Question

Whichis greater (1/2)^e or 1/e^2 if there is a given function sinx^(sinx), 0 < x < π?

(a) 1/e^2

(b) (1/2)^e

(c) Data not sufficient

(d) Varies as the value of x changes

I have been asked this question in examination.

Enquiry is from Second Order Derivative in portion Limits and Derivatives of Mathematics – Class 11

Answer 1

Right option is (b) (1/2)^e

To explain I would say: Let, f(x) = sinx^(sinx)

So, f(x) = e^sinx log sinx

So, on differentiating it we get,

f’(x) = sinx^(sinx) [cosx log sinx + cos x]

So, f’(x) = 0 when, x = sin^-1(1/e) or x = π/2

Also, f”(x) = sinx^(sinx)[cos^2x(1 + log sinx)^2 – sinx log sinx + (cos^2x/sinx) – sinx]

So that, f”(sin^-1(1/e)) = (e – 1/e)e^-1/e

And, f”(π/2) = -1

Hence, f(x) has local as well as global minimum at x = sin^-1(1/e) also have local and global maximum at x = π/2

Global minimum value of f(x) is (1/e)^1/e

It therefore follows that ,

sin π/6 ^(sinπ/6) > (1/e)^1/e

=>(1/2)^1/2 > (1/e)^1/e

=>(1/2)^e > 1/e^2