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latsm1x3m4CNbZCrVO1kmknb7se | maths | limits-continuity-and-differentiability | limits-of-algebric-function | The value of $$\mathop {\lim }\limits_{n \to \infty } {{[r] + [2r] + ... + [nr]} \over {{n^2}}}$$, where r is a non-zero real number and [r] denotes the greatest integer less than or equal to r, is equal to : | [{"identifier": "A", "content": "r"}, {"identifier": "B", "content": "$${r \\over 2}$$"}, {"identifier": "C", "content": "0"}, {"identifier": "D", "content": "2r"}] | ["B"] | null | We know,<br><br>(x $$-$$ 1) $$ \le $$ [x] < x<br><br>$$ \therefore $$ (r $$-$$ 1) $$ \le $$ [r] < r <br><br>(2r $$-$$ 1) $$ \le $$ [2r] < 2r <br><br>.<br><br>.<br><br>.<br><br>(nr $$-$$ 1) $$ \le $$ [nr] < nr<br><br>Adding<br><br>$${{n(n + 1)} \over 2}r - n \le [r] + [2r] + .......[nr] < {{n(n + 1)} \ove... | mcq | jee-main-2021-online-17th-march-evening-shift |
1krrqx6yp | maths | limits-continuity-and-differentiability | limits-of-algebric-function | If $$f:R \to R$$ is given by $$f(x) = x + 1$$, then the value of $$\mathop {\lim }\limits_{n \to \infty } {1 \over n}\left[ {f(0) + f\left( {{5 \over n}} \right) + f\left( {{{10} \over n}} \right) + ...... + f\left( {{{5(n - 1)} \over n}} \right)} \right]$$ is : | [{"identifier": "A", "content": "$${3 \\over 2}$$"}, {"identifier": "B", "content": "$${5 \\over 2}$$"}, {"identifier": "C", "content": "$${1 \\over 2}$$"}, {"identifier": "D", "content": "$${7 \\over 2}$$"}] | ["D"] | null | $$f(0) + f\left( {{5 \over n}} \right) + f\left( {{{10} \over n}} \right) + ...... + f\left( {{{5(n - 1)} \over n}} \right)$$<br><br>$$ \Rightarrow 1 + 1 + {5 \over n} + 1 + {{10} \over n} + .... + 1 + {{5(n - 1)} \over n}$$<br><br>$$ \Rightarrow n + {5 \over n}{{(n - 1)n} \over 2} = {{2n + 5n - 5} \over 2} = {{7n - 5}... | mcq | jee-main-2021-online-20th-july-evening-shift |
1ks0994hx | maths | limits-continuity-and-differentiability | limits-of-algebric-function | Let f : R $$\to$$ R be a function such that f(2) = 4 and f'(2) = 1. Then, the value of $$\mathop {\lim }\limits_{x \to 2} {{{x^2}f(2) - 4f(x)} \over {x - 2}}$$ is equal to : | [{"identifier": "A", "content": "4"}, {"identifier": "B", "content": "8"}, {"identifier": "C", "content": "16"}, {"identifier": "D", "content": "12"}] | ["D"] | null | This limit can be solved using L'Hopital's Rule, which states that for the limit of the form 0/0 or ±∞/±∞, the limit can be found by taking the derivative of the numerator and the derivative of the denominator separately.
<br/><br/>$$\mathop {\lim }\limits_{x \to 2} {{{x^2}f(2) - 4f(x)} \over {x - 2}}$$ is in the inde... | mcq | jee-main-2021-online-27th-july-morning-shift |
1ktd2oyng | maths | limits-continuity-and-differentiability | limits-of-algebric-function | $$\mathop {\lim }\limits_{x \to 2} \left( {\sum\limits_{n = 1}^9 {{x \over {n(n + 1){x^2} + 2(2n + 1)x + 4}}} } \right)$$ is equal to : | [{"identifier": "A", "content": "$${9 \\over {44}}$$"}, {"identifier": "B", "content": "$${5 \\over {24}}$$"}, {"identifier": "C", "content": "$${1 \\over 5}$$"}, {"identifier": "D", "content": "$${7 \\over {36}}$$"}] | ["A"] | null | $$S = \mathop {\lim }\limits_{x \to 2} \sum\limits_{n = 1}^9 {{x \over {n(n + 1){x^2} + 2(2n + 1)x + 4}}} $$<br><br>$$S = \sum\limits_{n = 1}^9 {{2 \over {4({n^2} + 3n + 2)}}} = {1 \over 2}\sum\limits_{n = 1}^9 {\left( {{1 \over {n + 1}} - {1 \over {n + 2}}} \right)} $$<br><br>$$S = {1 \over 2}\left( {{1 \over 2} - {1... | mcq | jee-main-2021-online-26th-august-evening-shift |
1ktg3wzp1 | maths | limits-continuity-and-differentiability | limits-of-algebric-function | If $$\mathop {\lim }\limits_{x \to \infty } \left( {\sqrt {{x^2} - x + 1} - ax} \right) = b$$, then the ordered pair (a, b) is : | [{"identifier": "A", "content": "$$\\left( {1,{1 \\over 2}} \\right)$$"}, {"identifier": "B", "content": "$$\\left( {1, - {1 \\over 2}} \\right)$$"}, {"identifier": "C", "content": "$$\\left( { - 1,{1 \\over 2}} \\right)$$"}, {"identifier": "D", "content": "$$\\left( { - 1, - {1 \\over 2}} \\right)$$"}] | ["B"] | null | $$\mathop {\lim }\limits_{x \to \infty } \left( {\sqrt {{x^2} - x + 1} } \right) - ax = b$$ ($$\infty$$ $$-$$ $$\infty$$)<br><br>Now, $$\mathop {\lim }\limits_{x \to \infty } {{({x^2} - x + 1 - {a^2}{x^2}}) \over {\sqrt {{x^2} - x + 1} + ax}} = b$$<br><br>$$ \Rightarrow \mathop {\lim }\limits_{x \to \infty } {{(1 - {a... | mcq | jee-main-2021-online-27th-august-evening-shift |
1ktoau6kf | maths | limits-continuity-and-differentiability | limits-of-algebric-function | Let $$f(x) = {x^6} + 2{x^4} + {x^3} + 2x + 3$$, x $$\in$$ R. Then the natural number n for which $$\mathop {\lim }\limits_{x \to 1} {{{x^n}f(1) - f(x)} \over {x - 1}} = 44$$ is __________. | [] | null | 7 | $$f(x) = {x^6} + 2{x^4} + {x^3} + 2x + 3$$<br><br>$$\mathop {\lim }\limits_{x \to 1} {{{x^n}f(1) - f(x)} \over {x - 1}} = 44$$<br><br>$$\mathop {\lim }\limits_{x \to 1} {{9{x^n} - ({x^6} + 2{x^4} + {x^3} + 2x + 3)} \over {x - 1}} = 44$$<br><br>$$\mathop {\lim }\limits_{x \to 1} {{9n{x^{n - 1}} - (6{x^5} + 8{x^3} + 3{x^... | integer | jee-main-2021-online-1st-september-evening-shift |
1l56u4jaa | maths | limits-continuity-and-differentiability | limits-of-algebric-function | <p>Let [t] denote the greatest integer $$\le$$ t and {t} denote the fractional part of t. The integral value of $$\alpha$$ for which the left hand limit of the function</p>
<p>$$f(x) = [1 + x] + {{{\alpha ^{2[x] + {\{x\}}}} + [x] - 1} \over {2[x] + \{ x\} }}$$ at x = 0 is equal to $$\alpha - {4 \over 3}$$, is ________... | [] | null | 3 | <p>$$f(x) = [1 + x] + {{{a^{2[x] + \{ x\} }} + [x] - 1} \over {2[x] + \{ x\} }}$$</p>
<p>$$\mathop {\lim }\limits_{x \to {0^ - }} f(x) = \alpha - {4 \over 3}$$</p>
<p>$$ \Rightarrow \mathop {\lim }\limits_{x \to {0^ - }} 1 + [x] + {{{\alpha ^{x + [x]}} + [x] - 1} \over {x + [x]}} = \alpha - {4 \over 3}$$</p>
<p>$$ \R... | integer | jee-main-2022-online-27th-june-evening-shift |
1l57o3waq | maths | limits-continuity-and-differentiability | limits-of-algebric-function | <p>Let a be an integer such that $$\mathop {\lim }\limits_{x \to 7} {{18 - [1 - x]} \over {[x - 3a]}}$$ exists, where [t] is greatest integer $$\le$$ t. Then a is equal to :</p> | [{"identifier": "A", "content": "$$-$$6"}, {"identifier": "B", "content": "$$-$$2"}, {"identifier": "C", "content": "2"}, {"identifier": "D", "content": "6"}] | ["A"] | null | <p>$$\mathop {\lim }\limits_{x \to 7} {{18 - [1 - x]} \over {[x - 3a]}}$$ exist & $$a \in I$$.</p>
<p>$$ = \mathop {\lim }\limits_{x \to 7} {{17 - [ - x]} \over {[x] - 3a}}$$ exist</p>
<p>$$RHL = \mathop {\lim }\limits_{x \to {7^ + }} {{17 - [ - x]} \over {[x] - 3a}} = {{25} \over {7 - 3a}}$$ $$\left[ {a \ne {7 \over 3... | mcq | jee-main-2022-online-27th-june-morning-shift |
1l5ai4x49 | maths | limits-continuity-and-differentiability | limits-of-algebric-function | <p>Let f(x) be a polynomial function such that $$f(x) + f'(x) + f''(x) = {x^5} + 64$$. Then, the value of $$\mathop {\lim }\limits_{x \to 1} {{f(x)} \over {x - 1}}$$ is equal to:</p> | [{"identifier": "A", "content": "$$-$$15"}, {"identifier": "B", "content": "$$-$$60"}, {"identifier": "C", "content": "60"}, {"identifier": "D", "content": "15"}] | ["A"] | null | <p>Given, $$f(x) + f'(x) + f''(x) = {x^5} + 64$$ .........(i)</p>
<p>$\Rightarrow f(x)$ is a polynomial in $x$ whose degree is 5.</p>
<p>Let $$f(x) = {x^5} + a{x^4} + b{x^3} + c{x^2} + dx + e$$</p>
<p>$$f'(x) = 5{x^4} + 4a{x^3} + 3b{x^2} + 2cx + d$$</p>
<p>$$f''(x) = 20{x^3} + 12a{x^2} + 6bx + 2c$$</p>
<p>On substituti... | mcq | jee-main-2022-online-25th-june-morning-shift |
1l6dv4a68 | maths | limits-continuity-and-differentiability | limits-of-algebric-function | <p>If $$\mathop {\lim }\limits_{n \to \infty } \left( {\sqrt {{n^2} - n - 1} + n\alpha + \beta } \right) = 0$$, then $$8(\alpha+\beta)$$ is equal to :
</p> | [{"identifier": "A", "content": "4"}, {"identifier": "B", "content": "$$-$$8"}, {"identifier": "C", "content": "$$-$$4"}, {"identifier": "D", "content": "8"}] | ["C"] | null | <p>$$\mathop {\lim }\limits_{n \to \alpha } \left( {\sqrt {{n^2} - n - 1} + n\alpha + \beta } \right) = 0$$</p>
<p>[ This limit will be zero when $$\alpha$$ < 0 as when $$\alpha$$ > 0 then overall limit will be $$\infty $$. ]</p>
<p>$$ \Rightarrow \mathop {\lim }\limits_{n \to \alpha } {{\left( {\sqrt {{n^2} - n - 1}... | mcq | jee-main-2022-online-25th-july-morning-shift |
ldoa27ax | maths | limits-continuity-and-differentiability | limits-of-algebric-function | $$
\lim\limits_{x \rightarrow \infty} \frac{(\sqrt{3 x+1}+\sqrt{3 x-1})^6+(\sqrt{3 x+1}-\sqrt{3 x-1})^6}{\left(x+\sqrt{x^2-1}\right)^6+\left(x-\sqrt{x^2-1}\right)^6} x^3
$$ | [{"identifier": "A", "content": "is equal to 9"}, {"identifier": "B", "content": "is equal to $\\frac{27}{2}$"}, {"identifier": "C", "content": "does not exist"}, {"identifier": "D", "content": "is equal to 27"}] | ["D"] | null | $\lim \limits_{x \rightarrow \infty} \frac{(\sqrt{3 x+1}+\sqrt{3 x-1})^{6}+(\sqrt{3 x+1}-\sqrt{3 x-1})^{6}}{\left(x+\sqrt{x^{2}-1}\right)^{6}+\left(x-\sqrt{x^{2}-1}\right)^{6}} x^{3}$
<br/><br/>$$
\begin{aligned}
& = \lim \limits_{x \rightarrow \infty} x^{3} \times\left\{\frac{x^{3}\left\{\left(\sqrt{3+\frac{1}{x}}+\s... | mcq | jee-main-2023-online-31st-january-evening-shift |
1ldv1f3qc | maths | limits-continuity-and-differentiability | limits-of-algebric-function | <p>The value of $$\mathop {\lim }\limits_{n \to \infty } {{1 + 2 - 3 + 4 + 5 - 6\, + \,.....\, + \,(3n - 2) + (3n - 1) - 3n} \over {\sqrt {2{n^4} + 4n + 3} - \sqrt {{n^4} + 5n + 4} }}$$ is :</p> | [{"identifier": "A", "content": "$${3 \\over {2\\sqrt 2 }}$$"}, {"identifier": "B", "content": "$${3 \\over 2}(\\sqrt 2 + 1)$$"}, {"identifier": "C", "content": "$$3(\\sqrt 2 + 1)$$"}, {"identifier": "D", "content": "$${{\\sqrt 2 + 1} \\over 2}$$"}] | ["B"] | null | $$
\begin{aligned}
& I=\lim _{n \rightarrow \infty} \frac{(1+2+3+\ldots+3 n)-2(3+6+9+. .+3 n)}{\sqrt{2 n^4+4 n+3}-\sqrt{n^4+5 n+4}} \\\\
& =\lim _{n \rightarrow \infty} \frac{\frac{3 n(3 n+1)}{2}-6 \frac{n(n+1)}{2}}{\left(\sqrt{2 n^4+4 n+3}-\sqrt{n^4+5 n+4}\right)} \\\\
& =\lim _{n \rightarrow \infty} \frac{3 n(n-1)\le... | mcq | jee-main-2023-online-25th-january-morning-shift |
1ldwwk58p | maths | limits-continuity-and-differentiability | limits-of-algebric-function | <p>The set of all values of $$a$$ for which $$\mathop {\lim }\limits_{x \to a} ([x - 5] - [2x + 2]) = 0$$, where [$$\alpha$$] denotes the greatest integer less than or equal to $$\alpha$$ is equal to</p> | [{"identifier": "A", "content": "$$[-7.5,-6.5]$$"}, {"identifier": "B", "content": "$$(-7.5,-6.5]$$"}, {"identifier": "C", "content": "$$[-7.5,-6.5)$$"}, {"identifier": "D", "content": "$$(-7.5,-6.5)$$"}] | ["D"] | null | <p>$$\mathop {\lim }\limits_{x \to a} \left( {[x - 5] - [2x + 2]} \right) = 0$$</p>
<p>$$ \Rightarrow \mathop {\lim }\limits_{x \to a} \left( {[x] - 5 - [2x] - 2} \right) = 0$$</p>
<p>$$ \Rightarrow \mathop {\lim }\limits_{x \to a} [x] - [2x] - 7 = 0$$</p>
<p>Case 1 :</p>
<p>If $$a \in \left[ {n,n + {1 \over 2}} \right... | mcq | jee-main-2023-online-24th-january-evening-shift |
1lh22m2xj | maths | limits-continuity-and-differentiability | limits-of-algebric-function | <p>Let $$a_{1}, a_{2}, a_{3}, \ldots, a_{\mathrm{n}}$$ be $$\mathrm{n}$$ positive consecutive terms of an arithmetic progression. If $$\mathrm{d} > 0$$ is its common difference, then</p>
<p>$$\lim_\limits{n \rightarrow \infty} \sqrt{\frac{d}{n}}\left(\frac{1}{\sqrt{a_{1}}+\sqrt{a_{2}}}+\frac{1}{\sqrt{a_{2}}+\sqrt{a_... | [{"identifier": "A", "content": "$$\\frac{1}{\\sqrt{d}}$$"}, {"identifier": "B", "content": "1"}, {"identifier": "C", "content": "0"}, {"identifier": "D", "content": "$$\\sqrt{d}$$"}] | ["B"] | null | $$\lim_\limits{n \rightarrow \infty} \sqrt{\frac{d}{n}}\left(\frac{1}{\sqrt{a_{1}}+\sqrt{a_{2}}}+\frac{1}{\sqrt{a_{2}}+\sqrt{a_{3}}}+\ldots \ldots \ldots+\frac{1}{\sqrt{a_{n-1}}+\sqrt{a_{n}}}\right)$$
<br/><br/>Now,
<br/><br/>$\begin{aligned} & \frac{1}{\sqrt{a_1}+\sqrt{a_2}}+\frac{1}{\sqrt{a_2}+\sqrt{a_3}}+\ldots+\fr... | mcq | jee-main-2023-online-6th-april-morning-shift |
lsamze84 | maths | limits-continuity-and-differentiability | limits-of-algebric-function | Let $f(x)=\left\{\begin{array}{l}x-1, x \text { is even, } \\ 2 x, \quad x \text { is odd, }\end{array} x \in \mathbf{N}\right.$.
