Let `a = cos^(-1) cos 20, b = cos^(-1) cos 30 and c = sin^(-1) sin (a + b)` then
The largest integer x for which `sin^(-1) (sin x) ge |x -(a + b + c)|` is

Options

A. 1
B. 2
C. 3
D. 4

Explanation

To solve the problem step by step, we will follow the logic presented in the video transcript and clarify each step.

### Step 1: Define the Variables
Let:
- \( a = \cos^{-1}(\cos 20^\circ) \)
- \( b = \cos^{-1}(\cos 30^\circ) \)
- \( c = \sin^{-1}(\sin(a + b)) \)

### Step 2: Calculate \( a \)
Using the property of the inverse cosine function:
\[ \cos^{-1}(\cos x) = x \quad \text{for } x \in [0, \pi] \]
Since \( 20^\circ \) is in the range \( [0, \pi] \):
\[ a = 20^\circ \]

### Step 3: Calculate \( b \)
Similarly, for \( b \):
\[ b = \cos^{-1}(\cos 30^\circ) = 30^\circ \]

### Step 4: Calculate \( a + b \)
Now, we can calculate \( a + b \):
\[ a + b = 20^\circ + 30^\circ = 50^\circ \]

### Step 5: Calculate \( c \)
Next, we need to find \( c \):
\[ c = \sin^{-1}(\sin(a + b)) = \sin^{-1}(\sin 50^\circ) \]
Since \( 50^\circ \) is in the range \( [-\frac{\pi}{2}, \frac{\pi}{2}] \), we have:
\[ c = 50^\circ \]

### Step 6: Calculate \( a + b + c \)
Now, we can calculate:
\[ a + b + c = 50^\circ + 50^\circ = 100^\circ \]

### Step 7: Set up the Inequality
We need to find the largest integer \( x \) such that:
\[ \sin^{-1}(\sin x) \geq |x - (a + b + c)| \]
Substituting \( a + b + c = 100^\circ \):
\[ \sin^{-1}(\sin x) \geq |x - 100^\circ| \]

### Step 8: Analyze \( \sin^{-1}(\sin x) \)
The function \( \sin^{-1}(\sin x) \) returns \( x \) when \( x \) is in the range \( [-\frac{\pi}{2}, \frac{\pi}{2}] \) and adjusts to keep the output in that range otherwise.

### Step 9: Determine the Range for \( x \)
To satisfy the inequality, we need to consider the ranges:
- For \( x \in [0, 180^\circ] \), \( \sin^{-1}(\sin x) = x \).
- For \( x \in [180^\circ, 360^\circ] \), \( \sin^{-1}(\sin x) = 180^\circ - x \).

### Step 10: Solve the Inequality
We analyze the inequality:
1. For \( x \in [0, 100^\circ] \):
\[ x \geq 100^\circ - x \implies 2x \geq 100^\circ \implies x \geq 50^\circ \]
2. For \( x \in [100^\circ, 180^\circ] \):
\[ x \geq x - 100^\circ \implies 0 \geq -100^\circ \quad \text{(always true)} \]
3. For \( x \in [180^\circ, 360^\circ] \):
\[ 180^\circ - x \geq x - 100^\circ \implies 280^\circ \geq 2x \implies x \leq 140^\circ \]

### Step 11: Combine Results
From the analysis:
- \( x \) must be in the range \( [50^\circ, 140^\circ] \).

### Step 12: Find the Largest Integer
The largest integer \( x \) in the range \( [50, 140] \) is:
\[ \boxed{140} \]