InterviewSolution
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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
Write the range of one branch of \(sin^{–1} x\), other than the principal branch. |
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Answer» \([\frac{\pi}2,\frac{3\pi}2]\) |
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| 2. |
Find the principal value of \(cos^{-1} [cos (– 680°)]\). |
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Answer» \(cos^{-1}[cos(-680^o)]\)\(=cos^{-1}[cos(680^o)]\) \([\because cos(-\theta)=cos\,\theta]\) \(=cos^{-1}[cos\,(720^o-40^o)]\) \(=cos^{-1}[cos\,(4\pi-40^o)]\)\(=cos^{-1}(cos\,40^o)\) \(=40^o\) or \(\frac{2\pi}9\) \([\because40^o=\frac{2\pi}9\in/0,\pi/]\) |
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| 3. |
Evaluate: \(tan (tan^{–1} (–4))\). |
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Answer» \(tan(tan^{-1}(-4))=-4\) \([\because\,tan(tan^{-1}x)=x\,if\,x\in R\,and\,-4\in R]\) |
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| 4. |
Find the value of \(tan^{-1}(\frac xy)-tan^{-1}(\frac{x-y}{x+y})\). |
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Answer» \(tan^{-1}(\frac xy)-tan^{-1}(\frac{x-y}{x+y})\) \(=tan^{-1}\left(\cfrac{\frac xy-\frac{x-y}{x+y}}{1+\frac xy.\frac{x-y}{x+y}}\right)\) \([Here \frac xy.\frac{x-y}{x+y}>-1]\) \(=tan^{-1}(\frac{x^2+xy-xy+y^2}{y(x+y)}\times\)\(\frac{y(x+y)}{xy+y^2+x^2-xy})\) \(=tan^{-1}(\frac{x^2+y^2}{x^2+y^2})\) \(=tan^{-1}(1)=\frac{\pi}4\) |
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| 5. |
Prove that cos^-1 X = 2sin^-1 √1-x/2 = 2cos^-1 √1+X/2 |
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Answer» RHS 2SIN√1-X/2 2SIN^-1(√1-COSTHETA/2) =2SIN^-1(SINTHETA/2) =2(SINTHETA/2) =2(THETA/2] LHS=RHS |
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