Let f: R rightarrow R be a function such that | vedclass

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

Question

(C) Given $f(x)=x^3+x^2 f^{\prime}(1)+x f^{\prime \prime}(2)+6$. Differentiating with respect to $x$,we get $f^{\prime}(x)=3 x^2+2 x f^{\prime}(1)+f^{

Vedclass · Accepted Answer

$-2$

Vedclass · Answer

(C) Given $f(x)=x^3+x^2 f^{\prime}(1)+x f^{\prime \prime}(2)+6$. Differentiating with respect to $x$,we get $f^{\prime}(x)=3 x^2+2 x f^{\prime}(1)+f^{\prime \prime}(2)$. Differentiating again,we get $f^{\prime \prime}(x)=6 x+2 f^{\prime}(1)$. Substituting $x=1$ in the expression for $f^{\prime}(x)$: $f^{\prime}(1)=3(1)^2+2(1) f^{\prime}(1)+f^{\prime \prime}(2) \Rightarrow f^{\prime}(1)+f^{\prime \prime}(2)=-3$ (Equation $I$). Substituting $x=2$ in the expression for $f^{\prime \prime}(x)$: $f^{\prime \prime}(2)=6(2)+2 f^{\prime}(1) \Rightarrow f^{\prime \prime}(2)=12+2 f^{\prime}(1)$ (Equation $II$). Substituting $f^{\prime \prime}(2)$ from Equation $II$ into Equation $I$: $f^{\prime}(1)+12+2 f^{\prime}(1)=-3 \Rightarrow 3 f^{\prime}(1)=-15 \Rightarrow f^{\prime}(1)=-5$. Now,find $f^{\prime \prime}(2)$ using Equation $II$: $f^{\prime \prime}(2)=12+2(-5)=12-10=2$. Finally,substitute $f^{\prime}(1)$ and $f^{\prime \prime}(2)$ into the original function $f(x)$ at $x=2$: $f(2)=2^3+2^2(-5)+2(2)+6 = 8-20+4+6 = -2$.

Let $f: R \rightarrow R$ be a function such that $f(x)=x^3+x^2 f^{\prime}(1)+x f^{\prime \prime}(2)+6, x \in R$,then $f(2)$ equals

Similar Questions

If $y = (1 + x^2) \tan^{-1} x - x$,then $\frac{dy}{dx}$ is

If $[x]$ represents the greatest integer function and $f(x) = x - [x] - \cos x$,then $f^{\prime}\left(\frac{\pi}{2}\right) = $

$\frac{d}{dx} \left( x^3 \tan^2 \frac{x}{2} \right) =$

Find the derivative: $\frac{d}{dx} \cosh^{-1}(\sec x) = $

If $\frac{d}{d x}\left[(x+1)\left(x^2+1\right)\left(x^4+1\right)\left(x^8+1\right)\right] = \left(15 x^p-16 x^q+1\right)(x-1)^{-2}$, then $(p, q)$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module