derivatives | maxima-and-minima | If S<sub>1</sub> and S<sub>2</sub> are respectively the sets of local
minimum and local maximum points of the function,
<br/><br/>ƒ(x) = 9x<sup>4</sup> + 12x<sup>3</sup> – 36x<sup>2</sup> + 25, x $$ \in $$ R,
then : | [{"identifier": "A", "content": "S<sub>1</sub> = {\u20131}; S<sub>2</sub> = {0, 2}"}, {"identifier": "B", "content": "S<sub>1</sub> = {\u20132}; S<sub>2</sub> = {0, 1}"}, {"identifier": "C", "content": "S<sub>1</sub> = {\u20132, 0}; S<sub>2</sub> = {1}"}, {"identifier": "D", "content": "S<sub>1</sub> = {\u20132, 1}; S<... | ["D"] | null | ƒ(x) = 9x<sup>4</sup> + 12x<sup>3</sup> – 36x<sup>2</sup> + 25
<br><br>ƒ'(x) = 36x<sup>3</sup> + 36x<sup>2</sup> – 72x
<br><br>ƒ'(x) = 36x(x<sup>2</sup> + x – 2)
<br><br>ƒ'(x) = 36x(x + 2)(x - 1)
<picture><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263290/exam_image... | mcq | jee-main-2019-online-8th-april-morning-slot |
tOS3ma3QAHGsURo5qS5KT | maths | application-of-derivatives | maxima-and-minima | The maximum value of the function f(x) = 3x<sup>3</sup> – 18x<sup>2</sup> + 27x – 40 on the set S = $$\left\{ {x\, \in R:{x^2} + 30 \le 11x} \right\}$$ is : | [{"identifier": "A", "content": "$$-$$ 222"}, {"identifier": "B", "content": "$$-$$ 122"}, {"identifier": "C", "content": "$$122$$ "}, {"identifier": "D", "content": "222"}] | ["C"] | null | S = {x $$ \in $$ R, x<sup>2</sup> + 30 $$-$$ 11x $$ \le $$ 0}
<br><br>= {x $$ \in $$ R, 5 $$ \le $$ x $$ \le $$ 6}
<br><br>Now f(x) = 3x<sup>3</sup> $$-$$ 18x<sup>2</sup> + 27x $$-$$ 40
<br><br>$$ \Rightarrow $$ f '(x) = 9(x $$-$$ 1)(x $$-$$ 3),
<br><br>which is positive in [5, 6]
<br><br>$$ \Rightarrow $$&n... | mcq | jee-main-2019-online-11th-january-morning-slot |
wQ3wgXwJYh85rfjWcJkJl | maths | application-of-derivatives | maxima-and-minima | The shortest distance between the point $$\left( {{3 \over 2},0} \right)$$ and the curve y = $$\sqrt x $$, (x > 0), is - | [{"identifier": "A", "content": "$${{\\sqrt 3 } \\over 2}$$"}, {"identifier": "B", "content": "$${5 \\over 4}$$"}, {"identifier": "C", "content": "$${3 \\over 2}$$"}, {"identifier": "D", "content": "$${{\\sqrt 5 } \\over 2}$$"}] | ["D"] | null | Let points $$\left( {{3 \over 2},0} \right),\left( {{t^2},t} \right),t > 0$$
<br><br>Distance = $$\sqrt {{t^2} + {{\left( {{t^2} - {3 \over 2}} \right)}^2}} $$
<br><br>= $$\sqrt {{t^4} - 2{t^2} + {9 \over 4}} = \sqrt {{{\left( {{t^2} - 1} \right)}^2} + {5 \over 4}} $$
<br><br>So minimum distance is $$\sqrt {{5 \ov... | mcq | jee-main-2019-online-10th-january-morning-slot |
U1DtFitPjt3mM5fLmUIkD | maths | application-of-derivatives | maxima-and-minima | The maximum volume (in cu.m) of the right circular cone having slant height 3 m is : | [{"identifier": "A", "content": "2$$\\sqrt3$$$$\\pi $$"}, {"identifier": "B", "content": "3$$\\sqrt3$$$$\\pi $$"}, {"identifier": "C", "content": "6$$\\pi $$"}, {"identifier": "D", "content": "$${4 \\over 3}\\pi $$"}] | ["A"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263605/exam_images/srfihf2ilic7xfbenv64.webp" style="max-width: 100%; height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2019 (Online) 9th January Morning Slot Mathematics - Application of Derivatives Question 156 English Explanati... | mcq | jee-main-2019-online-9th-january-morning-slot |
1lBNcIAtmid94BAZmo7k9k2k5hkbtc7 | maths | application-of-derivatives | maxima-and-minima | Let ƒ(x) be a polynomial of degree 3 such that
ƒ(–1) = 10, ƒ(1) = –6, ƒ(x) has a critical point
at x = –1 and ƒ'(x) has a critical point at x = 1.
