Test: MCQs (One or More Correct Option): Definite Integrals and Applications of Integrals | JEE Advanced - JEE MCQ

Differentiating both sides w.r.t. x,

[∵ x is an odd function]

Thus, putting value in equation (1) we get 

The two curves meet at

or (1 - m)³ = 27 ,
∴ m = -2 

But if m >1 then 1– m is – ive, then 

∴f (x) is a non constant twice differentiable function such that f (x) = f (1– x)  ⇒ f '(x) = – f ' (1 – x)    ...(1)

but given that 
Hence, f '(x) satisfies all conditions of Rolle's theorem for   So there exists at least one point   and at least one point 

Such that
f "(C₁) = 0 and f "(C₂) = 0 

The area bounded by the curve y = e^x and lines x = 0 and y = e is as shown in the graph.

Also required area

Adding equations (1) and (2), we get
    [as integrand is an even function]

We have

and f '(x) has finite  continuous

Which does not exist at the points where

∴ f '(x) is not differentiable.
∴ (a) is false but (b) is true

First of all let us draw a rough sketch of y = e–x.
At x = 0, y = 1 and at x = 1, y = 1/e

∴ is decreasing on (0, 1)
Hence its graph is as shown in figure given below

Now, S = area exclosed by curve = ABRO

and area of rectangle ORBM = 1/e 

Now S < area of rectangle APSO + area of rectangle CSRN

where ‘a’ can take any even
value.

f(x) = 7 tan⁸x + 7tan⁶x – 3tan⁴x – 3tan²x                
= (7tan⁴x – 3) (tan⁴x + tan²x)                
= (7tan⁶x – 3tan²x) sec²x

= 1/12

∴ Only (d) is the correct option.

∴ f is an incr easing function.

Hence (b) and (c) are the correct options.