If `f(x)={sin(pi/2)(x-[x]),x<5 5(b-1),x=5(a b^2|x^2-11 x+24|)/(x-3),x >5` is continuous at `x=5,a ,b in R` then ([.] denotes the greatest integer function) `a=(25)/(108), b=6/5` b. `a=6/(13), b=(17)/(29)` c. `a=1/2, b=(25)/(36)` d. `a=(23)/(100), b=6/5`

Options

A. `a=(25)/(108),b=(6)/(5)`
B. `a=(6)/(13),b=(17)/(29)`
C. `a=(1)/(2),b=(25)/(36)`
D. `a=(23),(100),b=(6)/(5)`

Explanation

To determine the values of \( a \) and \( b \) such that the function \( f(x) \) is continuous at \( x = 5 \), we need to ensure that the left-hand limit (LHL) and the right-hand limit (RHL) at \( x = 5 \) are equal to the function value at that point.

The function is defined as follows:

\[ f(x) = \begin{cases} \sin\left(\frac{\pi}{2}(x - [x])\right) & \text{if } x < 5 \\ 5(b - 1) & \text{if } x = 5 \\ \frac{a b^2 |x^2 - 11x + 24|}{x - 3} & \text{if } x > 5 \end{cases} \]

### Step 1: Calculate the Left-Hand Limit (LHL)

For \( x < 5 \):

\[ LHL = \lim_{x \to 5^-} f(x) = \lim_{x \to 5^-} \sin\left(\frac{\pi}{2}(x - [x])\right) \]

Since \( [x] = 4 \) when \( x < 5 \):

\[ LHL = \sin\left(\frac{\pi}{2}(5 - 4)\right) = \sin\left(\frac{\pi}{2}\right) = 1 \]

### Step 2: Calculate the Right-Hand Limit (RHL)

For \( x > 5 \):

\[ RHL = \lim_{x \to 5^+} f(x) = \lim_{x \to 5^+} \frac{a b^2 |x^2 - 11x + 24|}{x - 3} \]

First, we need to factor \( |x^2 - 11x + 24| \):

\[ x^2 - 11x + 24 = (x - 3)(x - 8) \]

Since \( x > 5 \), \( x - 3 > 0 \) and \( x - 8 < 0 \), thus:

\[ |x^2 - 11x + 24| = -(x - 3)(x - 8) = -(x^2 - 11x + 24) \]

So we can write:

\[ RHL = \lim_{x \to 5^+} \frac{a b^2 (-(x - 3)(x - 8))}{x - 3} \]

The \( (x - 3) \) terms cancel out:

\[ RHL = \lim_{x \to 5^+} -a b^2 (x - 8) = -a b^2 (5 - 8) = 3ab^2 \]

### Step 3: Set LHL equal to RHL

For continuity at \( x = 5 \):

\[ LHL = RHL \implies 1 = 3ab^2 \]

### Step 4: Set the function value at \( x = 5 \)

The function value at \( x = 5 \):

\[ f(5) = 5(b - 1) \]

### Step 5: Set the two equations

We have two equations:

1. \( 3ab^2 = 1 \)
2. \( 5(b - 1) = 1 \)

From the second equation:

\[ 5(b - 1) = 1 \implies b - 1 = \frac{1}{5} \implies b = \frac{6}{5} \]

### Step 6: Substitute \( b \) into the first equation

Substituting \( b = \frac{6}{5} \) into \( 3ab^2 = 1 \):

\[ 3a\left(\frac{6}{5}\right)^2 = 1 \implies 3a \cdot \frac{36}{25} = 1 \implies a = \frac{25}{108} \]

### Conclusion

The values of \( a \) and \( b \) are:

\[ a = \frac{25}{108}, \quad b = \frac{6}{5} \]