Let $\mathrm{A}=\left[\mathrm{a}_{\mathrm{ij}}\right]$ be a $3 \times 3$ matrix, where $\mathrm{a}_{\mathrm{ij}}=\left\{\begin{array}{cc}1, & \text { if } \mathrm{i}=\mathrm{j} \\ -\mathrm{x}, & \text { if }|\mathrm{i}-\mathrm{j}|=1 \\ 2 \mathrm{x}+1, & \text { otherwise }\end{array}\right.$ Let a function $f: R \rightarrow R$ be defined as $f(x)=\operatorname{det}(A)$. Then the sum of maximum and minimum values of $f$ on $R$ is equal to :

JEE Main · 2021 · 20 Jul 2021 · Shift 1 · Determinants

Let

A = [a_{ij}]

be a

3 \times 3

matrix, where

a_{ij} = ⎩ ⎨ ⎧ 1, - x, 2 x + 1, if i = j if ∣ i - j ∣ = 1 otherwise

Let a function

f : R \to R

be defined as

f (x) = det (A)

. Then the sum of maximum and minimum values of

f

R

is equal to :

Answer: (A)

The three rules fill the matrix as

A = 1 - x 2 x + 1 - x 1 - x 2 x + 1 - x 1

and expanding along the first row gives

f (x) = det A = (1 - x^{2}) + x (2 x^{2}) + (2 x + 1) (x^{2} - 2 x - 1) = 4 x^{3} - 4 x^{2} - 4 x

Then

f^{'} (x) = 12 x^{2} - 8 x - 4 = 4 (3 x + 1) (x - 1)

, so the stationary points are

x = - \frac{1}{3}

and

x = 1

. Since

f^{''} (x) = 24 x - 8

is negative at

x = - \frac{1}{3}

and positive at

x = 1

, the local maximum is

f (- \frac{1}{3}) = \frac{20}{27}

and the local minimum is

f (1) = - 4

. Their sum is

\frac{20}{27} - 4 = - \frac{88}{27}

JEE Main · 2021 · 20 Jul 2021 · Shift 1 · Single correct option