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BITSAT 2020
Exam
BITSAT
Year
2020
Subject
Mathematics
Chapter
Trigonometric Ratios & Identities

BITSAT 2020 · Mathematics · Trigonometric Ratios & Identities

The equation ^ 4 x-(k+2) ^ 2 x-(k+3)=0 possesses a solution if

The equation sin⁡4x−(k+2)sin⁡2x−(k+3)=0\sin ^{4} x-(k+2) \sin ^{2} x-(k+3)=0 possesses a solution if

Options

  1. Ak>−3k > -3
  2. Bk<−2k< -2
  3. C−3≤k≤−2-3 \leq \mathrm{k} \leq-2
  4. Dk\mathrm{k} is any positive integer

Correct answer

C. −3≤k≤−2-3 \leq \mathrm{k} \leq-2

Step-by-step solution

We have, sin⁡4x−(k+2)sin⁡2x−(k+3)=0⇒sin⁡2x=(k+2)±(k+2)2+4(k+3)2=(k+2)±(k+4)2⇒sin⁡2x=k+3(∵sin⁡2x=−1\sin ^{4} x-(k+2) \sin ^{2} x-(k+3)=0\Rightarrow \sin ^{2} x=\frac{(k+2) \pm \sqrt{(k+2)^{2}+4(k+3)}}{2}=\frac{(k+2) \pm(k+4)}{2}\Rightarrow \sin ^{2} x=k+3\left(\because \sin ^{2} x=-1\right. is not possible) Since 0≤sin⁡2x≤1∴0≤k+3≤10 \leq \sin ^{2} x \leq 1\therefore 0 \leq \mathrm{k}+3 \leq 1 or −3≤k≤−2-3 \leq \mathrm{k} \leq-2
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