<br/><br/> If for some $\mathrm{a} \in \mathbf{N}, f(f(f(\mathrm{a})))=21$, then $\lim\limits_{x \rightarrow \mathrm{a}^{-}}\left\{\frac{|x|^3}{\mathrm{a}}-\left[\frac{x}{\mathrm{a}}\right]\... | [{"identifier": "A", "content": "169"}, {"identifier": "B", "content": "121"}, {"identifier": "C", "content": "225"}, {"identifier": "D", "content": "144"}] | ["D"] | null | $f(x)=\left\{\begin{array}{l}x-1, x \text { is even, } \\ 2 x, \quad x \text { is odd, }\end{array} x \in \mathbf{N}\right.$
<br/><br/>Let $a$ is odd
<br/><br/>$$
\begin{aligned}
& \Rightarrow f(a)=2 a \\\\
& \Rightarrow f(f(a))=2 a-1 \\\\
& \Rightarrow f(f(f(a)))=2(2 a-1)
\end{aligned}
$$
<br/><br/>$2(2 a-1)=21$ Not p... | mcq | jee-main-2024-online-1st-february-evening-shift |
lv0vxdko | maths | limits-continuity-and-differentiability | limits-of-algebric-function | <p>If $$\lim _\limits{x \rightarrow 1} \frac{(5 x+1)^{1 / 3}-(x+5)^{1 / 3}}{(2 x+3)^{1 / 2}-(x+4)^{1 / 2}}=\frac{\mathrm{m} \sqrt{5}}{\mathrm{n}(2 \mathrm{n})^{2 / 3}}$$, where $$\operatorname{gcd}(\mathrm{m}, \mathrm{n})=1$$, then $$8 \mathrm{~m}+12 \mathrm{n}$$ is equal to _______.</p> | [] | null | 100 | <p>$$I=\lim _\limits{x \rightarrow 1} \frac{(5 x+1)^{1 / 3}-(+5)^{1 /}}{(2 x+3)^{1 / 2}-(x+4)^{1 / 2}}$$</p>
<p>From: $$\frac{0}{0}$$, using $$\mathrm{L}-\mathrm{H}$$ rule</p>
<p>$$\begin{aligned}
& I=\lim _{x \rightarrow 1} \frac{\frac{1}{3} \times 5(5 x+1)^{-2 / 3}--(+5)}{\frac{1}{2} \times 2(2 x+3)^{-1 / 2}-\frac{1}... | integer | jee-main-2024-online-4th-april-morning-shift |
lvb294dv | maths | limits-continuity-and-differentiability | limits-of-algebric-function | <p>$$\lim _\limits{n \rightarrow \infty} \frac{\left(1^2-1\right)(n-1)+\left(2^2-2\right)(n-2)+\cdots+\left((n-1)^2-(n-1)\right) \cdot 1}{\left(1^3+2^3+\cdots \cdots+n^3\right)-\left(1^2+2^2+\cdots \cdots+n^2\right)}$$ is equal to :</p> | [{"identifier": "A", "content": "$$\\frac{2}{3}$$\n"}, {"identifier": "B", "content": "$$\\frac{1}{2}$$\n"}, {"identifier": "C", "content": "$$\\frac{3}{4}$$\n"}, {"identifier": "D", "content": "$$\\frac{1}{3}$$"}] | ["D"] | null | <p>$$\mathop {\lim }\limits_{n \to \infty } {{({1^2} - 1)(n - 1) + ({2^2} - 2)(n - 2) + ... + \left( {{{(n - 1)}^2} - (n - 1)} \right) \times 1} \over {({1^3} + {2^3} + ... + {n^3}) - ({1^2} + {2^2} + ... + {n^2})}}$$</p>
<p>$$\begin{aligned}
& \text { Numerator }=\sum_{r=1}^{n-1}\left((r-1)^2-(r-1)\right)(n-r) \\
& =\... | mcq | jee-main-2024-online-6th-april-evening-shift |
IKRUfBV0w2RAAt7D | maths | limits-continuity-and-differentiability | limits-of-exponential-functions | $$\mathop {\lim }\limits_{x \to \infty } {\left( {{{{x^2} + 5x + 3} \over {{x^2} + x + 2}}} \right)^x}$$ | [{"identifier": "A", "content": "$${e^4}$$"}, {"identifier": "B", "content": "$${e^2}$$"}, {"identifier": "C", "content": "$${e^3}$$"}, {"identifier": "D", "content": "$$1$$"}] | ["A"] | null | $$\mathop {\lim }\limits_{x \to \infty } {\left( {{{{x^2} + 5x + 3} \over {{x^2} + x + 2}}} \right)^x}$$
<br><br>$$ = \mathop {\lim }\limits_{x \to \infty } {\left( {1 + {{4x + 1} \over {{x^2} + x + 2 }}} \right)^x}$$
<br><br>$$ = \mathop {\lim }\limits_{x \to \infty } {\left[ {{{\left( {1 + {{4x + 1} \over {{x^2} + x ... | mcq | aieee-2002 |
MCKjvmuPcTeZAzZD | maths | limits-continuity-and-differentiability | limits-of-exponential-functions | If $$\mathop {\lim }\limits_{x \to \infty } {\left( {1 + {a \over x} + {b \over {{x^2}}}} \right)^{2x}} = {e^2}$$, then the value of $$a$$ and $$b$$, are | [{"identifier": "A", "content": "$$a$$ = 1 and $$b$$ = 2"}, {"identifier": "B", "content": "$$a$$ = 1 and $$b$$ $$ \\in R$$"}, {"identifier": "C", "content": "$$a$$ $$ \\in R$$ and $$b$$ = 2"}, {"identifier": "D", "content": "$$a$$ $$ \\in R$$ and $$b$$ $$ \\in R$$"}] | ["B"] | null | We know that $$\mathop {\lim }\limits_{x \to \infty } \left( {1 + x{1 \over x}} \right) = e$$
<br><br>$$\therefore$$ $$\mathop {\lim }\limits_{x \to \infty } {\left( {1 + {a \over x} + {b \over {{x^2}}}} \right)^{2x}} = {e^2}$$
<br><br>$$ \Rightarrow \mathop {\lim }\limits_{x \to \infty } {\left[ {\left( {1 + {a \over ... | mcq | aieee-2004 |
qkqEIPW5oUqkmilbNPzuL | maths | limits-continuity-and-differentiability | limits-of-exponential-functions | If $$\mathop {\lim }\limits_{x \to \infty } {\left( {1 + {a \over x} - {4 \over {{x^2}}}} \right)^{2x}} = {e^3},$$ then 'a' is equal to : | [{"identifier": "A", "content": "2"}, {"identifier": "B", "content": "$${3 \\over 2}$$ "}, {"identifier": "C", "content": "$${2 \\over 3}$$"}, {"identifier": "D", "content": "$${1 \\over 2}$$"}] | ["B"] | null | Given,
<br><br>$$\mathop {\lim }\limits_{x \to \propto } {\left( {1 + {a \over x} - {4 \over {{x^2}}}} \right)^{2x}} = {e^3}$$
<br><br>So, $$\mathop {\lim }\limits_{x \to \propto } {\left( {1 + {a \over x} - {4 \over {{x^2}}}} \right)^{2x}}\,\left[ {{1^ \propto }\,\,\,form} \right]$$
<br><br>$$ = {e^{\ma... | mcq | jee-main-2016-online-9th-april-morning-slot |
6GVqfQiZB8zwXJZLZS7k9k2k5e4r5pg | maths | limits-continuity-and-differentiability | limits-of-exponential-functions | $$\mathop {\lim }\limits_{x \to 2} {{{3^x} + {3^{3 - x}} - 12} \over {{3^{ - x/2}} - {3^{1 - x}}}}$$ is equal to_______. | [] | null | 36 | $$\mathop {\lim }\limits_{x \to 2} {{{3^x} + {3^{3 - x}} - 12} \over {{3^{ - x/2}} - {3^{1 - x}}}}$$
<br><br>let 3<sup>x/2</sup> = t
<br><br>= $$\mathop {\lim }\limits_{t \to 3} {{{t^2} + {{27} \over {{t^2}}} - 12} \over {{1 \over t} - {3 \over {{t^2}}}}}$$
<br><br>= $$\mathop {\lim }\limits_{t \to 3} {{\left( {{t^2} -... | integer | jee-main-2020-online-7th-january-morning-slot |
6XlUY8FhC12Q5CmfPJ7k9k2k5gqyxk1 | maths | limits-continuity-and-differentiability | limits-of-exponential-functions | $$\mathop {\lim }\limits_{x \to 0} {\left( {{{3{x^2} + 2} \over {7{x^2} + 2}}} \right)^{{1 \over {{x^2}}}}}$$ is equal to | [{"identifier": "A", "content": "e"}, {"identifier": "B", "content": "e<sup>2</sup>"}, {"identifier": "C", "content": "$${1 \\over {{e^2}}}$$"}, {"identifier": "D", "content": "$${1 \\over e}$$"}] | ["C"] | null | Given $$\mathop {\lim }\limits_{x \to 0} {\left( {{{3{x^2} + 2} \over {7{x^2} + 2}}} \right)^{{1 \over {{x^2}}}}}$$
<br><br>Putting x = 0 we get 1<sup>$$\infty $$</sup> form.
<br><br>$$ \therefore $$ $${e^{\mathop {\lim }\limits_{x \to 0} {1 \over {{x^2}}}\left[ {{{3{x^2} + 2} \over {7{x^2} + 2}} - 1} \right]}}$$
<br><... | mcq | jee-main-2020-online-8th-january-morning-slot |
FBNwrp6wuHYjAyCO6Pjgy2xukezaisy8 | maths | limits-continuity-and-differentiability | limits-of-exponential-functions | $$\mathop {\lim }\limits_{x \to 0} {\left( {\tan \left( {{\pi \over 4} + x} \right)} \right)^{{1 \over x}}}$$ is equal to : | [{"identifier": "A", "content": "2"}, {"identifier": "B", "content": "1"}, {"identifier": "C", "content": "$$e$$"}, {"identifier": "D", "content": "$$e$$<sup>2</sup>"}] | ["D"] | null | $$\mathop {\lim }\limits_{x \to 0} {\left( {\tan \left( {{\pi \over 4} + x} \right)} \right)^{{1 \over x}}}$$
<br><br>This is 1<sup>$$\infty $$</sup> form.
<br><br>= $${e^{\mathop {\lim }\limits_{x \to 0} \left[ {\tan \left( {{\pi \over 4} + x} \right) - 1} \right] \times {1 \over x}}}$$
<br><br>= $${e^{\mathop {\lim... | mcq | jee-main-2020-online-2nd-september-evening-slot |
t00z1tGGAJXa8Rxka5jgy2xukfqbtvgj | maths | limits-continuity-and-differentiability | limits-of-exponential-functions | $$\mathop {\lim }\limits_{x \to 0} {{x\left( {{e^{\left( {\sqrt {1 + {x^2} + {x^4}} - 1} \right)/x}} - 1} \right)} \over {\sqrt {1 + {x^2} + {x^4}} - 1}}$$ | [{"identifier": "A", "content": "is equal to 0."}, {"identifier": "B", "content": "is equal to $$\\sqrt e $$."}, {"identifier": "C", "content": "is equal to 1."}, {"identifier": "D", "content": "does not exist."}] | ["C"] | null | $$\mathop {\lim }\limits_{x \to 0} {{x\left( {{e^{\left( {\sqrt {1 + {x^2} + {x^4}} - 1} \right)/x}} - 1} \right)} \over {\sqrt {1 + {x^2} + {x^4}} - 1}}$$
<br><br>= $$\mathop {\lim }\limits_{x \to 0} {{x\left[ {{e^{{{\left( {\sqrt {1 + {x^2} + {x^4}} - 1} \right)\left( {\sqrt {1 + {x^2} + {x^4}} + 1} \right)} \ove... | mcq | jee-main-2020-online-5th-september-evening-slot |
yESaXThRJN9Sg4L4Rs1kls4zfqx | maths | limits-continuity-and-differentiability | limits-of-exponential-functions | $$\mathop {\lim }\limits_{n \to \infty } {\left( {1 + {{1 + {1 \over 2} + ........ + {1 \over n}} \over {{n^2}}}} \right)^n}$$ is equal to : | [{"identifier": "A", "content": "$${{1 \\over 2}}$$"}, {"identifier": "B", "content": "1"}, {"identifier": "C", "content": "0"}, {"identifier": "D", "content": "$${{1 \\over e}}$$"}] | ["B"] | null | It is $${1^\infty }$$ form<br><br>$$L = {e^{\mathop {\lim }\limits_{n \to \infty } \left( {{{1 + {1 \over 2} + {1 \over 3} + ...{1 \over n}} \over n}} \right)}}$$<br><br>$$S = 1 + \left( {{1 \over 2} + {1 \over 3}} \right) + \left( {{1 \over 4} + {1 \over 5} + {1 \over 6} + {1 \over 7}} \right) + \left( {{1 \over 8} + ... | mcq | jee-main-2021-online-25th-february-morning-slot |
BJ4mdrETo7ewozsDTW1klta44r9 | maths | limits-continuity-and-differentiability | limits-of-exponential-functions | If $$\mathop {\lim }\limits_{x \to 0} {{ax - ({e^{4x}} - 1)} \over {ax({e^{4x}} - 1)}}$$ exists and is equal to b, then the value of a $$-$$ 2b is __________. | [] | null | 5 | $$\mathop {\lim }\limits_{x \to 0} {{ax - \left( {{e^{4x}} - 1} \right)} \over {ax\left( {{e^{4x}} - 1} \right)}}$$<br><br>Applying L' Hospital Rule<br><br>$$\mathop {\lim }\limits_{x \to 0} {{a - 4{e^{4x}}} \over {a\left( {{e^{4x}} - 1} \right) + ax\left( {4{e^{4x}}} \right)}}$$
<br><br>This is $${{a - 4} \over 0}$$.
