The total number of distinct x in [0, 1] for | vedclass

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

Question

(A) Let $f(x) = \int_0^x \frac{t^2}{1+t^4} dt - 2x + 1$ for $x \in [0, 1]$.Taking the derivative,$f'(x) = \frac{x^2}{1+x^4} - 2$.Since $x^4 + 1 \geq 2

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(A) Let $f(x) = \int_0^x \frac{t^2}{1+t^4} dt - 2x + 1$ for $x \in [0, 1]$.Taking the derivative,$f'(x) = \frac{x^2}{1+x^4} - 2$.Since $x^4 + 1 \geq 2x^2$ by $AM$-$GM$ inequality,we have $\frac{x^2}{1+x^4} \leq \frac{1}{2}$.Thus,$f'(x) = \frac{x^2}{1+x^4} - 2 \leq \frac{1}{2} - 2 = -\frac{3}{2} Since $f'(x) We evaluate the endpoints:$f(0) = \int_0^0 \frac{t^2}{1+t^4} dt - 2(0) + 1 = 1$.$f(1) = \int_0^1 \frac{t^2}{1+t^4} dt - 2(1) + 1 = \int_0^1 \frac{t^2}{1+t^4} dt - 1$.Since $\frac{t^2}{1+t^4} Since $f(0) > 0$ and $f(1) < 0$ and $f(x)$ is continuous and strictly decreasing,by the Intermediate Value Theorem,there exists exactly one root $x \in [0, 1]$ such that $f(x) = 0$.

The total number of distinct $x \in [0, 1]$ for which $\int_0^x \frac{t^2}{1+t^4} dt = 2x - 1$ is

Similar Questions

The value of $\int_{0}^{4}|x-1| dx$ is

Number of values of $x$ satisfying the equation $\int_{-1}^{x} (8t^2 + \frac{28}{3}t + 4) dt = \frac{(\frac{3}{2})x + 1}{\log_{(x+1)} \sqrt{x+1}}$.

All values of $a$ for which the inequality $\frac{1}{\sqrt{a}} \int_{1}^{a} (\frac{3}{2} \sqrt{x} + 1 - \frac{1}{\sqrt{x}}) dx < 4$ is satisfied, lie in the interval

$\int_0^{1.5} [x^2] dx$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module