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Quadratic Equation and Inequalities question

2019 · 10 Apr · Shift 1 · Q29
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  5. /2019 · 10 Apr · Shift 1 · Q29

Quadratic Equation and Inequalities question

2019 · 10 Apr · Shift 1 · Q29

JEE MainMathematicsQuadratic Equation and InequalitiesMCQ+4 / −1
All the pairs (x, y) that satisfy the inequality 2sin⁡2x−2sin⁡x+5.14sin⁡2y≤1{2^{\sqrt {{{\sin }^2}x - 2\sin x + 5} }}.{1 \over {{4^{{{\sin }^2}y}}}} \le 12sin2x−2sinx+5​.4sin2y1​≤1 also satisfy the equation
  1. A
    sin x = |sin y|
  2. B
    sin x = 2sin y
  3. C
    2 sin x = sin y
  4. D
    2 |sin x | = 3 sin y
View written solutionFree

Correct answer: A

  1. Rewrite the inequality in powers of 222

Given

2sin⁡2x−2sin⁡x+5⋅14sin⁡2y≤1.2^{\sqrt{\sin^2 x-2\sin x+5}}\cdot \frac{1}{4^{\sin^2 y}}\le 1.2sin2x−2sinx+5​⋅4sin2y1​≤1.

Since 4sin⁡2y=(22)sin⁡2y=22sin⁡2y4^{\sin^2 y}=(2^2)^{\sin^2 y}=2^{2\sin^2 y}4sin2y=(22)sin2y=22sin2y, the inequality becomes

2sin⁡2x−2sin⁡x+5−2sin⁡2y≤1.2^{\sqrt{\sin^2 x-2\sin x+5}-2\sin^2 y}\le 1.2sin2x−2sinx+5​−2sin2y≤1.

Now, because the base 2>12>12>1, we must have

sin⁡2x−2sin⁡x+5−2sin⁡2y≤0,\sqrt{\sin^2 x-2\sin x+5}-2\sin^2 y\le 0,sin2x−2sinx+5​−2sin2y≤0,

i.e.

sin⁡2x−2sin⁡x+5≤2sin⁡2y.\sqrt{\sin^2 x-2\sin x+5}\le 2\sin^2 y.sin2x−2sinx+5​≤2sin2y.
  1. Estimate both sides

Let t=sin⁡xt=\sin xt=sinx. Then −1≤t≤1-1\le t\le 1−1≤t≤1.

The expression inside the square root is

t2−2t+5=(t−1)2+4.t^2-2t+5=(t-1)^2+4.t2−2t+5=(t−1)2+4.

So

sin⁡2x−2sin⁡x+5=(sin⁡x−1)2+4≥2.\sqrt{\sin^2 x-2\sin x+5}=\sqrt{(\sin x-1)^2+4}\ge 2.sin2x−2sinx+5​=(sinx−1)2+4​≥2.

Also, since 0≤sin⁡2y≤10\le \sin^2 y\le 10≤sin2y≤1,

0≤2sin⁡2y≤2.0\le 2\sin^2 y\le 2.0≤2sin2y≤2.

Thus the inequality

(sin⁡x−1)2+4≤2sin⁡2y\sqrt{(\sin x-1)^2+4}\le 2\sin^2 y(sinx−1)2+4​≤2sin2y

can hold only if both sides are exactly 222.

So we must have

(sin⁡x−1)2+4=2\sqrt{(\sin x-1)^2+4}=2(sinx−1)2+4​=2

and

2sin⁡2y=2.2\sin^2 y=2.2sin2y=2.
  1. Solve these conditions

From

(sin⁡x−1)2+4=2,\sqrt{(\sin x-1)^2+4}=2,(sinx−1)2+4​=2,

squaring gives

(sin⁡x−1)2+4=4(\sin x-1)^2+4=4(sinx−1)2+4=4 (sin⁡x−1)2=0(\sin x-1)^2=0(sinx−1)2=0 sin⁡x=1.\sin x=1.sinx=1.

From

2sin⁡2y=2,2\sin^2 y=2,2sin2y=2,

we get

sin⁡2y=1\sin^2 y=1sin2y=1 ∣sin⁡y∣=1.|\sin y|=1.∣siny∣=1.

Therefore all solutions of the inequality satisfy

sin⁡x=1=∣sin⁡y∣.\sin x=1=|\sin y|.sinx=1=∣siny∣.

Hence

sin⁡x=∣sin⁡y∣.\sin x=|\sin y|.sinx=∣siny∣.
  1. Check the options

  • A: sin⁡x=∣sin⁡y∣\sin x=|\sin y|sinx=∣siny∣
    True for all solution pairs.

  • B: sin⁡x=2sin⁡y\sin x=2\sin ysinx=2siny
    Here sin⁡x=1\sin x=1sinx=1 and sin⁡y=±1\sin y=\pm 1siny=±1. This would require 1=±21=\pm 21=±2, impossible. False.

  • C: 2sin⁡x=sin⁡y2\sin x=\sin y2sinx=siny
    This would require 2=±12=\pm 12=±1, impossible. False.

  • D: 2∣sin⁡x∣=3sin⁡y2|\sin x|=3\sin y2∣sinx∣=3siny
    Since ∣sin⁡x∣=1|\sin x|=1∣sinx∣=1, this gives 2=3sin⁡y2=3\sin y2=3siny, so sin⁡y=23\sin y=\frac23siny=32​, not possible for our solutions where sin⁡y=±1\sin y=\pm1siny=±1. False.

  1. Final answer

The correct option is

A\boxed{A}A​

with relation

sin⁡x=∣sin⁡y∣.\boxed{\sin x=|\sin y|}.sinx=∣siny∣​.
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