Limit Continuity And Differentiability Ques 61
- Let $p(x)$ be a polynomial of degree $4$ having extremum at $x=1,2$ and $\lim _{x \rightarrow 0}\left[1+\frac{p(x)}{x^2}\right]=2$. Then, the value of $p(2)$ is $\qquad$
(2010)
Show Answer
Answer:
Correct Answer: 61.$p(2)=0$
Solution: Let $p(x)=a x^4+b x^3+c x^2+d x+e$
$\Rightarrow \quad p^{\prime}(x)=4 a x^3+3 b x^2+2 c x+d$
$\therefore \quad p^{\prime}(1)=4 a+3 b+2 c+d=0$ $\quad$ ……..(i)
and $p^{\prime}(2)=32 a+12 b+4 c+d=0$ $\quad$ ……..(ii)
Since, $\quad \lim _{x \rightarrow 0}\left(1+\frac{p(x)}{x^2}\right)=2 \quad$ [given]
$\therefore \quad \lim _{x \rightarrow 0} \frac{a x^4+b x^3+(c+1) x^2+d x+e}{x^2}=2$
$\Rightarrow \quad c+1=2, d=0, e=0$
$\Rightarrow \quad c=1$
From Eqs. (i) and (ii), we get
$4 a+3 b=-2$
$32a +12b=- 4$
$\Rightarrow \quad a=\frac{1}{4}$ and $b=-1$.
$\therefore \quad p(x)=\frac{x^4}{4}-x^3+x^2$
$\Rightarrow \quad p(2)=\frac{16}{4}-8+4$
$\Rightarrow \quad p(2)=0$
BETA
