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JEE Advanced 2022 Paper 1 Online
Numerical
+3
-0
Let $$\alpha$$ be a positive real number. Let $$f: \mathbb{R} \rightarrow \mathbb{R}$$ and $$g:(\alpha, \infty) \rightarrow \mathbb{R}$$ be the functions defined by
$$ f(x)=\sin \left(\frac{\pi x}{12}\right) \quad \text { and } \quad g(x)=\frac{2 \log _{\mathrm{e}}(\sqrt{x}-\sqrt{\alpha})}{\log _{\mathrm{e}}\left(e^{\sqrt{x}}-e^{\sqrt{\alpha}}\right)} . $$
Then the value of $$\lim \limits_{x \rightarrow \alpha^{+}} f(g(x))$$ is
$$ f(x)=\sin \left(\frac{\pi x}{12}\right) \quad \text { and } \quad g(x)=\frac{2 \log _{\mathrm{e}}(\sqrt{x}-\sqrt{\alpha})}{\log _{\mathrm{e}}\left(e^{\sqrt{x}}-e^{\sqrt{\alpha}}\right)} . $$
Then the value of $$\lim \limits_{x \rightarrow \alpha^{+}} f(g(x))$$ is
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2
JEE Advanced 2020 Paper 2 Offline
Numerical
+3
-1
Let the functions $$f:( - 1,1) \to R$$ and $$g:( - 1,1) \to ( - 1,1)$$ be defined by $$f(x) = |2x - 1| + |2x + 1|$$ and $$g(x) = x - [x]$$, where [x] denotes the greatest integer less than or equal to x. Let $$f\,o\,g:( - 1,1) \to R$$ be the composite function defined by $$(f\,o\,g)(x) = f(g(x))$$. Suppose c is the number of points in the interval ($$-$$1, 1) at which $$f\,o\,g$$ is NOT continuous, and suppose d is the number of points in the interval ($$-$$1, 1) at which $$f\,o\,g$$ is NOT differentiable. Then the value of c + d is ............
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3
JEE Advanced 2020 Paper 2 Offline
Numerical
+3
-1
The value of the limit
$$\mathop {\lim }\limits_{x \to {\pi \over 2}} {{4\sqrt 2 (\sin 3x + \sin x)} \over {\left( {2\sin 2x\sin {{3x} \over 2} + \cos {{5x} \over 2}} \right) - \left( {\sqrt 2 + \sqrt 2 \cos 2x + \cos {{3x} \over 2}} \right)}}$$
is ...........
$$\mathop {\lim }\limits_{x \to {\pi \over 2}} {{4\sqrt 2 (\sin 3x + \sin x)} \over {\left( {2\sin 2x\sin {{3x} \over 2} + \cos {{5x} \over 2}} \right) - \left( {\sqrt 2 + \sqrt 2 \cos 2x + \cos {{3x} \over 2}} \right)}}$$
is ...........
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4
JEE Advanced 2020 Paper 1 Offline
Numerical
+4
-0
let e denote the base of the natural logarithm. The value of the real number a for which the right hand limit
$$\mathop {\lim }\limits_{x \to {0^ + }} {{{{(1 - x)}^{1/x}} - {e^{ - 1}}} \over {{x^a}}}$$
is equal to a non-zero real number, is .............
$$\mathop {\lim }\limits_{x \to {0^ + }} {{{{(1 - x)}^{1/x}} - {e^{ - 1}}} \over {{x^a}}}$$
is equal to a non-zero real number, is .............
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