The least natural number 'n' for which `(n-2)x^2+8x+n+4>sin^(-1)(sin12)+cos^(-1)(cos12) AA x in R` is

Answer

To solve the problem, we need to find the least natural number \( n \) such that the inequality

\[ (n-2)x^2 + 8x + (n+4) > \sin^{-1}(\sin 12) + \cos^{-1}(\cos 12) \]

holds for all \( x \in \mathbb{R} \).

### Step 1: Simplify the Right Side

First, we need to evaluate \( \sin^{-1}(\sin 12) + \cos^{-1}(\cos 12) \).

- Since \( 12 \) radians is greater than \( \pi \) (approximately \( 3.14 \)), we can find the equivalent angle in the range of \( [0, 2\pi] \):
\[ 12 - 2\pi \approx 12 - 6.28 \approx 5.72 \text{ radians} \]

- Now, we can calculate:
\[ \sin^{-1}(\sin 12) = \sin^{-1}(\sin(5.72)) = 5.72 \quad \text{(since \( 5.72 \) is in the range of \( [-\frac{\pi}{2}, \frac{\pi}{2}] \))} \]
\[ \cos^{-1}(\cos 12) = 2\pi - 12 \quad \text{(since \( 12 \) is in the range of \( [\pi, 2\pi] \))} \]
\[ \cos^{-1}(\cos 12) = 2\pi - 12 \approx 6.28 - 12 = -5.72 \]

Thus,
\[ \sin^{-1}(\sin 12) + \cos^{-1}(\cos 12) = 5.72 - 5.72 = 0 \]

### Step 2: Set Up the Inequality

Now we need to ensure that:
\[ (n-2)x^2 + 8x + (n+4) > 0 \quad \text{for all } x \in \mathbb{R} \]

### Step 3: Conditions for the Quadratic

For a quadratic \( ax^2 + bx + c \) to be greater than zero for all \( x \):
1. The coefficient of \( x^2 \) (which is \( n-2 \)) must be positive:
\[ n - 2 > 0 \implies n > 2 \]

2. The discriminant must be less than zero:
\[ D = b^2 - 4ac < 0 \]
Here, \( a = n - 2 \), \( b = 8 \), and \( c = n + 4 \).

Therefore, the discriminant is:
\[ D = 8^2 - 4(n-2)(n+4) < 0 \]
\[ 64 - 4[(n-2)(n+4)] < 0 \]
\[ 64 - 4(n^2 + 2n - 8) < 0 \]
\[ 64 - 4n^2 - 8n + 32 < 0 \]
\[ -4n^2 - 8n + 96 < 0 \]
Dividing the entire inequality by -4 (and reversing the inequality):
\[ n^2 + 2n - 24 > 0 \]

### Step 4: Factor the Quadratic

Next, we factor the quadratic:
\[ (n + 6)(n - 4) > 0 \]

### Step 5: Find the Intervals

The critical points are \( n = -6 \) and \( n = 4 \). We analyze the sign of the product in the intervals:
1. \( n < -6 \) (both factors negative)
2. \( -6 < n < 4 \) (one factor positive, one negative)
3. \( n > 4 \) (both factors positive)

Thus, the solution to the inequality is:
\[ n < -6 \quad \text{or} \quad n > 4 \]

### Step 6: Find the Least Natural Number

Since we are looking for the least natural number \( n \), we take:
\[ n = 5 \]

### Final Answer

The least natural number \( n \) for which the inequality holds for all \( x \in \mathbb{R} \) is:
\[ \boxed{5} \]