1
JEE Main 2022 (Online) 26th June Morning Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

$$\mathop {\lim }\limits_{x \to {1 \over {\sqrt 2 }}} {{\sin ({{\cos }^{ - 1}}x) - x} \over {1 - \tan ({{\cos }^{ - 1}}x)}}$$ is equal to :

A
$$\sqrt 2 $$
B
$$ - \sqrt 2 $$
C
$${1 \over {\sqrt 2 }}$$
D
$$ - {1 \over {\sqrt 2 }}$$
2
JEE Main 2022 (Online) 26th June Morning Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

Let f, g : R $$\to$$ R be two real valued functions defined as $$f(x) = \left\{ {\matrix{ { - |x + 3|} & , & {x < 0} \cr {{e^x}} & , & {x \ge 0} \cr } } \right.$$ and $$g(x) = \left\{ {\matrix{ {{x^2} + {k_1}x} & , & {x < 0} \cr {4x + {k_2}} & , & {x \ge 0} \cr } } \right.$$, where k1 and k2 are real constants. If (gof) is differentiable at x = 0, then (gof) ($$-$$ 4) + (gof) (4) is equal to :

A
$$4({e^4} + 1)$$
B
$$2(2{e^4} + 1)$$
C
$$4{e^4}$$
D
$$2(2{e^4} - 1)$$
3
JEE Main 2022 (Online) 25th June Evening Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

$$\mathop {\lim }\limits_{x \to {\pi \over 2}} \left( {{{\tan }^2}x\left( {{{(2{{\sin }^2}x + 3\sin x + 4)}^{{1 \over 2}}} - {{({{\sin }^2}x + 6\sin x + 2)}^{{1 \over 2}}}} \right)} \right)$$ is equal to

A
$${1 \over {12}}$$
B
$$-$$$${1 \over {18}}$$
C
$$-$$$${1 \over {12}}$$
D
$${1 \over {6}}$$
4
JEE Main 2022 (Online) 25th June Morning Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

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:

A
$$-$$15
B
$$-$$60
C
60
D
15
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