Solved Objective Question on Area

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

(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

(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

(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.

(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.

(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

(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

(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

(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 .

(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

(D) a²sq. units

Solution

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

= 2

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

A = 2a² = 2a² = 2a² .

Question.13

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

(A) 2e

(B) 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

(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 .

(D) none of these

Solution

The required area =

= = .