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

NTA Abhyas 2020 · Mathematics · Trigonometric Equations

If the equation x^ 2 + 4 + 3 (ax + b) - 2x = 0 has atleast one real...

If the equation x2+4+3sin⁡(ax+b)−2x=0x^{2} + 4 + 3\sin \left(ax + b\right) - 2x = 0 has atleast one real solution, where aa, b∈[0,2π]b \in [ 0, 2\pi ], then one possible value of (a+b)( a + b ) can be equal to&

Options

  1. A7π2\dfrac{7\pi}{2}
  2. B5π2\dfrac{5\pi}{2}
  3. C9π2\dfrac{9\pi}{2}
  4. DNone of these

Correct answer

A. 7π2\dfrac{7\pi}{2}

Step-by-step solution

(x−1)2+3+3sin⁡(ax+b)=0{\left(x - 1\right)}^{2} + 3 + 3\sin \left(ax + b\right) = 0

(x−1)2+3=−3sin⁡(ax+b){\left(x - 1\right)}^{2} + 3 = - 3\sin ( ax + b )

L.H.S≥3,RHS∈[−3,3]L. H. S \ge 3, RHS \in [ - 3, 3 ] now sin⁡(ax+b)=−1\sin ( ax + b ) = - 1

∴x=1sin⁡(b+a)=−1∴ x = 1\sin \left(b + a\right) = - 1
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