1. The problem asks for the value of $k$ such that the function $$f(x) = \begin{cases} kx + 1, & x \leq \pi \\ \cos x, & x > \pi \end{cases}$$ is continuous at $x = \pi$. 2. For continuity at $x=\pi$, the left-hand limit and the right-hand limit at $\pi$ must be equal, and both must equal $f(\pi)$. 3. Calculate the left-hand limit: $$\lim_{x \to \pi^-} f(x) = k\pi + 1$$ 4. Calculate the right-hand limit: $$\lim_{x \to \pi^+} f(x) = \cos \pi = -1$$ 5. For continuity: $$k\pi + 1 = -1 \Rightarrow k\pi = -2 \Rightarrow k = \frac{-2}{\pi}$$ 6. So, the value of $k$ is $-\frac{2}{\pi}$ which corresponds to option (D). --- 7. The second problem asks for the number of discontinuities of the piecewise function: $$f(x) = \begin{cases} x + 2, & x < 0 \\ e^x, & 0 \leq x \leq 1 \\ 2 - x, & x > 1 \end{cases}$$ 8. Check continuity at the boundary points where the definition changes: $x=0$ and $x=1$. 9. At $x=0$: - Left limit: $\lim_{x \to 0^-} (x+2) = 0 + 2 = 2$ - Right limit: $\lim_{x \to 0^+} e^x = e^0 = 1$ - Since $2 \neq 1$, there is a discontinuity at $x=0$. 10. At $x=1$: - Left limit: $\lim_{x \to 1^-} e^x = e^1 = e$ - Right limit: $\lim_{x \to 1^+} (2 - x) = 2 - 1 = 1$ - Since $e \neq 1$, there is a discontinuity at $x=1$. 11. There are 2 discontinuities at $x=0$ and $x=1$. So the answer is (C) 2. --- 12. The third problem is about the function: $$f(x) = |x| - x$$ 13. Check continuity and differentiability at $x=0$. 14. Evaluate $f(0)$: $$f(0) = |0| - 0 = 0$$ 15. Check limits: - Left limit at $0$: $\lim_{x \to 0^-} (|x| - x) = \lim_{x \to 0^-} (-x - x) = \lim_{x \to 0^-} (-2x) = 0$ - Right limit at $0$: $\lim_{x \to 0^+} (x - x) = \lim_{x \to 0^+} 0 = 0$ - Both limits equal $f(0)$, so $f$ is continuous at 0. 16. Check differentiability by evaluating the left and right derivatives at $x=0$: - For $x<0$, $f(x) = -x - x = -2x$, so derivative is $f'(x) = -2$. - For $x>0$, $f(x) = x - x = 0$, so derivative is $f'(x) = 0$. 17. Left derivative at 0 is $-2$; right derivative at 0 is $0$. Since they are not equal, $f$ is not differentiable at $x=0$. 18. Therefore, $f$ is continuous but not differentiable at $x=0$. The answer is (B).