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ISI MStat 2016 Problem 1 | Area bounded by the curves | PSB Sample

This is a beautiful problem from ISI MStat 2016 Problem 1, PSB Sample, based on area bounded by the curves. We provide a detailed solution with the prerequisites mentioned explicitly.

Problem- ISI MStat 2016 Problem 1

In the diagram below, L(x) is a straight line that intersects the graph of a polynomial P(x) of degree 2 at the points A=(-1,0) and B=(5,12) . The area of the shaded region is 36 square units. Obtain the expression for P(x).

ISI MStat 2016 Problem 1 figure

Prerequisites

  • Area bounded by the curve
  • Polynomials of degree 2
  • Area of a triangle

Solution

Let P(x)=ax^2 +bx + c as given P(x) is of degree 2 .

Now from the figure we can see that L(x) intersect P(x) at points A=(-1,0) and B=(5,12) .

Hence we have P(-1)=0 and P(5)=12 , which gives ,

a-b+c=0 ---(1) and 25a+5b +c =12 ----(2)

Then ,

ISI MStat 2016 Problem 1 graph
Fig-1

See from Fig-1 we can say that Area of the shaded region = (Area bounded by the curve P(x) and x-axis )- (Area of the triangle ABC) - (Area bounded by the curve P(x) , x=5 and x=L )

= \int^{L}_{-1} P(x)\,dx - \frac{1}{2} \times (5+1) \times 12 -\int^{5}_{L} P(x)\,dx

=\int^{L}_{-1} P(x)\,dx -  \int^{5}_{L} P(x)\,dx -36

=\int^{5}_{-1} P(x)\,dx -36

Again it is given that area of the shaded region is 36 square units.

So, \int^{5}_{-1} P(x)\,dx -36 =36 \Rightarrow \int^{5}_{-1} P(x)\,dx =2 \times 36

\int^{5}_{-1} (ax^2+bx+c) \,dx  = 2 \times 36 . After integration we get ,

7a + 2b +c =12 ---(3)

Now we have three equations and three unknows

a-b+c=0

25a+5b +c =12

7a + 2b +c =12

Solving this three equations by elimination and substitution we get ,

a=-1 , b=6 , c=7

Therefore , the expression for P(x) is P(x)= -x^2+6x+7 .

Previous MStat Posts:

This is a beautiful problem from ISI MStat 2016 Problem 1, PSB Sample, based on area bounded by the curves. We provide a detailed solution with the prerequisites mentioned explicitly.

Problem- ISI MStat 2016 Problem 1

In the diagram below, L(x) is a straight line that intersects the graph of a polynomial P(x) of degree 2 at the points A=(-1,0) and B=(5,12) . The area of the shaded region is 36 square units. Obtain the expression for P(x).

ISI MStat 2016 Problem 1 figure

Prerequisites

  • Area bounded by the curve
  • Polynomials of degree 2
  • Area of a triangle

Solution

Let P(x)=ax^2 +bx + c as given P(x) is of degree 2 .

Now from the figure we can see that L(x) intersect P(x) at points A=(-1,0) and B=(5,12) .

Hence we have P(-1)=0 and P(5)=12 , which gives ,

a-b+c=0 ---(1) and 25a+5b +c =12 ----(2)

Then ,

ISI MStat 2016 Problem 1 graph
Fig-1

See from Fig-1 we can say that Area of the shaded region = (Area bounded by the curve P(x) and x-axis )- (Area of the triangle ABC) - (Area bounded by the curve P(x) , x=5 and x=L )

= \int^{L}_{-1} P(x)\,dx - \frac{1}{2} \times (5+1) \times 12 -\int^{5}_{L} P(x)\,dx

=\int^{L}_{-1} P(x)\,dx -  \int^{5}_{L} P(x)\,dx -36

=\int^{5}_{-1} P(x)\,dx -36

Again it is given that area of the shaded region is 36 square units.

So, \int^{5}_{-1} P(x)\,dx -36 =36 \Rightarrow \int^{5}_{-1} P(x)\,dx =2 \times 36

\int^{5}_{-1} (ax^2+bx+c) \,dx  = 2 \times 36 . After integration we get ,

7a + 2b +c =12 ---(3)

Now we have three equations and three unknows

a-b+c=0

25a+5b +c =12

7a + 2b +c =12

Solving this three equations by elimination and substitution we get ,

a=-1 , b=6 , c=7

Therefore , the expression for P(x) is P(x)= -x^2+6x+7 .

Previous MStat Posts:

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