Question

If $f(x) = \begin{cases} \frac{a+b\cos x+c\sin x}{x^2}; x > 0 \\ 9 ; x \leq 0 \end{cases}$ is continuous at x = 0, then the value of $\frac{|a|+|b|}{5}$

Answer 1

36/5

Answer 2

For the function $f(x)$ to be continuous at $x=0$, the following condition must be satisfied:

$\lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0)$

From the definition of the function:

1. $f(0) = 9$
2. $\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} 9 = 9$

Now, we must have $\lim_{x \to 0^+} f(x) = 9$.

$\lim_{x \to 0^+} \frac{a+b\cos x+c\sin x}{x^2} = 9$

Let $L = \lim_{x \to 0^+} \frac{a+b\cos x+c\sin x}{x^2}$.
As $x \to 0$, the denominator $x^2 \to 0$. For the limit $L$ to be a finite value (9), the numerator must also approach $0$. This implies the form $\frac{0}{0}$.

Step 1: Numerator must be 0 at $x=0$.
Substitute $x=0$ into the numerator:
$a+b\cos(0)+c\sin(0) = 0$
$a+b(1)+c(0) = 0$
$a+b = 0 \implies b = -a$

Now, substitute $b=-a$ into the limit expression:

$L = \lim_{x \to 0^+} \frac{a-a\cos x+c\sin x}{x^2} = \lim_{x \to 0^+} \frac{a(1-\cos x)+c\sin x}{x^2}$

This is still of the $\frac{0}{0}$ form. We can apply L'Hopital's Rule.

Step 2: Apply L'Hopital's Rule (first time).
Differentiate the numerator and the denominator with respect to $x$:

$L = \lim_{x \to 0^+} \frac{\frac{d}{dx}(a(1-\cos x)+c\sin x)}{\frac{d}{dx}(x^2)} = \lim_{x \to 0^+} \frac{a\sin x+c\cos x}{2x}$

Again, as $x \to 0$, the denominator $2x \to 0$. For $L$ to be finite, the new numerator must also approach $0$.
Substitute $x=0$ into the new numerator:
$a\sin(0)+c\cos(0) = 0$
$a(0)+c(1) = 0$
$c = 0$

Now, substitute $c=0$ into the limit expression:

$L = \lim_{x \to 0^+} \frac{a\sin x}{2x}$

This is still of the $\frac{0}{0}$ form. We can apply L'Hopital's Rule again.

Step 3: Apply L'Hopital's Rule (second time).
Differentiate the numerator and the denominator with respect to $x$:

$L = \lim_{x \to 0^+} \frac{\frac{d}{dx}(a\sin x)}{\frac{d}{dx}(2x)} = \lim_{x \to 0^+} \frac{a\cos x}{2}$

Now, substitute $x=0$:

$L = \frac{a\cos(0)}{2} = \frac{a(1)}{2} = \frac{a}{2}$

We are given that $L=9$.
So, $\frac{a}{2} = 9 \implies a = 18$.

Step 4: Find the value of $b$.
From Step 1, we found $b = -a$.
So, $b = -18$.

Step 5: Calculate the required value.
We need to find $\frac{|a|+|b|}{5}$.

$\frac{|a|+|b|}{5} = \frac{|18|+|-18|}{5} = \frac{18+18}{5} = \frac{36}{5}$