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Solved Objective Question on Area

Posted on - 04-01-2017

Math

IIT JEE

Question.1

The area bounded by y =, x-axis and the ordinates x = 0, x = p/4 is

(A) equal to p/4

(B) less than p/4

(C) greater than p/4

(D) less than .

Solution

Since £ 1


&⇒
£
&⇒
£ p/4

Also £ (in the similar way).

Question.2

The area bounded by the curve y = x3 , the x-axis and the ordinates x = -2 and x= 1 is

(A) –9

(B) –15/4

(C) 15/4

(D) 17/4

Solution

The bounded area is shown shaded in the fig.

The required Area =Area(OABO)+Area(OCDO)

==

= sq. units

Question.3

The maximum area of a rectangle whose two vertices lie on the x-axis and two on the curve y = 3 -|x|, -3 £ x £ 3 is

(A) 9

(B) 9/4

(C) 3

(D) None of these

Solution

The area bounded by the lines y = 3 -|x| -3 £ x£ 3 is shown in the fig. Area A(x) = 2x .(3-x).

\ A¢(x) = 2 (3-x) –2x

= 6 – 4x = 0
&⇒
x = 3/2

\ Maximum area of the rectangle occurs when x = 3/2

Maximum area

= 2.sq. units

Question.4

The curve x = 4-3y - y2 cuts the y-axis into two points P and Q. Then the area enclosed by the y-axis and the portion of the curve which lies between P and Q is .

(A) 20 sq. units .

(B) 18 sq. units.

(C) 17 sq. units .

(D) none of these. .

Solution

Required area =

=

= 20sq. units.

Question.5

If f(x) = a + bx + cx2, where c > 0 and b2 – 4ac < 0, then the area enclosed by the co-ordinate axes , the line x = 2 and the curve y = f(x) is given by

(A)

(B)

(C)

(D)

Solution

Area of OABL

=

=

= (2a+2b +c)

= . . . (1)

But, f(x) = a +bx+cx2

f(0) = a , f(1) = a+b+c

f(2) = a + 2b + 4c


&⇒
=

= .

Question.6

The curve y = x2 –7x +10 intersects the x-axis at the points A and B then the area bounded by the curve and the line AB is

(A) 4sq. units

(B) 4 sq. units.

(C) 6 sq. units .

(D) 2 sq. units.

Solution

The curve y = x2 – 7x +1 0 intersects x –axis at the points A (2, 0) and B( 5, 0) .

Hence the required area is

= –

= –

Question.7

The area of the region bounded by the curve y = and and the x-axis between the ordinates x= p/6 and x= p/3 is ;

(A) p/4

(B) p/2

(C) p/8

(D) none of these

Solution

The given curve is y =


&⇒
y =

Area bounded between the intervals x = p/6 and x = p/3 is given by

A =

Also A =

Adding 2A =
&⇒
A = = sq. units

Question.8

Let f(x) = minimum (tanx , cotx, 1/Ö3) " x ∈ [0, p/2] . Then area bounded by y = f(x) and the x-axis is ;.

(A)

(B)

(C)

(D)

Solution

We have f(x) = min(tanx, cotx, 1/Ö3) " x ∈

The area bounded is shown in the fig We have A = (p/6, 1/Ö3), B = (p/3, 1/Ö3)

The required area =


&⇒
+


&⇒

&⇒
log+

Question.9

The area bounded by the curve, y = f(x)= x4-2x3+x2+3, the x-axis and the ordinates corresponding to the minimum of function f(x) is

(A) 1

(B) 91/30

(C) 30/9

(D) 4

Solution

We have f(x) = x4-2x3 +x2+3

To find minimum

f ¢(x) = 4x3-6x2 +2x = 0


&⇒
2x (x-1)(2x-1) = 0
&⇒
x = 0, 1,1/2

f ¢¢(x) = 12 x2 –12x +2 = 2 (6x2 –6x+1)

f ¢¢(x) x=0 > 0
&⇒
Minimum exists at x = 0

f ¢¢(x)x=1/2 < 0
&⇒
Maximum exists at x = ½

f ¢¢(x)x = 1 > 0
&⇒
Minimum exists at x = 1


&⇒
The curve is bounded by the ordinates x = 0 and x = 1

\ Required Area =

= = sq. units.

Question.10

The area bounded by y = ln x, the x-axis and the ordinates x = 0 and x = 1 is

(A) 1

(B) 3/2

(C) -1

(D) none of these

Solution

The required area is shown shaded in the fig.

\ Area = –


&⇒

Question.11

The area enclosed by the parabola ay = 3(a2-x2) and the x- axis is

(A) 4a2 sq. units .

(B) a2 sq. units .

(C) 6a2 sq. units .

(D) 5a2 sq. units.

Solution

The parabola cuts the x-axis at points (–a, 0) and (a, 0).
A == 2

= 4a2 sq. units. .

Question.12

The whole area contained between the curve y2(a - x) = x2 (a + x) and the line x = a ( a > 0) is

(A) 2a2sq. units

(B) a2sq. units

(C) 2a2sq. units

(D) a2sq. units

Solution

The curve is symmetrical about the x-axis and cuts it at (-a, 0) and (0, 0)
Area = 2

= 2

= 2

Let x = asinq. So that dx = a cosq dq.

A = 2a2 = 2a2 = 2a2 .

Question.13

The area between the curves y = xex and y=x e-x and the line x= 1 is

(A) 2e

(B) e

(C) 2/e

(D) 1/e

Solution

The line x = 1 meets the curves in A(1, e) and B(1, 1/e). Both the curves pass through origin.

The required area

A ==

=

= == 2/e sq. units.

Question.14

The area in the first quadrant bounded by y = 4x2, x =0 y =1, and
y = 4 is

(A) 2 sq. units .

(B) 2sq. units

(C) 2sq. units

(D) 3 sq. units.

Solution

Required area =
= =

==2sq. units.

Question.15

The area bounded by y2 = - 4x, and its latus-rectum is

(A) Not defined

(B) 1 sq. units .

(C) 2/3 sq. units .

(D) none of these

Solution

The required area =

=

= = .

 
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