Solved Subjective Question on Area Set 2

Example.1

Find the area bounded by y = |x-1| and y = 3 - |x| .

Solution

Let y = f₁(x) =|x-1| and y = f₂(x) = 3 - |x|. Now .

f₁(x) = and f₂(x) =

The graph of y = f₁(x) and y = f₂(x) are as shown : We have to find the area ABCD. Here
A º (1, 0), Cº (0, 3). B is the point of intersection of y = x–1 and y = 3-x . .

So B º (2, 1) and D is the point of intersection of y = – x + 1 and y = 3 + x .
Hence D º (-1, 2) and N º (2, 0). .

The required area =

= 9 –

= 9 – 5 = 4 sq. units .

Note:

Obviously ABCD is a rectangle with AB =and AD = 2. Hence the required area = 2= 4 sq. units.

Example.2

Find the smaller area enclosed by y = f(x) , when f(x) is a polynomial of least degree satisfying and the circle x²+ y²=2 above the x- axis.

Solution

Since exists, = 0.

Thus f(x) = a₄x⁴+a₅x⁵+-------+ a_nxⁿ, a_n > 0 , n ≥ 4.

Since f(x) is of least degree, f(x) = a₄x⁴. Given that.

= e
&⇒ = e
&⇒ a₄ = 1
&⇒ f(x) = x⁴.

The graph of y = x⁴ and x²+ y²= 2 are shown below.
The point A º(1, 1). The required area =

= =sq. units

Example.3

Find the area bounded by y = f(x) and the curve y =satisfying the conditions f(x) . f(y) = f(xy) " x, y ∈ R and f¢(1) =2, f(1) =1 .

Solution

f¢(x) = =

= = =
&⇒ .

Integrating both sides, we get f(x) = cx². Since f(1) = 1
&⇒ c = 1.

So, f(x) = x² . Now
&⇒ x⁴ + x² –2 = 0
&⇒ x² =1
&⇒ x = ±1

Required area = 2

= 2= . sq. units.

Example.4

Find the area of the region bounded by y² + 4ax = 0 and (y² + 4a²)x + 8a³ = 0, a > 0.

Solution

y² = – 4ax is the parabola opening towards left and vertex at origin.

x = -is a curve existing on the negative side of x – axis only and symmetrical w.r.t. x – axis. The lowest value of x occurs when y = 0 (then x = -2a). The curve tends to approach the y – axis as x approaches 0.