1
JEE Main 2022 (Online) 26th July Morning Shift
+4
-1

Let $$f(x) = \left\{ {\matrix{ {{x^3} - {x^2} + 10x - 7,} & {x \le 1} \cr { - 2x + {{\log }_2}({b^2} - 4),} & {x > 1} \cr } } \right.$$.

Then the set of all values of b, for which f(x) has maximum value at x = 1, is :

A
($$-$$6, $$-$$2)
B
(2, 6)
C
$$[ - 6, - 2) \cup (2,6]$$
D
$$\left[ {-\sqrt 6 , - 2} \right) \cup \left( {2,\sqrt 6 } \right]$$
2
JEE Main 2022 (Online) 25th July Evening Shift
+4
-1

$$\lim\limits_{x \rightarrow \frac{\pi}{4}} \frac{8 \sqrt{2}-(\cos x+\sin x)^{7}}{\sqrt{2}-\sqrt{2} \sin 2 x}$$ is equal to

A
14
B
7
C
14$$\sqrt2$$
D
7$$\sqrt2$$
3
JEE Main 2022 (Online) 25th July Morning Shift
+4
-1

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 :

A
4
B
$$-$$8
C
$$-$$4
D
8
4
JEE Main 2022 (Online) 29th June Evening Shift
+4
-1

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:

A
$${{{\pi ^2}} \over 6}$$
B
$${{{\pi ^2}} \over 3}$$
C
$${{{\pi ^2}} \over 2}$$
D
$$\pi$$2
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