JEE Main 2021 — Mathematics Calculus

We have,

{a}_{ij}={\begin{matrix}1 & , & \mathrm{if}i=j \\ -x & , & \mathrm{if}|i-j|=1 \\ 2x+1 & , & \mathrm{otherwise}\end{matrix}

Then,

$$A=[\begin{matrix}1 & -x & 2x+1 \\ -x & 1 & -x \\ 2x+1 & -x & 1\end{matrix}]$$A=[1−x2x+1​−x1−x​2x+1−x1​]

$$\Rightarrow |A|=(1-{x}^{2})+x(-x+2{x}^{2}+x)+(2x+1)({x}^{2}-2x-1)$$⇒∣A∣=(1−x2)+x(−x+2x2+x)+(2x+1)(x2−2x−1)

$$\Rightarrow |A|=1-{x}^{2}+2{x}^{3}+(2{x}^{3}-4{x}^{2}-2x)+({x}^{2}-2x-1)$$⇒∣A∣=1−x2+2x3+(2x3−4x2−2x)+(x2−2x−1)

$$\Rightarrow |A|=f(x)=4{x}^{3}-4{x}^{2}-4x$$⇒∣A∣=f(x)=4x3−4x2−4x

$$\Rightarrow {f}^{'}(x)=4(3{x}^{2}-2x-1)$$⇒f′(x)=4(3x2−2x−1)

For maxima or minima, ${f}^{'}(x)=0$f′(x)=0

$$\Rightarrow 3{x}^{2}-2x-1=0$$⇒3x2−2x−1=0

$$\Rightarrow x=1,\frac{-1}{3}$$⇒x=1,3−1​

Now,

$${f}^{''}(x)=4(6x-2)$$f′′(x)=4(6x−2)

${f}^{''}(1)=4(6\cdot 1-2)=16>0$f′′(1)=4(6⋅1−2)=16>0( point of minima)

${f}^{''}(-\frac{1}{3})=4(6\cdot (-\frac{1}{3})-2)=-16<0$f′′(−31​)=4(6⋅(−31​)−2)=−16<0 ( point of maxima)

Hence, required sum is

$$f(1)+f(-\frac{1}{3})$$f(1)+f(−31​)

$$=-4+\frac{20}{27}=-\frac{88}{27}$$=−4+2720​=−2788​