If range of f(x)= cos -1(x2 ⋅ cos 1+ sin 1 ⋅ √1-x4) is [a, b], then (b-a) is equal to

Question

If range of f(x)= cos -1(x2 ⋅ cos 1+ sin 1 ⋅ √1-x4) is [a, b], then (b-a) is equal to (A) (π/2) (B) π (C) 6-(3 π/2) (D) 1

Tardigrade Admin · Accepted Answer

f ( x )= cos -1( x 2 ⋅ cos 1+ sin 1 ⋅ √1- x 4)=| cos -1( x 2)- cos -1( cos 1)|=| cos -1( x 2)-1| text (even function) text Domian of f ( x ) text is [-1,1] text Range of f ( x ) text is [0,1] ∴ b - a =1 Aliter: f(x)= begincases1-θ, θ ∈[0,1] θ-1, θ ∈[1, (π/2)] endcases Range of f(x) is [0,1].

Q. If range of $f (x) = cos^{- 1} (x^{2} \cdot cos 1 + sin 1 \cdot 1 - x^{4})$ is $[a, b]$ , then $(b - a)$ is equal to

$\frac{π}{2}$

$π$

$6 - \frac{3 π}{2}$

1

Q. If range of f(x)=cos−1(x2⋅cos1+sin1⋅1−x4​) is [a,b], then (b−a) is equal to

2π​

π

6−23π​

1

f(x)=cos−1(x2⋅cos1+sin1⋅1−x4​)=∣∣​cos−1(x2)−cos−1(cos1)∣∣​=∣∣​cos−1(x2)−1∣∣​ (even function) Domian of f(x) is [−1,1] Range of f(x) is [0,1] ∴b−a=1 Aliter: f(x)={1−θ,θ−1,​θ∈[0,1]θ∈[1,2π​]​ Range of f(x) is [0,1].

Q. If range of $f (x) = cos^{- 1} (x^{2} \cdot cos 1 + sin 1 \cdot 1 - x^{4})$ is $[a, b]$ , then $(b - a)$ is equal to

$\frac{π}{2}$

$π$

$6 - \frac{3 π}{2}$