Let f, g:(0, ∞) → ℝ be two functions defined by f(x)=∫₋ₓˣ(|t|-t²) e^(-t²) d t and g(x)=∫₀^(x²) t^(1 / 2) e⁻ᵗ d t. Then, the value of 9(f(√(log ₑ 9))+g(√(log ₑ 9))) is equal to :

Solution

Maths sir · 4 steps

The t² part of f and all of g are the same integral, so they cancel. Only an easy integral is left.

1. Use evenness. The integrand (|t|-t²)e^(-t²) is even, so f(x) = 2∫₀ˣ (t-t²)e^(-t²) dt. On [0, x] we have |t| = t, which splits f into 2∫₀ˣ t e^(-t²) dt - 2∫₀ˣ t² e^(-t²) dt.
2. Rewrite g. In g, put t = u², so dt = 2u du and the limits 0 to x² become 0 to x. Then g(x) = ∫₀ˣ u e^(-u²)· 2u du = 2∫₀ˣ u² e^(-u²) du, which is exactly the second part of f.

   f(x)+g(x) = 2∫₀ˣ t e^(-t²) dt = [-e^(-t²)]₀ˣ = 1 - e^(-x²)

3. Put in x. With x = √(logₑ 9), we get x² = logₑ 9, so e^(-x²) = 1/9. That gives f + g = 1 - 1/9 = 8/9.
4. So. 9(f+g) = 9 × 8/9 = 8, which is option (c).

The idea to keep. When two integrals look hard on their own but you are asked for their sum, substitute so both use the same variable and limits. The hard parts usually cancel.

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