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Question

Mathematics Question on Inverse Trigonometric Functions

Let
f(x)=2cos−1⁡x+4cot−1⁡x−3x2−2x+10,X∈[−1,1]f(x)=2cos^{−1}⁡x+4cot^{−1}⁡x−3x^2−2x+10,X∈[−1,1]
If [a, b] is the range of the function, f then 4a – b is equal to :

A

11

B

11–π

C

11+π

D

15-π

Answer

11–π

Explanation

Solution

The correct asnwer is (B) : 11–π
f(x)=2cos−1⁡x+4cot−1⁡x−3x2−2x+10∀x∈[−1,1]f(x)=2cos^{−1}⁡x+4cot^{−1}⁡x−3x^2−2x+10 ∀x∈[−1,1]
⇒f‘(x)=−2(1−x2−41+x2−6x−2<0∀x∈[−1,1]⇒f‘(x)=−\frac{2}{\sqrt{(1−x^2}}−\frac{4}{1+x^2}−6x−2<0 ∀x∈[−1,1]
So f(x) is decreasing function and range of f(x) is
[f(1), f(-1)], which is [π + 5, 5π + 9]
Now 4a – b = 4(π + 5) – (5π + 9)
= 11 – π