Then ƒ(x) has a local minima at x = _______. | [] | null | 3 | Let f(x) = ax<sup>3</sup>
+ bx<sup>2</sup>
+ cx + d
<br><br>Given f(-1) = 10, f(1) = -6
<br><br>$$ \therefore $$ -a + b - c + d = 10 ....(i)
<br><br>and a + b + c + d = -6 ......(ii)
<br><br>adding (i) + (ii)
<br><br>2(b + d) = 4
<br><br>$$ \Rightarrow $$ b + d = 2 ....(iii)
<br><br>f'(x) = 3ax<sup>2</sup> + 2bx + c
... | integer | jee-main-2020-online-8th-january-evening-slot |
bWXZpKvMYXQpzsJPLljgy2xukg38i6xk | maths | application-of-derivatives | maxima-and-minima | The set of all real values of $$\lambda $$ for which the
function<br/><br/>
$$f(x) = \left( {1 - {{\cos }^2}x} \right)\left( {\lambda + \sin x} \right),x \in \left( { - {\pi \over 2},{\pi \over 2}} \right)$$<br/><br/>has exactly one maxima and exactly one
minima, is :
| [{"identifier": "A", "content": "$$\\left( { - {3 \\over 2},{3 \\over 2}} \\right) - \\left\\{ 0 \\right\\}$$"}, {"identifier": "B", "content": "$$\\left( { - {3 \\over 2},{3 \\over 2}} \\right)$$"}, {"identifier": "C", "content": "$$\\left( { - {1 \\over 2},{1 \\over 2}} \\right) - \\left\\{ 0 \\right\\}$$"}, {"identi... | ["A"] | null | $$f(x) = \left( {1 - {{\cos }^2}x} \right)\left( {\lambda + \sin x} \right)$$
<br><br>$$ \Rightarrow $$ f(x) = sin<sup>2</sup> x($$\lambda $$ + sinx) ....(1)
<br><br>$$ \therefore $$ f'(x) = 2sinx cosx ($$\lambda $$ +sinx) + sin<sup>2</sup>x (cosx)
<br><br>$$ \Rightarrow $$ f'(x) = sin2x($${{2\lambda + 3\sin x} \o... | mcq | jee-main-2020-online-6th-september-evening-slot |
cMYWcwhlFTJFfWbA6Ejgy2xukfw0x33e | maths | application-of-derivatives | maxima-and-minima | The position of a moving car at time t is <br/>given by f(t) = at<sup>2</sup> + bt + c, t > 0, where a, b and c are real
numbers greater than 1. Then the average speed of the car over the time interval [t<sub>1</sub>
, t<sub>2</sub>
] is
attained at the point :
| [{"identifier": "A", "content": "$${{\\left( {{t_1} + {t_2}} \\right)} \\over 2}$$"}, {"identifier": "B", "content": "$${{\\left( {{t_2} - {t_1}} \\right)} \\over 2}$$"}, {"identifier": "C", "content": "2a(t<sub>1</sub>\n + t<sub>2</sub>) + b"}, {"identifier": "D", "content": "a(t<sub>2</sub>\n\u2013 t<sub>1</sub>) + b... | ["A"] | null | V<sub>av</sub> = $${{f\left( {{t_2}} \right) - f\left( {{t_1}} \right)} \over {{t_2} - {t_1}}}$$ = f'(t)
<br><br>$$ \Rightarrow $$ $${{a\left( {t_2^2 - t_1^2} \right) - b\left( {{t_2} - {t_1}} \right)} \over {{t_2} - {t_1}}}$$ = 2$$a$$t + b
<br><br>$$ \Rightarrow $$ a(t<sub>2</sub> + t<sub>1</sub>) + b = 2at + b
<br><b... | mcq | jee-main-2020-online-6th-september-morning-slot |
ZG73oFifL0rz2mngI3jgy2xukfqc6v02 | maths | application-of-derivatives | maxima-and-minima | If x = 1 is a critical point of the function
<br/>f(x) = (3x<sup>2</sup>
+ ax – 2 – a)e<sup>x</sup>
, then : | [{"identifier": "A", "content": "x = 1 is a local maxima and x = $$ - {2 \\over 3}$$ is a\nlocal minima of f."}, {"identifier": "B", "content": "x = 1 and x = $$ - {2 \\over 3}$$ are local maxima of f."}, {"identifier": "C", "content": "x = 1 and x = $$ - {2 \\over 3}$$ are local minima of f."}, {"identifier": "D", "co... | ["D"] | null | f(x) = (3x<sup>2</sup>
+ ax – 2 – a)e<sup>x</sup>
<br><br>$$ \therefore $$ f'(x) = e<sup>x</sup>(6x + a) + (3x<sup>2</sup>
+ ax – 2 – a)e<sup>x</sup>
<br><br> = e<sup>x</sup>(3x<sup>2</sup> + x(6 + a) – 2)
<br><br>f '(x) = 0 at x = 1
<br><br>$$ \Rightarrow $$ 3 + (6 + a) – 2 = 0
<br><br>$$ \Rightarrow $$ a = -7
<br><... | mcq | jee-main-2020-online-5th-september-evening-slot |
hOgXPc5U67INRRRyO9jgy2xukfg6uw39 | maths | application-of-derivatives | maxima-and-minima | If the point P on the curve, 4x<sup>2</sup> + 5y<sup>2</sup> = 20 is <br/>farthest from the point Q(0, -4), then PQ<sup>2</sup> is equal to:
| [{"identifier": "A", "content": "36"}, {"identifier": "B", "content": "48"}, {"identifier": "C", "content": "21"}, {"identifier": "D", "content": "29"}] | ["A"] | null | Given ellipse is $${{{x^2}} \over 5} + {{{y^2}} \over 4} = 1$$<br><br>Let point P is $$(\sqrt 5 \cos \theta ,\,2\sin \theta )$$<br><br>$${(PQ)^2}=5{\cos ^2}\theta + {(2\sin \theta + 4)^2}$$
<br><br>$$ \Rightarrow $$ (PQ)<sup>2</sup> = $$5{\cos ^2}\theta + 4{\sin ^2}\theta + 16\sin \theta + 16$$
<br><br>$$ \Rightar... | mcq | jee-main-2020-online-5th-september-morning-slot |
HBKNx1NQFioVfmrFstjgy2xukfahe3zt | maths | application-of-derivatives | maxima-and-minima | The area (in sq. units) of the largest rectangle ABCD whose vertices A and B lie on the x-axis and vertices C and D lie on the parabola, y = x<sup>2</sup>–1 below the x-axis, is : | [{"identifier": "A", "content": "$${1 \\over {3\\sqrt 3 }}$$"}, {"identifier": "B", "content": "$${2 \\over {3\\sqrt 3 }}$$"}, {"identifier": "C", "content": "$${4 \\over {3\\sqrt 3 }}$$"}, {"identifier": "D", "content": "$${4 \\over 3}$$"}] | ["C"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264037/exam_images/uk75haohyuzity3go0py.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2020 (Online) 4th September Evening Slot Mathematics - Application of Derivatives Question 119 English Explanati... | mcq | jee-main-2020-online-4th-september-evening-slot |
Bss5ARx998ScyyRnSIjgy2xukewssjfx | maths | application-of-derivatives | maxima-and-minima | If p(x) be a polynomial of degree three that has
a local maximum value 8 at x = 1 and a local
minimum value 4 at x = 2; then p(0) is equal to : | [{"identifier": "A", "content": "6"}, {"identifier": "B", "content": "12"}, {"identifier": "C", "content": "-12"}, {"identifier": "D", "content": "-24"}] | ["C"] | null | Since p(x) has relative extreme at
<br><br>x = 1 & 2
<br><br>so p'(x) = 0 at x = 1 & 2
<br><br>$$ \therefore $$ Let p'(x) = A(x – 1) (x – 2)
<br><br>$$ \Rightarrow $$ p(x) = $$\int {A\left( {{x^2} - 3x + 2} \right)dx} $$
<br><br>$$ \Rightarrow $$ p(x) = $${A\left( {{{{x^2}} \over 3} - 3{x^2} + 2x} \right)}$$ + ... | mcq | jee-main-2020-online-2nd-september-morning-slot |
mmwaTQKip2yYQJU89z7k9k2k5fo9w02 | maths | application-of-derivatives | maxima-and-minima | Let ƒ(x) be a polynomial of degree 5 such that x = ±1 are its critical points.
