If the value of the integral ∫₋₁¹ (cos α x)/(1+3ˣ) d x is 2/π.Then, a value of α is

Solution

Maths sir · 4 steps

Swap x with -x and add the two integrals, and the awkward 1+3ˣ in the denominator disappears.

1. Flip the sign. The limits are -1 and 1, so ∫₋₁¹ f(x) dx = ∫₋₁¹ f(-x) dx. Since cos(-α x) = cos α x and 1/(1+3⁻ˣ) = 3ˣ/(3ˣ+1), the same integral is I = ∫₋₁¹ (3ˣ cos α x)/(1+3ˣ) dx.
2. Add both forms. Add the original I to this one. The numerators combine to (1+3ˣ)cos α x, which cancels the denominator.

   2I = ∫₋₁¹ cos α x dx = [(sin α x)/α]₋₁¹ = (2 sin α)/α

3. Match the value. So I = (sin α)/α, and this must equal 2/π. For α = π/2 it gives 1/(π/2) = 2/π. The other options give 2√2/π, 3√3/2π and 3/π.
4. So. A value of α is π/2, option (a).

The idea to keep. On symmetric limits, a denominator like 1 + aˣ is a signal. Replace x by -x and add the two integrals. The aˣ terms cancel and the integrand becomes plain.

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