... | integer | jee-main-2021-online-25th-february-evening-slot |
1krq1tmci | maths | limits-continuity-and-differentiability | limits-of-exponential-functions | If the value of $$\mathop {\lim }\limits_{x \to 0} {(2 - \cos x\sqrt {\cos 2x} )^{\left( {{{x + 2} \over {{x^2}}}} \right)}}$$ is equal to e<sup>a</sup>, then a is equal to __________. | [] | null | 3 | $$\mathop {\lim }\limits_{x \to 0} {(2 - \cos x\sqrt {\cos 2x} )^{{{x + 2} \over {{x^2}}}}}$$<br><br>form : 1<sup>$$\infty$$</sup> $$ = {e^{\mathop {\lim }\limits_{x \to 0} \left( {{{1 - \cos x\sqrt {\cos 2x} } \over {{x^2}}}} \right) \times (x + 2)}}$$<br><br>Now, $$\mathop {\lim }\limits_{x \to 0} {{1 - \cos x\sqrt {... | integer | jee-main-2021-online-20th-july-morning-shift |
1kten7puo | maths | limits-continuity-and-differentiability | limits-of-exponential-functions | If $$\alpha$$, $$\beta$$ are the distinct roots of x<sup>2</sup> + bx + c = 0, then <br/><br/>$$\mathop {\lim }\limits_{x \to \beta } {{{e^{2({x^2} + bx + c)}} - 1 - 2({x^2} + bx + c)} \over {{{(x - \beta )}^2}}}$$ is equal to : | [{"identifier": "A", "content": "b<sup>2</sup> + 4c"}, {"identifier": "B", "content": "2(b<sup>2</sup> + 4c)"}, {"identifier": "C", "content": "2(b<sup>2</sup> $$-$$ 4c)"}, {"identifier": "D", "content": "b<sup>2</sup> $$-$$ 4c"}] | ["C"] | null | $$\mathop {\lim }\limits_{x \to \beta } {{{e^{2({x^2} + bx + c)}} - 1 - 2({x^2} + bx + c)} \over {{{(x - \beta )}^2}}}$$<br><br>$$ = \mathop {\lim }\limits_{x \to \beta } {{1\left( {1 + {{2({x^2} + bx + c)} \over {1!}} + {{{2^2}{{({x^2} + bx + c)}^2}} \over {2!}} + ...} \right) - 1 - 2({x^2} + bx + c)} \over {{{(x - \b... | mcq | jee-main-2021-online-27th-august-morning-shift |
1l6hy7n8m | maths | limits-continuity-and-differentiability | limits-of-exponential-functions | <p>Let $$\beta=\mathop {\lim }\limits_{x \to 0} \frac{\alpha x-\left(e^{3 x}-1\right)}{\alpha x\left(e^{3 x}-1\right)}$$ for some $$\alpha \in \mathbb{R}$$. Then the value of $$\alpha+\beta$$ is :</p> | [{"identifier": "A", "content": "$$\\frac{14}{5}$$"}, {"identifier": "B", "content": "$$\\frac{3}{2}$$"}, {"identifier": "C", "content": "$$\\frac{5}{2}$$"}, {"identifier": "D", "content": "$$\\frac{7}{2}$$"}] | ["C"] | null | <p>$$\beta = \mathop {\lim }\limits_{x \to 0} {{\alpha x - ({e^{3x}} - 1)} \over {\alpha x({e^{3x}} - 1)}},\,\alpha \in R$$</p>
<p>$$ = \mathop {\lim }\limits_{x \to 0} {{{\alpha \over 3} - \left( {{{{e^{3x}} - 1} \over {3x}}} \right)} \over {\alpha x\left( {{{{e^{3x - 1}}} \over {3x}}} \right)}}$$</p>
<p>So, $$\alp... | mcq | jee-main-2022-online-26th-july-evening-shift |
1l6m6yoet | maths | limits-continuity-and-differentiability | limits-of-exponential-functions | <p>$$\lim\limits_{x \rightarrow 0}\left(\frac{(x+2 \cos x)^{3}+2(x+2 \cos x)^{2}+3 \sin (x+2 \cos x)}{(x+2)^{3}+2(x+2)^{2}+3 \sin (x+2)}\right)^{\frac{100}{x}}$$ is equal to ___________.</p> | [] | null | 1 | <p>Let $$x + 2\cos x = a$$</p>
<p>$$x + 2 = b$$</p>
<p>as $$x \to 0$$, $$a \to 2$$ and $$b \to 2$$</p>
<p>$$\mathop {\lim }\limits_{x \to 0} {\left( {{{{a^3} + 2{a^2} + 3\sin a} \over {{b^3} + 2{b^2} + 3\sin b}}} \right)^{{{100} \over x}}}$$</p>
<p>$$ = {e^{\mathop {\lim }\limits_{x \to 0} \,.\,{{100} \over x}\,.\,{{({... | integer | jee-main-2022-online-28th-july-morning-shift |
1ldyaxaot | maths | limits-continuity-and-differentiability | limits-of-exponential-functions | <p>$$\mathop {\lim }\limits_{t \to 0} {\left( {{1^{{1 \over {{{\sin }^2}t}}}} + {2^{{1 \over {{{\sin }^2}t}}}}\, + \,...\, + \,{n^{{1 \over {{{\sin }^2}t}}}}} \right)^{{{\sin }^2}t}}$$ is equal to</p> | [{"identifier": "A", "content": "$${{n(n + 1)} \\over 2}$$"}, {"identifier": "B", "content": "n"}, {"identifier": "C", "content": "n$$^2$$ + n"}, {"identifier": "D", "content": "n$$^2$$"}] | ["B"] | null | $$
\begin{aligned}
& \lim _{t \rightarrow 0}\left(1^{\operatorname{cosec}^2 t}+2^{\operatorname{cosec}^2 t}+\ldots \ldots . .+n^{\operatorname{cosec}^2 t}\right)^{\sin ^2 t} \\\\
& =\lim _{t \rightarrow 0} n\left(\left(\frac{1}{n}\right)^{\operatorname{cosec}^2 t}+\left(\frac{2}{n}\right)^{\operatorname{cosec}^2 t}+\ld... | mcq | jee-main-2023-online-24th-january-morning-shift |
jaoe38c1lsd53rxf | maths | limits-continuity-and-differentiability | limits-of-exponential-functions | <p>If $$\lim _\limits{x \rightarrow 0} \frac{a x^2 e^x-b \log _e(1+x)+c x e^{-x}}{x^2 \sin x}=1$$, then $$16\left(a^2+b^2+c^2\right)$$ is equal to ________.</p> | [] | null | 81 | <p>$$\mathop {\lim }\limits_{x \to 0} {{a{x^2}\left( {1 + x + {{{x^2}} \over {2!}} + {{{x^3}} \over {3!}} + ....} \right) - b\left( {x - {{{x^2}} \over 2} + {{{x^3}} \over 3} - .....} \right) + cx\left( {1 - x + {{{x^2}} \over {x!}} - {{{x^3}} \over {3!}} + .....} \right)} \over {{x^3}\,.\,{{\sin x} \over x}}}$$</p>
<p... | integer | jee-main-2024-online-31st-january-evening-shift |
jaoe38c1lse4rgly | maths | limits-continuity-and-differentiability | limits-of-exponential-functions | <p>$$\lim _\limits{x \rightarrow 0} \frac{e^{2|\sin x|}-2|\sin x|-1}{x^2}$$</p> | [{"identifier": "A", "content": "is equal to 1\n"}, {"identifier": "B", "content": "does not exist\n"}, {"identifier": "C", "content": "is equal to $$-1$$\n"}, {"identifier": "D", "content": "is equal to 2"}] | ["D"] | null | <p>$$\begin{aligned}
& \lim _{x \rightarrow 0} \frac{e^{2|\sin x|}-2|\sin x|-1}{x^2} \\
& \lim _{x \rightarrow 0} \frac{e^{2|\sin x|}-2|\sin x|-1}{|\sin x|^2} \times \frac{\sin ^2 x}{x^2}
\end{aligned}$$</p>
<p>Let $$|\sin \mathrm{x}|=\mathrm{t}$$</p>
<p>$$\begin{aligned}
& \lim _{t \rightarrow 0} \frac{e^{2 t}-2 t-1}{... | mcq | jee-main-2024-online-31st-january-morning-shift |
luxwdeld | maths | limits-continuity-and-differentiability | limits-of-exponential-functions | <p>$$\lim _\limits{x \rightarrow 0} \frac{e-(1+2 x)^{\frac{1}{2 x}}}{x}$$ is equal to</p> | [{"identifier": "A", "content": "$$\\frac{-2}{e}$$\n"}, {"identifier": "B", "content": "$$e-e^2$$"}, {"identifier": "C", "content": "0"}, {"identifier": "D", "content": "$$e$$"}] | ["D"] | null | <p>$$\lim _\limits{x \rightarrow 0} \frac{e-(1+2 x)^{\frac{1}{2 x}}}{x}$$</p>
<p>Using expansion</p>
<p>$$\begin{aligned}
& =\lim _\limits{x \rightarrow 0} \frac{e-e\left[1-\frac{2 x}{2}+\frac{11 \times 4 x^2}{24}+\ldots\right]}{x} \\
& =\lim _\limits{x \rightarrow 0}\left(e-\frac{11 x}{6} e+\ldots\right)=e
\end{aligne... | mcq | jee-main-2024-online-9th-april-evening-shift |
lv3vec6m | maths | limits-continuity-and-differentiability | limits-of-exponential-functions | <p>If $$\alpha=\lim _\limits{x \rightarrow 0^{+}}\left(\frac{\mathrm{e}^{\sqrt{\tan x}}-\mathrm{e}^{\sqrt{x}}}{\sqrt{\tan x}-\sqrt{x}}\right)$$ and $$\beta=\lim _\limits{x \rightarrow 0}(1+\sin x)^{\frac{1}{2} \cot x}$$ are the roots of the quadratic equation $$\mathrm{a} x^2+\mathrm{b} x-\sqrt{\mathrm{e}}=0$$, then $$... | [] | null | 6 | <p>$$\begin{aligned}
\alpha & =\lim _{x \rightarrow 0^{+}} \frac{e^{\sqrt{\tan x}}-e^{\sqrt{x}}}{(\sqrt{\tan x}-\sqrt{x})} \\
& =\lim _{x \rightarrow 0} \frac{e^{\sqrt{x}}\left(e^{\sqrt{\tan x}-\sqrt{x}}-1\right)}{(\sqrt{\tan x}-\sqrt{x})}=1 \\
\beta & =\lim _{x \rightarrow 0}(1+\sin x)^{\frac{1}{2} \cot x}=\lim _{x \r... | integer | jee-main-2024-online-8th-april-evening-shift |
DZUSj4MdIWEgezAl | maths | limits-continuity-and-differentiability | limits-of-logarithmic-functions | $$\mathop {\lim }\limits_{x \to 0} {{\log {x^n} - \left[ x \right]} \over {\left[ x \right]}}$$, $$n \in N$$, ( [x] denotes the greatest integer less than or equal to x ) | [{"identifier": "A", "content": "has value $$ -1$$"}, {"identifier": "B", "content": "has value $$0$$"}, {"identifier": "C", "content": "has value $$1$$"}, {"identifier": "D", "content": "does not exist"}] | ["D"] | null | Since $$\mathop {\lim }\limits_{x \to 0} \left[ x \right]$$ does not exist, hence the required limit does not exist. | mcq | aieee-2002 |
gJ8W4ffbWbmqeEai | maths | limits-continuity-and-differentiability | limits-of-logarithmic-functions | If $$\mathop {\lim }\limits_{x \to 0} {{\log \left( {3 + x} \right) - \log \left( {3 - x} \right)} \over x}$$ = k, the value of k is | [{"identifier": "A", "content": "$$ - {2 \\over 3}$$"}, {"identifier": "B", "content": "0"}, {"identifier": "C", "content": "$$ - {1 \\over 3}$$"}, {"identifier": "D", "content": "$${2 \\over 3}$$"}] | ["D"] | null | $$\mathop {\lim }\limits_{x \to 0} {{\log \left( {3 + x} \right) - \log \left( {3 - x} \right)} \over x} = K$$
<br><br>(by $$L'$$ Hospital rule)
<br><br>$$\mathop {\lim }\limits_{x \to 0} {{{1 \over {3 + x}} - {{ - 1} \over {3 - x}}} \over 1} = K$$
<br><br>$$\therefore$$ $${2 \over 3} = K$$ | mcq | aieee-2003 |
1l6ggon0k | maths | limits-continuity-and-differentiability | limits-of-logarithmic-functions | <p>If the function $$f(x) = \left\{ {\matrix{
{{{{{\log }_e}(1 - x + {x^2}) + {{\log }_e}(1 + x + {x^2})} \over {\sec x - \cos x}}} & , & {x \in \left( {{{ - \pi } \over 2},{\pi \over 2}} \right) - \{ 0\} } \cr
k & , & {x = 0} \cr
} } \right.$$ is continuous at x = 0, then k is equal to:</p... | [{"identifier": "A", "content": "1"}, {"identifier": "B", "content": "$$-$$1"}, {"identifier": "C", "content": "e"}, {"identifier": "D", "content": "0"}] | ["A"] | null | <p>$$f(x) = \left\{ {\matrix{
{{{{{\log }_e}(1 - x + {x^2}) + {{\log }_e}(1 + x + {x^2})} \over {\sec x - \cos x}}} & , & {x \in \left( {{{ - \pi } \over 2},{\pi \over 2}} \right) - \{ 0\} } \cr
k & , & {x = 0} \cr
} } \right.$$</p>
<p>for continuity at $$x = 0$$</p>
<p>$$\mathop {\lim }\limits_{x \to 0} f... | mcq | jee-main-2022-online-26th-july-morning-shift |
C3FEi2ihx23NKCcI | maths | limits-continuity-and-differentiability | limits-of-trigonometric-functions | $$\mathop {\lim }\limits_{x \to 0} {{\sqrt {1 - \cos 2x} } \over {\sqrt 2 x}}$$ is | [{"identifier": "A", "content": "$$1$$"}, {"identifier": "B", "content": "$$-1$$"}, {"identifier": "C", "content": "zero"}, {"identifier": "D", "content": "does not exist"}] | ["D"] | null | $$\lim {{\sqrt {1 - \cos \,2x} } \over {\sqrt 2 x}} \Rightarrow \lim {{\sqrt {1 - \left( {1 - 2\,{{\sin }^2}\,x} \right)} } \over {\sqrt 2 x}}$$
<br><br>$$\mathop {\lim }\limits_{x \to 0} {{\sqrt {2\,{{\sin }^2}\,x} } \over {\sqrt {2x} }} \Rightarrow \mathop {\lim }\limits_{x \to 0} {{\left| {\sin x} \right|} \over x}... | mcq | aieee-2002 |
