If , , then is
- A.3
- B.
- C.1/3
- D.None of these
Show answer and solution
Correct option: A
Solution
Differentiate both sides with respect to x. The left side, by Leibniz rule (with upper limit constant 1 and lower limit sin x), gives:
-f(sin x)·cos x = -cos x
So f(sin x) = 1 for all x ∈ (0, π/2).
Wait — that gives f(sin x) = 1, meaning f is identically 1, so f(1/√3) = 1. But the official answer is 3.
Let me re-examine. Differentiating ∫sin x1 t²f(t) dt = 1 - sin x:
d/dx[∫sin x1 t²f(t) dt] = -sin²(x)·f(sin x)·cos x = -cos x
So sin²(x)·f(sin x) = 1, giving f(sin x) = 1/sin²(x).
Now set sin x = 1/√3, so f(1/√3) = 1/(1/√3)² = 1/(1/3) = 3.
The key step is correctly applying the chain rule: the derivative of the lower limit sin x brings a factor of cos x AND the integrand evaluated at sin x is sin²(x)·f(sin x), so sin²(x)·f(sin x)·cos x = cos x, yielding f(1/√3) = 3.
-f(sin x)·cos x = -cos x
So f(sin x) = 1 for all x ∈ (0, π/2).
Wait — that gives f(sin x) = 1, meaning f is identically 1, so f(1/√3) = 1. But the official answer is 3.
Let me re-examine. Differentiating ∫sin x1 t²f(t) dt = 1 - sin x:
d/dx[∫sin x1 t²f(t) dt] = -sin²(x)·f(sin x)·cos x = -cos x
So sin²(x)·f(sin x) = 1, giving f(sin x) = 1/sin²(x).
Now set sin x = 1/√3, so f(1/√3) = 1/(1/√3)² = 1/(1/3) = 3.
The key step is correctly applying the chain rule: the derivative of the lower limit sin x brings a factor of cos x AND the integrand evaluated at sin x is sin²(x)·f(sin x), so sin²(x)·f(sin x)·cos x = cos x, yielding f(1/√3) = 3.
[AIEEE 2005]