<br/><br/>If $$\mathop {\lim }\limits_{x \to 0} \left( {2 + {{f\left( x \right)} \over {{x^3}}}} \right) = 4$$, then which one of the following is not true? | [{"identifier": "A", "content": "\u0192(1) - 4\u0192(-1) = 4."}, {"identifier": "B", "content": "x = 1 is a point of minima and x = -1 is a point of maxima of \u0192."}, {"identifier": "C", "content": "x = 1 is a point of maxima and x = -1 is a point of minimum of \u0192."}, {"identifier": "D", "content": "\u0192 is an... | ["B"] | null | let f(x) = ax<sup>5</sup> + bx<sup>4</sup> + cx<sup>3</sup> + dx<sup>2</sup> + ex + f
<br><br>Given $$\mathop {\lim }\limits_{x \to 0} \left( {2 + {{f\left( x \right)} \over {{x^3}}}} \right) = 4$$
<br><br>$$ \Rightarrow $$ $$\mathop {\lim }\limits_{x \to 0} \left( {2 + {{a{x^5} + b{x^4} + c{x^3} + d{x^2} + ex + f} \ov... | mcq | jee-main-2020-online-7th-january-evening-slot |
XCWdLnNu3Bl74gtzgZ1klri0zmo | maths | application-of-derivatives | maxima-and-minima | The minimum value of $$\alpha $$ for which the <br/>equation $${4 \over {\sin x}} + {1 \over {1 - \sin x}} = \alpha $$
has at least one
solution in $$\left( {0,{\pi \over 2}} \right)$$ is ....... | [] | null | 9 | $$f(x) = {4 \over {\sin x}} + {1 \over {1 - \sin x}}$$<br><br>Let sinx = t $$ \because $$ $$x \in \left( {0,{\pi \over 2}} \right) \Rightarrow 0 < t < 1$$<br><br>$$f(t) = {4 \over t} + {1 \over {1 - t}}$$<br><br>$$f'(t) = {{ - 4} \over {{t^2}}} + {1 \over {{{(1 - t)}^2}}}$$<br><br>$$ = {{{t^2} - 4{{(1 - t)}^2}} ... | integer | jee-main-2021-online-24th-february-morning-slot |
8ssmAtzXibYG3PMlMr1kls5x94e | maths | application-of-derivatives | maxima-and-minima | Let f(x) be a polynomial of degree 6 in x, in which the coefficient of x<sup>6</sup> is unity and it has extrema at x = $$-$$1 and x = 1. If $$\mathop {\lim }\limits_{x \to 0} {{f(x)} \over {{x^3}}} = 1$$, then $$5.f(2)$$ is equal to _________. | [] | null | 144 | $$f(x) = {x^6} + a{x^5} + b{x^4} + {x^3}$$<br><br>$$\therefore$$ $$f'(x) = 6{x^5} + 5a{x^4} + 4b{x^3} + 3{x^2}$$<br><br>Roots 1 & $$-$$1<br><br>$$ \therefore $$ $$6 + 5z + 4b + 3 = 0$$ & $$ - 6 + 5a - 4b + 3 = 0$$ solving<br><br>$$a = - {3 \over 5}$$<br><br>$$b = - {3 \over 2}$$<br><br>$$ \therefore $$ $$f(x)... | integer | jee-main-2021-online-25th-february-morning-slot |
cPe9u55dm7F2RKDxWx1kluh1ma0 | maths | application-of-derivatives | maxima-and-minima | The maximum slope of the curve $$y = {1 \over 2}{x^4} - 5{x^3} + 18{x^2} - 19x$$ occurs at the point : | [{"identifier": "A", "content": "$$\\left( {3,{{21} \\over 2}} \\right)$$"}, {"identifier": "B", "content": "(0, 0)"}, {"identifier": "C", "content": "(2, 9)"}, {"identifier": "D", "content": "(2, 2)"}] | ["D"] | null | Given, $$y = {1 \over 2}{x^4} - 5{x^3} + 18{x^2} - 19x$$
<br><br>$${{dy} \over {dx}} = {1 \over 2} \times 4{x^3} - 15{x^2} + 36x - 19$$<br><br>$$ \Rightarrow $$ Slope M = $$2{x^3} - 15{x^2} + 36x - 19$$<br><br>At max of slope $${{dM} \over {dx}} = 0$$<br><br>$$ \therefore $$ $${{dM} \over {dx}} = 6{x^2} - 30x + 36 = 0$... | mcq | jee-main-2021-online-26th-february-morning-slot |
MNEisjfUr6CRd4ILoq1kmiz0uri | maths | application-of-derivatives | maxima-and-minima | The maximum value of <br/><br/>$$f(x) = \left| {\matrix{
{{{\sin }^2}x} & {1 + {{\cos }^2}x} & {\cos 2x} \cr
{1 + {{\sin }^2}x} & {{{\cos }^2}x} & {\cos 2x} \cr
{{{\sin }^2}x} & {{{\cos }^2}x} & {\sin 2x} \cr
} } \right|,x \in R$$ is : | [{"identifier": "A", "content": "$$\\sqrt 5 $$"}, {"identifier": "B", "content": "$${3 \\over 4}$$"}, {"identifier": "C", "content": "5"}, {"identifier": "D", "content": "$$\\sqrt 7 $$"}] | ["A"] | null | $$f(x) = \left| {\matrix{
{{{\sin }^2}x} & {1 + {{\cos }^2}x} & {\cos 2x} \cr
{1 + {{\sin }^2}x} & {{{\cos }^2}x} & {\cos 2x} \cr
{{{\sin }^2}x} & {{{\cos }^2}x} & {\sin 2x} \cr
} } \right|$$
<br><br>$${C_1} \to {C_1} + {C_2}$$<br><br>= $$\left| {\matrix{
2 & {1 + {{\cos ... | mcq | jee-main-2021-online-16th-march-evening-shift |
axb7TtMHVL2HxjaxWs1kmko64sd | maths | application-of-derivatives | maxima-and-minima | Let f : [$$-$$1, 1] $$ \to $$ R be defined as f(x) = ax<sup>2</sup> + bx + c for all x$$\in$$[$$-$$1, 1], where a, b, c$$\in$$R such that f($$-$$1) = 2, f'($$-$$1) = 1 for x$$\in$$($$-$$1, 1) the maximum value of f ''(x) is $${{1 \over 2}}$$. If f(x) $$ \le $$ $$\alpha$$, x$$\in$$[$$-$$1, 1], then the least value of $$... | [] | null | 5 | $$f(x) = a{x^2} + bx + c$$<br><br>$$f'(x) = 2ax + b,$$<br><br>$$f''(x) = 2a$$<br><br>Given, $$f''( - 1) = {1 \over 2}$$<br><br>$$ \Rightarrow a = {1 \over 4}$$<br><br>$$f'( - 1) = 1 \Rightarrow b - 2a = 1$$<br><br>$$ \Rightarrow b = {3 \over 2}$$<br><br>$$f( - 1) = a - b + c = 2$$<br><br>$$ \Rightarrow c = {{13} \over ... | integer | jee-main-2021-online-17th-march-evening-shift |
