Question

Let p(x) be a polynomial of degree 4 having extremum at x = 1 ,2 and
$\displaystyle\sf \lim_{x\to 0}\left( 1+\dfrac{p(x)}{x^2}\right) = 2$

Then the value of p(2) is ? ​

Answer 1

SOLUTION:-

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Answer 2

We are given that ;

${\quad \longrightarrow \displaystyle \sf \lim_{x\to 0}\bigg\{1+\dfrac{p(x)}{x^2}\bigg\}=2}$

Where p(x) is a polynomial of degree 4 , it will help us later, but let's do some manipulations first ;

Can be further written as ;

${:\implies \quad \displaystyle \sf 1+\lim_{x\to 0}\bigg\{\dfrac{p(x)}{x^2}\bigg\}=2}$

${:\implies \quad \displaystyle \sf \lim_{x\to 0}\bigg\{\dfrac{p(x)}{x^2}\bigg\}=1}$

So, here p(x) is polynomial of degree 4, so it will be a biquadratic polynomial, so we will write p(x) in the form of general biquadratic polynomial, so p(x) = ax⁴ + bx³ + cx² + dx + e

Now, first find p(x)/x² ;

${:\implies \quad \sf \dfrac{p(x)}{x^2}=\dfrac{ax^{4}+bx^{3}+cx^{2}+dx+e}{x^2}}$

${:\implies \quad \sf \dfrac{p(x)}{x^2}=ax^{2}+bx+c+\dfrac{d}{x}+\dfrac{e}{x^2}}$

So now, as ${\bf{x\to 0}}$ , for any values of a, b, c, d and e, the RHS will approach ∞ iff d ≠ e ≠ 0, as the denominator of d and e will be approaching 0 and so the whole limit will be ∞ , but we want the limit to be approaching 1, so when if d = e = 0, the denominator of d and e will be approaching 0 (not absolutely 0), and if d = e = 0, we will have the limit be approaching ax²+ bx + c for x approaching 0 being the limit 1 , and for any values of a, b and c . So now we have ;

${:\implies \quad \displaystyle \bf \lim_{x\to 0}\dfrac{p(x)}{x^{2}}=\begin{cases}\bf \infty \:, iff\: d\neq e\neq 0 \\ \\ \bf 1\:, iff\: d=e=0\end{cases}}$

So, now we had to consider the second case, for which the limit is approaching 1, for d = e = 0, so the limand here will just be ax² + bx + c

Now, we so have ;

${:\implies \quad \displaystyle \sf \lim_{x\to 0}ax^{2}+bx+c=1}$

Putting the limit we will have ;

${:\implies \quad \sf c=1}$

So, now as p(x) have extremum at 1 and 2, so p'(x) = 0, for x = 1, 2 , so now finding p'(x)

${:\implies \quad \sf p(x)=ax^{4}+bx^{3}+x^{2}\quad \qquad \{\because c=1\: and\: d=e=0\}}$

So, differentiating both sides wr.t.x ;

${:\implies \quad \sf p^{\prime}(x)=4ax^{3}+3bx^{2}+2x}$

Now, p'(1) and p'(2) must be 0

${:\implies \quad \sf p^{\prime}(1)=4a+3b+2=0}$

${:\implies \quad \sf p^{\prime}(2)=32a+12b+4=0}$

So, now we have ;

${\quad \longrightarrow \displaystyle \begin{cases}\bf 4a+3b=-2 \\ \\ \bf 32a+12b=-4\end{cases}}$

On multiplying first equation by 8 on both sides we can thus obtain ;

${\quad \longrightarrow \displaystyle \begin{cases}\bf 32a+24b=-16 \\ \\ \bf 32a+12b=-4\end{cases}}$

On solving both the equations we will be having ;

${\quad \longrightarrow \displaystyle \begin{cases}\bf a=\dfrac{1}{4} \\ \\ \bf b=-1\end{cases}}$

So , now as d = e = 0, c = 1, a = (1/4), b = -1, so putting all the values in p(x) we can obtain p(x) as ;

${:\implies \quad \sf p(x)=\dfrac{1}{4}(x)^{4}-(x)^{3}+x^{2}}$

Now, at x = 2 ;

${:\implies \quad \sf p(2)=\dfrac{1}{4}(2)^{4}-(2)^{3}+2^{2}}$

${:\implies \quad \sf p(2)=4-8+4=8-8}$

${:\implies \quad \bf \therefore \quad \underline{\underline{p(2)=0}}}$

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