Let f(x) and g(x) be twice differentiable fun | vedclass

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

Question

(B) Given $f''(x) = g''(x)$. Integrating both sides with respect to $x$, we get $f'(x) = g'(x) + C_1$.From the given conditions, $f'(1) = 4$ and $2g'(

Vedclass · Accepted Answer

$40$

Vedclass · Answer

(B) Given $f''(x) = g''(x)$. Integrating both sides with respect to $x$, we get $f'(x) = g'(x) + C_1$.From the given conditions, $f'(1) = 4$ and $2g'(1) = 4 \implies g'(1) = 2$.Substituting these into the derivative equation: $4 = 2 + C_1 \implies C_1 = 2$.Thus, $f'(x) - g'(x) = 2$. Integrating again, we get $f(x) - g(x) = 2x + C_2$.Given $g(2) = 9$ and $3f(2) = 9 \implies f(2) = 3$.Substituting $x = 2$ into the equation $f(x) - g(x) = 2x + C_2$: $3 - 9 = 2(2) + C_2 \implies -6 = 4 + C_2 \implies C_2 = -10$.Therefore, $f(x) - g(x) = 2x - 10$.For $x = 25$, $f(25) - g(25) = 2(25) - 10 = 50 - 10 = 40$.

Let $f(x)$ and $g(x)$ be twice differentiable functions satisfying $f''(x) = g''(x)$ for all $x \in R$, $f'(1) = 2g'(1) = 4$ and $g(2) = 3f(2) = 9$. Then $f(25) - g(25)$ is equal to:

Similar Questions

$\frac{d}{d x}\left[\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)^2\right]=$ . . . . . .

If $y = \cos (\sin {x^2}),$ then at $x = \sqrt {\frac{\pi }{2}} , \frac{dy}{dx} = $

If $\frac{d}{d x}\left(\frac{1+x^2+x^4}{1+x+x^2}\right)=a x+b$,then $(a, b)=$

Find the derivative of $\tan (2 x+3)$

Let $f$ and $g$ be differentiable functions satisfying $g'(a) = 2$,$g(a) = b$,and $f \circ g = I$ (identity function). Then $f'(b)$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module