jQEkUesKCe4vxz8X | maths | limits-continuity-and-differentiability | limits-of-trigonometric-functions | $$\mathop {\lim }\limits_{x \to {\pi \over 2}} {{\left[ {1 - \tan \left( {{x \over 2}} \right)} \right]\left[ {1 - \sin x} \right]} \over {\left[ {1 + \tan \left( {{x \over 2}} \right)} \right]{{\left[ {\pi - 2x} \right]}^3}}}$$ is | [{"identifier": "A", "content": "$$\\infty $$"}, {"identifier": "B", "content": "$${1 \\over 8}$$"}, {"identifier": "C", "content": "0"}, {"identifier": "D", "content": "$${1 \\over 32}$$"}] | ["D"] | null | $$\mathop {\lim }\limits_{x \to {\pi \over 2}} {{\tan \left( {{\pi \over 4} - {x \over 2}} \right).\left( {1 - \sin x} \right)} \over {{{\left( {\pi - 2x} \right)}^3}}}$$
<br><br>Let $$x = {\pi \over 2} + x;\,\,y \to 0$$
<br><br>$$=$$ $$\mathop {\lim }\limits_{y \to 0} {{\tan \left( { - {y \over 2}} \right).\left( ... | mcq | aieee-2003 |
NJEFjGuNFNugq1vY | maths | limits-continuity-and-differentiability | limits-of-trigonometric-functions | Let $$\alpha$$ and $$\beta$$ be the distinct roots of $$a{x^2} + bx + c = 0$$, then
<br/><br/>$$\mathop {\lim }\limits_{x \to \alpha } {{1 - \cos \left( {a{x^2} + bx + c} \right)} \over {{{\left( {x - \alpha } \right)}^2}}}$$ is equal to | [{"identifier": "A", "content": "$${{{a^2}{{\\left( {\\alpha - \\beta } \\right)}^2}} \\over 2}$$"}, {"identifier": "B", "content": "0"}, {"identifier": "C", "content": "$$ - {{{a^2}{{\\left( {\\alpha - \\beta } \\right)}^2}} \\over 2}$$"}, {"identifier": "D", "content": "$${{{{\\left( {\\alpha - \\beta } \\right)}^... | ["A"] | null | Given limit $$ = \mathop {\lim }\limits_{x \to \alpha } {{1 - \cos \,a\left( {x - \alpha } \right)\left( {x - \beta } \right)} \over {{{\left( {x - \alpha } \right)}^2}}}$$
<br><br>$$ = \mathop {\lim }\limits_{x \to \alpha } {{2{{\sin }^2}\left( {a{{\left( {x - \alpha } \right)\left( {x - \beta } \right)} \over 2}} \ri... | mcq | aieee-2005 |
PuCPbTskQL6G76Mb | maths | limits-continuity-and-differentiability | limits-of-trigonometric-functions | $$\mathop {\lim }\limits_{x \to 2} \left( {{{\sqrt {1 - \cos \left\{ {2(x - 2)} \right\}} } \over {x - 2}}} \right)$$ | [{"identifier": "A", "content": "Equals $$\\sqrt 2 $$"}, {"identifier": "B", "content": "Equals $$-\\sqrt 2 $$"}, {"identifier": "C", "content": "Equals $${1 \\over {\\sqrt 2 }}$$"}, {"identifier": "D", "content": "does not exist"}] | ["D"] | null | $$\mathop {\lim }\limits_{x \to 2} {{\sqrt {1 - \cos \left\{ {2\left( {x - 2} \right)} \right\}} } \over {x - 2}}$$
<br><br>$$ = \mathop {\lim }\limits_{x \to 2} {{\sqrt 2 \left| {\sin \left( {x - 2} \right)} \right|} \over {x - 2}}$$
<br><br>$$L.H.L. = \mathop {\lim }\limits_{x \to {2^ - }} {{\sqrt 2 \sin \left( {x - ... | mcq | aieee-2011 |
K2B3XCHyhhQxaDGa | maths | limits-continuity-and-differentiability | limits-of-trigonometric-functions | $$\mathop {\lim }\limits_{x \to 0} {{\left( {1 - \cos 2x} \right)\left( {3 + \cos x} \right)} \over {x\tan 4x}}$$ is equal to | [{"identifier": "A", "content": "$$ - {1 \\over 4}$$"}, {"identifier": "B", "content": "$${1 \\over 2}$$"}, {"identifier": "C", "content": "1"}, {"identifier": "D", "content": "2"}] | ["D"] | null | Multiply and divide by $$x$$ in the given expression, we get
<br><br>$$\mathop {\lim }\limits_{x \to 0} {{\left( {1 - \cos 2x} \right)} \over {{x_2}}}{{\left( {3 + \cos x} \right)} \over 1}.{x \over {\tan \,4x}}$$
<br><br>$$ = \mathop {\lim }\limits_{x \to 0} {{2{{\sin }^2}x} \over {{x^2}}}.{{3 + \cos x} \over 1}.{x \... | mcq | jee-main-2013-offline |
wmvCA4BOiiEDsFrU | maths | limits-continuity-and-differentiability | limits-of-trigonometric-functions | $$\mathop {\lim }\limits_{x \to 0} {{\sin \left( {\pi {{\cos }^2}x} \right)} \over {{x^2}}}$$ is equal to : | [{"identifier": "A", "content": "$$ - \\pi $$"}, {"identifier": "B", "content": "$$ \\pi $$"}, {"identifier": "C", "content": "$${\\pi \\over 2}$$"}, {"identifier": "D", "content": "1"}] | ["B"] | null | Consider $$\mathop {\lim }\limits_{x \to 0} {{\sin \left( {\pi {{\cos }^2}x} \right)} \over {{x^2}}}$$
<br><br>$$ = \mathop {\lim }\limits_{x \to 0} {{\sin \left[ {\pi \left( {1 - {{\sin }^2}x} \right)} \right]} \over {{x^2}}}$$
<br><br>$$ = \mathop {\lim }\limits_{x \to 0} \sin {{\left( {\pi - \pi {{\sin }^2}x} \rig... | mcq | jee-main-2014-offline |
R1Gk6O2WAxXViPUh | maths | limits-continuity-and-differentiability | limits-of-trigonometric-functions | $$\mathop {\lim }\limits_{x \to 0} {{\left( {1 - \cos 2x} \right)\left( {3 + \cos x} \right)} \over {x\tan 4x}}$$ is equal to | [{"identifier": "A", "content": "2"}, {"identifier": "B", "content": "$${1 \\over 2}$$"}, {"identifier": "C", "content": "4"}, {"identifier": "D", "content": "3"}] | ["A"] | null | Multiply and divide by $$x$$ in the given expression, we get
<br><br>$$\mathop {\lim }\limits_{x \to 0} {{\left( {1 - \cos 2x} \right)} \over {{x^2}}}{{\left( {3 + \cos x} \right)} \over 1}.{x \over {\tan \,4x}}$$
<br><br>$$ = \mathop {\lim }\limits_{x \to 0} {{2{{\sin }^2}x} \over {{x^2}}}.{{3 + \cos x} \over 1}.{x \o... | mcq | jee-main-2015-offline |
nFuvyJlrfDgLfn04 | maths | limits-continuity-and-differentiability | limits-of-trigonometric-functions | Let $$p = \mathop {\lim }\limits_{x \to {0^ + }} {\left( {1 + {{\tan }^2}\sqrt x } \right)^{{1 \over {2x}}}}$$ then $$log$$ $$p$$ is equal to : | [{"identifier": "A", "content": "$${1 \\over 2}$$ "}, {"identifier": "B", "content": "$${1 \\over 4}$$ "}, {"identifier": "C", "content": "$$2$$ "}, {"identifier": "D", "content": "$$1$$ "}] | ["A"] | null | $$\ln \,P = \mathop {\lim }\limits_{x \to {0^ + }} {1 \over {2x}}\ln \left( {1 + {{\tan }^2}\sqrt x } \right)$$
<br><br>$$\mathop {\lim }\limits_{x \to {0^ + }} {1 \over x}\ln \left( {\sec \sqrt x \,\,\,\,\,\,} \right)$$
<br><br>Applying $$L$$ Hospital's rule :
<br><br>$$ = \mathop {\lim }\limits_{x \to {0^ + }} {{\sec... | mcq | jee-main-2016-offline |
59WFqj1zr1H1BHbGnna5X | maths | limits-continuity-and-differentiability | limits-of-trigonometric-functions | $$\mathop {\lim }\limits_{x \to 0} \,{{{{\left( {1 - \cos 2x} \right)}^2}} \over {2x\,\tan x\, - x\tan 2x}}$$ is : | [{"identifier": "A", "content": "$$-$$ 2"}, {"identifier": "B", "content": "$$-$$ $${1 \\over 2}$$"}, {"identifier": "C", "content": "$${1 \\over 2}$$"}, {"identifier": "D", "content": "2"}] | ["A"] | null | $$\mathop {\lim }\limits_{x \to 0} {{{{\left( {1 - \cos 2x} \right)}^2}} \over {2x\tan x - x\tan 2x}}$$
<br><br>$$ = \mathop {\lim }\limits_{x \to 0} {{{{\left( {2{{\sin }^2}x} \right)}^2}} \over {2x\left( {x + {{{x^3}} \over 3} + {{2{x^5}} \over {15}} + ....} \right) - x\left( {2x + {{{2^3}{x^3}} \over 3} + 2.{{{2^5}{... | mcq | jee-main-2016-online-10th-april-morning-slot |
YSGWJySyjCgT1nDb | maths | limits-continuity-and-differentiability | limits-of-trigonometric-functions | $$\mathop {\lim }\limits_{x \to {\pi \over 2}} {{\cot x - \cos x} \over {{{\left( {\pi - 2x} \right)}^3}}}$$ equals | [{"identifier": "A", "content": "$${1 \\over {16}}$$"}, {"identifier": "B", "content": "$${1 \\over 8}$$"}, {"identifier": "C", "content": "$${1 \\over {4}}$$"}, {"identifier": "D", "content": "$${1 \\over {24}}$$"}] | ["A"] | null | $$\mathop {\lim }\limits_{x \to {\pi \over 2}} {{\cot x - \cos x} \over {{{\left( {\pi - 2x} \right)}^3}}}$$
<br><br>= $$\mathop {\lim }\limits_{x \to {\pi \over 2}} {1 \over 8}.{{\cos x\left( {1 - \sin x} \right)} \over {\sin x{{\left( {{\pi \over 2} - x} \right)}^3}}}$$
<br><br>Put $${{\pi \over 2} - x}$$ = t
<b... | mcq | jee-main-2017-offline |
JcX96L117UGDKvIsATAfR | maths | limits-continuity-and-differentiability | limits-of-trigonometric-functions | $$\mathop {\lim }\limits_{x \to 0} {{x\tan 2x - 2x\tan x} \over {{{\left( {1 - \cos 2x} \right)}^2}}}$$ equals : | [{"identifier": "A", "content": "$${1 \\over 4}$$"}, {"identifier": "B", "content": "1"}, {"identifier": "C", "content": "$${1 \\over 2}$$"}, {"identifier": "D", "content": "$$-$$ $${1 \\over 2}$$"}] | ["C"] | null | Let, L = $$\mathop {\lim }\limits_{x \to 0} {{\left( {x\tan 2x - 2x\tan x} \right)} \over {{{\left( {1 - \cos 2x} \right)}^2}}}$$ = $$\mathop {\lim }\limits_{x \to 0} K$$ (say)
<br><br>$$ \Rightarrow $$ K = $${{x\left[ {{{2\tan x} \over {1 - {{\left( {\tan x} \right)}^2}}}} \right] - 2x\tan x} \over {{{\left( {1 - \lef... | mcq | jee-main-2018-online-15th-april-evening-slot |
M9bWcFHcnYf7eCFffvkjG | maths | limits-continuity-and-differentiability | limits-of-trigonometric-functions | $$\mathop {\lim }\limits_{x \to 0} {{x\cot \left( {4x} \right)} \over {{{\sin }^2}x{{\cot }^2}\left( {2x} \right)}}$$ is equal to : | [{"identifier": "A", "content": "0"}, {"identifier": "B", "content": "4"}, {"identifier": "C", "content": "1"}, {"identifier": "D", "content": "2"}] | ["C"] | null | $$\mathop {\lim }\limits_{x \to 0} {{x{{\tan }^2}2x} \over {\tan 4x{{\sin }^2}x}} = \mathop {\lim }\limits_{x \to 0} {{x\left( {{{{{\tan }^2}2x} \over {4{x^2}}}} \right)4{x^2}} \over {\left( {{{\tan 4x} \over {4x}}} \right)4x\left( {{{{{\sin }^2}x} \over {{x^2}}}} \right){x^2}}} = 1$$ | mcq | jee-main-2019-online-11th-january-evening-slot |
VrGyCUrha6FdMeKKOG3rsa0w2w9jxb3fiw6 | maths | limits-continuity-and-differentiability | limits-of-trigonometric-functions | $$\mathop {\lim }\limits_{x \to 0} {{x + 2\sin x} \over {\sqrt {{x^2} + 2\sin x + 1} - \sqrt {{{\sin }^2}x - x + 1} }}$$ is : | [{"identifier": "A", "content": "6"}, {"identifier": "B", "content": "1"}, {"identifier": "C", "content": "3"}, {"identifier": "D", "content": "2"}] | ["D"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267071/exam_images/hbbibexjhoou7gayvkwp.webp" style="max-width: 100%; height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2019 (Online) 12th April Evening Slot Mathematics - Limits, Continuity and Differentiability Question 157 Engl... | mcq | jee-main-2019-online-12th-april-evening-slot |
B0OUXwsIJuKbg7GgFHrzw | maths | limits-continuity-and-differentiability | limits-of-trigonometric-functions | $$\mathop {\lim }\limits_{x \to 0} {{{{\sin }^2}x} \over {\sqrt 2 - \sqrt {1 + \cos x} }}$$ equals: | [{"identifier": "A", "content": "$$ \\sqrt 2$$"}, {"identifier": "B", "content": "$$2 \\sqrt 2$$"}, {"identifier": "C", "content": "4"}, {"identifier": "D", "content": "$$4 \\sqrt 2$$"}] | ["D"] | null | $$\mathop {\lim }\limits_{x \to 0} {{{{\sin }^2}x} \over {\sqrt 2 - \sqrt {1 + \cos x} }}$$
