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Question: The area of the region above the x-axis bounded by the curve \[y = \;tanx\], \[0 \leqslant x \leqsla...

The area of the region above the x-axis bounded by the curve y=  tanxy = \;tanx, 0x  π2  0 \leqslant x \leqslant \;\dfrac{\pi }{2}\; and the tangent to the curve at x=  π4x = \;\dfrac{\pi }{4} is
(A) 12(log212)\dfrac{1}{2}\left( {\log 2 - \dfrac{1}{2}} \right)
(B) 12(log2+12)\dfrac{1}{2}\left( {\log 2 + \dfrac{1}{2}} \right)
(C) 12(1log2)\dfrac{1}{2}\left( {1 - \log 2} \right)
(D) 12(1log2)\dfrac{1}{2}\left( {1 - \log 2} \right)

Explanation

Solution

Here we will first find the equation of the tangent using the slope point form i.e.
yy1=dydx(xx1)y - {y_1} = \dfrac{{dy}}{{dx}}\left( {x - {x_1}} \right) and then we will find the coordinate where the tangent cuts the x axis and then we will finally find the area of the shaded region.

Complete step-by-step answer:

First of all we will find the equation of the tangent which is passing through (π4,1)\left( {\dfrac{\pi }{4},1} \right)
Now we know that for any curve tangent is passing through a point (x1,y1)\left( {{x_1},{y_1}} \right), then the equation of tangent using slope point form is given by:-
yy1=dydx(xx1)y - {y_1} = \dfrac{{dy}}{{dx}}\left( {x - {x_1}} \right)…………………………………..(1)
Hence we need to find dydx\dfrac{{dy}}{{dx}} at x=  π4x = \;\dfrac{\pi }{4}
Now since y=  tanxy = \;tanx
Differentiating it both the sides we get:-
dydx=ddx(tanx)\dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\left( {\tan x} \right)
Now as we know that:-
ddx(tanx)=sec2x\dfrac{d}{{dx}}\left( {\tan x} \right) = {\sec ^2}x
Hence putting in the value we get:-
dydx=sec2x\dfrac{{dy}}{{dx}} = {\sec ^2}x
Now since we have calculate its value at x=  π4x = \;\dfrac{\pi }{4}
Therefore substituting x=  π4x = \;\dfrac{\pi }{4} in above equation we get:-
dydx=sec2(  π4)\dfrac{{dy}}{{dx}} = {\sec ^2}\left( {\;\dfrac{\pi }{4}} \right)
Now since we know that:-
sec(  π4)=2\sec \left( {\;\dfrac{\pi }{4}} \right) = \sqrt 2
Hence putting the value in above equation we get:-
dydx=(2)2\dfrac{{dy}}{{dx}} = {\left( {\sqrt 2 } \right)^2}
Solving it further we get:-
dydx=2\dfrac{{dy}}{{dx}} = 2
Now putting this value and the values of passing points in equation 1 we get:-
y1=2(xπ4)y - 1 = 2\left( {x - \dfrac{\pi }{4}} \right)
Simplifying it we get:-
y1=2xπ2y - 1 = 2x - \dfrac{\pi }{2}
Therefore the equation of tangent is:-
y1=2xπ2y - 1 = 2x - \dfrac{\pi }{2}
Now we will find the point where the tangent is touching the x axis.
Therefore, this implies y=0y = 0
Hence putting y=0y = 0 in the equation of tangent we get:-
01=2xπ20 - 1 = 2x - \dfrac{\pi }{2}
Now evaluating for x we get:-
2x=π212x = \dfrac{\pi }{2} - 1
Dividing the equation by 2 we get:-
x=π412x = \dfrac{\pi }{4} - \dfrac{1}{2}
Hence the point where tangents cuts the x axis is (π412,0)\left( {\dfrac{\pi }{4} - \dfrac{1}{2},0} \right)
Now we will find the area of the shaded region.
Area of shaded region=area under the curve y=  tanxy = \;tanx- area under the tangent
Therefore, we get:-
required area=0π4tanxdxarea of triangle ABC{\text{required area}} = \int_0^{\dfrac{\pi }{4}} {\tan x dx - {\text{area of triangle ABC}}} ……………………….(2)
Now we know that area of triangle is given by:-
area=12×base×heightarea = \dfrac{1}{2} \times base \times height
In triangle ABC

base=π4(π412) base=π4π4+12 base=12  base = \dfrac{\pi }{4} - \left( {\dfrac{\pi }{4} - \dfrac{1}{2}} \right) \\\ \Rightarrow base = \dfrac{\pi }{4} - \dfrac{\pi }{4} + \dfrac{1}{2} \\\ \Rightarrow base = \dfrac{1}{2} \\\

Also,
height=1height = 1
Hence putting these values in formula of area we get:-
area=12×12×1area = \dfrac{1}{2} \times \dfrac{1}{2} \times 1
Solving it further we get:-
area=14area = \dfrac{1}{4}
Also, we know that:-
tanxdx=log(secx)+C\int {\tan x dx = \log \left( {\sec x} \right)} + C
Hence putting the respective values in equation 2 we get:-
required area=[logsecx]0π414{\text{required area}} = \left[ {\log \sec x} \right]_0^{\dfrac{\pi }{4}} - \dfrac{1}{4}
Solving it further and putting the limits we get:-
required area=(logsec(π4)logsec(0))14{\text{required area}} = \left( {\log \sec \left( {\dfrac{\pi }{4}} \right) - \log \sec \left( 0 \right)} \right) - \dfrac{1}{4}
Now we know that:-
sec(  π4)=2\sec \left( {\;\dfrac{\pi }{4}} \right) = \sqrt 2
sec0=1\sec 0 = 1
Hence putting these values we get:-
required area=[log2log1]14{\text{required area}} = \left[ {\log \sqrt 2 - \log 1} \right] - \dfrac{1}{4}
We know that:-
log1=0\log 1 = 0
Therefore putting the value we get:-
required area=[log20]14{\text{required area}} = \left[ {\log \sqrt 2 - 0} \right] - \dfrac{1}{4}
We know that:-
logxn=nlogx\log {x^n} = n\log x
Hence applying this property we get:-

required area=[log(2)120]14 required area=12log214  {\text{required area}} = \left[ {\log {{\left( 2 \right)}^{\dfrac{1}{2}}} - 0} \right] - \dfrac{1}{4} \\\ \Rightarrow {\text{required area}} = \dfrac{1}{2}\log 2 - \dfrac{1}{4} \\\

Simplifying it we get:-
required area=12[log212]{\text{required area}} = \dfrac{1}{2}\left[ {\log 2 - \dfrac{1}{2}} \right]sq units

Hence option A is the correct option.

Note: Students should note that area under the curve is the area between the two curves.
Students might make mistake in integration so, they should keep in mind the following formula:
tanxdx=log(secx)+C\int {\tan x dx = \log \left( {\sec x} \right)} + C
Also, the limits should be substituted carefully.