Step 1: Analyze the points where discontinuity might occur.
The given piecewise function transitions at \( x = -3 \) and \( x = 3 \). These are the potential points where discontinuities could arise.
Step 2: Verify continuity at \( x = -3 \).
At \( x = -3 \), calculate the left-hand and right-hand limits: \[ \text{Left-hand limit (LHL)} = |x| + 3 = |-3| + 3 = 3 + 3 = 6, \] \[ \text{Right-hand limit (RHL)} = -2x = -2(-3) = 6. \] The function value at \( x = -3 \) is: \[ f(-3) = |x| + 3 = |-3| + 3 = 6. \] Since the left-hand limit, right-hand limit, and the function's value at \( x = -3 \) are all equal, the function is continuous at \( x = -3 \).
Step 3: Verify continuity at \( x = 3 \).
At \( x = 3 \), calculate the left-hand and right-hand limits: \[ \text{Left-hand limit (LHL)} = -2x = -2(3) = -6, \] \[ \text{Right-hand limit (RHL)} = 6x + 2 = 6(3) + 2 = 18 + 2 = 20. \] Since the left-hand limit and right-hand limit are not equal, the function is discontinuous at \( x = 3 \).
Step 4: Final Conclusion.
There is exactly one point of discontinuity, which occurs at \( x = 3 \). Therefore, the total number of points of discontinuity is: \[ \boxed{1}. \]
Let $ f(x) + 2f\left( \frac{1}{x} \right) = x^2 + 5 $ and $ 2g(x) - 3g\left( \frac{1}{2} \right) = x, \, x>0. \, \text{If} \, \alpha = \int_{1}^{2} f(x) \, dx, \, \beta = \int_{1}^{2} g(x) \, dx, \text{ then the value of } 9\alpha + \beta \text{ is:}$
If $$ \int \frac{\left( \sqrt{1 + x^2} + x \right)^{10}}{\left( \sqrt{1 + x^2} - x \right)^9} \, dx = \frac{1}{m} \left( \left( \sqrt{1 + x^2} + x \right)^n \left( n\sqrt{1 + x^2} - x \right) \right) + C, $$ $\text{where } m, n \in \mathbb{N} \text{ and }$ $C \text{ is the constant of integration, then } m + n$ $\text{ is equal to:}$
Analyse the characters of William Douglas from ‘Deep Water’ and Mukesh from ‘Lost Spring’ in terms of their determination and will power in pursuing their goals.