Question

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

Answer 1

Hint: Here we will first find the equation of the tangent using the slope point form i.e.
\[y - {y_1} = \dfrac{{dy}}{{dx}}\left( {x - {x_1}} \right)\] 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 \[\left( {\dfrac{\pi }{4},1} \right)\]
Now we know that for any curve tangent is passing through a point \[\left( {{x_1},{y_1}} \right)\], then the equation of tangent using slope point form is given by:-
\[y - {y_1} = \dfrac{{dy}}{{dx}}\left( {x - {x_1}} \right)\]…………………………………..(1)
Hence we need to find \[\dfrac{{dy}}{{dx}}\] at \[x = \;\dfrac{\pi }{4}\]
Now since \[y = \;tanx\]
Differentiating it both the sides we get:-
\[\dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\left( {\tan x} \right)\]
Now as we know that:-
\[\dfrac{d}{{dx}}\left( {\tan x} \right) = {\sec ^2}x\]
Hence putting in the value we get:-
\[\dfrac{{dy}}{{dx}} = {\sec ^2}x\]
Now since we have calculate its value at \[x = \;\dfrac{\pi }{4}\]
Therefore substituting \[x = \;\dfrac{\pi }{4}\] in above equation we get:-
\[\dfrac{{dy}}{{dx}} = {\sec ^2}\left( {\;\dfrac{\pi }{4}} \right)\]
Now since we know that:-
\[\sec \left( {\;\dfrac{\pi }{4}} \right) = \sqrt 2 \]
Hence putting the value in above equation we get:-
\[\dfrac{{dy}}{{dx}} = {\left( {\sqrt 2 } \right)^2}\]
Solving it further we get:-
\[\dfrac{{dy}}{{dx}} = 2\]
Now putting this value and the values of passing points in equation 1 we get:-
\[y - 1 = 2\left( {x - \dfrac{\pi }{4}} \right)\]
Simplifying it we get:-
\[y - 1 = 2x - \dfrac{\pi }{2}\]
Therefore the equation of tangent is:-
\[y - 1 = 2x - \dfrac{\pi }{2}\]
Now we will find the point where the tangent is touching the x axis.
Therefore, this implies \[y = 0\]
Hence putting \[y = 0\] in the equation of tangent we get:-
\[0 - 1 = 2x - \dfrac{\pi }{2}\]
Now evaluating for x we get:-
\[2x = \dfrac{\pi }{2} - 1\]
Dividing the equation by 2 we get:-
\[x = \dfrac{\pi }{4} - \dfrac{1}{2}\]
Hence the point where tangents cuts the x axis is \[\left( {\dfrac{\pi }{4} - \dfrac{1}{2},0} \right)\]
Now we will find the area of the shaded region.
Area of shaded region=area under the curve \[y = \;tanx\]- area under the tangent
Therefore, we get:-
\[{\text{required area}} = \int_0^{\dfrac{\pi }{4}} {\tan x dx - {\text{area of triangle ABC}}} \]……………………….(2)
Now we know that area of triangle is given by:-
\[area = \dfrac{1}{2} \times base \times height\]
In triangle ABC
\[
  base = \dfrac{\pi }{4} - \left( {\dfrac{\pi }{4} - \dfrac{1}{2}} \right) \\
   \Rightarrow base = \dfrac{\pi }{4} - \dfrac{\pi }{4} + \dfrac{1}{2} \\
   \Rightarrow base = \dfrac{1}{2} \\
 \]
Also,
\[height = 1\]
Hence putting these values in formula of area we get:-
\[area = \dfrac{1}{2} \times \dfrac{1}{2} \times 1\]
Solving it further we get:-
\[area = \dfrac{1}{4}\]
Also, we know that:-
\[\int {\tan x dx = \log \left( {\sec x} \right)} + C\]
Hence putting the respective values in equation 2 we get:-
\[{\text{required area}} = \left[ {\log \sec x} \right]_0^{\dfrac{\pi }{4}} - \dfrac{1}{4}\]
Solving it further and putting the limits we get:-
\[{\text{required area}} = \left( {\log \sec \left( {\dfrac{\pi }{4}} \right) - \log \sec \left( 0 \right)} \right) - \dfrac{1}{4}\]
Now we know that:-
\[\sec \left( {\;\dfrac{\pi }{4}} \right) = \sqrt 2 \]
\[\sec 0 = 1\]
Hence putting these values we get:-
\[{\text{required area}} = \left[ {\log \sqrt 2 - \log 1} \right] - \dfrac{1}{4}\]
We know that:-
\[\log 1 = 0\]
Therefore putting the value we get:-
\[{\text{required area}} = \left[ {\log \sqrt 2 - 0} \right] - \dfrac{1}{4}\]
We know that:-
\[\log {x^n} = n\log x\]
Hence applying this property we get:-
\[
  {\text{required area}} = \left[ {\log {{\left( 2 \right)}^{\dfrac{1}{2}}} - 0} \right] - \dfrac{1}{4} \\
   \Rightarrow {\text{required area}} = \dfrac{1}{2}\log 2 - \dfrac{1}{4} \\
 \]
Simplifying it we get:-
\[{\text{required area}} = \dfrac{1}{2}\left[ {\log 2 - \dfrac{1}{2}} \right]\]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:
\[\int {\tan x dx = \log \left( {\sec x} \right)} + C\]
Also, the limits should be substituted carefully.