The area of the region bounded by the curve y=tanx, the tangent to the curve at x=π4 and the x-axis is
A
14(loge4−1)
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B
14(loge2−1)
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C
12(loge4−1)
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D
12(loge2−1)
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Solution
The correct option is A14(loge4−1)
Slope of the tangent at x=π4 is sec2π4=2
Equation of the tangent at A is, y−1=2(x−π4) ∴ Coordinates of B are (π4−12,0) BC=12
Required area = Area of OACO− Area of ΔABC =π/4∫0tanxdx−12×AC×BC =[logesecx]π/40−12×1×12=(loge√2−0)−14=14(loge4−1)