1krpwaqea | maths | application-of-derivatives | maxima-and-minima | Let $$A = [{a_{ij}}]$$ be a 3 $$\times$$ 3 matrix, where $${a_{ij}} = \left\{ {\matrix{
1 & , & {if\,i = j} \cr
{ - x} & , & {if\,\left| {i - j} \right| = 1} \cr
{2x + 1} & , & {otherwise.} \cr
} } \right.$$<br/><br/>Let a function f : R $$\to$$ R be defined as f(x) = det(A). Th... | [{"identifier": "A", "content": "$$ - {{20} \\over {27}}$$"}, {"identifier": "B", "content": "$${{88} \\over {27}}$$"}, {"identifier": "C", "content": "$${{20} \\over {27}}$$"}, {"identifier": "D", "content": "$$ - {{88} \\over {27}}$$"}] | ["D"] | null | $$A = \left[ {\matrix{
1 & { - x} & {2x + 1} \cr
{ - x} & 1 & { - x} \cr
{2x + 1} & { - x} & 1 \cr
} } \right]$$<br><br>$$\left| A \right| = 4{x^3} - 4{x^2} - 4x = f(x)$$<br><br>$$f'(x) = 4(3{x^2} - 2x - 1) = 0$$<br><br>$$ \Rightarrow x = 1;x = {{ - 1} \over 3}$$<br><br>$$\there... | mcq | jee-main-2021-online-20th-july-morning-shift |
1krpzb31f | maths | application-of-derivatives | maxima-and-minima | Let 'a' be a real number such that the function f(x) = ax<sup>2</sup> + 6x $$-$$ 15, x $$\in$$ R is increasing in $$\left( { - \infty ,{3 \over 4}} \right)$$ and decreasing in $$\left( {{3 \over 4},\infty } \right)$$. Then the function g(x) = ax<sup>2</sup> $$-$$ 6x + 15, x$$\in$$R has a : | [{"identifier": "A", "content": "local maximum at x = $$-$$ $${{3 \\over 4}}$$"}, {"identifier": "B", "content": "local minimum at x = $$-$$$${{3 \\over 4}}$$"}, {"identifier": "C", "content": "local maximum at x = $${{3 \\over 4}}$$"}, {"identifier": "D", "content": "local minimum at x = $${{3 \\over 4}}$$"}] | ["A"] | null | $${{ - B} \over {2A}} = {3 \over 4}$$<br><br>$$ \Rightarrow {{ - (6)} \over {2a}} = {3 \over 4}$$<br><br>$$ \Rightarrow a = {{ - 6 \times 4} \over 6} \Rightarrow a = - 4$$<br><br>$$\therefore$$ $$g(x) = 4{x^2} - 6x + 15$$<br><br>Local max. at $$x = {{ - B} \over {2A}} = - {{( - 6)} \over {2( - 4)}}$$<br><br>$$ = {{ -... | mcq | jee-main-2021-online-20th-july-morning-shift |
1krrsddj5 | maths | application-of-derivatives | maxima-and-minima | The sum of all the local minimum values of the twice differentiable function f : R $$\to$$ R defined by $$f(x) = {x^3} - 3{x^2} - {{3f''(2)} \over 2}x + f''(1)$$ is : | [{"identifier": "A", "content": "$$-$$22"}, {"identifier": "B", "content": "5"}, {"identifier": "C", "content": "$$-$$27"}, {"identifier": "D", "content": "0"}] | ["C"] | null | $$f(x) = {x^3} - 3{x^2} - {{3f''(2)} \over 2}x + f''(1)$$ ..... (i)<br><br>$$f(x) = 3{x^2} - 6x - {3 \over 2}f''(2)$$ ..... (ii)<br><br>$$f''(x) = 6x - 6$$ ..... (iii)<br><br>Now, is 3<sup>rd</sup> equation<br><br>$$f''(2) = 12 - 6 = 6$$<br><br>$$f''(11 = 0)$$<br><br>Use (ii)<br><br>$$f'(x) = 3{x^2} - 6x - {3 \over 2}f... | mcq | jee-main-2021-online-20th-july-evening-shift |
1ktbihwcy | maths | application-of-derivatives | maxima-and-minima | A wire of length 36 m is cut into two pieces, one of the pieces is bent to form a square and the other is bent to form a circle. If the sum of the areas of the two figures is minimum, and the circumference of the circle is k (meter), then $$\left( {{4 \over \pi } + 1} \right)k$$ is equal to _____________. | [] | null | 36 | Let x + y = 36<br><br>x is perimeter of square and y is perimeter of circle side of square = x/4<br><br>radius of circle = $${y \over {2\pi }}$$<br><br>Sum Areas = $${\left( {{x \over 4}} \right)^2} + \pi {\left( {{y \over {2\pi }}} \right)^2}$$<br><br>$$ = {{{x^2}} \over {16}} + {{{{(36 - x)}^2}} \over {4\pi }}$$<br><... | integer | jee-main-2021-online-26th-august-morning-shift |
1ktcydszh | maths | application-of-derivatives | maxima-and-minima | The local maximum value of the function $$f(x) = {\left( {{2 \over x}} \right)^{{x^2}}}$$, x > 0, is | [{"identifier": "A", "content": "$${\\left( {2\\sqrt e } \\right)^{{1 \\over e}}}$$"}, {"identifier": "B", "content": "$${\\left( {{4 \\over {\\sqrt e }}} \\right)^{{e \\over 4}}}$$"}, {"identifier": "C", "content": "$${(e)^{{2 \\over e}}}$$"}, {"identifier": "D", "content": "1"}] | ["C"] | null | $$f(x) = {\left( {{2 \over x}} \right)^{{x^2}}}$$ ; x > 0<br><br>$$\ln f(x) = {x^2}(\ln 2 - \ln x)$$<br><br>$$f'(x) = f(x)\{ - x + (\ln 2 - \ln x)2x\} $$<br><br>$$f'(x) = \underbrace {f(x)}_ + \,.\,\underbrace x_ + \underbrace {(2\ln 2 - 2\ln x - 1)}_{g(x)}$$<br><br>$$g(x) = 2{\ln ^2} - 2\ln x - 1$$<br><br>$$ = \ln... | mcq | jee-main-2021-online-26th-august-evening-shift |
