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1

JEE Main 2021 (Online) 27th August Evening Shift

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
$$3 - 2\sqrt 2$$

Explanation

Let g(x) = sin x + cos x = $$\sqrt 2$$ sin$$\left( {x + {\pi \over 4}} \right)$$

g(x)$$\in$$ $$\left[ {1,\sqrt 2 } \right]$$ for x$$\in$$ [0, $$\pi$$/2]

f(x) = tan$$-$$1 (sin x + cos x) $$\in$$ $$\left[ {{\pi \over 4},{{\tan }^{ - 1}}\sqrt 2 } \right]$$

tan$$({\tan ^{ - 1}}\sqrt 2 - {\pi \over 4}) = {{\sqrt 2 - 1} \over {1 + \sqrt 2 }} \times {{\sqrt 2 - 1} \over {\sqrt 2 - 1}} = 3 - 2\sqrt 2$$
2

JEE Main 2021 (Online) 27th August Morning Shift

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
$$\sin \left( {{{4a} \over \pi }} \right)$$

Explanation

Given $$a = {({\sin ^{ - 1}}x)^2} - {({\cos ^{ - 1}}x)^2}$$

$$= ({\sin ^{ - 1}}x + {\cos ^{ - 1}}x)({\sin ^{ - 1}}x - {\cos ^{ - 1}}x)$$

$$= {\pi \over 2}\left( {{\pi \over 2} - 2{{\cos }^{ - 1}}x} \right)$$

$$\Rightarrow 2{\cos ^{ - 1}}x = {\pi \over 2} - {{2a} \over \pi }$$

$$\Rightarrow {\cos ^{ - 1}}(2{x^2} - 1) = {\pi \over 2} - {{2a} \over \pi }$$

$$\Rightarrow 2{x^2} - 1 = \cos \left( {{\pi \over 2} - {{2a} \over \pi }} \right)$$
3

JEE Main 2021 (Online) 26th August Evening Shift

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
$${{51} \over {50}}$$

Explanation

$$\sum\limits_{r = 1}^{50} {{{\tan }^{ - 1}}\left( {{2 \over {4{r^2}}}} \right) = \sum\limits_{r = 1}^{50} {{{\tan }^{ - 1}}\left( {{{(2r + 1) - (2r - 1)} \over {1 + (2r + 1)(2r - 1)}}} \right)} }$$

= $$\sum\limits_{r = 1}^{50} {{{\tan }^{ - 1}}(2r + 1) - {{\tan }^{ - 1}}(2r - 1)}$$

= $${\tan ^{ - 1}}(101) - {\tan ^{ - 1}}1 = {\tan ^{ - 1}}{{50} \over {51}}$$
4

JEE Main 2021 (Online) 26th August Morning Shift

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
$${(1 + x)^2}f'(x) - 2{(f(x))^2} = 0$$

Explanation

$$f(x) = \cos \left( {2{{\tan }^{ - 1}}\sin \left( {{{\cot }^{ - 1}}\sqrt {{{1 - x} \over x}} } \right)} \right)$$

$${\cot ^{ - 1}}\sqrt {{{1 - x} \over x}} = {\sin ^{ - 1}}\sqrt x$$

or $$f(x) = \cos (2{\tan ^{ - 1}}\sqrt x )$$

$$= \cos {\tan ^{ - 1}}\left( {{{2\sqrt x } \over {1 - x}}} \right)$$

$$f(x) = {{1 - x} \over {1 + x}}$$

Now, $$f'(x) = {{ - 2} \over {{{(1 + x)}^2}}}$$

or $$f'(x){(1 - x)^2} = - 2{\left( {{{1 - x} \over {1 + x}}} \right)^2}$$

or $${(1 - x)^2}f'(x) + 2{(f(x))^2} = 0$$.

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