Home
Class 12
MATHS
Given f(x) = tan(sqrt(pi^2/16 - x^2)) an...

Given `f(x) = tan(sqrt(pi^2/16 - x^2))` and A = R - [0,1]

A

Range of f (x) is A

B

Range of f(x) is A'

C

Maximum value of f(x)=1

D

Minimum value of `(f(x))^(-1)=1`

Text Solution

Verified by Experts

The correct Answer is:
B, C
Promotional Banner

Similar Questions

Explore conceptually related problems

STATEMENT-1: The values of f(x)=3sin(sqrt((pi^(2))/(16)-x^(2)) lie in the interval [0,(3)/(sqrt(3))]. and STATEMENT-2The domain of definition of the function f(x)=3sin(sqrt((pi^(2))/(16)-x^(2)) is [-(pi)/(4),(pi)/(4)]

If f(x) = tan^-1 {1/x (sqrt(1+x^2))} then f'(0) is

If f (x) = tan ^(-1)sqrt((1 + sin x )/(1 - sin x)), 0 le x le (pi)/(2) then f' ((pi)/(6)) =?

Find the simplest value of f(x)=tan^(-1)((sqrt(1+x^(2))-1)/(x)),x in R-{0}

If f(x) = tan^(-1)(sqrt((1+sinx)/(1-sinx))), 0 lt x lt pi/2 , then f'(pi/6) is

f(x)=2x-tan^(-1)x-ln(x+sqrt(1+x^(2)));x in R then

The inverse of the function of f:R to R given by f(x)=log_(a) (x+sqrt(x^(2)+1)(a gt 0, a ne 1) is

If f (x) = sqrt(cos ec ^(2) x - 2 sin x cos x - (1)/(tan ^(2) x )) x in ((7pi)/(4), 2pi ) then f' ((11 pi)/(6))=