<br><br>= $$\mathop {\lim }\limits_{x \to 0} \left( {{{{{\sin }^2}x} \over {\sqrt 2 - \sqrt {1 + \cos x} }}} \right)\left( {{{\sqrt 2 + \sqrt {1 + \cos x} } \over {\sqrt 2 + \sqrt {1 + \cos x} }}} \right)$$
<br><br>= $$\math... | mcq | jee-main-2019-online-8th-april-morning-slot |
HYXw2jZ1xzLywGWiYZrVQ | maths | limits-continuity-and-differentiability | limits-of-trigonometric-functions | $$\mathop {\lim }\limits_{x \to {1^ - }} {{\sqrt \pi - \sqrt {2{{\sin }^{ - 1}}x} } \over {\sqrt {1 - x} }}$$ is equal to : | [{"identifier": "A", "content": "$$\\sqrt {{2 \\over \\pi }} $$"}, {"identifier": "B", "content": "$${1 \\over {\\sqrt {2\\pi } }}$$"}, {"identifier": "C", "content": "$$\\sqrt {{\\pi \\over 2}} $$"}, {"identifier": "D", "content": "$$\\sqrt \\pi $$"}] | ["A"] | null | $$\mathop {\lim }\limits_{x \to {1^ - }} {{\sqrt \pi - \sqrt {2{{\sin }^{ - 1}}x} } \over {\sqrt {1 - x} }} \times {{\sqrt \pi + \sqrt {2{{\sin }^{ - 1}}} } \over {\sqrt \pi + \sqrt {2{{\sin }^{ - 1}}x} }}$$
<br><br>$$\mathop {\lim }\limits_{x \to {1^ - }} {{2\left( {{\pi \over 2} - {{\sin }^{ - 1}}x} \right)} \... | mcq | jee-main-2019-online-12th-january-evening-slot |
ZhCzqfek1CZLC206tdsJx | maths | limits-continuity-and-differentiability | limits-of-trigonometric-functions | $$\mathop {\lim }\limits_{x \to \pi /4} {{{{\cot }^3}x - \tan x} \over {\cos \left( {x + {\pi \over 4}} \right)}}$$ is : | [{"identifier": "A", "content": "$$8\\sqrt 2 $$"}, {"identifier": "B", "content": "4"}, {"identifier": "C", "content": "$$4\\sqrt 2 $$"}, {"identifier": "D", "content": "8"}] | ["D"] | null | $$\mathop {\lim }\limits_{x \to \pi /4} {{{{\cot }^3}x - \tan x} \over {\cos \left( {x + {\pi \over 4}} \right)}}$$
<br><br>= $$\mathop {\lim }\limits_{x \to {\pi \over 4}} {{{1 \over {{{\tan }^3}x}} - \tan x} \over {\cos \left( {x + {\pi \over 4}} \right)}}$$
<br><br>= $$\mathop {\lim }\limits_{x \to {\pi \over 4}... | mcq | jee-main-2019-online-12th-january-morning-slot |
MnT35fZqunH4z67MqKLbB | maths | limits-continuity-and-differentiability | limits-of-trigonometric-functions | Let [x] denote the greatest integer less than or equal to x. Then $$\mathop {\lim }\limits_{x \to 0} {{\tan \left( {\pi {{\sin }^2}x} \right) + {{\left( {\left| x \right| - \sin \left( {x\left[ x \right]} \right)} \right)}^2}} \over {{x^2}}}$$ | [{"identifier": "A", "content": "equals $$\\pi $$ + 1"}, {"identifier": "B", "content": "equals 0"}, {"identifier": "C", "content": "does not exist "}, {"identifier": "D", "content": "equals $$\\pi $$"}] | ["C"] | null | R.H.L. $$=$$ $$\mathop {\lim }\limits_{x \to {0^ + }} {{\tan \left( {\pi {{\sin }^2}x} \right) + {{\left( {\left| x \right| - \sin \left( {x\left[ x \right]} \right)} \right)}^2}} \over {{x^2}}}$$
<br><br>(as x $$ \to $$ 0<sup>+</sup> $$ \Rightarrow $$ [x] $$=$$ 0)
<br><br>$$=$$ $$\mathop {\lim }\limit... | mcq | jee-main-2019-online-11th-january-morning-slot |
EvdzNUS9OvijXzGchQf93 | maths | limits-continuity-and-differentiability | limits-of-trigonometric-functions | For each t $$ \in $$ R , let [t] be the greatest integer less than or equal to t
<br/><br/>Then $$\mathop {\lim }\limits_{x \to 1^ + } {{\left( {1 - \left| x \right| + \sin \left| {1 - x} \right|} \right)\sin \left( {{\pi \over 2}\left[ {1 - x} \right]} \right)} \over {\left| {1 - x} \right|.\left[ {1 - x} \right]}}$... | [{"identifier": "A", "content": "equals $$-$$ 1 "}, {"identifier": "B", "content": "equals 1"}, {"identifier": "C", "content": "equals 0"}, {"identifier": "D", "content": "does not exist"}] | ["C"] | null | $$\mathop {\lim }\limits_{x \to {1^ + }} {{\left( {1 - \left| x \right| + \sin \left| {1 - x} \right|} \right)\sin \left( {{\pi \over 2}\left[ {1 - x} \right]} \right)} \over {\left| {1 - x} \right|\left[ {1 - x} \right]}}$$
<br><br>$$=$$ $$\mathop {\lim }\limits_{x \to {1^ + }} {{\left( {1 - x} \right) + \sin \left( ... | mcq | jee-main-2019-online-10th-january-morning-slot |
Mjg8JDXWkk9XV53FaN3r3 | maths | limits-continuity-and-differentiability | limits-of-trigonometric-functions | For each x$$ \in $$<b>R</b>, let [x] be the greatest integer less than or equal to x.
<br/><br/>Then $$\mathop {\lim }\limits_{x \to {0^ - }} \,\,{{x\left( {\left[ x \right] + \left| x \right|} \right)\sin \left[ x \right]} \over {\left| x \right|}}$$ is equal to : | [{"identifier": "A", "content": "$$-$$ sin 1"}, {"identifier": "B", "content": "1"}, {"identifier": "C", "content": "sin 1"}, {"identifier": "D", "content": "0"}] | ["A"] | null | $$\mathop {\lim }\limits_{x \to {0^ - }} {{x\left( {\left[ x \right] + \left| x \right|} \right)\sin \left[ x \right]} \over {\left| x \right|}}$$
<br><br>$$ = \mathop {\lim }\limits_{h \to 0} {{\left( {0 - h} \right)\left( {\left[ {0 - h} \right] + \left| {0 - h} \right|} \right)\sin \left[ {0 - h} \right]} \over {\le... | mcq | jee-main-2019-online-9th-january-evening-slot |
2x9c5HPz6lbLnD5rcAjgy2xukf0z9feu | maths | limits-continuity-and-differentiability | limits-of-trigonometric-functions | If $$\mathop {\lim }\limits_{x \to 0} \left\{ {{1 \over {{x^8}}}\left( {1 - \cos {{{x^2}} \over 2} - \cos {{{x^2}} \over 4} + \cos {{{x^2}} \over 2}\cos {{{x^2}} \over 4}} \right)} \right\}$$ = 2<sup>-k</sup>
<br/><br/>then the value of k is _______ . | [] | null | 8 | $$\mathop {\lim }\limits_{x \to 0} \left\{ {{1 \over {{x^8}}}\left( {1 - \cos {{{x^2}} \over 2} - \cos {{{x^2}} \over 4} + \cos {{{x^2}} \over 2}\cos {{{x^2}} \over 4}} \right)} \right\} = {2^{ - k}}$$
<br><br>$$ \Rightarrow $$ $$\mathop {\lim }\limits_{x \to 0} {{\left( {1 - \cos {{{x^2}} \over 2}} \right)} \over {4{{... | integer | jee-main-2020-online-3rd-september-morning-slot |
CbTrUE9Q2rPhXV8fxzjgy2xukfjjo939 | maths | limits-continuity-and-differentiability | limits-of-trigonometric-functions | If $$\alpha $$ is positive root of the equation, p(x) = x<sup>2</sup> - x - 2 = 0, then<br/><br/>
$$\mathop {\lim }\limits_{x \to {\alpha ^ + }} {{\sqrt {1 - \cos \left( {p\left( x \right)} \right)} } \over {x + \alpha - 4}}$$ is equal to : | [{"identifier": "A", "content": "$${1 \\over \\sqrt2}$$"}, {"identifier": "B", "content": "$${1 \\over 2}$$"}, {"identifier": "C", "content": "$${3 \\over \\sqrt2}$$"}, {"identifier": "D", "content": "$${3 \\over 2}$$"}] | ["C"] | null | $${x^2} - x - 2 = 0$$<br><br>roots are 2 & $$-$$1 $$ \Rightarrow $$ $$\alpha $$ = 2 (given $$\alpha$$ is positive)<br><br>Now $$ \mathop {\lim }\limits_{x \to {2^ + }} {{\sqrt {1 - \cos ({x^2} - x - 2)} } \over {(x - 2)}}$$<br><br>$$ = \mathop {\lim }\limits_{x \to {2^ + }} {{\sqrt {2{{\sin }^2}{{({x^2} - x - 2)} \... | mcq | jee-main-2020-online-5th-september-morning-slot |
K9MKlxYQfb5W0pceJy1klrisk9y | maths | limits-continuity-and-differentiability | limits-of-trigonometric-functions | $$\mathop {\lim }\limits_{n \to \infty } \tan \left\{ {\sum\limits_{r = 1}^n {{{\tan }^{ - 1}}\left( {{1 \over {1 + r + {r^2}}}} \right)} } \right\}$$ is equal to ______. | [] | null | 1 | $${\tan ^{ - 1}}\left( {{1 \over {1 + r + {r^2}}}} \right)$$<br><br>$$ = {\tan ^{ - 1}}\left( {{{r + 1 - r} \over {1 + r(r + 1)}}} \right)$$<br><br>$$ = {\tan ^{ - 1}}(r + 1) - {\tan ^{ - 1}}r$$<br><br>$$ \therefore $$ $$\sum\limits_{r = 1}^n {\left( {{{\tan }^{ - 1}}(r + 1) - {{\tan }^{ - 1}}(r)} \right)} $$<br><br>$$... | integer | jee-main-2021-online-24th-february-morning-slot |
raNEv8qHb2QGtIhf9v1klug6ks1 | maths | limits-continuity-and-differentiability | limits-of-trigonometric-functions | The value of $$\mathop {\lim }\limits_{h \to 0} 2\left\{ {{{\sqrt 3 \sin \left( {{\pi \over 6} + h} \right) - \cos \left( {{\pi \over 6} + h} \right)} \over {\sqrt 3 h\left( {\sqrt 3 \cosh - \sinh } \right)}}} \right\}$$ is : | [{"identifier": "A", "content": "$${4 \\over 3}$$"}, {"identifier": "B", "content": "$${2 \\over 3}$$"}, {"identifier": "C", "content": "$${3 \\over 4}$$"}, {"identifier": "D", "content": "$${2 \\over {\\sqrt 3 }}$$"}] | ["A"] | null | Let L = $$\mathop {\lim }\limits_{h \to 0} 2\left\{ {{{\sqrt 3 \sin \left( {{\pi \over 6} + h} \right) - \cos \left( {{\pi \over 6} + h} \right)} \over {\sqrt 3 h\left( {\sqrt 3 \cosh - \sinh } \right)}}} \right\}$$
<br><br>$$ \Rightarrow $$ L = $$\mathop {\lim }\limits_{h \to 0} 2 \times 2\left\{ {{{{{\sqrt 3 } \ov... | mcq | jee-main-2021-online-26th-february-morning-slot |
2ZjqG1divjp9n1jn1B1kmhvs8fg | maths | limits-continuity-and-differentiability | limits-of-trigonometric-functions | Let $${S_k} = \sum\limits_{r = 1}^k {{{\tan }^{ - 1}}\left( {{{{6^r}} \over {{2^{2r + 1}} + {3^{2r + 1}}}}} \right)} $$. Then $$\mathop {\lim }\limits_{k \to \infty } {S_k}$$ is equal to : | [{"identifier": "A", "content": "$${\\cot ^{ - 1}}\\left( {{3 \\over 2}} \\right)$$"}, {"identifier": "B", "content": "$${\\pi \\over 2}$$"}, {"identifier": "C", "content": "tan<sup>$$-$$1</sup> (3)"}, {"identifier": "D", "content": "$${\\tan ^{ - 1}}\\left( {{3 \\over 2}} \\right)$$"}] | ["A"] | null | S<sub>k</sub> = $$\sum\limits_{r = 1}^k {{{\tan }^{ - 1}}} \left( {{{{6^r}(3 - 2)} \over {\left( {1 + {{\left( {{3 \over 2}} \right)}^{2r + 1}}} \right){2^{2r + 1}}}}} \right)$$<br><br>= $$\sum\limits_{r = 1}^k {{{\tan }^{ - 1}}} \left( {{{{2^r}\,.\,{3^{r + 1}} - {3^r}{2^{r + 1}}} \over {\left( {1 + {{\left( {{3 \ove... | mcq | jee-main-2021-online-16th-march-morning-shift |
gr60nFENdvz6ErSCTo1kmhzpkqe | maths | limits-continuity-and-differentiability | limits-of-trigonometric-functions | If $$\mathop {\lim }\limits_{x \to 0} {{a{e^x} - b\cos x + c{e^{ - x}}} \over {x\sin x}} = 2$$, then a + b + c is equal to ____________. | [] | null | 4 | $$\mathop {\lim }\limits_{x \to 0} {{\left\{ {a\left( {1 + x + {{{x^2}} \over {2!}} + .....} \right) - b\left( {1 - {{{x^2}} \over {2!}} + {{{x^4}} \over {4!}}......} \right) + c\left( {1 - x + {{{x^2}} \over {2!}}......} \right)} \right\}} \over {x\left( {x - {{{x^3}} \over {3!}} + .....} \right)}} = 2$$<br><br>$$ \th... | integer | jee-main-2021-online-16th-march-morning-shift |
