Let f(x) = int_{1}^{x} sqrt{2 - t^2} dt. Then | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Given $f(x) = \int_{1}^{x} \sqrt{2 - t^2} dt$.Using the Leibniz integral rule,we differentiate $f(x)$ with respect to $x$:$f'(x) = \frac{d}{dx} \i

Vedclass · Accepted Answer

$\pm 1$

Vedclass · Answer

(A) Given $f(x) = \int_{1}^{x} \sqrt{2 - t^2} dt$.Using the Leibniz integral rule,we differentiate $f(x)$ with respect to $x$:$f'(x) = \frac{d}{dx} \int_{1}^{x} \sqrt{2 - t^2} dt = \sqrt{2 - x^2}$.The given equation is $x^2 - f'(x) = 0$,which implies $x^2 = f'(x)$.Substituting $f'(x)$,we get $x^2 = \sqrt{2 - x^2}$.Squaring both sides,we obtain $x^4 = 2 - x^2$,which simplifies to $x^4 + x^2 - 2 = 0$.Let $u = x^2$. Then the equation becomes $u^2 + u - 2 = 0$.Factoring the quadratic,we get $(u + 2)(u - 1) = 0$.Since $u = x^2$,we have $x^2 = -2$ (which gives no real roots) or $x^2 = 1$.Thus,$x = \pm 1$.

Let $f(x) = \int_{1}^{x} \sqrt{2 - t^2} dt$. Then the real roots of the equation $x^2 - f'(x) = 0$ are

Similar Questions

$\int_0^1 x^{3/2} \sqrt{1-x} \, dx$ is equal to

The points of extremum of $\int_0^{x^2} \frac{t^2-5t+4}{2+e^t} dt$ are

$\int_{-2 \pi}^{2 \pi} \sin ^4(2 x) \cos ^6(2 x) d x=$

The derivative of $F(x) = \int_{x^2}^{x^3} \frac{1}{\log t} \, dt$,$(x > 0)$ is

$\int_0^{\pi /2} \sin^5 x \, dx = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module