Let f be a real-valued function defined on th | vedclass

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

Question

(B) Given $e^{-x} f(x) = 2 + \int_0^x \sqrt{t^4 + 1} \, dt \dots (i)$At $x = 0$,$e^0 f(0) = 2 + \int_0^0 \sqrt{t^4 + 1} \, dt \implies f(0) = 2$.Since

Vedclass · Accepted Answer

$1/3$

Vedclass · Answer

(B) Given $e^{-x} f(x) = 2 + \int_0^x \sqrt{t^4 + 1} \, dt \dots (i)$At $x = 0$,$e^0 f(0) = 2 + \int_0^0 \sqrt{t^4 + 1} \, dt \implies f(0) = 2$.Since $f(0) = 2$,it follows that $f^{-1}(2) = 0$.We know that $(f^{-1})'(y) = \frac{1}{f'(f^{-1}(y))}$.For $y = 2$,$(f^{-1})'(2) = \frac{1}{f'(f^{-1}(2))} = \frac{1}{f'(0)}$.Differentiating equation $(i)$ with respect to $x$ using the product rule and the Fundamental Theorem of Calculus:$-e^{-x} f(x) + e^{-x} f'(x) = \sqrt{x^4 + 1}$.At $x = 0$:$-e^0 f(0) + e^0 f'(0) = \sqrt{0^4 + 1}$$-1(2) + 1(f'(0)) = 1$$-2 + f'(0) = 1 \implies f'(0) = 3$.Therefore,$(f^{-1})'(2) = \frac{1}{f'(0)} = \frac{1}{3}$.

Let $f$ be a real-valued function defined on the interval $(-1, 1)$ such that $e^{-x} f(x) = 2 + \int_0^x \sqrt{t^4 + 1} \, dt$,for all $x \in (-1, 1)$ and let $f^{-1}$ be the inverse function of $f$. Then $(f^{-1})'(2)$ is equal to

Similar Questions

If $[\cdot]$ denotes the greatest integer function and if $f:(5,10) \rightarrow(7,12)$ is a function defined by $f(x)=x+2\left[\frac{x}{5}\right]$,then

If $f(x) = \frac{x}{1 + x}$,then ${f^{-1}}(x)$ is equal to

If $f(x) = \frac{a^x - a^{-x}}{a^x + a^{-x}}$,where $a$ and $x$ satisfy the necessary conditions,then $f^{-1}(x) =$

Let $S = \{a, b, c\}$ and $T = \{1, 2, 3\}$. Find $F^{-1}$ of the following function $F$ from $S$ to $T$,if it exists: $F = \{(a, 2), (b, 1), (c, 1)\}$.

If a function $f:(-1,1) \rightarrow B(\subseteq R)$ is defined as $f(x)=x+x^2+x^3+\ldots \infty$, then in order to have the inverse function of $f$, $B$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module