Question

Let \(f(x) =   \begin{cases}     3x+6,       & if\ x\geq c \\       x^2-3x-1,       & if\ x\lt c   \end{cases}\), where x∈R and c is a constant. The values of c for which f is continuous on R are

• A. -7, 1
• B. 1, 3
• C. -1, 7
• D. -1, 6
• E. 2, -3

Answer 1

The Correct Option is C

For $f(x)$ to be continuous at $x = c$, the following condition must hold: \[ \lim_{x \to c^-} f(x) = \lim_{x \to c^+} f(x) = f(c) \] That is, the left-hand limit must equal the right-hand limit at $x = c$. For $x \geq c$, we have $f(x) = 3x + 6$ and for $x < c$, we have $f(x) = x^2 - 3x - 1$. At $x = c$, we need: \[ \lim_{x \to c^-} f(x) = \lim_{x \to c^+} f(x) \] Substituting the functions for the limits: \[ \lim_{x \to c^-} (x^2 - 3x - 1) = \lim_{x \to c^+} (3x + 6) \] \[ c^2 - 3c - 1 = 3c + 6 \] Now, solve for $c$: \[ c^2 - 3c - 1 = 3c + 6 \] \[ c^2 - 6c - 7 = 0 \] Solving this quadratic equation using the quadratic formula: \[ c = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(1)(-7)}}{2(1)} = \frac{6 \pm \sqrt{36 + 28}}{2} = \frac{6 \pm \sqrt{64}}{2} \] \[ c = \frac{6 \pm 8}{2} \] Thus, the two possible values for $c$ are: \[ c = \frac{6 + 8}{2} = 7 \quad \text{or} \quad c = \frac{6 - 8}{2} = -1 \]

The correct option is (C) : \(-1, 7\)

Answer 2

• For \(f(x)\) to be continuous at \(x=c\), the left-hand limit and the right-hand limit must be equal at \(x=c\), and they must also be equal to the function's value at \(x=c\).
• The right-hand limit is: \[\displaystyle\lim_{x\rightarrow c^+} f(x) = \displaystyle\lim_{x\rightarrow c^+} (3x+6) = 3c+6\]
• The left-hand limit is: \[\displaystyle\lim_{x\rightarrow c^-} f(x) = \displaystyle\lim_{x\rightarrow c^-} (x^2-3x-1) = c^2-3c-1\]
• For continuity, we need: \[3c+6 = c^2-3c-1\]
• Rearrange the equation: \[c^2 - 6c - 7 = 0\]
• Factor the quadratic: \[(c-7)(c+1) = 0\]
• Solve for \(c\): \[c = 7 \text{ or } c = -1\]
• Therefore, the values of \(c\) for which \(f\) is continuous on R are -1 and 7.