1kteodvui | maths | application-of-derivatives | maxima-and-minima | A wire of length 20 m is to be cut into two pieces. One of the pieces is to be made into a square and the other into a regular hexagon. Then the length of the side (in meters) of the hexagon, so that the combined area of the square and the hexagon is minimum, is : | [{"identifier": "A", "content": "$${5 \\over {2 + \\sqrt 3 }}$$"}, {"identifier": "B", "content": "$${{10} \\over {2 + 3\\sqrt 3 }}$$"}, {"identifier": "C", "content": "$${5 \\over {3 + \\sqrt 3 }}$$"}, {"identifier": "D", "content": "$${{10} \\over {3 + 2\\sqrt 3 }}$$"}] | ["D"] | null | Let the wire is cut into two pieces of length x and 20 $$-$$ x.<br><br><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263573/exam_images/qgzb6vevzrjgj6qtkfoq.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 27th August Morning Shift M... | mcq | jee-main-2021-online-27th-august-morning-shift |
1kteoog7u | maths | application-of-derivatives | maxima-and-minima | The number of distinct real roots of the equation 3x<sup>4</sup> + 4x<sup>3</sup> $$-$$ 12x<sup>2</sup> + 4 = 0 is _____________. | [] | null | 4 | 3x<sup>4</sup> + 4x<sup>3</sup> $$-$$ 12x<sup>2</sup> + 4 = 0<br><br>So, let f(x) = 3x<sup>4</sup> + 4x<sup>3</sup> $$-$$ 12x<sup>2</sup> + 4<br><br>$$\therefore$$ f'(x) = 12x(x<sup>2</sup> + x $$-$$ 2)<br><br>= 12x (x + 2) (x $$-$$ 1)<br><br>$$ \therefore $$ f'(x) = 12x<sup>3</sup> + 12x<sup>2</sup> – 24x = 12x(x + 2)... | integer | jee-main-2021-online-27th-august-morning-shift |
1ktg2hnod | maths | application-of-derivatives | maxima-and-minima | A box open from top is made from a rectangular sheet of dimension a $$\times$$ b by cutting squares each of side x from each of the four corners and folding up the flaps. If the volume of the box is maximum, then x is equal to : | [{"identifier": "A", "content": "$${{a + b - \\sqrt {{a^2} + {b^2} - ab} } \\over {12}}$$"}, {"identifier": "B", "content": "$${{a + b - \\sqrt {{a^2} + {b^2} + ab} } \\over 6}$$"}, {"identifier": "C", "content": "$${{a + b - \\sqrt {{a^2} + {b^2} - ab} } \\over 6}$$"}, {"identifier": "D", "content": "$${{a + b + \\sqr... | ["C"] | null | <picture><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266717/exam_images/wemjsa96uqnlahmrdgwc.webp"><source media="(max-width: 500px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264426/exam_images/yprti8h0wfn66ietnppr.webp"><img src="https://res.c... | mcq | jee-main-2021-online-27th-august-evening-shift |
1ktio607h | maths | application-of-derivatives | maxima-and-minima | The number of real roots of the equation <br/><br/>$${e^{4x}} + 2{e^{3x}} - {e^x} - 6 = 0$$ is : | [{"identifier": "A", "content": "2"}, {"identifier": "B", "content": "4"}, {"identifier": "C", "content": "1"}, {"identifier": "D", "content": "0"}] | ["C"] | null | Let $${e^x} = t > 0$$<br><br>$$f(t) = {t^4} + 2{t^3} - t - 6 = 0$$<br><br>$$f'(t) = 4{t^3} + 6{t^2} - 1$$<br><br><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266142/exam_images/xzvjyoinhhpqhttdoa57.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE M... | mcq | jee-main-2021-online-31st-august-morning-shift |
1ktkevgzw | maths | application-of-derivatives | maxima-and-minima | Let f(x) be a cubic polynomial with f(1) = $$-$$10, f($$-$$1) = 6, and has a local minima at x = 1, and f'(x) has a local minima at x = $$-$$1. Then f(3) is equal to ____________. | [] | null | 22 | <p>Let f(x) = ax<sup>3</sup> + bx<sup>2</sup> + cx + d</p>
<p>f'(x) = 3ax<sup>2</sup> + 2bx + c $$\Rightarrow$$ f''(x) = 6ax + 2b</p>
<p>f'(x) has local minima at x = $$-$$1, so</p>
<p>$$\because$$ f''($$-$$1) = 0 $$\Rightarrow$$ $$-$$6a + 2b = 0 $$\Rightarrow$$ b = 3a ..... (i)</p>
<p>f(x) has local minima at x = 1</p... | integer | jee-main-2021-online-31st-august-evening-shift |
1l545dst9 | maths | application-of-derivatives | maxima-and-minima | <p>A wire of length 22 m is to be cut into two pieces. One of the pieces is to be made into a square and the other into an equilateral triangle. Then, the length of the side of the equilateral triangle, so that the combined area of the square and the equilateral triangle is minimum, is :</p> | [{"identifier": "A", "content": "$${{22} \\over {9 + 4\\sqrt 3 }}$$"}, {"identifier": "B", "content": "$${{66} \\over {9 + 4\\sqrt 3 }}$$"}, {"identifier": "C", "content": "$${{22} \\over {4 + 9\\sqrt 3 }}$$"}, {"identifier": "D", "content": "$${{66} \\over {4 + 9\\sqrt 3 }}$$"}] | ["B"] | null | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l5nil7kj/2ee06cba-de30-46ff-9d0f-1870cf0f243a/6bf44540-04d1-11ed-93b8-936002ac8631/file-1l5nil7kk.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l5nil7kj/2ee06cba-de30-46ff-9d0f-1870cf0f243a/6bf44540-04d1-11ed-93b8-936002ac8631... | mcq | jee-main-2022-online-29th-june-morning-shift |
1l589itwu | maths | application-of-derivatives | maxima-and-minima | <p>The sum of the absolute minimum and the absolute maximum values of the <br/><br/>function f(x) = |3x $$-$$ x<sup>2</sup> + 2| $$-$$ x in the interval [$$-$$1, 2] is :</p> | [{"identifier": "A", "content": "$${{\\sqrt {17} + 3} \\over 2}$$"}, {"identifier": "B", "content": "$${{\\sqrt {17} + 5} \\over 2}$$"}, {"identifier": "C", "content": "5"}, {"identifier": "D", "content": "$${{9 - \\sqrt {17} } \\over 2}$$"}] | ["A"] | null | $f(x)=\left|x^2-3 x-2\right|-x \forall x \in[-1,2]$<br><br>
