Let g(t)=∫ limits-π / 2π / 2 cos ((π/4) t+f(x)) d x, where f(x)= log e(x+√x2+1), x ∈ R. Then which one of the following is correct?

Question

Let g(t)=∫ limits-π / 2π / 2 cos ((π/4) t+f(x)) d x, where f(x)= log e(x+√x2+1), x ∈ R. Then which one of the following is correct? (A) g(1)=g(0) (B) √2 g(1)=g(0) (C) g(1)=√2 g(0) (D) g(1)+g(0)=0

Tardigrade Admin · Accepted Answer

g(t)=∫ limits-π / 2π / 2( cos (π/4) t+f(x)) d x g(t)=π cos (π/4) t+∫ limits-π / 2π / 2 f(x) d x g(t)=π cos (π/4) t g(1)=(π/√2), g(0)= pi

Q. Let $g (t) = - π /2 \int π /2 cos (\frac{π}{4} t + f (x)) d x$ ,
where $f (x) = lo g_{e} (x + x^{2} + 1), x \in R$ . Then which one of the following is correct?

$g (1) = g (0)$

$2 g (1) = g (0)$

$g (1) = 2 g (0)$

$g (1) + g (0) = 0$

$g (t) = - π /2 \int π /2 (cos \frac{π}{4} t + f (x)) d x$ $g (t) = π cos \frac{π}{4} t + - π /2 \int π /2 f (x) d x$ $g (t) = π cos \frac{π}{4} t$ $g (1) = \frac{π}{2}, g (0) = p i$

Q. Let g(t)=−π/2∫π/2​cos(4π​t+f(x))dx, where f(x)=loge​(x+x2+1​),x∈R. Then which one of the following is correct?

g(1)=g(0)

2​g(1)=g(0)

g(1)=2​g(0)

g(1)+g(0)=0