Key Concepts and Formulas

• Differentiability implies Continuity: If a function is differentiable at a point, it must also be continuous at that point.
• Condition for Differentiability: A function $f(x)$ is differentiable at $x=c$ if the left-hand derivative (LHD) and the right-hand derivative (RHD) at $x=c$ are equal and finite.
  • LHD at $x=c$: $f'(c^-) = \lim_{h \to 0^-} \frac{f(c+h) - f(c)}{h}$
  • RHD at $x=c$: $f'(c^+) = \lim_{h \to 0^+} \frac{f(c+h) - f(c)}{h}$
• Derivative of Inverse Cosine: The derivative of $\cos^{-1}(u)$ is $-\frac{1}{\sqrt{1-u^2}} \frac{du}{dx}$.

Step-by-Step Solution

Step 1: Apply the continuity condition at x = 1. Since the function $f(x)$ is differentiable at $x=1$, it must be continuous at $x=1$. This means the left-hand limit, the right-hand limit, and the function value at $x=1$ must be equal. $\lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x) = f(1)$ For $x < 1$, $f(x) = -x$. So, $\lim_{x \to 1^-} f(x) = -1$. For $1 \le x \le 2$, $f(x) = a + \cos^{-1}(x+b)$. So, $\lim_{x \to 1^+} f(x) = a + \cos^{-1}(1+b)$. Also, $f(1) = a + \cos^{-1}(1+b)$. Equating these values, we get: $-1 = a + \cos^{-1}(1+b) \quad (\text{Equation 1})$

Step 2: Calculate the Left-Hand Derivative (LHD) at x = 1. The LHD is the derivative of the function for $x < 1$ evaluated at $x=1$. For $x < 1$, $f(x) = -x$. The derivative of $f(x)$ for $x < 1$ is $f'(x) = -1$. Therefore, the LHD at $x=1$ is: $f'(1^-) = -1$

Step 3: Calculate the Right-Hand Derivative (RHD) at x = 1. The RHD is the derivative of the function for $1 \le x \le 2$ evaluated at $x=1$. For $1 \le x \le 2$, $f(x) = a + \cos^{-1}(x+b)$. We need to find the derivative of $f(x)$ with respect to $x$: $f'(x) = \frac{d}{dx} (a + \cos^{-1}(x+b))$ Since $a$ is a constant, $\frac{d}{dx}(a) = 0$. Using the derivative formula for $\cos^{-1}(u)$ where $u = x+b$, we have $\frac{du}{dx} = 1$. $\frac{d}{dx} (\cos^{-1}(x+b)) = -\frac{1}{\sqrt{1-(x+b)^2}} \cdot 1 = -\frac{1}{\sqrt{1-(x+b)^2}}$ So, the derivative of $f(x)$ for $1 \le x \le 2$ is: $f'(x) = -\frac{1}{\sqrt{1-(x+b)^2}}$ Now, we evaluate this at $x=1$ to find the RHD: $f'(1^+) = -\frac{1}{\sqrt{1-(1+b)^2}}$

Step 4: Apply the condition that LHD = RHD. Since $f(x)$ is differentiable at $x=1$, we must have $f'(1^-) = f'(1^+)$. $-1 = -\frac{1}{\sqrt{1-(1+b)^2}}$ Multiplying both sides by $-1$: $1 = \frac{1}{\sqrt{1-(1+b)^2}}$ This implies: $\sqrt{1-(1+b)^2} = 1$ Squaring both sides: $1-(1+b)^2 = 1$ $-(1+b)^2 = 0$ $(1+b)^2 = 0$ $1+b = 0$ $b = -1$

Step 5: Substitute the value of b back into Equation 1 to find a. From Equation 1, we have: $-1 = a + \cos^{-1}(1+b)$ Substitute $b=-1$: $-1 = a + \cos^{-1}(1+(-1))$ $-1 = a + \cos^{-1}(0)$ The principal value of $\cos^{-1}(0)$ is $\frac{\pi}{2}$. $-1 = a + \frac{\pi}{2}$ Solving for $a$: $a = -1 - \frac{\pi}{2}$ $a = -\frac{2}{2} - \frac{\pi}{2} = -\frac{2+\pi}{2}$

Step 6: Calculate the value of ab. We have $a = -\frac{\pi+2}{2}$ and $b = -1$. $ab = \left(-\frac{\pi+2}{2}\right) \cdot (-1)$ $ab = \frac{\pi+2}{2}$

Common Mistakes & Tips

• Forgetting the Continuity Condition: Differentiability at a point implies continuity. Always start by ensuring the function is continuous at the point of differentiability.
• Incorrect Derivative Formula: Be careful with the derivative of inverse trigonometric functions, especially the sign and the term under the square root.
• Principal Values: When dealing with inverse trigonometric functions, remember to use their principal values. For $\cos^{-1}(0)$, the principal value is $\frac{\pi}{2}$.

Summary

The problem requires the function to be differentiable at $x=1$. This implies two conditions: continuity at $x=1$ and equality of the left-hand derivative and the right-hand derivative at $x=1$. By applying the continuity condition, we obtained an equation relating $a$ and $b$. By calculating the LHD and RHD and equating them, we found the value of $b$. Substituting this value back into the continuity equation allowed us to find the value of $a$. Finally, we computed the product $ab$.

The final answer is $\boxed{\frac{\pi + 2}{2}}$.