$\Rightarrow f(x)=\left\{\begin{array}{l}x^2-4 x-2 \text { if }-1 \leq x<\frac{3-\sqrt{17}}{2} \\ -x^2+2 x+2 \text { if } \frac{3-\sqrt{17}}{2} \leq x \leq 2\end{array}\right.$<br><br>
<img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lc6nga5... | mcq | jee-main-2022-online-26th-june-morning-shift |
1l58f6wgh | maths | application-of-derivatives | maxima-and-minima | <p>Consider a cuboid of sides 2x, 4x and 5x and a closed hemisphere of radius r. If the sum of their surface areas is a constant k, then the ratio x : r, for which the sum of their volumes is maximum, is :</p> | [{"identifier": "A", "content": "2 : 5"}, {"identifier": "B", "content": "19 : 45"}, {"identifier": "C", "content": "3 : 8"}, {"identifier": "D", "content": "19 : 15"}] | ["B"] | null | <p>$$\because$$ $${s_1} + {s_2} = k$$</p>
<p>$$76{x^2} + 3\pi {r^2} = k$$</p>
<p>$$\therefore$$ $$152x{{dx} \over {dr}} + 6\pi r = 0$$</p>
<p>$$\therefore$$ $${{dx} \over {dr}} = {{ - 6\pi r} \over {152x}}$$</p>
<p>Now $$V = 40{x^3} + {2 \over 3}\pi {r^3}$$</p>
<p>$$\therefore$$ $${{dv} \over {dr}} = 120{x^2}\,.\,{{dx}... | mcq | jee-main-2022-online-26th-june-evening-shift |
1l59lf1ub | maths | application-of-derivatives | maxima-and-minima | <p>Let $$f(x) = |(x - 1)({x^2} - 2x - 3)| + x - 3,\,x \in R$$. If m and M are respectively the number of points of local minimum and local maximum of f in the interval (0, 4), then m + M is equal to ____________.</p> | [] | null | 3 | <p>$$f(x) = \left| {(x - 1)(x + 1)(x - 3)} \right| + (x - 3)$$</p>
<p>$$f(x) = \left\{ {\matrix{
{(x - 3)({x^2})} & {3 \le x \le 4} \cr
{(x - 3)(2 - {x^2})} & {1 \le x < 3} \cr
{(x - 3)({x^2})} & {0 < x < 1} \cr
} } \right.$$</p>
<p>$$f'(x) = \left\{ {\matrix{
{3{x^2} - 6x} & {3 < x < 4} \cr
{ ... | integer | jee-main-2022-online-25th-june-evening-shift |
1l5c1bifq | maths | application-of-derivatives | maxima-and-minima | <p>The sum of absolute maximum and absolute minimum values of the function $$f(x) = |2{x^2} + 3x - 2| + \sin x\cos x$$ in the interval [0, 1] is :</p> | [{"identifier": "A", "content": "$$3 + {{\\sin (1){{\\cos }^2}\\left( {{1 \\over 2}} \\right)} \\over 2}$$"}, {"identifier": "B", "content": "$$3 + {1 \\over 2}(1 + 2\\cos (1))\\sin (1)$$"}, {"identifier": "C", "content": "$$5 + {1 \\over 2}(\\sin (1) + \\sin (2))$$"}, {"identifier": "D", "content": "$$2 + \\sin \\left... | ["B"] | null | $f(x)=|(2 x-1)(x+2)|+\frac{\sin 2 x}{2}$
<br/><br/>
$0 \leq x<\frac{1}{2} \quad f(x)=(1-2 x)(x+2)+\frac{\sin 2 x}{2}$
<br/><br/>
$f^{\prime}(x)=-4 x-3+\cos 2 x<0$
<br/><br/>
For $x \geq \frac{1}{2}: \quad f^{\prime}(x)=4 x+3+\cos 2 x>0$
<br/><br/>
So, minima occurs at $x=\frac{1}{2}$
<br/><br/>
$$
\begin{aligned}
\left... | mcq | jee-main-2022-online-24th-june-morning-shift |
1l5vzjzdv | maths | application-of-derivatives | maxima-and-minima | <p>If xy<sup>4</sup> attains maximum value at the point (x, y) on the line passing through the points (50 + $$\alpha$$, 0) and (0, 50 + $$\alpha$$), $$\alpha$$ > 0, then (x, y) also lies on the line :</p> | [{"identifier": "A", "content": "y = 4x"}, {"identifier": "B", "content": "x = 4y"}, {"identifier": "C", "content": "y = 4x + $$\\alpha$$"}, {"identifier": "D", "content": "x = 4y $$-$$ $$\\alpha$$"}] | ["A"] | null | <p>Equation of line passing through the point (50 + $$\alpha$$, 0) and (0, 50 + $$\alpha$$) is</p>
<p>$$y - 0 = {{50 + \alpha - 0} \over {0 - (50 + \alpha )}}\left( {x - (50 + \alpha )} \right)$$</p>
<p>$$ \Rightarrow y = - 1\left( {x - (50 + \alpha )} \right)$$</p>
<p>$$ \Rightarrow y = (50 + \alpha ) - x$$</p>
<p>$... | mcq | jee-main-2022-online-30th-june-morning-shift |
1l5vzpx90 | maths | application-of-derivatives | maxima-and-minima | <p>Let $$f(x) = 4{x^3} - 11{x^2} + 8x - 5,\,x \in R$$. Then f :</p> | [{"identifier": "A", "content": "has a local minina at $$x = {1 \\over 2}$$"}, {"identifier": "B", "content": "has a local minima at $$x = {3 \\over 4}$$"}, {"identifier": "C", "content": "is increasing in $$\\left( {{1 \\over 2},{3 \\over 4}} \\right)$$"}, {"identifier": "D", "content": "is decreasing in $$\\left( {{1... | ["D"] | null | <p>Given,</p>
<p>$$f(x) = 4{x^3} - 11{x^2} + 8x - 5$$</p>
<p>$$\therefore$$ $$f'(x) = 12{x^2} - 22x + 8$$</p>
<p>$$ = 2(6{x^2} - 11x + 4)$$</p>
<p>$$ = 2(6{x^2} - 8x - 3x + 4)$$</p>
<p>$$ = 2\left[ {2x(3x - 4) - 1(3x - 4)} \right]$$</p>
<p>$$ = 2\left[ {(3x - 4)(2x - 1)} \right]$$</p>
<p><img src="https://app-content.c... | mcq | jee-main-2022-online-30th-june-morning-shift |
1l6dv7vz2 | maths | application-of-derivatives | maxima-and-minima | <p>If the absolute maximum value of the function $$f(x)=\left(x^{2}-2 x+7\right) \mathrm{e}^{\left(4 x^{3}-12 x^{2}-180 x+31\right)}$$ in the interval $$[-3,0]$$ is $$f(\alpha)$$, then :
