JEE Main 2021 — Mathematics Calculus

Let $f(x)={x}^{6}+a{x}^{5}+b{x}^{4}+c{x}^{3}+d{x}^{2}+ex+f$f(x)=x6+ax5+bx4+cx3+dx2+ex+fas $\underset{x\rightarrow 0}{\mathrm{lim}}\frac{f(x)}{{x}^{3}}=1$x→0lim​x3f(x)​=1 non-zero finite

So, $d=e=f=0$d=e=f=0

and $f(x)={x}^{3}({x}^{3}+a{x}^{2}+bx+c)$f(x)=x3(x3+ax2+bx+c)

Hence, $\underset{x\rightarrow 0}{\mathrm{lim}}\frac{f(x)}{{x}^{3}}=c=1$x→0lim​x3f(x)​=c=1

Now, as $f(x)={x}^{6}+a{x}^{5}+b{x}^{4}+{x}^{3}$f(x)=x6+ax5+bx4+x3

and ${f}^{'}(x)=0$f′(x)=0 at $x=1$x=1 and $x=-1$x=−1

i.e., ${f}^{'}(x)=6{x}^{5}+5a{x}^{4}+4b{x}^{3}+3{x}^{2}$f′(x)=6x5+5ax4+4bx3+3x2

$$f'(1)=0$$f′(1)=0

$$\Rightarrow 6+5a+4b+3=0$$⇒6+5a+4b+3=0

$$\Rightarrow 5a+4b=-9$$⇒5a+4b=−9

and ${f}^{'}(-1)=0$f′(−1)=0

$$\Rightarrow -6+5a-4b+3=0$$⇒−6+5a−4b+3=0

$$\Rightarrow 5a-4b=3$$⇒5a−4b=3

Solving both we get,

$$a=\frac{-6}{10}=\frac{-3}{5};b=\frac{-3}{2}$$a=10−6​=5−3​;b=2−3​

$$\therefore f(x)={x}^{6}-\frac{3}{5}{x}^{5}-\frac{3}{2}{x}^{4}+{x}^{3}$$∴f(x)=x6−53​x5−23​x4+x3

$$\therefore 5f(2)=5[64-\frac{3}{5}\cdot 32-\frac{3}{2}\cdot 16+8]$$∴5f(2)=5[64−53​⋅32−23​⋅16+8]

$$=320-96-120+40$$=320−96−120+40

$$=144$$=144