f(x) and g(x) are two differentiable function | vedclass

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

Question

(D) Given $f''(x) - g''(x) = 0$.Integrating with respect to $x$,we get $f'(x) - g'(x) = c$,where $c$ is a constant.At $x = 1$,$f'(1) - g'(1) = c \impl

Vedclass · Accepted Answer

$-5$

Vedclass · Answer

(D) Given $f''(x) - g''(x) = 0$.Integrating with respect to $x$,we get $f'(x) - g'(x) = c$,where $c$ is a constant.At $x = 1$,$f'(1) - g'(1) = c \implies 2 - 4 = c \implies c = -2$.Thus,$f'(x) - g'(x) = -2$.Integrating again with respect to $x$,we get $f(x) - g(x) = -2x + c_1$,where $c_1$ is a constant.At $x = 2$,$f(2) - g(2) = -2(2) + c_1 \implies 3 - 9 = -4 + c_1 \implies -6 = -4 + c_1 \implies c_1 = -2$.Therefore,$f(x) - g(x) = -2x - 2$.At $x = 3/2$,$f(3/2) - g(3/2) = -2(3/2) - 2 = -3 - 2 = -5$.

$f(x)$ and $g(x)$ are two differentiable functions on $[0, 2]$ such that $f''(x) - g''(x) = 0$,$f'(1) = 2$,$g'(1) = 4$,$f(2) = 3$,and $g(2) = 9$. Then $f(x) - g(x)$ at $x = 3/2$ is:

Similar Questions

If $y = x^{x^{x...\infty}},$ then $x (1 - y \log x) \frac{dy}{dx} =$

If $\log _{10}\left(\frac{x^2-y^2}{x^2+y^2}\right)=2$,then $\frac{dy}{dx} = \dots$

Let $x^{k}+y^{k}=a^{k}$ where $a, k > 0$. If $\frac{dy}{dx}+\left(\frac{y}{x}\right)^{\frac{1}{3}}=0$,then the value of $k$ is:

If $\cos ^{-1}\left(\frac{x^2-y^2}{x^2+y^2}\right)=k$ (a constant), then $\frac{d y}{d x}$ is equal to

If $x \sqrt{1+y}+y \sqrt{1+x}=0$ for $-1 < x < 1$,prove that $\frac{dy}{dx} = -\frac{1}{(1+x)^2}$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module