If f(x)= sin x+ cos x, g(x)=x2-1, then g(f(x)) is invertible in the domain

Question

If f(x)= sin x+ cos x, g(x)=x2-1, then g(f(x)) is invertible in the domain (A) [0, (π/2)] (B) [-(π/4), (π/4)] (C) [-(π/2), (π/2)] (D) [0, π]

Tardigrade Admin · Accepted Answer

f(x)= sin x+ cos x, g(x)=x2-1 ⇒ g(f(x))=( sin x+ cos x)2-1= sin 2 x Clearly g(f(x)) is invertible in -(π/2) ≤ 2 x ≤ (π/2) [∵ sin θ is invertible when -π / 2 ≤ θ ≤ π / 2] ⇒ -(π/4) ≤ x ≤ (π/4)

Q. If $f (x) = sin x + cos x, g (x) = x^{2} - 1$ , then $g (f (x))$ is invertible in the domain

$[0, \frac{π}{2}]$

$[- \frac{π}{4}, \frac{π}{4}]$

$[- \frac{π}{2}, \frac{π}{2}]$

$[0, π]$

$f (x) = sin x + cos x, g (x) = x^{2} - 1$
$\Rightarrow g (f (x)) = (sin x + cos x)^{2} - 1 = sin 2 x$
Clearly $g (f (x))$ is invertible in
$- \frac{π}{2} \leq 2 x \leq \frac{π}{2}$
$[∵ sin θ$ is invertible when $- π /2 \leq θ \leq π /2]$
$\Rightarrow - \frac{π}{4} \leq x \leq \frac{π}{4}$

Q. If f(x)=sinx+cosx,g(x)=x2−1, then g(f(x)) is invertible in the domain

[0,2π​]

[−4π​,4π​]

[−2π​,2π​]

[0,π]

f(x)=sinx+cosx,g(x)=x2−1 ⇒g(f(x))=(sinx+cosx)2−1=sin2x Clearly g(f(x)) is invertible in −2π​≤2x≤2π​ [∵sinθ is invertible when −π/2≤θ≤π/2] ⇒−4π​≤x≤4π​

Q. If $f (x) = sin x + cos x, g (x) = x^{2} - 1$ , then $g (f (x))$ is invertible in the domain

$[0, \frac{π}{2}]$

$[- \frac{π}{4}, \frac{π}{4}]$

$[- \frac{π}{2}, \frac{π}{2}]$

$[0, π]$