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BITSAT 2018 Slot 2 Solved Paper

Section: Mathematics

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Question : 110 of 150

Marks: +1, -0

If

x^{2} = \begin{vmatrix} \sin \theta & \cos \theta & 0 \\ -\cos \theta & \sin \theta & 1 \\ \sin \theta & \cos \theta & 2 \end{vmatrix}

, then the value of

4 x^{2} + x \sin \frac{3\pi}{2} + 5

is

$13 - \sqrt{2}$
$13 + \sqrt{2}$
$\sqrt{2} - 13$
Both (a) and (b)

Solution:

Given,

x^{2} = \begin{vmatrix} \sin \theta & \cos \theta & 0 \\ -\cos \theta & \sin \theta & 1 \\ \sin \theta & \cos \theta & 2 \end{vmatrix}

On expanding

x^{2} = \begin{vmatrix} \sin \theta & \cos \theta & 0 \\ -\cos \theta & \sin \theta & 1 \\ \sin \theta & \cos \theta & 2 \end{vmatrix}

along

C_3

, we get

x^{2} = 0 - 1 \begin{vmatrix} \sin \theta & \cos \theta \\ \sin \theta & \cos \theta \end{vmatrix}

+ 2 \begin{vmatrix} \sin \theta & \cos \theta \\ -\cos \theta & \sin \theta \end{vmatrix}

= -1 (\sin \theta \cos \theta - \sin \theta \cos \theta) + 2 (\sin^2 \theta + \cos^2 \theta)

= -1 \times 0 + 2 \times 1

⇒

x^{2} = 2 \Rightarrow x = \pm \sqrt{2}

If

x = \sqrt{2}

, then

4 x^{2} + x \sin \frac{3\pi}{2} + 5 = 4 \times (\sqrt{2})^{2}

- \sqrt{2} + 5

= 8 - \sqrt{2} + 5 = (13 - \sqrt{2})

If

x = -\sqrt{2}

, then

4 x^{2} + x \sin \frac{3\pi}{2} + 5

= 4 \times (-\sqrt{2})^{2} + \sqrt{2} + 5

= 8 + \sqrt{2} + 5 = (13 + \sqrt{2})

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