QJEEM15Ma9fda: JEE Main 2015 Mathematics - Class 12 JEEM15Ma9fda Solution: Let be a polynomial of degree four having extreme values at and …

Question

Verified Solution • 2025–26 Session moderate

Let

f(x)

be a polynomial of degree four having extreme values
at

x=1

and

x=2

. If

\mathop {\lim }\limits_{x \to 0} \left[ {1 + {{f\left( x \right)} \over {{x^2}}}} \right] = 3

, then f

(2)

is equal to :

Accepted Answer

\mathop {\lim }\limits_{x \to 0} \left[ {1 + {{f\left( x \right)} \over {{x^2}}}} \right] = 3 \Rightarrow \mathop {Lim}\limits_{x \to 0} {{f\left( x \right)} \over {{x^2}}} = 2

So,

f(x)

contains terms in

x{}^2,{x^3}

and

{x^4}

Let

f\left( x \right) = {a_1}{x^2} + {a_2}{x^3} + {a_3}{x^4}

Since

\mathop {\lim }\limits_{x \to 0} {{f\left( x \right)} \over {{x^2}}} = 2 \Rightarrow {a_1} = 2

Hence,

f\left( x \right) = 2{x^2} + {a_2}{x^3} + {a_3}{x^4}

f'\left( x \right) = 4x + 3{a_2}{x^2} + 4{a_3}{x^3}

As given:

f'\left( 1 \right) = 0

and

f'\left( 2 \right) = 0

Hence,

4 + 3{a_2} + 4{a_3} = 0\,\,\,\,\,\,\,\,\,\,...\left( 1 \right)

and

8 + 12{a_2} + 32{a_3} = 0\,\,\,\,\,...\left( 1 \right)

By

4x\left( {eq1} \right) - eq\left( 2 \right),

we get

16 + 12{a_2} + 16{a_3} - \left( {8 + 12{a_2} + 32{a_3}} \right) = 0

\Rightarrow 8 - 16{a_3} = 0 \Rightarrow {a_3} = 1/2

and by eqn.

\left( 1 \right),4 + 3{a_2} + 4/2 = 0 \Rightarrow {a_2} = - 2

\Rightarrow f\left( x \right) = 2{x^2} - 2{x^3} + {1 \over 2}{x^4}

f\left( 2 \right) = 2 \times 4 - 2 \times 8 + {1 \over 2} \times 16 = 0

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