JEE Mains · Maths · STD 12 - 8. Application and integration
The area of the region above the \(x-\) axis bounded by the curve \(y\, = tan\, x\), \(0 \leq x \leq \frac{\pi }{2}\) and the tangent to the curve at \(x\, = \frac{\pi}{4}\) is
- A \(\frac{1}{2}\left( {\log \,2 - \frac{1}{2}} \right)\)
- B \(\frac{1}{2}\left( {\log \,2 + \frac{1}{2}} \right)\)
- C \(\frac{1}{2}\left( {1 - \log \,2} \right)\)
- D \(\frac{1}{2}\left( {1 + \log \,2} \right)\)
Answer & Solution
Correct Answer
(A) \(\frac{1}{2}\left( {\log \,2 - \frac{1}{2}} \right)\)
Step-by-step Solution
Detailed explanation
The given curve is \(y=\tan x \quad \ldots(1)\) when \(x=\frac{\pi}{4}, y=1\) Equation of tangent at \(P\) is \(y-1=\left(\sec ^{2} \frac{\pi}{4}\right)\left(x-\frac{\pi}{4}\right)\) or \(y=2 x+1-\frac{\pi}{2}\) .....\((2)\) Area of shaded region \(=\) area of…
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