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tarinbansal

CHECK THIS OUT ALSO-

In the ques-

I₁ = pi²/4 + root 2

I₂ = [tan^-1(1/e)]² + 2e/root(e² +1)

I₃ = [tan^-1(e)]² + 2/root(e² +1)

The answer given is I₂<I₁<I₃

The solution-

Consider f(x)=[tan^-1(x)]² + 2/root(x² +1)

f '(x)>0 for all x belonging to (0,infinity)

so f(1/e)<f(1)<f(e)

U think it is correct?

DHYAAN SE DEKHO.

In I₂ , they hav considered f(1), tan^-1 1 =pi/4 NOT pi/2

It means I₂ = pi²/16 + .....

I think they shud be sent to study in class X to learn that tan 45=1 NOT tan 90.

The correct ans shud be D.