Originality!!

Calculus Level 5

The equation of the curve y y = f ( x ) f(x) .The tangents at :-

(1, f ( 1 ) f(1) ),(2, f ( 2 ) f(2) ) ,and(3, f ( 3 ) f(3) ) makes angle π 6 \dfrac{\pi}{6} , π 3 \dfrac{\pi}{3} and π 4 \dfrac{\pi}{4} respectively with the positive direction of x x - axis.Then the value of

2 3 \displaystyle\int_{2}^{3} f ( x ) f'(x) f " ( x ) f"(x) d x dx + 1 3 \displaystyle\int_{1}^{3} f " ( x ) f"(x) d x dx can be expressed as

1 a \dfrac{-1}{a} find the value of a a .


The answer is 1.7.

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4 solutions

Chew-Seong Cheong
Dec 24, 2014

It is given that:

f ( 1 ) = tan π 6 = 1 3 f ( 2 ) = tan π 3 = 3 f ( 3 ) = tan π 4 = 1 f'(1) = \tan {\frac {\pi}{6}} = \frac {1}{\sqrt{3}} \quad \Rightarrow f'(2) = \tan {\frac {\pi}{3}} = \sqrt{3} \quad \Rightarrow f'(3) = \tan {\frac {\pi}{4}} = 1

and we need to solve the integral below:

2 3 f ( x ) f ( x ) d x + 1 3 f ( x ) d x \displaystyle \int _2 ^3 {f'(x)f '' (x)\space dx} + \int _1 ^3 {f''(x)\space dx}

Now consider:

d d x ( f ( x ) f ( x ) ) = f ( x ) f ( x ) + f ( x ) f ( x ) = 2 f ( x ) f ( x ) \frac {d}{dx} \left( f'(x)f'(x) \right) = f'(x)f''(x) + f''(x)f'(x) = 2 f'(x)f''(x)

f ( x ) f ( x ) d x = 1 2 f ( x ) f ( x ) + C \Rightarrow \displaystyle \int {f'(x)f '' (x)\space dx} = \frac {1}{2} f'(x)f'(x) + C

It is obvious that: f ( x ) d x = f ( x ) + C \displaystyle \int {f''(x)\space dx} = f'(x) + C

Therefore,

2 3 f ( x ) f ( x ) d x + 1 3 f ( x ) d x = [ 1 2 f ( x ) f ( x ) ] x = 2 x = 3 + [ f ( x ) ] x = 1 x = 3 \displaystyle \int _2 ^3 {f'(x)f '' (x)\space dx} + \int _1 ^3 {f''(x)\space dx} = \left[ \frac {1}{2} f'(x)f'(x) \right] _{x=2} ^{x=3} + \left[ f'(x) \right] _{x=1} ^{x=3}

= 1 2 ( 1 3 ) + 1 1 3 = 1 3 a = 3 = \dfrac {1}{2} (1-3) + 1 - \dfrac {1}{\sqrt{3}} = \dfrac {-1}{\sqrt{3}} \quad \Rightarrow a = \boxed {\sqrt{3}}

Nice solution sir.

Parth Lohomi - 6 years, 5 months ago

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tailored the original question?? hun?

A Former Brilliant Member - 6 years, 1 month ago
Keshav Tiwari
Nov 30, 2014

@Parth Lohomi I follow you already . ;)

Thanks!! Just wanna increase the number of followers!!

Parth Lohomi - 6 years, 6 months ago
Kunal Gupta
Dec 26, 2014

Well not a level 5 I suppose!!!!

Akash Doshi
Dec 7, 2014

If we look at the first integrand, we see an f ( x ) f^{ ' }\left( x \right) and an f ( x ) f^{ '' }\left( x \right) . This should remind us of chain rule, and we should think about how we would differentiate a function to get f ( x ) f^{ ' }\left( x \right) and f ( x ) f^{ '' }\left( x \right) . We should realize that we need to differentiate f ( x ) f^{ ' }\left( x \right) to some power in order to obtain the f ( x ) f^{ '' }\left( x \right) , and then we realize that power has to be 2.

The indefinite integral of f ( x ) f ( x ) = 1 2 f ( x ) 2 \int { f'\left( x \right) f''\left( x \right) = } \frac { 1 }{ 2 } { f'\left( x \right) }^{ 2 }

Just by simple integration, we see that the indefinite integral of the other integrand is f ( x ) f^{ ' }\left( x \right) .

We have bounds given to us, so we now need to find the definite values of these indefinite integrals. If the tangent at (1, f(1)) makes an angle of pi/6 with the +x axis, we know the slope, and therefore the quantity f ( 1 ) f^{ ' }(1) , is arctan(pi/6)= 1 3 \frac { 1 }{ \sqrt { 3 } } . The same principle can be applied to the other two points, and we find f ( 2 ) = arctan π 3 = 3 f^{ ' }(2)\quad =\arctan { \frac { \pi }{ 3 } }=\quad \sqrt { 3 } and f ( 3 ) = arctan π 4 = 1 f^{ ' }(3)\quad =\arctan { \frac { \pi }{ 4 } } =\quad 1 .

Then we go to our integrals and plug in our values:

2 3 f ( x ) f ( x ) = 1 2 ( f ( 3 ) ) 2 1 2 ( f ( 2 ) ) 2 \int _{ 2 }^{ 3 }{ f'\left( x \right) f''\left( x \right) } =\frac { 1 }{ 2 } { \left( f'\left( 3 \right) \right) }^{ 2 }-\frac { 1 }{ 2 } { \left( f'\left( 2 \right) \right) }^{ 2 }

and

1 3 f ( x ) = f ( 3 ) f ( 1 ) \int _{ 1 }^{ 3 }{ f''\left( x \right) } =f'\left( 3 \right) \quad -\quad f'\left( 1 \right)

Plugging in our values and adding the two integrals we get it to come out to 1 3 \frac { -1 }{ \sqrt { 3 } } , so our answer is a = 3 1.732 \sqrt { 3 } \approx \quad 1.732 .

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