Search any question & find its solution
Question:
Answered & Verified by Expert
Let $f: \mathbf{R} \rightarrow \mathbf{R}$ be a function such that
$\mathrm{f}(\mathrm{x})=\mathrm{x}^{3}+\mathrm{x}^{2} \mathrm{f}^{\prime}(1)+\mathrm{xf}^{\prime}(2)+\mathrm{f}^{\prime \prime}(3)$
for $\mathrm{x} \in \mathbf{R}$
What is $\mathrm{f}(1)$ equal to?
Options:
$\mathrm{f}(\mathrm{x})=\mathrm{x}^{3}+\mathrm{x}^{2} \mathrm{f}^{\prime}(1)+\mathrm{xf}^{\prime}(2)+\mathrm{f}^{\prime \prime}(3)$
for $\mathrm{x} \in \mathbf{R}$
What is $\mathrm{f}(1)$ equal to?
Solution:
2727 Upvotes
Verified Answer
The correct answer is:
4
$f(x)=x^{3}+x^{2} f^{\prime}(1)+x f^{\prime \prime}(2)+f^{\prime \prime \prime}(3)$ ...(1)
$f^{\prime}(x)=3 x^{2}+2 x f^{\prime}(1)+f^{\prime \prime}(2)$ ...(2)
$f^{\prime \prime}(x)=6 x+2 f^{\prime}(1)$ ...(3)
$f^{\prime \prime \prime}(x)=6$ ...(4)
$f^{\prime}(1)=3+2 f^{\prime}(1)+f^{\prime \prime}(2)$ ...(5)
$f^{\prime}(2)=12+2 f^{\prime}(1)$ ...(6)
Using $(6)$ in $(5)$, we get
$f^{\prime}(1)=3+2 f^{\prime}(1)+12+2 f^{\prime \prime}(1)$
$-3 f^{\prime}(1)=15$
$f^{\prime}(1)=-5$
Using this value in eqn (6) we get
$f^{\prime}(2)=12+2 \times(-5)$
$f^{\prime \prime}(2)=2$
Using $x=3$ in eqn (4),
$f^{\prime \prime \prime}(3)=6$
Putting value of $f^{\prime}(1)+f^{\prime \prime}(2)$ and $\mathrm{f}^{\prime \prime \prime}(3)$ in eqn (1)
We get
$\begin{aligned} f(x) &=x^{3}+x^{2}(-5)+x(2)+6 \\ &=x^{3}-5 x^{2}+2 x+6 \end{aligned}$
Putting $\mathrm{x}=1$ $f(1)=(1)^{3}-5(1)^{2}+2(1)+6$
$f(1)=4$
$f^{\prime}(x)=3 x^{2}+2 x f^{\prime}(1)+f^{\prime \prime}(2)$ ...(2)
$f^{\prime \prime}(x)=6 x+2 f^{\prime}(1)$ ...(3)
$f^{\prime \prime \prime}(x)=6$ ...(4)
$f^{\prime}(1)=3+2 f^{\prime}(1)+f^{\prime \prime}(2)$ ...(5)
$f^{\prime}(2)=12+2 f^{\prime}(1)$ ...(6)
Using $(6)$ in $(5)$, we get
$f^{\prime}(1)=3+2 f^{\prime}(1)+12+2 f^{\prime \prime}(1)$
$-3 f^{\prime}(1)=15$
$f^{\prime}(1)=-5$
Using this value in eqn (6) we get
$f^{\prime}(2)=12+2 \times(-5)$
$f^{\prime \prime}(2)=2$
Using $x=3$ in eqn (4),
$f^{\prime \prime \prime}(3)=6$
Putting value of $f^{\prime}(1)+f^{\prime \prime}(2)$ and $\mathrm{f}^{\prime \prime \prime}(3)$ in eqn (1)
We get
$\begin{aligned} f(x) &=x^{3}+x^{2}(-5)+x(2)+6 \\ &=x^{3}-5 x^{2}+2 x+6 \end{aligned}$
Putting $\mathrm{x}=1$ $f(1)=(1)^{3}-5(1)^{2}+2(1)+6$
$f(1)=4$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.