Limit Continuity And Differentiability Ques 61

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 $b=-1$.

$\therefore \quad p(x)=\frac{x^4}{4}-x^3+x^2$

$\Rightarrow \quad p(2)=\frac{16}{4}-8+4$

$\Rightarrow \quad p(2)=0$