1
JEE Main 2021 (Online) 27th August Evening Shift
+4
-1
Let M and m respectively be the maximum and minimum values of the function
f(x) = tan$$-$$1 (sin x + cos x) in $$\left[ {0,{\pi \over 2}} \right]$$, then the value of tan(M $$-$$ m) is equal to :
A
$$2 + \sqrt 3$$
B
$$2 - \sqrt 3$$
C
$$3 + 2\sqrt 2$$
D
$$3 - 2\sqrt 2$$
2
JEE Main 2021 (Online) 27th August Morning Shift
+4
-1
If $${({\sin ^{ - 1}}x)^2} - {({\cos ^{ - 1}}x)^2} = a$$; 0 < x < 1, a $$\ne$$ 0, then the value of 2x2 $$-$$ 1 is :
A
$$\cos \left( {{{4a} \over \pi }} \right)$$
B
$$\sin \left( {{{2a} \over \pi }} \right)$$
C
$$\cos \left( {{{2a} \over \pi }} \right)$$
D
$$\sin \left( {{{4a} \over \pi }} \right)$$
3
JEE Main 2021 (Online) 26th August Evening Shift
+4
-1
If $$\sum\limits_{r = 1}^{50} {{{\tan }^{ - 1}}{1 \over {2{r^2}}} = p}$$, then the value of tan p is :
A
$${{101} \over {102}}$$
B
$${{50} \over {51}}$$
C
100
D
$${{51} \over {50}}$$
4
JEE Main 2021 (Online) 26th August Morning Shift
+4
-1
Let $$f(x) = \cos \left( {2{{\tan }^{ - 1}}\sin \left( {{{\cot }^{ - 1}}\sqrt {{{1 - x} \over x}} } \right)} \right)$$, 0 < x < 1. Then :
A
$${(1 - x)^2}f'(x) - 2{(f(x))^2} = 0$$
B
$${(1 + x)^2}f'(x) + 2{(f(x))^2} = 0$$
C
$${(1 - x)^2}f'(x) + 2{(f(x))^2} = 0$$
D
$${(1 + x)^2}f'(x) - 2{(f(x))^2} = 0$$
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