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1

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

The set of all values of K > $$-$$1, for which the equation $${(3{x^2} + 4x + 3)^2} - (k + 1)(3{x^2} + 4x + 3)(3{x^2} + 4x + 2) + k{(3{x^2} + 4x + 2)^2} = 0$$ has real roots, is :
A
$$\left( {1,{5 \over 2}} \right]$$
B
[2, 3)
C
$$\left[ { - {1 \over 2},1} \right)$$
D
$$\left( {{1 \over 2},{3 \over 2}} \right] - \{ 1\}$$

## Explanation

$${(3{x^2} + 4x + 3)^2} - (k + 1)(3{x^2} + 4x + 3)(3{x^2} + 4x + 2) + k{(3{x^2} + 4x + 2)^2} = 0$$

Let $$3{x^2} + 4x + 3 = a$$

and $$3{x^2} + 4x + 2 = b \Rightarrow b = a - 1$$

Given equation becomes

$$\Rightarrow {a^2} - (k + 1)ab + k{b^2} = 0$$

$$\Rightarrow a(a - kb) - b(a - kb) = 0$$

$$\Rightarrow (a - kb)(a - b) = 0 \Rightarrow a = kb$$ or a = b (reject) $$\because$$ a = kb

$$\Rightarrow 3{x^2} + 4x + 3 = k(3{x^2} + 4x + 2)$$

$$\Rightarrow 3(k - 1){x^2} + 4(k - 1)x + (2k - 3) = 0$$ for real roots

D $$\ge$$ 0

$$\Rightarrow 16{(k - 1)^2} - 4(3(k - 1))(2k - 3) \ge 0$$

$$\Rightarrow 4(k - 1)\{ 4(k - 1) - 3(2k - 3)\} \ge 0$$

$$\Rightarrow 4(k - 1)\{ - 2k + 5\} \ge 0$$

$$\Rightarrow - 4(k - 1)\{ 2k - 5\} \ge 0$$

$$\Rightarrow (k - 1)(2k - 5) \le 0$$

$$\therefore$$ $$k \in \left[ {1,{5 \over 2}} \right]$$

$$\therefore$$ $$k \ne 1$$

$$\therefore$$ $$k \in \left( {1,{5 \over 2}} \right]$$
2

### JEE Main 2021 (Online) 27th July Morning Shift

Let $$\alpha$$, $$\beta$$ be two roots of the

equation x2 + (20)1/4x + (5)1/2 = 0. Then $$\alpha$$8 + $$\beta$$8 is equal to
A
10
B
100
C
50
D
160

## Explanation

x2 + (20)1/4x + (5)1/2 = 0

$$\Rightarrow$$ x2 + $$\sqrt 5$$ = - (20)1/4x

Squaring both sides, we get

$${\left( {{x^2} + \sqrt 5 } \right)^2} = \sqrt {20} {x^2}$$

$$\Rightarrow$$ x4 = $$-$$5 $$\Rightarrow$$ x8 = 25

$$\Rightarrow$$ $$\alpha$$8 + $$\beta$$8 = 50
3

### JEE Main 2021 (Online) 25th July Evening Shift

The number of real solutions of the equation, x2 $$-$$ |x| $$-$$ 12 = 0 is :
A
2
B
3
C
1
D
4

## Explanation

|x|2 $$-$$ |x| $$-$$ 12 = 0

$$\Rightarrow$$ (|x| + 3)(|x| $$-$$ 4) = 0

$$\Rightarrow$$ |x| = 4

$$\Rightarrow$$ x = $$\pm$$2
4

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

Let $$\alpha = \mathop {\max }\limits_{x \in R} \{ {8^{2\sin 3x}}{.4^{4\cos 3x}}\}$$ and $$\beta = \mathop {\min }\limits_{x \in R} \{ {8^{2\sin 3x}}{.4^{4\cos 3x}}\}$$. If $$8{x^2} + bx + c = 0$$ is a quadratic equation whose roots are $$\alpha$$1/5 and $$\beta$$1/5, then the value of c $$-$$ b is equal to :
A
42
B
47
C
43
D
50

## Explanation

$$\alpha = \mathop {\max }\limits_{x \in R} \{ {8^{2\sin 3x}}{.4^{4\cos 3x}}\}$$

$$= \max \{ {2^{6\sin 3x}}{.2^{8\cos 3x}}\}$$

$$= max\{ {2^{6\sin 3x + 8\cos 3x}}\}$$

and $$\beta = \min \{ {8^{2\sin 3x}}{.4^{4\cos 3x}}\} = \min \{ {2^{6\sin 3x + 8\cos 3x}}\}$$

Now range of $$6sin3x + 8cos3x$$

$$= \left[ { - \sqrt {{6^2} + {8^2}} , + \sqrt {{6^2} + {8^2}} } \right] = [ - 10,10]$$

$$\alpha$$ = 210 & $$\beta$$ = 2$$-$$10

So, $$\alpha$$1/5 = 22 = 4

$$\Rightarrow$$ $$\beta$$1/5 = 2$$-$$2 = 1/4

quadratic 8x2 + bx + c = 0

$$- {b \over 8} = {{17} \over 4}$$ $$\Rightarrow$$ b = -34

$${c \over 8} = 1$$ $$\Rightarrow$$ c = 8

$$\therefore$$ c – b = 8 + 34 = 42

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