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Let $f(x)$ be a polynomial of degree four having extreme values at $x=1\; and\; x=2.$

if $\lim_{x\rightarrow 0}\left [ 1+\frac{f(x)}{x^{2}} \right ]\; =\; 3,$

then $f(2)$ is equal to :

Option 1)
-8

Option 2)
-4

Option 3)
0

Option 4)
4

Answers (1)

As we learnt in

Evaluation of limits : (algebraic limits) : (Method of direct substitution) -

$\lim_{x\rightarrow a}f(x)\:defines\:by\:direct\:x=a$

$ex: \:\lim_{x\rightarrow 1}\:\frac{x^{2}+x+1}{x^{2}+x-1}=3$

- wherein

Means at x = a f(x) defined.

Let $f(x)=ax^{4}+bx^{3}+cx^{2}+dx+e$

$\frac{f(x)}{x^{2}}=ax^{2}+bx+c+\left ( \frac{dx+e}{x^{2}} \right )$

for finite value d=0, e=0

$\lim_{n\rightarrow 0}\left [ 1+ax^{2} +bx+c\right ]=3$

$\therefore 1+c=3$

$\Rightarrow$ c=2

Now

$f{}'(x)=4ax^{3}+3bx^{2}+2cx+d$

$f'(1)=4a+3b+2c=0 \left [ \because d=0 \right ]$

$f{}'(2)=32a+12b+2c=0$

So that $a=\frac{1}{2}$

b=-2

c=2

So $f(x)=\frac{x^{4}}{2}-2x^{3}+2x^{2}$

$f(2)=\frac{16}{2}-2\times 8+2\times 4$

=8-16+8=0

Option 1)

-8

This option is incorrect

Option 2)

-4

This option is incorrect

Option 3)

0

This option is correct

Option 4)

4

This option is incorrect

Posted by

Sabhrant Ambastha

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