question_answer

Area of the region bounded by the curve \[y=\tan x,\]tangent drawn to the curve at \[x=\frac{\pi }{4}\] and the \[x-\]axis is:

A) \[\log \sqrt{2}\,\text{sq}\,\text{unit}\]

B)         \[\left( \log \sqrt{2}+\frac{1}{4} \right)\,\text{sq}\,\text{unit}\]

C)         \[\left( \log \sqrt{2}-\frac{1}{4} \right)\text{sq}\,\text{unit}\]

D)         \[\frac{1}{4}\text{sq}\,\text{unit}\]

Correct Answer: C

Solution :

Given curve is At           \[x=\frac{\pi }{4}\]                 \[y=\tan \frac{\pi }{4}=1\] On differentiating Eq. (i), w.r.t. \[x,\]we get \[\frac{dy}{dx}={{\sec }^{2}}x\Rightarrow {{\left( \frac{dy}{dx} \right)}_{x=\pi /4}}={{\sec }^{2}}\frac{\pi }{4}=2\] Equation of tangent at \[P\left( \frac{\pi }{4},1 \right)\]is \[y-1=2\left( x-\frac{\pi }{4} \right)\]                 \[\Rightarrow \]               \[y=2x+1-\frac{\pi }{2}\] It meets \[x-\]axis at \[T\left( \frac{\pi -2}{4},0 \right).\] \[\therefore \] Required area = area of curve OTP - area of \[\Delta TNP\] \[=\int_{0}^{\pi /4}{\tan x\,dx-\frac{1}{2}TN.PN}\] \[=[\log \sec x]_{0}^{\pi /4}-\frac{1}{2}.\frac{1}{2}.1\] \[=\log \sqrt{2}-0-\frac{1}{4}\] \[=\left( \log \sqrt{2}-\frac{1}{4} \right)\text{sq}\,\text{unit}\]