Let y=f(x) be a thrice differentiable function in (-5,5). Let the tangents to the curve y=f(x) at (1, f(1)) and (3, f(3)) make angles π / 6 and π / 4, respectively with positive x-axis. If 27 ∫1³((f′(t))²+1) f^(′ ′)(t) d t=α+β √3 where α, β are integers, then the value of α+β equals

Solution

Maths sir · 4 steps

f''(t) dt is just d(f'(t)), so put u = f'(t) and the integral becomes a polynomial.

1. Read the slopes. The slope of a tangent is the tan of its angle with the positive x-axis. So f'(1) = tan π/6 = 1/√3 and f'(3) = tan π/4 = 1.
2. Substitute. Let u = f'(t), so du = f''(t) dt. The limits change from t = 1, 3 to u = 1/√3, 1.

   ∫₁³((f'(t))²+1)f''(t) dt = ∫_(1/√3)¹(u²+1) du = [u³/3+u]_(1/√3)¹

3. Evaluate. At u = 1 the bracket is 4/3. At u = 1/√3 it is 1/9√3 + 1/√3 = 10/9√3 = 10√3/27. So the integral is 4/3 - 10√3/27.
4. So. Multiplying by 27 gives 36 - 10√3, so α = 36, β = -10 and α + β = 26, option (a).

The idea to keep. When an integrand has g(f'(t)) times f''(t), substitute u = f'(t). Tangent angles then give you the new limits directly as tan θ.

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