JEE Main 2014 · 19 Apr · Shift 1 · Mathematics Q76

← JEE Main 2014 · 19 Apr · Shift 1

MathematicsSingle correct+4 / -1Question 76 of 90

The area of the region above the $x$ -axis bounded by the curve $y=\tan x$ , $0\le x\le\dfrac\pi2$ and the tangent to the curve at $x=\dfrac\pi4$ is :

\dfrac12\left(\log2-\dfrac12\right)

\dfrac12\left(\log2+\dfrac12\right)

\dfrac12(1-\log2)

\dfrac12(1+\log2)

Answer and worked solution

Official answer: A

The tangent at $\left(\frac\pi4,1\right)$ is $y=2x-\frac\pi2+1$ , meeting the $x$ -axis at $x=\frac\pi4-\frac12$ .

The region is the area under $\tan x$ from 0 to $\frac\pi4$ minus the triangle under the tangent (base $\frac12$ , height 1):
$\frac12\ln2-\frac14=\frac12\left(\log2-\frac12\right).$

Topics:Area between curves Tangents and normals

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JEE Main 2014 · 19 Apr · Shift 1, Mathematics question 76