owmdXqRf8g9VE2IoDB1kmja76au | maths | limits-continuity-and-differentiability | limits-of-trigonometric-functions | The value of <br/>$$\mathop {\lim }\limits_{x \to {0^ + }} {{{{\cos }^{ - 1}}(x - {{[x]}^2}).{{\sin }^{ - 1}}(x - {{[x]}^2})} \over {x - {x^3}}}$$, where [ x ] denotes the greatest integer $$ \le $$ x is : | [{"identifier": "A", "content": "$$\\pi$$"}, {"identifier": "B", "content": "$${\\pi \\over 4}$$"}, {"identifier": "C", "content": "$${\\pi \\over 2}$$"}, {"identifier": "D", "content": "0"}] | ["C"] | null | $$\mathop {\lim }\limits_{x \to {0^ + }} {{{{\cos }^{ - 1}}\left( {x - {{[x]}^2}} \right).{{\sin }^{ - 1}}\left( {x - {{[x]}^2}} \right)} \over {x - {x^3}}}$$<br><br>$$ = \mathop {\lim }\limits_{x \to {0^ + }} {{{{\cos }^{ - 1}}x} \over {1 - {x^2}}}.{{{{\sin }^{ - 1}}x} \over x}$$<br><br>$$ = {\cos ^{ - 1}}0 = {\pi \o... | mcq | jee-main-2021-online-17th-march-morning-shift |
7EgukTgHs5JcAZ2Ac41kmkn3tcl | maths | limits-continuity-and-differentiability | limits-of-trigonometric-functions | The value of the limit <br/><br/>$$\mathop {\lim }\limits_{\theta \to 0} {{\tan (\pi {{\cos }^2}\theta )} \over {\sin (2\pi {{\sin }^2}\theta )}}$$ is equal to : | [{"identifier": "A", "content": "0"}, {"identifier": "B", "content": "$$-$$$${1 \\over 2}$$"}, {"identifier": "C", "content": "$${1 \\over 4}$$"}, {"identifier": "D", "content": "$$-$$$${1 \\over 4}$$"}] | ["B"] | null | Given,<br><br>$$\mathop {\lim }\limits_{\theta \to 0} {{\tan (\pi {{\cos }^2}\theta )} \over {\sin (2\pi {{\sin }^2}\theta )}}$$<br><br>$$ = \mathop {\lim }\limits_{\theta \to 0} {{\tan (\pi - \pi {{\sin }^2}\theta )} \over {\sin (2\pi {{\sin }^2}\theta )}}$$ $$ \therefore $$ $$\left( {{{\cos }^2}\theta = 1 - {{\si... | mcq | jee-main-2021-online-17th-march-evening-shift |
CFfZTluMCLEQbMjgF01kmlihjxq | maths | limits-continuity-and-differentiability | limits-of-trigonometric-functions | If $$\mathop {\lim }\limits_{x \to 0} {{{{\sin }^{ - 1}}x - {{\tan }^{ - 1}}x} \over {3{x^3}}}$$ is equal to L, then the value of (6L + 1) is | [{"identifier": "A", "content": "$${1 \\over 6}$$"}, {"identifier": "B", "content": "$${1 \\over 2}$$"}, {"identifier": "C", "content": "6"}, {"identifier": "D", "content": "2"}] | ["D"] | null | $$L = \mathop {\lim }\limits_{x \to 0} {{\left( {x + {{{x^3}} \over 6} + .....} \right) - \left( {x - {{{x^3}} \over 3}.....} \right)} \over {3{x^3}}}$$<br><br>$$L = {1 \over 3}\left( {{1 \over 6} + {1 \over 3}} \right) = {1 \over 6}$$<br><br>$$ \therefore $$ $$ 6L + 1 = 6.{1 \over 6} + 1 = 2$$ | mcq | jee-main-2021-online-18th-march-morning-shift |
1krxkdox3 | maths | limits-continuity-and-differentiability | limits-of-trigonometric-functions | The value of <br/><br/>$$\mathop {\lim }\limits_{x \to 0} \left( {{x \over {\root 8 \of {1 - \sin x} - \root 8 \of {1 + \sin x} }}} \right)$$ is equal to : | [{"identifier": "A", "content": "0"}, {"identifier": "B", "content": "4"}, {"identifier": "C", "content": "$$-$$4"}, {"identifier": "D", "content": "$$-$$1"}] | ["C"] | null | $$\mathop {\lim }\limits_{x \to 0} \left( {{x \over {\root 8 \of {1 - \sin x} - \root 8 \of {1 + \sin x} }}} \right)$$
<br><br>= $$\mathop {\lim }\limits_{x \to 0} {x \over {\left[ {{{\left( {1 - \sin x} \right)}^{{1 \over 8}}} - {{\left( {1 + \sin x} \right)}^{{1 \over 8}}}} \right]}} \times \left[ {{{{{\left( {1 - \... | mcq | jee-main-2021-online-27th-july-evening-shift |
1ktir1jy2 | maths | limits-continuity-and-differentiability | limits-of-trigonometric-functions | $$\mathop {\lim }\limits_{x \to 0} {{{{\sin }^2}\left( {\pi {{\cos }^4}x} \right)} \over {{x^4}}}$$ is equal to : | [{"identifier": "A", "content": "$${\\pi ^2}$$"}, {"identifier": "B", "content": "$$2{\\pi ^2}$$"}, {"identifier": "C", "content": "$$4{\\pi ^2}$$"}, {"identifier": "D", "content": "$$4\\pi $$"}] | ["C"] | null | $$\mathop {\lim }\limits_{x \to 0} {{{{\sin }^2}\left( {\pi {{\cos }^4}x} \right)} \over {{x^4}}}$$<br><br>= $$\mathop {\lim }\limits_{x \to 0} {{1 - \cos \left( {2\pi {{\cos }^4}x} \right)} \over {2{x^4}}}$$<br><br>= $$\mathop {\lim }\limits_{x \to 0} {{1 - \cos \left( {2\pi - 2\pi {{\cos }^4}x} \right)} \over {{{\le... | mcq | jee-main-2021-online-31st-august-morning-shift |
1ktk6iv5f | maths | limits-continuity-and-differentiability | limits-of-trigonometric-functions | If $$\alpha = \mathop {\lim }\limits_{x \to {\pi \over 4}} {{{{\tan }^3}x - \tan x} \over {\cos \left( {x + {\pi \over 4}} \right)}}$$ and $$\beta = \mathop {\lim }\limits_{x \to 0 } {(\cos x)^{\cot x}}$$ are the roots of the equation, ax<sup>2</sup> + bx $$-$$ 4 = 0, then the ordered pair (a, b) is : | [{"identifier": "A", "content": "(1, $$-$$3)"}, {"identifier": "B", "content": "($$-$$1, 3)"}, {"identifier": "C", "content": "($$-$$1, $$-$$3)"}, {"identifier": "D", "content": "(1, 3)"}] | ["D"] | null | $$\alpha = \mathop {\lim }\limits_{x \to {\pi \over 4}} {{{{\tan }^3}x - \tan x} \over {\cos \left( {x + {\pi \over 4}} \right)}};{0 \over 0}$$ form<br><br>Using L Hospital rule<br><br>$$\alpha = \mathop {\lim }\limits_{x \to {\pi \over 4}} {{3{{\tan }^2}x{{\sec }^2}x - {{\sec }^2}x} \over { - \sin \left( {x + {\p... | mcq | jee-main-2021-online-31st-august-evening-shift |
1l54ar2pq | maths | limits-continuity-and-differentiability | limits-of-trigonometric-functions | <p>The value of $$\mathop {\lim }\limits_{x \to 1} {{({x^2} - 1){{\sin }^2}(\pi x)} \over {{x^4} - 2{x^3} + 2x - 1}}$$ is equal to:</p> | [{"identifier": "A", "content": "$${{{\\pi ^2}} \\over 6}$$"}, {"identifier": "B", "content": "$${{{\\pi ^2}} \\over 3}$$"}, {"identifier": "C", "content": "$${{{\\pi ^2}} \\over 2}$$"}, {"identifier": "D", "content": "$$\\pi$$<sup>2</sup>"}] | ["D"] | null | <p>$$\mathop {\lim }\limits_{x \to 1} {{({x^2} - 1){{\sin }^2}(\pi x)} \over {{x^4} - 2{x^2} + 2x - 1}}$$</p>
<p>$$ = \mathop {\lim }\limits_{x \to 1} {{({x^2} - 1)si{n^2}(\pi x)} \over {({x^2} - 1){{(x - 1)}^2}}}$$</p>
<p>$$ = \mathop {\lim }\limits_{x \to 1} {{{{\sin }^2}(\pi x)} \over {{{(x - 1)}^2}}}$$</p>
<p>Let $... | mcq | jee-main-2022-online-29th-june-evening-shift |
1l55iq9zy | maths | limits-continuity-and-differentiability | limits-of-trigonometric-functions | <p>The value of <br/><br/>$$\mathop {\lim }\limits_{n \to \infty } 6\tan \left\{ {\sum\limits_{r = 1}^n {{{\tan }^{ - 1}}\left( {{1 \over {{r^2} + 3r + 3}}} \right)} } \right\}$$ is equal to :</p> | [{"identifier": "A", "content": "1"}, {"identifier": "B", "content": "2"}, {"identifier": "C", "content": "3"}, {"identifier": "D", "content": "6"}] | ["C"] | null | <p>$$\mathop {\lim }\limits_{n \to \infty } 6\tan \left\{ {\sum\limits_{r = 1}^n {{{\tan }^{ - 1}}\left( {{1 \over {{r^2} + 3r + 3}}} \right)} } \right\}$$</p>
<p>$$ = \mathop {\lim }\limits_{n \to \infty } 6\tan \left\{ {\sum\limits_{r = 1}^n {{{\tan }^{ - 1}}\left( {{{(r + 2) - (r + 1)} \over {1 + (r + 2)(r + 1)}}} \... | mcq | jee-main-2022-online-28th-june-evening-shift |
1l55j1jkd | maths | limits-continuity-and-differentiability | limits-of-trigonometric-functions | <p>If $$\mathop {\lim }\limits_{x \to 1} {{\sin (3{x^2} - 4x + 1) - {x^2} + 1} \over {2{x^3} - 7{x^2} + ax + b}} = - 2$$, then the value of (a $$-$$ b) is equal to ___________.</p> | [] | null | 11 | $$
\begin{aligned}
& \lim _{x \rightarrow 1} \frac{\left(\frac{\sin \left(3 x^{2}-4 x+1\right)}{3 x^{2}-4 x+1}\right)\left(3 x^{2}-4 x+1\right)-x^{2}+1}{2 x^{3}-7 x^{2}+a x+b}=-2 \\\\
\Rightarrow & \lim _{x \rightarrow 1} \frac{3 x^{2}-4 x+1-x^{2}+1}{2 x^{3}-7 x^{2}+a x+b}=-2 \\\\
\Rightarrow & \lim _{x \rightarrow 1} ... | integer | jee-main-2022-online-28th-june-evening-shift |
1l589fknh | maths | limits-continuity-and-differentiability | limits-of-trigonometric-functions | <p>$$\mathop {\lim }\limits_{x \to {1 \over {\sqrt 2 }}} {{\sin ({{\cos }^{ - 1}}x) - x} \over {1 - \tan ({{\cos }^{ - 1}}x)}}$$ is equal to :</p> | [{"identifier": "A", "content": "$$\\sqrt 2 $$"}, {"identifier": "B", "content": "$$ - \\sqrt 2 $$"}, {"identifier": "C", "content": "$${1 \\over {\\sqrt 2 }}$$"}, {"identifier": "D", "content": "$$ - {1 \\over {\\sqrt 2 }}$$"}] | ["D"] | null | <p>$$\mathop {\lim }\limits_{x \to {1 \over {\sqrt 2 }}} {{\sin ({{\cos }^{ - 1}}x) - x} \over {1 - \tan ({{\cos }^{ - 1}}x)}}$$</p>
<p>Let $${\cos ^{ - 1}}x = t$$</p>
<p>$$ \Rightarrow x = \cos t$$</p>
<p>When $$x \to {1 \over {\sqrt 2 }}$$, then $$t \to {\cos ^{ - 1}}\left( {{1 \over {\sqrt 2 }}} \right) \to {\pi \o... | mcq | jee-main-2022-online-26th-june-morning-shift |
1l58f4816 | maths | limits-continuity-and-differentiability | limits-of-trigonometric-functions | <p>$$\mathop {\lim }\limits_{x \to 0} {{\cos (\sin x) - \cos x} \over {{x^4}}}$$ is equal to :</p> | [{"identifier": "A", "content": "$${1 \\over 3}$$"}, {"identifier": "B", "content": "$${1 \\over 4}$$"}, {"identifier": "C", "content": "$${1 \\over 6}$$"}, {"identifier": "D", "content": "$${1 \\over 12}$$"}] | ["C"] | null | <p>$$\mathop {\lim }\limits_{x \to 0} {{\cos (\sin x) - \cos x} \over {{x^4}}} = \mathop {\lim }\limits_{x \to 0} {{2\sin (x + \sin x)\,.\,\sin \left( {{{x - \sin x} \over 2}} \right)} \over {{x^4}}}$$</p>
<p>$$ = \mathop {\lim }\limits_{x \to 0} 2\,.\,\left( {{{\left( {{{x + \sin x} \over 2}} \right)\left( {{{x - \sin... | mcq | jee-main-2022-online-26th-june-evening-shift |
1l59jt81c | maths | limits-continuity-and-differentiability | limits-of-trigonometric-functions | <p>$$\mathop {\lim }\limits_{x \to {\pi \over 2}} \left( {{{\tan }^2}x\left( {{{(2{{\sin }^2}x + 3\sin x + 4)}^{{1 \over 2}}} - {{({{\sin }^2}x + 6\sin x + 2)}^{{1 \over 2}}}} \right)} \right)$$ is equal to</p> | [{"identifier": "A", "content": "$${1 \\over {12}}$$"}, {"identifier": "B", "content": "$$-$$$${1 \\over {18}}$$"}, {"identifier": "C", "content": "$$-$$$${1 \\over {12}}$$"}, {"identifier": "D", "content": "$${1 \\over {6}}$$"}] | ["A"] | null | <p>$$\mathop {\lim }\limits_{x \to {\pi \over 2}} {\tan ^2}x\left\{ {\sqrt {2{{\sin }^2}x + 3\sin x + 4} - \sqrt {{{\sin }^2}x + 6\sin x + 2} } \right\}$$</p>
<p>$$ = \mathop {\lim }\limits_{x \to {\pi \over 2}} {{{{\tan }^2}x({{\sin }^2}x - 3\sin x + 2)} \over {\sqrt {2{{\sin }^2}x + 3\sin x + 4} + \sqrt {{{\sin }... | mcq | jee-main-2022-online-25th-june-evening-shift |
