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Solved Subjective Question on Area Set 4

Posted on - 04-01-2017

Math

IIT JEE

Example.1

Find the area of the figure bounded by the parabola (y-2)2 = x-1, the tangent to it at the point with ordinate 3, and the x-axis.

Solution

The equation (y – 2)2 = x – 1 represents a parabola having vertex at A(1, 2) and axis as y =2 . The point P on this parabola with ordinate 3 has abscissa.

x = 1+ (y-2)2 = 2, then P º (2, 3)

(y - 2)2 = x -1


&⇒
2(y-2)= 1
&⇒

Thus equation of tangent at P is given by

y - 3 =


&⇒
x – 2y +4 = 0

The region, of which area is to be found is shown by shaded portion. We have to find the Area(EOBAPCE). C is the point of intersection of x-2y +4 =0 and y-axis

&⇒
C º (0, 2)
. Obviously .

D º (0, 3).

The required area

= area(OBAPDO) – area (DPCD) + area(DEOC)

=

= sq. units

Example.2

Let f(x) = maximum {x2, (1-x)2, 2x(1-x)}. Determine the area of the region bounded by the curve y = f(x), x-axis , x = 0 and x=1.

Solution     

First we draw the graph of y = x2, y = (1 - x)2 and y = 2x(1 - x) for

0 £ x £ 1. Obviously all these are parabolas.

Now graph of y = f(x) is shown by dark curves. Hence C is the point of intersection of
y = x2 and y = (1-x)2
. So
C º (1/2, ¼)
. B is the point of intersection of y =(1 – x)2,
y =2x(1-x)
. So B º(1/3, 4/9).

Similarly D º (2/3, 4/9)

The required area = Area (OPBDQAO)

=

=

= sq. units

Note: If f(x) = minimum {x2, (1-x)2, 2x(1-x)}. , then the required area will be OCAO. .

Example.3    

Find the area between the curves y = x2 + x –2 and y = 2x, for which |x2 + x –2| + | 2x | = |x2 + 3x –2| is satisfied.

Solution

y = x2 + x –2, y = 2x

| x2 + x –2 | + | 2x | = | x2 + 3x –2 |

This means we have to find area in the region where x2 + x –2 and 2x have same sign.

Thus, the required region is OABO and ECD.

\ x2 +x –2 = 2x


&⇒
x2 –x –2 = 0
&⇒
x = -1, x = 2

\ required area is,

A=Area of OABO+ Area of ECD

\

=

=

=

=

= =

Example.4    

Let f(x) = min{ex, 3/2, 1+ e-x | 0 £ x £ 1 }. Find the area bounded by
y = f(x) , x-axis, y-axis and the line x = 1
. .

Solution                 

f(x) =

Let A be the required area as shown in fig.

A = =

= = 2+ .

Example.5

Find a function ‘f’, (x4 – 4x2 £ f(x) £ 2x2 – x3) such that the area bounded by y = f(x), y =x4 –4x2, the y-axis and the line x=t, (0 £ t £ 2) is k times the area bounded by y = f(x), y = 2x2 – x3, y-axis and line x =t ( 0£ t £ 2) . .

Solution     

According to the given condition

Differentiate both sides w.r.t t , we get f(t) –(t4 – 4t2) = k(2t2 –t3 –f(t)).


&⇒
(1+k) f(t) = k 2t2 – kt3 + t4 – 4t2
&⇒
f(t) =

Hence, the required function ‘f’ is given by

f(x) = (x4 – kx3 +2(k – 2)x2) .

 
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