If `f(x)=sin^(-1)x and lim_(xto1//2+)f(3x-4x^3)=a-3lim_(xto1//2+)f(x)`, then [a] is equal to {where [] denotes G.I.F}

Answer

To solve the problem step by step, we start with the given function and limit expression.

### Step 1: Define the function and limit
We have the function:
\[ f(x) = \sin^{-1}(x) \]
We need to evaluate:
\[ \lim_{x \to \frac{1}{2}^+} f(3x - 4x^3) = a - 3 \lim_{x \to \frac{1}{2}^+} f(x) \]

### Step 2: Evaluate the limit on the right side
First, we calculate:
\[ \lim_{x \to \frac{1}{2}^+} f(x) = \lim_{x \to \frac{1}{2}^+} \sin^{-1}(x) \]
As \( x \) approaches \( \frac{1}{2} \):
\[ \lim_{x \to \frac{1}{2}^+} \sin^{-1}(x) = \sin^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{6} \]

### Step 3: Evaluate the limit on the left side
Next, we need to evaluate:
\[ \lim_{x \to \frac{1}{2}^+} f(3x - 4x^3) = \lim_{x \to \frac{1}{2}^+} \sin^{-1}(3x - 4x^3) \]
Calculating \( 3x - 4x^3 \) as \( x \) approaches \( \frac{1}{2} \):
\[ 3\left(\frac{1}{2}\right) - 4\left(\frac{1}{2}\right)^3 = \frac{3}{2} - 4 \cdot \frac{1}{8} = \frac{3}{2} - \frac{1}{2} = 1 \]
Thus,
\[ \lim_{x \to \frac{1}{2}^+} \sin^{-1}(3x - 4x^3) = \sin^{-1}(1) = \frac{\pi}{2} \]

### Step 4: Set up the equation
Now we have:
\[ \frac{\pi}{2} = a - 3 \cdot \frac{\pi}{6} \]
Calculating \( 3 \cdot \frac{\pi}{6} \):
\[ 3 \cdot \frac{\pi}{6} = \frac{3\pi}{6} = \frac{\pi}{2} \]
Substituting this back into the equation gives:
\[ \frac{\pi}{2} = a - \frac{\pi}{2} \]

### Step 5: Solve for \( a \)
Adding \( \frac{\pi}{2} \) to both sides:
\[ \frac{\pi}{2} + \frac{\pi}{2} = a \]
Thus,
\[ a = \pi \]

### Step 6: Find the greatest integer value of \( a \)
The value of \( \pi \) is approximately \( 3.14 \). Therefore, the greatest integer value of \( a \) is:
\[ \lfloor a \rfloor = \lfloor \pi \rfloor = 3 \]

### Final Answer
Thus, the greatest integer value of \( a \) is:
\[ \boxed{3} \]