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NTA Abhyas 2020
Exam
NTA Abhyas
Year
2020
Subject
Mathematics
Chapter
Trigonometric Equations

NTA Abhyas 2020 · Mathematics · Trigonometric Equations

If | ^ 2 x + 10 - x^ 2 | = |9 - x^ 2 | + 2sin^ 2 x + cos^ 2 x, then...

If ∣sin⁡2x+10−x2∣=∣9−x2∣+2sin2x+cos2x,\left|\sin^{2}x + 10 - x^{2}\right| = \left|9 - x^{2}\right| + 2sin^{2}x + cos^{2}x, then xx lies in

Options

  1. A[−8,8]\left[- 8,8\right]
  2. B[−3,3]\left[- 3,3\right]
  3. C[−17,17]\left[- \sqrt{17}, \sqrt{17}\right]
  4. D[−21,21]\left[- \sqrt{21}, \sqrt{21}\right]

Correct answer

B. [−3,3]\left[- 3,3\right]

Step-by-step solution

\left|\sin^{2}x + 10 - x^{2}\right| =\left|9 - x^{2}\right|+sin2x+(sin2x+cos2x)+ sin^{2}x + \left(sin^{2}x + cos^{2}x\right)
⇒∣sin⁡2x+1+9−x2∣=∣9−x2∣+sin2x+1⇒ \left|\sin^{2}x + 1 + 9 - x^{2}\right| =\left|9 - x^{2}\right| + sin^{2}x + 1
\left|x + y + z\right|=∣x∣=\left|x\right| + \left|y\right|+∣z∣+ \left|z\right|, if x,y,zx, y, z have the same sign
⇒sin2x(9−x2)≥0⇒ sin^{2}x \left(9 - x^{2}\right)≥ 0
⇒(9−x2)≥0⇒ \left(9 - x^{2}\right)≥ 0
⇒x∈[−3,3]⇒ x \in \left[-3,3\right]
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