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1

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

Numerical
Let S = {1, 2, 3, 4, 5, 6, 9}. Then the number of elements in the set T = {A $$\subseteq$$ S : A $$\ne$$ $$\phi$$ and the sum of all the elements of A is not a multiple of 3} is _______________.

## Explanation

3n type $$\to$$ 3, 6, 9 = P

3n $$-$$ 1 type $$\to$$ 2, 5 = Q

3n $$-$$ 2 type $$\to$$ 1, 4 = R

number of subset of S containing one element which are not divisible by 3 = $${}^2$$C1 + $${}^2$$C1 = 4

number of subset of S containing two numbers whose some is not divisible by 3

= $${}^3$$C1 $$\times$$ $${}^2$$C1 + $${}^3$$C1 $$\times$$ $${}^2$$C1 + $${}^2$$C2 + $${}^2$$C2 = 14

number of subsets containing 3 elements whose sum is not divisible by 3

= $${}^3$$C2 $$\times$$ $${}^4$$C1 + ($${}^2$$C2 $$\times$$ $${}^2$$C1)2 + $${}^3$$C1($${}^2$$C2 + $${}^2$$C2) = 22

number of subsets containing 4 elements whose sum is not divisible by 3

= $${}^3$$C3 $$\times$$ $${}^4$$C1 + $${}^3$$C2($${}^2$$C2 + $${}^2$$C2) + ($${}^3$$C1$${}^2$$C1 $$\times$$ $${}^2$$C2)2

= 4 + 6 + 12 = 22

number of subsets of S containing 5 elements whose sum is not divisible by 3.

= $${}^3$$C3($${}^2$$C2 + $${}^2$$C2) + ($${}^3$$C2$${}^2$$C1 $$\times$$ $${}^2$$C2) $$\times$$ 2 = 2 + 12 = 14

number of subsets of S containing 6 elements whose sum is not divisible by 3 = 4

$$\Rightarrow$$ Total subsets of Set A whose sum of digits is not divisible by 3 = 4 + 14 + 22 + 22 + 14 + 4 = 80.
2

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

Numerical
Let z1 and z2 be two complex numbers such that $$\arg ({z_1} - {z_2}) = {\pi \over 4}$$ and z1, z2 satisfy the equation | z $$-$$ 3 | = Re(z). Then the imaginary part of z1 + z2 is equal to ___________.

## Explanation

Let z1 = x1 + iy ; z2 = x2 + iy2

z1 $$-$$ z2 = (x1 $$-$$ x2) + i(y1 $$-$$ y2)

$$\therefore$$ $$\arg ({z_1} - {z_2}) = {\pi \over 4}$$ $$\Rightarrow$$ $${\tan ^{ - 1}}\left( {{{{y_1} - {y_2}} \over {{x_1} - {x_2}}}} \right) = {\pi \over 4}$$

$${y_1} - {y_2} = {x_1} - {x_2}$$ ....... (1)

$$|{z_1} - 3|\, = {\mathop{\rm Re}\nolimits} ({z_1}) \Rightarrow {({x_1} - 3)^2} + {y_1}^2 = {x_1}^2$$ .... (2)

$$|{z_2} - 3|\, = {\mathop{\rm Re}\nolimits} ({z_2}) \Rightarrow {({x_2} - 3)^2} + {y_2}^2 = {x_2}^2$$ .... (3)

sub (2) & (3)

$${({x_1} - 3)^2} - {({x_2} - 3)^2} + {y_1}^2 - {y_2}^2 = {x_1}^2 - {x_2}^2$$

$$({x_1} - {x_2})({x_1} + {x_2} - 6) + ({y_1} - {y_2})({y_1} + {y_2})$$

$$= ({x_1} - {x_2})({x_1} + {x_2})$$

$${x_1} + {x_2} - 6 + {y_1} + {y_2} = {x_1} + {x_2} \Rightarrow {y_1} + {y_2} = 6$$
3

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

Numerical
The probability distribution of random variable X is given by :

X 1 2 3 4 5
P(X) K 2K 2K 3K K

Let p = P(1 < X < 4 | X < 3). If 5p = $$\lambda$$K, then $$\lambda$$ equal to ___________.

## Explanation

$$\sum {P(X) = 1 \Rightarrow k + 2k + 3} k + k = 1$$

$$\Rightarrow k = {1 \over 9}$$

Now, $$p = P\left( {{{kx < 4} \over {X < 3}}} \right) = {{P(X = 2)} \over {P(X < 3)}} = {{{{2k} \over {9k}}} \over {{k \over {9k}} + {{2k} \over {9k}}}} = {2 \over 3}$$

$$\Rightarrow p = {2 \over 3}$$

Now, $$5p = \lambda k$$

$$\Rightarrow (5)\left( {{2 \over 3}} \right) = \lambda (1/9)$$

$$\Rightarrow \lambda = 30$$
4

### 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)

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

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

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

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

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