Question

The values of $a, b, c$ for which the function $f(x) = \begin{cases} \sin((a + 1)x) + \sin x, & x<0 \\ c, & x = 0 \\ \frac{(\sqrt{x + bx^2}) - \sqrt{x}}{bx^{1/2}}, & x > 0 \end{cases}$ is continuous at $x = 0$, are

• A. a=3/2, b=3/2, c=1/2
• B. a= - 3/2, c= 3/2, b is arbitrary non- zero real number
• C. a=-5/2, b=-3/2, c=3/2
• D. a=-2, bεR -{0}, c=0

Answer 1

The Correct Option is D

We are given the piecewise function: \[ f(x) = \begin{cases} \sin((a + 1)x) + \sin x, & x<0 \\ c, & x = 0 \\ \frac{(\sqrt{x + bx^2}) - \sqrt{x}}{bx^{1/2}}, & x>0 \end{cases} \] and we need to find the values of \(a\), \(b\), and \(c\) for which \(f(x)\) is continuous at \(x = 0\). Step 1: Continuity at \(x = 0\) For \(f(x)\) to be continuous at \(x = 0\), the following must hold: \[ \lim_{x \to 0^-} f(x) = f(0) = \lim_{x \to 0^+} f(x) \] Step 2: Left-hand limit (\(x \to 0^-\)) For \(x<0\), the function is: \[ f(x) = \sin((a + 1)x) + \sin x \] We need to find the limit as \(x \to 0^-\). Using the small angle approximation \(\sin x \approx x\) as \(x \to 0\): \[ \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \left( \sin((a + 1)x) + \sin x \right) = (a + 1) \cdot 0 + 0 = 0 \] Thus, the left-hand limit is 0. Step 3: Right-hand limit (\(x \to 0^+\)) For \(x>0\), the function is: \[ f(x) = \frac{(\sqrt{x + bx^2}) - \sqrt{x}}{bx^{1/2}} \] We simplify this expression by rationalizing the numerator: \[ \frac{(\sqrt{x + bx^2}) - \sqrt{x}}{bx^{1/2}} = \frac{\left( \sqrt{x + bx^2} - \sqrt{x} \right) \cdot \left( \sqrt{x + bx^2} + \sqrt{x} \right)}{bx^{1/2} \left( \sqrt{x + bx^2} + \sqrt{x} \right)} \] \[ = \frac{(x + bx^2) - x}{bx^{1/2} \left( \sqrt{x + bx^2} + \sqrt{x} \right)} = \frac{bx^2}{bx^{1/2} \left( \sqrt{x + bx^2} + \sqrt{x} \right)} \] \[ = \frac{x^{3/2}}{\sqrt{x + bx^2} + \sqrt{x}} \] As \(x \to 0\), \(\sqrt{x + bx^2} \to \sqrt{x}\), so the limit becomes: \[ \lim_{x \to 0^+} \frac{x^{3/2}}{2\sqrt{x}} = \lim_{x \to 0^+} \frac{x}{2} = 0 \] Thus, the right-hand limit is also 0. Step 4: Continuity condition at \(x = 0\) For the function to be continuous at \(x = 0\), we must have: \[ f(0) = \lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = 0 \] Therefore, we must have \(c = 0\). Step 5: Finding the values of \(a\) and \(b\) The left-hand and right-hand limits are both 0, and the function at \(x = 0\) is \(c = 0\), so there is no restriction on \(a\) and \(b\) for continuity as long as \(c = 0\). Final Answer: The correct choice is: \[ \boxed{a = -2, b \in \mathbb{R} \setminus \{0\}, c = 0} \]