1l5w0zq08 | maths | limits-continuity-and-differentiability | limits-of-trigonometric-functions | <p>Suppose $$\mathop {\lim }\limits_{x \to 0} {{F(x)} \over {{x^3}}}$$ exists and is equal to L, where</p>
<p>$$F(x) = \left| {\matrix{
{a + \sin {x \over 2}} & { - b\cos x} & 0 \cr
{ - b\cos x} & 0 & {a + \sin {x \over 2}} \cr
0 & {a + \sin {x \over 2}} & { - b\cos x} \cr
} } \... | [] | null | 14 | <p>Given,</p>
<p>$$F(x) = \left| {\matrix{
{a + \sin {x \over 2}} & { - b\cos x} & 0 \cr
{ - b\cos x} & 0 & {a + \sin {x \over 2}} \cr
0 & {a + \sin {x \over 2}} & { - b\cos x} \cr
} } \right|$$</p>
<p>$$ = \left( {a + \sin {x \over 2}} \right)\left( { - {{\left( {a + \sin {x \over 2}} \right)}^2}} \ri... | integer | jee-main-2022-online-30th-june-morning-shift |
1l6f0xruh | maths | limits-continuity-and-differentiability | limits-of-trigonometric-functions | <p>$$\lim\limits_{x \rightarrow \frac{\pi}{4}} \frac{8 \sqrt{2}-(\cos x+\sin x)^{7}}{\sqrt{2}-\sqrt{2} \sin 2 x}$$ is equal to </p> | [{"identifier": "A", "content": "14"}, {"identifier": "B", "content": "7"}, {"identifier": "C", "content": "14$$\\sqrt2$$"}, {"identifier": "D", "content": "7$$\\sqrt2$$"}] | ["A"] | null | <p>$$\mathop {\lim }\limits_{x \to {\pi \over 4}} {{8\sqrt 2 - {{(\cos x + \sin x)}^7}} \over {\sqrt 2 - \sqrt 2 \sin 2x}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {{0 \over 0}\,\mathrm{form}} \right)$$</p>
<p>$$ = \mathop {\lim }\limits_{x \to {\pi \over 4}} {{ - 7{{(\cos x + \sin x)}^6}( - \sin x + \cos x)} \over { - 2\... | mcq | jee-main-2022-online-25th-july-evening-shift |
1l6p17ava | maths | limits-continuity-and-differentiability | limits-of-trigonometric-functions | <p>If $$\lim\limits_{x \rightarrow 0} \frac{\alpha \mathrm{e}^{x}+\beta \mathrm{e}^{-x}+\gamma \sin x}{x \sin ^{2} x}=\frac{2}{3}$$, where $$\alpha, \beta, \gamma \in \mathbf{R}$$, then which of the following is NOT correct?</p> | [{"identifier": "A", "content": "$$\\alpha^{2}+\\beta^{2}+\\gamma^{2}=6$$"}, {"identifier": "B", "content": "$$\\alpha \\beta+\\beta \\gamma+\\gamma \\alpha+1=0$$"}, {"identifier": "C", "content": "$$\\alpha\\beta^{2}+\\beta \\gamma^{2}+\\gamma \\alpha^{2}+3=0$$"}, {"identifier": "D", "content": "$$\\alpha^{2}-\\beta^{... | ["C"] | null | <p>$$\mathop {\lim }\limits_{x \to 0} {{\alpha {e^x} + \beta {e^{ - x}} + \gamma \sin x} \over {x{{\sin }^2}x}} = {2 \over 3}$$</p>
<p>$$ \Rightarrow \alpha + \beta = 0$$ (to make indeterminant form) ...... (i)</p>
<p>Now,</p>
<p>$$\mathop {\lim }\limits_{x \to 0} {{\alpha {e^x} - \beta {e^{ - x}} + \gamma \cos x} \o... | mcq | jee-main-2022-online-29th-july-morning-shift |
ldqw6aac | maths | limits-continuity-and-differentiability | limits-of-trigonometric-functions | Let $f, g$ and $h$ be the real valued functions defined on $\mathbb{R}$ as
<br/><br/>$f(x)=\left\{\begin{array}{cc}\frac{x}{|x|}, & x \neq 0 \\ 1, & x=0\end{array}\right.$
<br/><br/>$g(x)=\left\{\begin{array}{cc}\frac{\sin (x+1)}{(x+1)}, & x \neq-1 \\ 1, & x=-1\end{array}\right.$
<br/><br/>and $h(x)=2[... | [{"identifier": "A", "content": "1"}, {"identifier": "B", "content": "$-1$"}, {"identifier": "C", "content": "$\\sin (1)$"}, {"identifier": "D", "content": "0"}] | ["A"] | null | <p>$$f(x) = {\mathop{\rm sgn}} (x)$$</p>
<p>$$h(x) = 2[x] - {\mathop{\rm sgn}} (x)$$</p>
<p>If $$x \to {1^ + }$$ then $$h(x - 1) = 2[x] - 2 - {\mathop{\rm sgn}} (x - 1)$$</p>
<p>$$ = 0 - 1 = - 1$$</p>
<p>& if $$x \to {1^ - }$$ then $$h(x - 1) = 2[x] - 2 - {\mathop{\rm sgn}} (x - 1)$$</p>
<p>$$ = - 2 + 1 = - 1$$</p>
... | mcq | jee-main-2023-online-30th-january-evening-shift |
1ldswip6v | maths | limits-continuity-and-differentiability | limits-of-trigonometric-functions | <p>Let $$x=2$$ be a root of the equation $$x^2+px+q=0$$ and $$f(x) = \left\{ {\matrix{
{{{1 - \cos ({x^2} - 4px + {q^2} + 8q + 16)} \over {{{(x - 2p)}^4}}},} & {x \ne 2p} \cr
{0,} & {x = 2p} \cr
} } \right.$$</p>
<p>Then $$\mathop {\lim }\limits_{x \to 2{p^ + }} [f(x)]$$, where $$\left[ . \right]$$ ... | [{"identifier": "A", "content": "2"}, {"identifier": "B", "content": "1"}, {"identifier": "C", "content": "0"}, {"identifier": "D", "content": "$$-1$$"}] | ["C"] | null | $\lim \limits_{x \rightarrow 2^{+}}\left(\frac{1-\cos \left(x^{2}-4 p x+q^{2}+8 q+16\right)}{\left(x^{2}-4 p x+q^{2}+8 q+16\right)^{2}}\right)\left(\frac{\left(x^{2}-4 p x+q^{2}+8 q+16\right)^{2}}{(x-2 p)^{2}}\right)$
<br/><br/>
$\lim \limits_{h \rightarrow 0} \frac{1}{2}\left(\frac{(2 p+h)^{2}-4 p(2 p+h)+q^{2}+82+16}{... | mcq | jee-main-2023-online-29th-january-morning-shift |
1lgow7xsy | maths | limits-continuity-and-differentiability | limits-of-trigonometric-functions | <p>If $$\lim_\limits{x \rightarrow 0} \frac{e^{a x}-\cos (b x)-\frac{cx e^{-c x}}{2}}{1-\cos (2 x)}=17$$, then $$5 a^{2}+b^{2}$$ is equal to</p> | [{"identifier": "A", "content": "64"}, {"identifier": "B", "content": "68"}, {"identifier": "C", "content": "72"}, {"identifier": "D", "content": "76"}] | ["B"] | null | <p>The Taylor series for $e^x$, $\cos x$, and $e^{-x}$ around $x=0$ are:</p>
<p>$$e^{ax} = 1 + ax + \frac{(ax)^2}{2!} + \ldots,$$
<br/><br/>$$\cos(bx) = 1 - \frac{(bx)^2}{2!} + \ldots,$$
<br/><br/>$$e^{-cx} = 1 - cx + \frac{(cx)^2}{2!} - \ldots.$$</p>
<p>Substituting these into the limit and simplifying:</p>
<p>$$\lim\... | mcq | jee-main-2023-online-13th-april-evening-shift |
1lgylxsd8 | maths | limits-continuity-and-differentiability | limits-of-trigonometric-functions | <p>If $$\alpha > \beta > 0$$ are the roots of the equation $$a x^{2}+b x+1=0$$, and $$\lim_\limits{x \rightarrow \frac{1}{\alpha}}\left(\frac{1-\cos \left(x^{2}+b x+a\right)}{2(1-\alpha x)^{2}}\right)^{\frac{1}{2}}=\frac{1}{k}\left(\frac{1}{\beta}-\frac{1}{\alpha}\right), \text { then } \mathrm{k} \text { is equa... | [{"identifier": "A", "content": "$$2 \\beta$$"}, {"identifier": "B", "content": "$$\\beta$$"}, {"identifier": "C", "content": "$$\\alpha$$"}, {"identifier": "D", "content": "$$2 \\alpha$$"}] | ["D"] | null | Since, $\alpha, \beta$ are roots of $a x^2+b x+1=0$
<br/><br/>Replace $x \rightarrow \frac{1}{x}$
<br/><br/>$$
\frac{a}{x^2}+\frac{b}{x}+1=0 \Rightarrow x^2+b x+a=0
$$
<br/><br/>So, $\frac{1}{\alpha}, \frac{1}{\beta}$ are the roots.
<br/><br/>Now, $\lim\limits_{x \rightarrow \frac{1}{\alpha}}\left[\frac{1-\cos \left(x^... | mcq | jee-main-2023-online-8th-april-evening-shift |
1lgzy95yj | maths | limits-continuity-and-differentiability | limits-of-trigonometric-functions | <p>$$\lim_\limits{x \rightarrow 0}\left(\left(\frac{\left(1-\cos ^{2}(3 x)\right.}{\cos ^{3}(4 x)}\right)\left(\frac{\sin ^{3}(4 x)}{\left(\log _{e}(2 x+1)\right)^{5}}\right)\right)$$ is equal to _____________.</p> | [{"identifier": "A", "content": "15"}, {"identifier": "B", "content": "18"}, {"identifier": "C", "content": "9"}, {"identifier": "D", "content": "24"}] | ["B"] | null | $$
\begin{aligned}
& \lim _{x \rightarrow 0}\left[\left(\frac{1-\cos ^2(3 x)}{\cos ^3(4 x)}\right)\left(\frac{\sin ^3(4 x)}{\left(\log _e(2 x+1)\right)^5}\right)\right] \\\\
& =\lim _{x \rightarrow 0}\left[\frac{1-\cos ^2(3 x)}{9 x^2} \times \frac{9 x^2}{\cos ^3(4 x)}\right] \times \frac{\frac{\sin ^3 4 x}{(4 x)^3} \t... | mcq | jee-main-2023-online-8th-april-morning-shift |
lsaq12m0 | maths | limits-continuity-and-differentiability | limits-of-trigonometric-functions | Let $\{x\}$ denote the fractional part of $x$ and $f(x)=\frac{\cos ^{-1}\left(1-\{x\}^2\right) \sin ^{-1}(1-\{x\})}{\{x\}-\{x\}^3}, x \neq 0$. If $\mathrm{L}$ and $\mathrm{R}$ respectively denotes the left hand limit and the right hand limit of $f(x)$ at $x=0$, then $\frac{32}{\pi^2}\left(\mathrm{~L}^2+\mathrm{R}^2\rig... | [] | null | 18 | Finding right hand limit
<br/><br/>R = $$
\begin{gathered}
\lim _{x \rightarrow 0^{+}} f(x)=\lim _{h \rightarrow 0} f(0+h)
=\lim _{h \rightarrow 0} f(h)
\end{gathered}
$$
<br/><br/>$\begin{aligned} & =\lim _{h \rightarrow 0} \frac{\cos ^{-1}\left(1-h^2\right) \sin ^{-1}(1-h)}{h\left(1-h^2\right)} \\\\ & =\lim _{\mathrm... | integer | jee-main-2024-online-1st-february-morning-shift |
lsbkqsz3 | maths | limits-continuity-and-differentiability | limits-of-trigonometric-functions | If $\mathrm{a}=\lim\limits_{x \rightarrow 0} \frac{\sqrt{1+\sqrt{1+x^4}}-\sqrt{2}}{x^4}$ and $\mathrm{b}=\lim\limits _{x \rightarrow 0} \frac{\sin ^2 x}{\sqrt{2}-\sqrt{1+\cos x}}$, then the value of $a b^3$ is : | [{"identifier": "A", "content": "36"}, {"identifier": "B", "content": "25"}, {"identifier": "C", "content": "32"}, {"identifier": "D", "content": "30"}] | ["C"] | null | <p>$$\begin{aligned}
a= & \lim _{x \rightarrow 0} \frac{\sqrt{1+\sqrt{1+x^4}}-\sqrt{2}}{x^4} \\
& =\lim _{x \rightarrow 0} \frac{\sqrt{1+x^4}-1}{x^4\left(\sqrt{1+\sqrt{1+x^4}}+\sqrt{2}\right)} \\
& =\lim _{x \rightarrow 0} \frac{x^4}{x^4\left(\sqrt{1+\sqrt{1+x^4}}+\sqrt{2}\right)\left(\sqrt{1+x^4}+1\right)}
\end{aligne... | mcq | jee-main-2024-online-27th-january-morning-shift |
jaoe38c1lscnb2gp | maths | limits-continuity-and-differentiability | limits-of-trigonometric-functions | <p>$$\text { If } \lim _\limits{x \rightarrow 0} \frac{3+\alpha \sin x+\beta \cos x+\log _e(1-x)}{3 \tan ^2 x}=\frac{1}{3} \text {, then } 2 \alpha-\beta \text { is equal to : }$$</p> | [{"identifier": "A", "content": "2"}, {"identifier": "B", "content": "1"}, {"identifier": "C", "content": "5"}, {"identifier": "D", "content": "7"}] | ["C"] | null | <p>$$\lim _\limits{x \rightarrow 0} \frac{3+\alpha \sin x+\beta \cos x+\log _e(1-x)}{3 \tan ^2 x}=\frac{1}{3}$$</p>
<p>$$\Rightarrow \lim _\limits{x \rightarrow 0} \frac{3+\alpha\left[x-\frac{x^3}{3 !}+\ldots .\right]+\beta\left[1-\frac{x^2}{2 !}+\frac{x^4}{4 !} \ldots .\right]+\left(-x-\frac{x^2}{2}-\frac{x^3}{3} \ldo... | mcq | jee-main-2024-online-27th-january-evening-shift |
lv5grwgp | maths | limits-continuity-and-differentiability | limits-of-trigonometric-functions | <p>The value of $$\lim _\limits{x \rightarrow 0} 2\left(\frac{1-\cos x \sqrt{\cos 2 x} \sqrt[3]{\cos 3 x} \ldots \ldots . \sqrt[10]{\cos 10 x}}{x^2}\right)$$ is __________.</p> | [] | null | 55 | <p>$$\mathop {\lim }\limits_{x \to 0} 2\left( {{{1 - \cos x{{(\cos 2x)}^{{1 \over 2}}}{{(\cos 3x)}^{{1 \over 3}}}\,...\,{{(\cos 10x)}^{{1 \over {10}}}}} \over {{x^2}}}} \right)$$ $$\left(\frac{0}{0} \text { form }\right)$$</p>
<p>Using L' hospital</p>
