Given: The piecewise function:
\( f(x) = \begin{cases} -2x, & -2 < x < -1, \\ -x, & -1 \leq x < 0, \\ 0, & 0 \leq x < 1, \\ x - 1, & 1 \leq x < 2. \end{cases} \)
Clearly, \( f(x) \) is discontinuous at \( x = -1 \). It is also non-differentiable at this point.
Thus, \( m = 1 \).
Differentiate \( f(x) \):
\( f'(x) = \begin{cases} -2, & -2 < x < -1, \\ -1, & -1 < x < 0, \\ 0, & 0 < x < 1, \\ 1, & 1 < x < 2. \end{cases} \)
\( f(x) \) is non-differentiable at \( x = -1, 0, 1 \).
The absolute value \( |f(x)| \) remains the same.
Thus, \( n = 3 \).
\( m + n = 1 + 3 = 4 \).
Final Answer: \( m + n = 4 \).
Let $ P_n = \alpha^n + \beta^n $, $ n \in \mathbb{N} $. If $ P_{10} = 123,\ P_9 = 76,\ P_8 = 47 $ and $ P_1 = 1 $, then the quadratic equation having roots $ \alpha $ and $ \frac{1}{\beta} $ is:
Let $ f(x) = \begin{cases} (1+ax)^{1/x} & , x<0 \\1+b & , x = 0 \\\frac{(x+4)^{1/2} - 2}{(x+c)^{1/3} - 2} & , x>0 \end{cases} $ be continuous at x = 0. Then $ e^a bc $ is equal to
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