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

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