<p>$$2 \lim _\limits{x \rightarrow 0} \frac{\sin x(\cos 2 x)^{\frac{... | integer | jee-main-2024-online-8th-april-morning-shift |
lv9s20md | maths | limits-continuity-and-differentiability | limits-of-trigonometric-functions | <p>Let $$\mathrm{a}>0$$ be a root of the equation $$2 x^2+x-2=0$$. If $$\lim _\limits{x \rightarrow \frac{1}{a}} \frac{16\left(1-\cos \left(2+x-2 x^2\right)\right)}{(1-a x)^2}=\alpha+\beta \sqrt{17}$$, where $$\alpha, \beta \in Z$$, then $$\alpha+\beta$$ is equal to _________.</p> | [] | null | 170 | <p>$$\because 2 x^2+x-2=0$$ has two roots where
$$a=\frac{\sqrt{17}-1}{4}$$ and another root is $$\frac{-\sqrt{17}-1}{4}$$</p>
<p>And $$2+x-2 x^2=-2\left(x-\frac{1}{a}\right)\left(x+\frac{4}{\sqrt{17}+1}\right)$$</p>
<p>Now $$\lim _\limits{x \rightarrow \frac{1}{a}} \frac{16\left(1-\cos \left(2+x-2 x^2\right)\right)}{... | integer | jee-main-2024-online-5th-april-evening-shift |
1lguw56u8 | maths | logarithm | logarithm-inequalities | <p>The number of integral solutions $$x$$ of $$\log _{\left(x+\frac{7}{2}\right)}\left(\frac{x-7}{2 x-3}\right)^{2} \geq 0$$ is :</p> | [{"identifier": "A", "content": "8"}, {"identifier": "B", "content": "7"}, {"identifier": "C", "content": "5"}, {"identifier": "D", "content": "6"}] | ["D"] | null | $$\log _{\left(x+\frac{7}{2}\right)}\left(\frac{x-7}{2 x-3}\right)^{2} \geq 0$$
<br/><br/><b>Domain :</b>
<br/><br/>$$
\begin{aligned}
& x+\frac{7}{2}>0 \\\\
& x>\frac{-7}{2} \\\\
& x+\frac{7}{2} \neq 1 \\\\
& x \neq \frac{-5}{2} \\\\
& \frac{x-7}{2 x-3} \neq 0 \\\\
& x \neq 7 \\\\
& x \neq \frac{3}{2}
\end{aligned}
$$... | mcq | jee-main-2023-online-11th-april-morning-shift |
L7moS1Zt7WzMgKYdTL7k9k2k5itzmge | maths | logarithm | logarithmic-equations | The number of distinct solutions of the equation<br/>
$${\log _{{1 \over 2}}}\left| {\sin x} \right| = 2 - {\log _{{1 \over 2}}}\left| {\cos x} \right|$$ in the interval
[0, 2$$\pi $$], is ____. | [] | null | 8 | $${\log _{{1 \over 2}}}\left| {\sin x} \right| = 2 - {\log _{{1 \over 2}}}\left| {\cos x} \right|$$
<br><br>$$ \Rightarrow $$ $${\log _{{1 \over 2}}}\left| {\sin x} \right|$$ + $${\log _{{1 \over 2}}}\left| {\cos x} \right|$$ = 2
<br><br>$$ \Rightarrow $$ $${\log _{{1 \over 2}}}\left( {\left| {\sin x\cos x} \right|} \r... | integer | jee-main-2020-online-9th-january-morning-slot |
Eqw2IGWkAs3wRXidEM1kluhm49z | maths | logarithm | logarithmic-equations | The number of solutions of the equation log<sub>4</sub>(x $$-$$ 1) = log<sub>2</sub>(x $$-$$ 3) is _________. | [] | null | 1 | $${\log _4}(x - 1) = {\log _2}(x - 3)$$<br><br>$$ \Rightarrow {1 \over 2}{\log _2}(x - 1) = {\log _2}(x - 3)$$<br><br>$$ \Rightarrow {\log _2}{(x - 1)^{1/2}} = {\log _2}(x - 3)$$<br><br>$$ \Rightarrow {(x - 1)^{1/2}} = {\log _2}(x - 3)$$<br><br>$$ \Rightarrow {(x - 1)^{1/2}} = x - 3$$<br><br>$$ \Rightarrow x - 1 = {x^2... | integer | jee-main-2021-online-26th-february-morning-slot |
1krrvdewm | maths | logarithm | logarithmic-equations | The number of solutions of the equation <br/><br/>$${\log _{(x + 1)}}(2{x^2} + 7x + 5) + {\log _{(2x + 5)}}{(x + 1)^2} - 4 = 0$$, x > 0, is : | [] | null | 1 | $${\log _{(x + 1)}}(2{x^2} + 7x + 5) + {\log _{(2x + 5)}}{(x + 1)^2} - 4 = 0$$<br><br>$${\log _{(x + 1)}}(2x + 5)(x + 1) + 2{\log _{(2x + 5)}}(x + 1) = 4$$<br><br>$${\log _{(x + 1)}}(2x + 5) + 1 + 2{\log _{(2x + 5)}}(x + 1) = 4$$<br><br>Put $${\log _{(x + 1)}}(2x + 5) = t$$<br><br>$$t + {2 \over t} = 3 \Rightarrow {t^2... | integer | jee-main-2021-online-20th-july-evening-shift |
1ldr78hpn | maths | logarithm | logarithmic-equations | <p>If the solution of the equation $$\log _{\cos x} \cot x+4 \log _{\sin x} \tan x=1, x \in\left(0, \frac{\pi}{2}\right)$$, is $$\sin ^{-1}\left(\frac{\alpha+\sqrt{\beta}}{2}\right)$$, where $$\alpha$$, $$\beta$$ are integers, then $$\alpha+\beta$$ is equal to :</p> | [{"identifier": "A", "content": "3"}, {"identifier": "B", "content": "6"}, {"identifier": "C", "content": "4"}, {"identifier": "D", "content": "5"}] | ["C"] | null | <p>$${\log _{\cos x}}\cot x + 4{\log _{\sin x}}\tan x = 1$$</p>
<p>$$ \Rightarrow {\log _{\cos x}}\cot x - 4{\log _{\sin x}}\cot x = 1$$</p>
<p>$$ \Rightarrow 1 - {\log _{\cos x}}\sin x - 4 - 4{\log _{\sin x}}\cos x = 1$$</p>
<p>Let $${\log _{\cos x}}\sin x = t$$</p>
<p>$$t + {4 \over t} = 4$$</p>
<p>$$ \Rightarrow t =... | mcq | jee-main-2023-online-30th-january-morning-shift |
1ldv30uxy | maths | logarithm | logarithmic-equations | <p>Let $$S = \left\{ {\alpha :{{\log }_2}({9^{2\alpha - 4}} + 13) - {{\log }_2}\left( {{5 \over 2}.\,{3^{2\alpha - 4}} + 1} \right) = 2} \right\}$$. Then the maximum value of $$\beta$$ for which the equation $${x^2} - 2{\left( {\sum\limits_{\alpha \in s} \alpha } \right)^2}x + \sum\limits_{\alpha \in s} {{{(\alpha... | [] | null | 25 | $$
\begin{aligned}
& \log _2\left(9^{2 \alpha-4}+13\right)-\log _2\left(\frac{5}{2} \cdot 3^{2 \alpha-4}+1\right)=2 \\\\
& \Rightarrow \frac{9^{2 \alpha-4}+13}{\frac{5}{2} 3^{2 \alpha-4}+1}=4 \\\\
& \Rightarrow \alpha=2 \quad \text { or } \quad 3 \\\\
& \sum_{\alpha \in \mathrm{S}} \alpha=5 \text { and } \sum_{\alpha \... | integer | jee-main-2023-online-25th-january-morning-shift |
1lgxwj4h5 | maths | logarithm | logarithmic-equations | <p>Let a, b, c be three distinct positive real numbers such that $${(2a)^{{{\log }_e}a}} = {(bc)^{{{\log }_e}b}}$$ and $${b^{{{\log }_e}2}} = {a^{{{\log }_e}c}}$$.<br/><br/>Then, 6a + 5bc is equal to ___________.</p> | [] | null | 8 | Given, $(2 a)^{\ln a}=(b c)^{\ln b}$, where $2 a>0, b c>0$
<br/><br/>$$
\Rightarrow \ln a(\ln 2+\ln a)=\ln b(\ln b+\ln c)
$$ ..........(i)
<br/><br/>and
$(b)^{\ln 2}=(a)^{\ln c}$
<br/><br/>$$
\Rightarrow \ln 2 \cdot \ln b=\ln c \cdot \ln a
$$ ..........(ii)
<br/><br/>Now, let $\ln a=x, \ln b=y$
<br/><br/>$$
\ln 2=p, \l... | integer | jee-main-2023-online-10th-april-morning-shift |
3YLjdLeFsjhnI2wc | maths | mathematical-induction | mathematical-induction | If $${a_n} = \sqrt {7 + \sqrt {7 + \sqrt {7 + .......} } } $$ having $$n$$ radical signs then by methods of mathematical induction which is true | [{"identifier": "A", "content": "$${a_n} > 7\\,\\,\\forall \\,\\,n \\ge 1$$ "}, {"identifier": "B", "content": "$${a_n} < 7\\,\\,\\forall \\,\\,n \\ge 1$$ "}, {"identifier": "C", "content": "$${a_n} < 4\\,\\,\\forall \\,\\,n \\ge 1$$ "}, {"identifier": "D", "content": "$${a_n} > 3\\,\\,\\forall \\,\\,n \\ge... | ["D"] | null | Given $${a_n} = \sqrt {7 + \sqrt {7 + \sqrt {7 + .......} } } $$
<br><br>$$\therefore$$ $${a_n} = \sqrt {7 + {a_n}} $$
<br><br>$$ \Rightarrow $$ $$a_n^2 = 7 + {a_n}$$
<br><br>$$ \Rightarrow $$ $$a_n^2 - {a_n} - 7 = 0$$
<br><br>$$ \Rightarrow {a_n} = {{1 \pm \sqrt {1 - 4 \times 1 \times - 7} } \over 2}$$
<br><br>$$ \Ri... | mcq | aieee-2002 |
hj54pq327GNyXr3n | maths | mathematical-induction | mathematical-induction | Let $$S(K)$$ $$ = 1 + 3 + 5... + \left( {2K - 1} \right) = 3 + {K^2}.$$ Then which of the following is true | [{"identifier": "A", "content": "Principle of mathematical induction can be used to prove the formula"}, {"identifier": "B", "content": "$$S\\left( K \\right) \\Rightarrow S\\left( {K + 1} \\right)$$ "}, {"identifier": "C", "content": "$$S\\left( K \\right) \\ne S\\left( {K + 1} \\right)$$ "}, {"identifier": "D", "cont... | ["B"] | null | Given $$S(K)$$ $$ = 1 + 3 + 5... + \left( {2K - 1} \right) = 3 + {K^2}$$
<br><br>When k = 1, S(1): 1 = 3 + 1,
<br><br>L.H.S of S(k) $$ \ne $$ R.H.S of S(k)
<br><br>So S(1) is not true.
<br><br>As S(1) is not true so principle of mathematical induction can not be used.
<br><br>S(K+1) = 1 + 3 + 5... + (2K - 1) + (2K + 1... | mcq | aieee-2004 |
9WbUqXiIlcooNReg | maths | mathematical-induction | mathematical-induction | If $$A = \left[ {\matrix{
1 & 0 \cr
1 & 1 \cr
} } \right]$$ and $$I = \left[ {\matrix{
1 & 0 \cr
0 & 1 \cr
} } \right],$$ then which one of the following holds for all $$n \ge 1,$$ by the principle of mathematical induction? | [{"identifier": "A", "content": "$${A^n} = nA - \\left( {n - 1} \\right){\\rm I}$$ "}, {"identifier": "B", "content": "$${A^n} = {2^{n - 1}}A - \\left( {n - 1} \\right){\\rm I}$$ "}, {"identifier": "C", "content": "$${A^n} = nA + \\left( {n - 1} \\right){\\rm I}$$"}, {"identifier": "D", "content": "$${A^n} = {2^{n - 1}... | ["A"] | null | Given $$A = \left[ {\matrix{
1 & 0 \cr
1 & 1 \cr
} } \right]$$
<br><br>$$\therefore$$ $$A \times A$$ = $${A^2}$$ = $$\left[ {\matrix{
1 & 0 \cr
2 & 1 \cr
} } \right]$$
<br><br>and $${A^3}$$ = $${A^2} \times A$$ = $$\left[ {\matrix{
1 & 0 \cr
3 & 1 \cr
} } \rig... | mcq | aieee-2005 |
x3tbE32XVPE9RRjV | maths | mathematical-reasoning | logical-connectives | The statement $$p \to \left( {q \to p} \right)$$ is equivalent to | [{"identifier": "A", "content": "$$p \\to \\left( {p \\leftrightarrow q} \\right)$$"}, {"identifier": "B", "content": "$$p \\to \\left( {p \\to q} \\right)$$"}, {"identifier": "C", "content": "$$p \\to \\left( {p \\vee q} \\right)$$"}, {"identifier": "D", "content": "$$p \\to \\left( {p \\wedge q} \\right)$$"}] | ["C"] | null | <p><style type="text/css">
.tg {border-collapse:collapse;border-spacing:0;}
.tg td{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px;
overflow:hidden;padding:10px 5px;word-break:normal;}
.tg th{border-color:black;border-style:solid;border-width:1px;font-family:Arial,... | mcq | aieee-2008 |
U4V9b6lrngYsbSHC | maths | mathematical-reasoning | logical-connectives | <b>Statement-1 :</b> $$ \sim \left( {p \leftrightarrow \sim q} \right)$$ is equivalent to $${p \leftrightarrow q}$$.
<br/><b>Statement-2 :</b> $$ \sim \left( {p \leftrightarrow \sim q} \right)$$ is a tautology. | [{"identifier": "A", "content": "Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1"}, {"identifier": "B", "content": "Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1"}, {"identifier": "C", "content": "Statement-1 is true, St... | ["C"] | null | <img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l3fdb05p/c9d4c926-25a7-47f2-b2d6-27e5c7d3dd72/0e89a3d0-d8be-11ec-bed7-8f98fee06619/file-1l3fdb05q.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l3fdb05p/c9d4c926-25a7-47f2-b2d6-27e5c7d3dd72/0e89a3d0-d8be-11ec-bed7-8f98fee06619/fi... | mcq | aieee-2009 |
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