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

NTA Abhyas 2020 · Mathematics · Trigonometric Equations

If x = 2nπ + tan^ - 1 (p)/(q) and y = r is a solution of the...

If x=2nπ+tan−1pqx = 2n\pi + tan^{- 1}\dfrac{p}{q} and y=ry = r is a solution of the equation 12sinx+5cosx=2y2−8y+21,12sin x + 5cos x = 2y^{2} - 8y + 21, then the value of k,k, such that& p2+q2+kr2=\sqrt{p^{2} + q^{2} + kr^{2}} =15,isequaltois equal to

Options

  1. A55
  2. B1414
  3. C314\dfrac{31}{4}
  4. D−22- 22

Correct answer

B. 1414

Step-by-step solution

12sin x + 5cos x = 2y^{2} - 8y + 21<br/><br />12sin x + 5cos xliesbetween[−13,13]lies between \left[- 13,13\right]
Also, 2y2−8y+21=2(y2−4y+4)+13=2(y−2)2+132y^{2} - 8y + 21 = 2\left(y^{2} - 4y + 4\right) + 13 = 2{\left(y - 2\right)}^{2} + 13
So, possibility of equality is when LHS=RHS=13LHS = RHS = 13
RHS=13RHS = 13, when y=2⇒r=2.y = 2 ⇒ r = 2 .
12sinx+5cosx=13⇒(sinx)1213+(cosx)513=112sin x + 5cos x = 13 ⇒ \left(sin x\right)\dfrac{12}{13} + \left(cos x\right)\dfrac{5}{13} = 1
⇒sin⁡(x+cos−11213)=1⇒x+cos−11213=(4n+1)π2⇒ \sin \left(x + cos^{- 1}\dfrac{12}{13}\right) = 1 ⇒ x + cos^{- 1}\dfrac{12}{13} = \left(4n + 1\right)\dfrac{\pi}{2}
⇒x=2nπ+(π2−cos−11213)=2nπ+sin−11213⇒ x = 2n\pi + \left(\dfrac{\pi}{2} - cos^{- 1}\dfrac{12}{13}\right) = 2n\pi + sin^{- 1}\dfrac{12}{13}
x=2nπ+tan−1125⇒p=12,q=5x = 2n\pi + tan^{- 1}\dfrac{12}{5} ⇒ p = 12, q = 5
⇒122+52+k⋅22=15⇒k=225−144−254=14⇒ \sqrt{1{2}^{2} + {5}^{2} + k ⋅ {2}^{2}} = 15 ⇒ k = \dfrac{225 - 144 - 25}{4} = 14
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