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+(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+(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\{(p(x))/(x^2)\bigg\}=2}`

`{:\implies \quad \displaystyle \sf \lim_(x\to 0)\bigg\{(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 (p(x))/(x^2)=(ax^(4)+bx^(3)+cx^(2)+dx+e)/(x^2)}`

`{:\implies \quad \sf (p(x))/(x^2)=ax^(2)+bx+c+(d)/(x)+(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)(p(x))/(x^(2))=\begin{cases}\bf \infty \:, iff\: d\\eq e\\eq 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=(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)=(1)/(4)(x)^(4)-(x)^(3)+x^(2)}`

Now, at x = 2 ;

`{:\implies \quad \sf p(2)=(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}}}`

This is the required answer