Polynomial of 1919

Calculus Level 3

If f ( x ) f(x) is a polynomial such that f ( x ) + f ′ ( x ) + f ′ ′ ( x ) + f ′ ′ ′ ( x ) = x 3 f (x) + f'(x) + f''(x) + f'''(x) = x^3 , g ( x ) = ∫ f ( x ) x 3 d x g(x) = \displaystyle \int \frac {f(x)}{x^3} dx and g ( 1 ) = 1 g(1) = 1 , find g ( e ) g(e) .

e + 3 e + 3 1 − e 1 - e e − 3 e - 3 1 + e 1 + e

Priyanshu Mishra by Priyanshu Mishra , 20, India

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2 solutions

Let f ( x ) = x 3 + a x 2 + b x + c f(x)=x^3+ax^2+bx+c , then

f ( x ) = x 3 + a x 2 + b x + c f ′ ( x ) = 3 x 2 + 2 a x + b f ′ ′ ( x ) = 6 x + 2 a f ′ ′ ′ ( x ) = 6 \begin{aligned} f(x) & = x^3 + ax^2 + bx + c \\ f'(x) & = 3x^2 + 2ax + b \\ f''(x) & = 6x + 2a \\ f'''(x) & = 6 \end{aligned}

Equating the coefficients of f ( x ) + f ′ ( x ) + f ′ ′ ( x ) + f ′ ′ ′ ( x ) = x 3 f(x) + f'(x) + f''(x) + f'''(x) = x^3 , we have 3 + a = 0 3+a=0 ⟹ a = − 3 \implies a = - 3 , b + 2 a + 6 = 0 b+2a+6=0 ⟹ b = 0 \implies b=0 , and c + b + 2 a + 6 = 0 c+b+2a+6=0 ⟹ c = 0 \implies c=0 . Therefore, f ( x ) = x 3 − 3 x 2 f(x) = x^3 - 3x^2 and

g ( x ) = ∫ f ( x ) x 3 d x = ∫ x 3 − 3 x 2 x 3 d x = ∫ ( 1 − 3 x ) d x = x − 3 ln ⁡ x + C where C is the constant of integration. g ( 1 ) = 1 − 3 ln ⁡ 1 + C = 1 ⟹ C = 0 ⟹ g ( x ) = x − 3 ln ⁡ x g ( e ) = e − 3 \begin{aligned} g(x) & = \int \frac {f(x)}{x^3} dx = \int \frac {x^3-3x^2}{x^3} dx= \int \left(1-\frac 3x\right) dx = x-3\ln x + C & \small \color{#3D99F6} \text{where }C \text{ is the constant of integration.} \\ g(1) & = 1 - 3\ln 1 + C = 1 \quad \small \color{#3D99F6} \implies C = 0 \\ \implies g(x) & = x - 3\ln x \\ g(e) & = \boxed{e - 3} \end{aligned}

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@Priyanshu Mishra , since you like to contribute problems, it would be advantageous if you learn up how to use LaTex probably. It is easier to enter LaTex codes just freehand than using the buttons of the editor. You also learn up LaTex codes faster.

For example, you can just enter \ (f(x) + f'(x) + f''(x) + f'''(x) = x^3 \ ) f ( x ) + f ′ ( x ) + f ′ ′ ( x ) + f ′ ′ ′ ( x ) = x 3 f(x) + f'(x) + f''(x) + f'''(x) = x^3 and it will turn out fine (of course there is no space between \ ( and \ )). Just showing you the different codings and results \int \frac {f(x)}{x^3} dx ∫ f ( x ) x 3 d x \int \frac {f(x)}{x^3} dx , \int \dfrac {f(x)}{x^3} dx ∫ f ( x ) x 3 d x \int \dfrac {f(x)}{x^3} dx and \displaystyle \int \frac {f(x)}{x^3} dx ∫ f ( x ) x 3 d x \displaystyle \int \frac {f(x)}{x^3} dx .

You can see the LaTex codes by placing your mouse cursor on top of the formulas. You can also choose Toggle LaTex under the pull-down menu ( ⋯ \cdots more) under the answer section. You can also learn LaTex codes from the LaTex editor .

Chew-Seong Cheong - 3 years, 4 months ago

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Thanks sir for suggesting. I will keep this in mind .

Priyanshu Mishra - 3 years, 4 months ago
Aaghaz Mahajan
Feb 8, 2018

Let the polynomial be of degree n. Solving the first equation gives us n=3. So, the polynomial is ax^3+bx^2+cx+d. Putting this in first equation leads us to polynomial being x^3-3x^2. Now the integral can be evaluated easily.....

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