JEE Main 2015 — Mathematics Calculus

Let $f(x)=a{x}^{4}+b{x}^{3}+c{x}^{2}+dx+e$f(x)=ax4+bx3+cx2+dx+e

Then $\underset{x\rightarrow 0}{\mathrm{lim}}[1+\frac{f(x)}{{x}^{2}}]=3$x→0lim​[1+x2f(x)​]=3

$$\Rightarrow \underset{x\rightarrow 0}{\mathrm{lim}}[1+a{x}^{2}+bx+c+\frac{d}{x}+\frac{e}{{x}^{2}}]=3$$⇒x→0lim​[1+ax2+bx+c+xd​+x2e​]=3

This limit exists when $d=e=0$d=e=0.

So, $\underset{x\rightarrow 0}{\mathrm{lim}}[1+a{x}^{2}+bx+c]=3$x→0lim​[1+ax2+bx+c]=3

$$\Rightarrow c+1=3$$⇒c+1=3

$$\Rightarrow c=2$$⇒c=2

It is given, $x=1$x=1 and $x=2$x=2 are the solution of ${f}^{'}(x)=0$f′(x)=0.

$$\Rightarrow {f}^{'}(x)=4a{x}^{3}+3b{x}^{2}+2cx$$⇒f′(x)=4ax3+3bx2+2cx

$$\Rightarrow x(4a{x}^{2}+3bx+2c)=0$$⇒x(4ax2+3bx+2c)=0

Given $1,2$1,2 are the roots of the quadratic equation.

$$\Rightarrow \text{Sum of roots}=\frac{-3b}{4a}=1+2=3$$⇒Sum of roots=4a−3b​=1+2=3

$$\Rightarrow b=-4a$$⇒b=−4a

Product of roots $=\frac{2c}{4a}=1\cdot 2=2$=4a2c​=1⋅2=2

$$\Rightarrow a=\frac{c}{4}$$⇒a=4c​

$$\Rightarrow a=\frac{1}{2}\text{, }b=-2$$⇒a=21​, b=−2

$$\therefore f(x)=\frac{1}{2}{x}^{4}-2{x}^{3}+2{x}^{2}$$∴f(x)=21​x4−2x3+2x2

$$\Rightarrow f(2)=8-16+8$$⇒f(2)=8−16+8

$$\Rightarrow f(2)=0$$⇒f(2)=0