</p> | [{"identifier": "A", "content": "$$\\alpha=0$$"}, {"identifier": "B", "content": "$$ \\alpha=-3$$"}, {"identifier": "C", "content": "$$\\alpha \\in(-1,0)$$"}, {"identifier": "D", "content": "$$\\alpha \\in(-3,-1]$$"}] | ["B"] | null | Given, $f(x)=\underbrace{\left(x^{2}-2 x+7\right)}_{f_{1}(x)} \underbrace{e^{\left(4 x^{3}-12 x^{2}-180 x+31\right)}}_{f_{2}(x)}$
<br/><br/>
$f_{1}(x)=x^{2}-2 x+7$
<br/><br/>
$f_{1}^{\prime}(x)=2 x-2$, so $f(x)$ is decreasing in $[-3,0]$ and positive also
<br/><br/>
$f_{2}(x)=e^{4 x^{3}-12 x^{2}-180 x+31}$
<br/><br/>
$... | mcq | jee-main-2022-online-25th-july-morning-shift |
1l6dvag5k | maths | application-of-derivatives | maxima-and-minima | <p>The curve $$y(x)=a x^{3}+b x^{2}+c x+5$$ touches the $$x$$-axis at the point $$\mathrm{P}(-2,0)$$ and cuts the $$y$$-axis at the point $$Q$$, where $$y^{\prime}$$ is equal to 3 . Then the local maximum value of $$y(x)$$ is:
</p> | [{"identifier": "A", "content": "$$\\frac{27}{4}$$"}, {"identifier": "B", "content": "$$\\frac{29}{4}$$"}, {"identifier": "C", "content": "$$\\frac{37}{4}$$"}, {"identifier": "D", "content": "$$\\frac{9}{2}$$"}] | ["A"] | null | $f(x)=y=a x^{3}+b x^{2}+c x+5 \quad \ldots$ (i)
<br/><br/>
$$
\frac{d y}{d x}=3 a x^{2}+2 b x+c \quad \ldots (ii)
$$
<br/><br/>
Touches $x$-axis at $P(-2,0)$
<br/><br/>
$\left.\Rightarrow y\right|_{x=-2}=0 \Rightarrow-8 a+4 b-2 c+5=0 \quad \ldots ...(iii)$
<br/><br/>
Touches $x$-axis at $P(-2,0)$ also implies
<br/><br/... | mcq | jee-main-2022-online-25th-july-morning-shift |
1l6f3c9hf | maths | application-of-derivatives | maxima-and-minima | <p>The sum of the maximum and minimum values of the function $$f(x)=|5 x-7|+\left[x^{2}+2 x\right]$$ in the interval $$\left[\frac{5}{4}, 2\right]$$, where $$[t]$$ is the greatest integer $$\leq t$$, is ______________.</p> | [] | null | 15 | <p>$$f(x) = |5x - 7| + [{x^2} + 2x]$$</p>
<p>$$ = |5x - 7| + [{(x + 1)^2}] - 1$$</p>
<p>Critical points of</p>
<p>$$f(x) = {7 \over 5},\sqrt 5 - 1,\,\sqrt 6 - 1,\,\sqrt 7 - 1,\,\sqrt 8 - 1,\,2$$</p>
<p>$$\therefore$$ Maximum or minimum value of $$f(x)$$ occur at critical points or boundary points</p>
<p>$$\therefor... | integer | jee-main-2022-online-25th-july-evening-shift |
1l6hy2szh | maths | application-of-derivatives | maxima-and-minima | <p>If the maximum value of $$a$$, for which the function $$f_{a}(x)=\tan ^{-1} 2 x-3 a x+7$$ is non-decreasing in $$\left(-\frac{\pi}{6}, \frac{\pi}{6}\right)$$, is $$\bar{a}$$, then $$f_{\bar{a}}\left(\frac{\pi}{8}\right)$$ is equal to :</p> | [{"identifier": "A", "content": "$$\n8-\\frac{9 \\pi}{4\\left(9+\\pi^{2}\\right)}\n$$"}, {"identifier": "B", "content": "$$8-\\frac{4 \\pi}{9\\left(4+\\pi^{2}\\right)}$$"}, {"identifier": "C", "content": "$$8\\left(\\frac{1+\\pi^{2}}{9+\\pi^{2}}\\right)$$"}, {"identifier": "D", "content": "$$8-\\frac{\\pi}{4}$$"}] | ["A"] | null | $\text {Given, }$
<br/><br/>$$
\begin{aligned}
f_a(x) & =\tan ^{-1} 2 x-3 a x+7 \\\\
f_a^{\prime}(x) & =\frac{2}{1+4 x^2}-3 a
\end{aligned}
$$
<br/><br/>As the function $f_a^{\prime}(x)$ is non-decreasing
<br/><br/>$$
\begin{aligned}
& \text { in }\left(-\frac{\pi}{6}, \frac{\pi}{6}\right), \\\\
& f_a^{\prime}(x) \geq... | mcq | jee-main-2022-online-26th-july-evening-shift |
1l6m63rp3 | maths | application-of-derivatives | maxima-and-minima | <p>If the minimum value of $$f(x)=\frac{5 x^{2}}{2}+\frac{\alpha}{x^{5}}, x>0$$, is 14 , then the value of $$\alpha$$ is equal to :</p> | [{"identifier": "A", "content": "32"}, {"identifier": "B", "content": "64"}, {"identifier": "C", "content": "128"}, {"identifier": "D", "content": "256"}] | ["C"] | null | <p>$$f(x) = {{5{x^2}} \over 2} + {\alpha \over {{x^5}}}\,\,\{ x > 0\} $$</p>
<p>$$f'(x) = 5x - {{5\alpha } \over {{x^6}}} = 0$$</p>
<p>$$ \Rightarrow x = {(\alpha )^{{1 \over 7}}}$$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l7rb2qy2/d1f9b6e9-ed06-4643-80c4-d9d8e53ea216/9bffa4a0-2e7f-11... | mcq | jee-main-2022-online-28th-july-morning-shift |
1l6p31rwv | maths | application-of-derivatives | maxima-and-minima | <p>Let $$f(x)=3^{\left(x^{2}-2\right)^{3}+4}, x \in \mathrm{R}$$. Then which of the following statements are true?</p>
<p>$$\mathrm{P}: x=0$$ is a point of local minima of $$f$$</p>
<p>$$\mathrm{Q}: x=\sqrt{2}$$ is a point of inflection of $$f$$</p>
<p>$$R: f^{\prime}$$ is increasing for $$x>\sqrt{2}$$</p> | [{"identifier": "A", "content": "Only P and Q"}, {"identifier": "B", "content": "Only P and R"}, {"identifier": "C", "content": "Only Q and R"}, {"identifier": "D", "content": "All P, Q and R"}] | ["D"] | null | <p>$$f(x) = {3^{{{({x^2} - 2)}^3} + 4}},\,x \in R$$</p>
<p>$$f(x) = {81.3^{{{({x^2} - 2)}^3}}}$$</p>
<p>$$f'(x) = {81.3^{{{({x^2} - 2)}^3}}}\ln 2.3({x^2} - 2)2x$$</p>
