Will you compare coefficients?

Calculus Level 5

$\begin{aligned} C_1: f\left( x \right) &=x^6+Ax^5+Bx^4+Cx^3+Dx^2+Ex+F \\ C_2: g\left( x \right) &=Px+Q \end{aligned}$

Consider the curves $C_1$ and $C_2$ where $A,B,C,D,E,F,P,Q \in \mathbb{R}$ .

Now it is given that $f\left( x \right)$ touches $g\left( x \right)$ at $x=1,2,3$ . Let $\mathcal{A}$ be the area bounded between the two curves $C_1$ and $C_2$ .

If $\mathcal{A}$ can be expressed in the form $\frac{m}{n}$ , find $m+n$ .

Details And Assumptions

$m,n$ are co-prime integers.

The answer is 121.

1 solution

Ronak Agarwal
Oct 1, 2014

We write $h(x)=f(x)-g(x)$ , which is a sextic polynomial function.

Now given that $f(x)$ TOUCHES $g(x)$ at $x=1,2,3$

$\Rightarrow x=1,2,3$ are double repeated roots of $h(x)$

Since $h(x)$ is a sextic and the leading power coefficient of $h(x)$ is $1$ hence we write :

$h(x)=((x-1)(x-2)(x-3))^2$

To find the area between ${C}_{1}$ and ${C}_{2}$ we will integrate $h(x)$ from $x=1$ to $x=3$

$A=\int _{ 1 }^{ 3 }{ ({ (x-1)(x-2)(x-3) })^{ 2 }dx } =\frac { 16 }{ 105 }$

Just extending the solution of Ronak Agarwal.

It is the tedious job to calculate the last integral. But it can be simplified.

Make the following substitution:- $x-2=t$ Hence the integral becomes:- $\int_{-1}^{1}(t+1)^{2} t^{2} (t-1)^{2}dt$ $\int_{-1}^{1} (t^{2}-1)^{2}t^{2}dt$ Using properties of integrals:- $2\int_{0}^{1} (t^{2}-1)^{2}t^{2}dt$ Now it can be integrated easily.

Prakhar Gupta - 6 years, 4 months ago

IN LAST STEP IT SHOULD BE 2*TIMES THE INTEGRAL

Atul Solanki - 5 years, 8 months ago

Thanks, I've corrected that.

Prakhar Gupta - 5 years, 7 months ago

@Ronak Agarwal Why are 1, 2, 3 double repeated roots?? How do you know they are repeated??

Aaghaz Mahajan - 3 years, 3 months ago

since the two curves touch each other at x = 1,2,3. Therefore f(x) - g(x) will touch the x-axis at those particular points and hence repeated roots. @Aaghaz Mahajan

Ankit Kumar Jain - 3 years ago

@Ankit Kumar Jain Ok thanks!! but, I had got that later on, after improving my Calculus skills..... :) But, Thanks anyways!!!

Aaghaz Mahajan - 3 years ago

Will you compare coefficients?

The answer is 121.

1 solution

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