If `f(x) ={{:( (a cos x+bx sin x+ce^(x)-2x)/(x^(2)), xne 0),( 0, x=0):}` is differentiable at `x=0`, then (a) `a+b+c=2` (b) `a+b=-4` (c) `f'(0)=1/3` (d) `a-c=4`

Options

A. `a+b+c=2`
B. `a+b=-4`
C. `f'(0)=1//3`
D. `a-c=4`

Answer

To determine the values of \(a\), \(b\), and \(c\) such that the function

\[ f(x) = \begin{cases} \frac{a \cos x + b x \sin x + c e^x - 2x}{x^2} & \text{if } x \neq 0 \\ 0 & \text{if } x = 0 \end{cases} \]

is differentiable at \(x = 0\), we will follow these steps:

### Step 1: Check the limit as \(x\) approaches 0

For \(f(x)\) to be differentiable at \(x = 0\), the limit

\[ \lim_{x \to 0} \frac{f(x) - f(0)}{x} = \lim_{x \to 0} \frac{f(x)}{x} \]

must exist. Since \(f(0) = 0\), we need to compute:

\[ \lim_{x \to 0} \frac{a \cos x + b x \sin x + c e^x - 2x}{x^2} \]

### Step 2: Apply L'Hôpital's Rule

Since both the numerator and denominator approach 0 as \(x \to 0\), we can apply L'Hôpital's Rule:

\[ \lim_{x \to 0} \frac{f(x)}{x} = \lim_{x \to 0} \frac{-a \sin x + b(\sin x + x \cos x) + c e^x - 2}{2x} \]

### Step 3: Evaluate the limit

As \(x \to 0\):

- \(\sin x \to 0\)
- \(\cos x \to 1\)
- \(e^x \to 1\)

Substituting these into the limit gives:

\[ \lim_{x \to 0} \frac{-a(0) + b(0 + 0) + c(1) - 2}{0} = \frac{c - 2}{0} \]

For this limit to exist, the numerator must also approach 0, thus:

\[ c - 2 = 0 \implies c = 2 \]

### Step 4: Substitute \(c\) back into the limit

Now substituting \(c = 2\) back into the limit:

\[ \lim_{x \to 0} \frac{a \cos x + b x \sin x + 2 e^x - 2x}{x^2} \]

### Step 5: Apply L'Hôpital's Rule again

We apply L'Hôpital's Rule again:

\[ \lim_{x \to 0} \frac{-a \sin x + b(\sin x + x \cos x) + 2 e^x - 2}{2x} \]

As \(x \to 0\):

\[ \lim_{x \to 0} \frac{-a(0) + b(0 + 0) + 2(1) - 2}{0} = \frac{2 - 2}{0} = 0 \]

### Step 6: Set up the equation for \(a\) and \(b\)

Now we need to differentiate again:

\[ \lim_{x \to 0} \frac{-a \cos x + b(\cos x + \cos x - x \sin x) + 2 e^x}{6x} \]

As \(x \to 0\):

\[ \lim_{x \to 0} \frac{-a(1) + b(1 + 1) + 2(1)}{6} = \frac{-a + 2b + 2}{6} \]

Setting this equal to 0 gives:

\[ -a + 2b + 2 = 0 \implies a = 2b + 2 \]

### Step 7: Find \(a + b\)

We also have the earlier limit condition:

\[ \lim_{x \to 0} \frac{a + c}{0} = 0 \implies a + c = 0 \implies a + 2 = 0 \implies a = -2 \]

Now substituting \(a = -2\) into \(a = 2b + 2\):

\[ -2 = 2b + 2 \implies 2b = -4 \implies b = -2 \]

### Summary of values

We have found:

- \(a = -2\)
- \(b = -2\)
- \(c = 2\)

### Step 8: Verify the options

1. \(a + b + c = -2 - 2 + 2 = -2\) (not correct)
2. \(a + b = -2 - 2 = -4\) (correct)
3. \(f'(0) = \frac{1}{3}\) (not correct)
4. \(a - c = -2 - 2 = -4\) (not correct)

### Final Answer

The correct option is:

(b) \(a + b = -4\)