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

Solution

Maths sir · 4 steps

Spot that f(x) is odd, expand the cosine, and the sine part dies on the symmetric interval.

1. f is odd. f(-x) = ln(√(x²+1) - x) = ln 1/(x + √(x²+1)) = -f(x). So cos f(x) is even and sin f(x) is odd.
2. Expand the cosine. Use cos(A + B) = cos A cos B - sin A sin B with A = (π t)/4, which does not depend on x. On [-π/2, π/2] the odd part sin f(x) integrates to 0.

   g(t) = cos(π t)/4∫_(-π/2)^(π/2)cos f(x) dx = I cos(π t)/4

3. I is not zero. |f(x)| ≤ ln(π/2 + √(π²/4 + 1)) ≈ 1.23 < π/2, so cos f(x) > 0 and I > 0. That means comparing values of g is the same as comparing values of cos(π t)/4.
4. So. g(0) = I and g(1) = I/√2. That gives √2 g(1) = g(0), which is option (b).

The idea to keep. On a symmetric interval, first check whether the function inside is odd. Then split the integrand into its even and odd parts, and drop the odd part straight away.

Answer from the NTA final answer key. Solution by PrepWiser.