JEE Main 2018 — Mathematics Calculus

$\because f(x)$∵f(x) has extremum values at $x=1$x=1 and $x=2$x=2 $\because f^{\prime}(1)=0$∵f′(1)=0 and $f^{\prime}(2)=0$f′(2)=0 As, $f(x)$f(x) is a polynomial of degree 4 . Suppose $f(x)=A x^4+B x^3+C x^2+D x+E$f(x)=Ax4+Bx3+Cx2+Dx+E $$\begin{aligned} &\because \lim _{x \rightarrow 0}\left(\frac{f(x)}{x^2}+1\right)=3 \\ &\Rightarrow \lim _{x \rightarrow 0}\left(\frac{A x^4+B x^3+C x^2+D x+E}{x^2}+1\right)=3 \\ &\Rightarrow \lim _{x \rightarrow 0}\left(A x^2+B x+C+\frac{D}{x}+\frac{E}{x^2}+1\right)=3 \end{aligned}$$​∵x→0lim​(x2f(x)​+1)=3⇒x→0lim​(x2Ax4+Bx3+Cx2+Dx+E​+1)=3⇒x→0lim​(Ax2+Bx+C+xD​+x2E​+1)=3​ As limit has finite value, so $D=0$D=0 and $E=0$E=0 Now $A(0)^2+B(0)+C+0+0+1=3$A(0)2+B(0)+C+0+0+1=3 $$\begin{aligned} &\Rightarrow c+1=3 \Rightarrow c=2 \\ &f^{\prime}(x)=4 A x^3+3 B x^2+2 C x+D \\ &f^{\prime}(1)=0 \Rightarrow 4 A(1)+3 B(1)+2 C(1)+D=0 \end{aligned}$$​⇒c+1=3⇒c=2f′(x)=4Ax3+3Bx2+2Cx+Df′(1)=0⇒4A(1)+3B(1)+2C(1)+D=0​ $$\begin{aligned} & \Rightarrow 4 A+3 B=-4 \\ f^{\prime}(2)=0 & \Rightarrow 4 A(8)+3 B(4)+2 C(2)+D=0 \\ & \Rightarrow 8 A+3 B=-2 \end{aligned}$$f′(2)=0​⇒4A+3B=−4⇒4A(8)+3B(4)+2C(2)+D=0⇒8A+3B=−2​ From equations (1) and (2), we get $A=\frac{1}{2}$A=21​ and $B=-2$B=−2 So, $f(x)=\frac{x^4}{2}-2 x^3+2 x^2$f(x)=2x4​−2x3+2x2 Therefore, $f(-1)=\frac{(-1)^4}{2}-2(-1)^3+2(-1)^2$f(−1)=2(−1)4​−2(−1)3+2(−1)2 $$=\frac{1}{2}+2+2=\frac{9}{2}$$=21​+2+2=29​ Hence $f(-1)=\frac{9}{2}$f(−1)=29​