<p>$$ = (486\ln 2)\left( {{3^{{{({x^2} - 2)}^3}}}({x^2} - 2)x} \right)$$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l7sso9mz... | mcq | jee-main-2022-online-29th-july-morning-shift |
1ldo4vr0v | maths | application-of-derivatives | maxima-and-minima | <p>The sum of the absolute maximum and minimum values of the function $$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)$$. To do this,... | mcq | jee-main-2023-online-1st-february-evening-shift |
1ldpsc9p1 | maths | application-of-derivatives | maxima-and-minima | <p>A wire of length $$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}$$ is equal to :</p> | [{"identifier": "A", "content": "6 : 1"}, {"identifier": "B", "content": "1 : 6"}, {"identifier": "C", "content": "4 : 1"}, {"identifier": "D", "content": "3 : 1"}] | ["A"] | null | $ \ell_{1}+\ell_{2}=20 \Rightarrow \frac{\mathrm{d} \ell_{2}}{\mathrm{~d} \ell_{1}}=-1$
<br/><br/>$\mathrm{A}_{1}=\left(\frac{\ell_{1}}{4}\right)^{2}$ and $\mathrm{A}_{2}=\pi\left(\frac{\ell_{2}}{2 \pi}\right)^{2}$
<br/><br/>Let $\mathrm{S}=2 \mathrm{~A}_{1}+3 \mathrm{~A}_{2}=\frac{\ell_{1}^{2}}{8}+\frac{3 \ell_{2}^{... | mcq | jee-main-2023-online-31st-january-morning-shift |
ldqwo36h | maths | application-of-derivatives | maxima-and-minima | If the functions $f(x)=\frac{x^3}{3}+2 b x+\frac{a x^2}{2}$
<br/><br/>and $g(x)=\frac{x^3}{3}+a x+b x^2, a \neq 2 b$ <br/><br/>have a common extreme point, then $a+2 b+7$ is equal to : | [{"identifier": "A", "content": "6"}, {"identifier": "B", "content": "$\\frac{3}{2}$"}, {"identifier": "C", "content": "3"}, {"identifier": "D", "content": "4"}] | ["A"] | null | <p>$$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$$</p> | mcq | jee-main-2023-online-30th-january-evening-shift |
1ldu57hdx | maths | application-of-derivatives | maxima-and-minima | <p>Let the function $$f(x) = 2{x^3} + (2p - 7){x^2} + 3(2p - 9)x - 6$$ have a maxima for some value of $$x < 0$$ and a minima for some value of $$x > 0$$. Then, the set of all values of p is</p> | [{"identifier": "A", "content": "$$\\left( { - {9 \\over 2},{9 \\over 2}} \\right)$$"}, {"identifier": "B", "content": "$$\\left( {{9 \\over 2},\\infty } \\right)$$"}, {"identifier": "C", "content": "$$\\left( {0,{9 \\over 2}} \\right)$$"}, {"identifier": "D", "content": "$$\\left( { - \\infty ,{9 \\over 2}} \\right)$$... | ["D"] | null | $f^{\prime}(x)=6 x^{2}+2 x(2 p-7)+3(2 p-9)$
<br/><br/>
$x_{1}<0, x_{2}>0$
<br/><br/>
$\Rightarrow f^{\prime}(0)<0$
<br/><br/>
$\Rightarrow p<\frac{9}{2}$<br/><br/>
$$p \in \left( { - \infty ,{9 \over 2}} \right)$$ | mcq | jee-main-2023-online-25th-january-evening-shift |
1ldv1bkyt | maths | application-of-derivatives | maxima-and-minima | <p>Let $$x=2$$ be a local minima of the function $$f(x)=2x^4-18x^2+8x+12,x\in(-4,4)$$. If M is local maximum value of the function $$f$$ in ($$-4,4)$$, then M =</p> | [{"identifier": "A", "content": "$$18\\sqrt6-\\frac{33}{2}$$"}, {"identifier": "B", "content": "$$18\\sqrt6-\\frac{31}{2}$$"}, {"identifier": "C", "content": "$$12\\sqrt6-\\frac{33}{2}$$"}, {"identifier": "D", "content": "$$12\\sqrt6-\\frac{31}{2}$$"}] | ["C"] | null | $$
\begin{aligned}
& f(x)=8 x^3-36 x+8 \\\\
& =4\left(2 x^3-9 x+2\right) \\\\
& =4(x-2)\left(2 x^2+4 x-1\right) \\\\
& =4(x-2)\left(x-\frac{-2+\sqrt{6}}{2}\right)\left(x-\frac{-2 \sqrt{6}}{2}\right)
\end{aligned}
$$<br/><br/>
Local maxima occurs at $x=\frac{-2+\sqrt{6}}{2}=x_0$<br/><br/>
$$
f\left(x_0\right)=12 \sqrt{6... | mcq | jee-main-2023-online-25th-january-morning-shift |
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 |
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/>$$
f(x) = x - \sin{2x} + \frac{1}{3}\sin{3x}... | mcq | jee-main-2023-online-13th-april-morning-shift |
1lgrg9l7x | maths | application-of-derivatives | maxima-and-minima | <p>If the local maximum value of the function $$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 |
1lgxt9s39 | maths | application-of-derivatives | maxima-and-minima | <p>A square piece of tin of side 30 cm is to be made into a box without top by cutting a square from each corner and folding up the flaps to form a box. If the volume of the box is maximum, then its surface area (in cm$$^2$$) is equal to :</p> | [{"identifier": "A", "content": "1025"}, {"identifier": "B", "content": "900"}, {"identifier": "C", "content": "800"}, {"identifier": "D", "content": "675"}] | ["C"] | null | Given, length of square sheet of side is $30 \mathrm{~cm}$.
<br><br><img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lnbdiknq/edf7bbaa-df8b-4c8e-bede-daa23c7634e0/12049450-6280-11ee-95f7-df7702d42956/file-6y3zli1lnbdiknr.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/6y3zli... | mcq | jee-main-2023-online-10th-april-morning-shift |
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