Big Problem but, Small solution!!

Calculus Level 4

Let $f$ be a function on $\mathbb{R} \to \mathbb{R}$ such that $f(1) = 1$ . If the y-intercept of the tangent at any point $\text{P}(x,y)$ on curve $y = f(x)$ is equal to the cube of abcissa of $\text{P}$ . Then find $f(-3)$ .

And now if , $\displaystyle\int_{0}^{ f(-3) - 9} 4 \cos\frac{x}{2} . \cos{x} . \sin\frac{21x}{2} dx = \text{k}$

Find the value of $\text{k} + 0.111$

The answer is 0.111.

1 solution

Abhijeeth Babu
Sep 19, 2016

Let the function be $f(x)=y$ such that $f'(x)=\frac{f(x)-x^3}{x} \Rightarrow \dfrac{\text{dy}}{\text{d}x} =\frac{y-x^3}{x}$ .

Which is a first order linear differential equation, and can be solved to $y=cx-\frac{x^3}{2}$ . $\\$
Using the condition $f(1)=1$ we solve for $c=\frac{3}{2}$ . $\\$

Thus $f(x)=\frac{3}{2}x-\frac{x^3}{2} \Rightarrow f(-3)=9$ . $\\$

Substituting $f(-3)=9$ in $I=\displaystyle\int_{0}^{f(-3)+9} 4\cdot cos(\frac{x}{2})\cdot cos(x)\cdot sin(\frac{21x}{2}) \text{d}x \\$

$I=\displaystyle\int_{0}^{-9+9} 4\cdot cos(\frac{x}{2})\cdot cos(x)\cdot sin(\frac{21x}{2})\text{d}x \\$ $\Rightarrow I=\displaystyle\int_{0}^{0} 4\cdot cos(\frac{x}{2})\cdot cos(x)\cdot sin(\frac{21x}{2}) \text{d}x =0 \\$ $\Rightarrow k=0 \\$ $\Rightarrow k+0.111=0.111 \\$ $\square$

Big Problem but, Small solution!!

The answer is 0.111.

1 solution

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