Let f(x) = int_1^x √(2 - t^2) dt. Then, the real roots of the equation x^2 - f'(x) = 0 are

Let f(x) = int_1^x √(2 - t^2) dt. Then, the real roots of… — JEE Main 2002 Mathematics PYQ

AIEEE 2002

Let

f(x) = \int_{1}^{x} \sqrt{2 - t^2}\, dt

. Then, the real roots of the equation

x^2 - f'(x) = 0

are

A.
$\pm 1$
B.
$\pm \dfrac{1}{\sqrt{2}}$
C.
$\pm \dfrac{1}{2}$
D.
$0$ and $1$

Show answer and solution

Correct option: A

Solution

f'(x) = √(2 - x²) by the Fundamental Theorem of Calculus. Setting x² = f'(x) gives x² = √(2 - x²). Squaring: x⁴ = 2 - x², so x⁴ + x² - 2 = 0, which factors as (x² - 1)(x² + 2) = 0, giving x² = 1, i.e., x = ±1. Both values satisfy the original equation (checking: 1 = √(2-1) = 1 ✓), so the real roots are ±1.

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