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1

### JEE Main 2021 (Online) 27th August Evening Shift

Numerical
Let S be the sum of all solutions (in radians) of the equation $${\sin ^4}\theta + {\cos ^4}\theta - \sin \theta \cos \theta = 0$$ in [0, 4$$\pi$$]. Then $${{8S} \over \pi }$$ is equal to ____________.

## Explanation

Given equation

$${\sin ^4}\theta + {\cos ^4}\theta - \sin \theta \cos \theta = 0$$

$$\Rightarrow 1 - {\sin ^2}\theta {\cos ^2}\theta - \sin \theta \cos \theta = 0$$

$$\Rightarrow 2 - {(\sin 2\theta )^2} - \sin 2\theta = 0$$

$$\Rightarrow {(\sin 2\theta )^2} + (\sin 2\theta ) - 2 = 0$$

$$\Rightarrow (\sin 2\theta + 2)(\sin 2\theta - 1) = 0$$

$$\Rightarrow \sin 2\theta = 1$$ or $$\sin 2\theta = - 2$$ (Not Possible)

$$\Rightarrow 2\theta = {\pi \over 2},{{5\pi } \over 2},{{9\pi } \over 2},{{13\pi } \over 2}$$

$$\Rightarrow \theta = {\pi \over 4},{{5\pi } \over 4},{{9\pi } \over 4},{{13\pi } \over 4}$$

$$\Rightarrow S = {\pi \over 4} + {{5\pi } \over 4} + {{9\pi } \over 4} + {{13\pi } \over 4} = 7\pi$$

$$\Rightarrow {{8S} \over \pi } = {{8 \times 7\pi } \over \pi } = 56.00$$
2

### JEE Main 2021 (Online) 18th March Morning Shift

Numerical
The number of solutions of the equation

$$|\cot x| = \cot x + {1 \over {\sin x}}$$ in the interval [ 0, 2$$\pi$$ ] is

## Explanation

Case I : When cot x > 0, $$x \in \left[ {0,{\pi \over 2}} \right] \cup \left[ {\pi ,{{3\pi } \over 2}} \right]$$

$$\cot x = \cot x + {1 \over {\sin x}} \Rightarrow$$ not possible

Case II : When cot x < 0, $$x \in \left[ {{\pi \over 2},\pi } \right] \cup \left[ {{{3\pi } \over 2},2\pi } \right]$$

$$- \cot x = \cot x + {1 \over {\sin x}}$$

$$\Rightarrow {{ - 2\cos x} \over {\sin x}} = {1 \over {\sin x}}$$

$$\Rightarrow \cos x = {{ - 1} \over 2}$$

$$\Rightarrow x = {{2\pi } \over 3},{{4\pi } \over 3}$$(Rejected)

One solution.
3

### JEE Main 2021 (Online) 26th February Morning Shift

Numerical
The number of integral values of 'k' for which the equation $$3\sin x + 4\cos x = k + 1$$ has a solution, k$$\in$$R is ___________.

## Explanation

We know,

$$- \sqrt {{a^2} + {b^2}} \le a\cos x + b\sin x \le \sqrt {{a^2} + {b^2}}$$

$$\therefore$$ $$- \sqrt {{3^2} + {4^2}} \le 3\cos x + 4\sin x \le \sqrt {{3^2} + {4^2}}$$

$$- 5 \le k + 1 \le 5$$

$$- 6 \le k \le 4$$

$$\therefore$$ Set of integers = $$- 6, - 5, - 4, - 3, - 2, - 1,0,1,2,3,4$$ = Total 11 intergers.
4

### JEE Main 2021 (Online) 26th February Morning Shift

Numerical
If $$\sqrt 3 ({\cos ^2}x) = (\sqrt 3 - 1)\cos x + 1$$, the number of solutions of the given equation when $$x \in \left[ {0,{\pi \over 2}} \right]$$ is __________.

## Explanation

$$\sqrt 3 ({\cos ^2}x) = (\sqrt 3 - 1)\cos x + 1$$

$$\Rightarrow$$ $$\sqrt 3 {\cos ^2}x - \sqrt 3 \cos x + \cos x - 1 = 0$$

$$\Rightarrow \sqrt 3 \cos x(\cos x - 1) + (\cos x - 1) = 0$$

$$\Rightarrow (\cos x - 1)(\sqrt 3 \cos x + 1) = 0$$

$$\cos x = 1$$

$$\Rightarrow x = 0$$ $$[as x \in \left[ {0,{\pi \over 2}} \right]$$]

and $$\cos x = - {1 \over {\sqrt 3 }}$$ (not possible in $$x \in \left[ {0,{\pi \over 2}} \right]$$]

$$\therefore$$ Number of solution = 1

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