Let m and M be respectively the minimum and maximum value values of | cases ^ 2 x & 1 + ^ 2 x & 2x 1 + ^ 2 x & ^ 2 x & 2x ^ 2 x & ^ 2 x & 1 + 2x cases | Then the ordered pair (m, M) is equal to:

Correct answer: B. (- 3, - 1) D = | cases ^ 2 x & 1 + ^ 2 x & 2x 1 + ^ 2 x & ^ 2 x & 2x ^ 2 x & ^ 2 x & 1 + 2x cases | C _ 1 → C _ 1 + C _ 2 array ll D = 2 1 + ^ 2 x 2x 2 & ^ 2 x & 2x 1 & ^ 2 x & 1 + 2x array array ll R _ 1 → R _ 1 - 2 R _ 3 and R _ 2 → R _ 2 - 2 R _ 3 array D = | cases 0 & ^ 2 x & - ( 2 + 2x ) 0 & - ^ 2 x & - ( 2 + 2x ) 1 & ^ 2 x & 1 + 2x cases | ⇒ D = - 2 - 2x As range of 2x ∈ [- 1, 1] , thus m = - 3, M = - 1

JEE Main 20206 SeptemberMorning Shift

Exam: JEE Main
Year: 2020
Date: 6 September
Shift: Morning Shift
Subject: Mathematics
Chapter: Determinants

JEE Main 2020 · 6 September · Morning Shift · Mathematics · Determinants

Let m and M be respectively the minimum and maximum value values of...

Let m and M be respectively the minimum and maximum value values of

\left|

\begin{cases}\cos^{2}x & 1 + \sin^{2}x & \sin 2x \\ 1 + \cos^{2}x & \sin^{2}x & \sin 2x \\ \cos^{2}x & \sin^{2}x & 1 + \sin 2x\end{cases}

\right|

Then the ordered pair (m, M) is equal to:

Options

A $\left(3, 3\right)$
B $\left(- 3, - 1\right)$
C $\left(4, 1\right)$
D $\left(1, 3\right)$

Correct answer

B. $\left(- 3, - 1\right)$

Step-by-step solution

$D = \left|$ \begin{cases}\cos^{2}x & 1 + \sin^{2}x & \sin 2x \\ 1 + \cos^{2}x & \sin^{2}x & \sin 2x \\ \cos^{2}x & \sin^{2}x & 1 + \sin 2x\end{cases} $\right|$

${C}_{1} \to {C}_{1} + {C}_{2}$

$\begin{array}{ll}D = 2 1 + \sin^{2}x \sin 2x \\ 2 & \sin^{2}x & \sin 2x \\ 1 & \sin^{2}x & 1 + \sin 2x\end{array}$

$\begin{array}{ll}{R}_{1} \to {R}_{1} - 2{R}_{3}\text{and}{R}_{2} \to {R}_{2} - 2{R}_{3}\end{array}$

$D = \left|$ \begin{cases}0 & \cos^{2}x & - ( 2 + \sin 2x ) \\ 0 & - \sin^{2}x & - ( 2 + \sin 2x ) \\ 1 & \sin^{2}x & 1 + \sin 2x\end{cases} $\right| ⇒ D = - 2 - \sin 2x$

As range of $\sin 2x \in \left[- 1, 1\right]$ , thus

$m = - 3, M = - 1$

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