Standard Problems in Calculus(Part 2)

Calculus Level 5

We have a continous and differentiblel curve $y=f(x)$ passing through $(1,0)$

Let the slope of the tangent at the point $(x,f(x))$ be ${m}_{1}$ and the slope of the line joining the point and the origin be ${m}_{2}$ .

Now, the ratio of $\left| log({ m }_{ 1 }+{ m }_{ 2 }) \right|$ and $\left| log(x) \right|$ is $2:1$

There are exactly two functions satisfying this property let them be ${f}_{1}(x)$ and ${f}_{2}(x)$

Find $\int _{ 1 }^{ e }{ ({ f }_{ 1 }(x)+{ f }_{ 2 }(x))dx }$

1 solution

Mvs Saketh
Sep 14, 2014

Easy one, the trick lies in the modulus function, So while the ratio of the modulus must be $2$ , the ratio of the insides can be $2$ or $-2$

Hence case 1)

we get $log(m1+m2)=2 log(x)$

now,slope of curve is $\frac{dy}{dx}$ and slope of line joining origin and point is $y/x$

hence we have ${y}^{'} + \frac{y}{x}= {x}^{2}$

multiplying through out by $xdx$ and integrating

and using boundary conditions we get :

$\frac { { x }^{ 3 } }{ 4 } -\frac { 1 }{ 4x } =y$

and integrating within limits we get

$\frac { { e }^{ 4 }-5 }{ 16 }$

Now taking $-2$ as ratio and similarly proceeding,

we get

$xy=ln(x)$

which on integrating yields $\frac{1}{2}$

adding them both,we get ans