Q245 / 270 · NDA 2026 (Phase 1) · Mathematics
Passage: Let $f(x)=\begin{cases} ax(x-1), & x<1 \\ x-1, & 1\leq x\leq3 \\ px^2+qx+2, & x>3 \end{cases}$
Given that $f(x)$ is continuous for all $x$ but not differentiable at $x=1$. Further $f'(x)$ is continuous at $x=3$.
Question: What is the value of $p$ ?
Given that $f(x)$ is continuous for all $x$ but not differentiable at $x=1$. Further $f'(x)$ is continuous at $x=3$.
Question: What is the value of $p$ ?
- A$-1$
- B$-\dfrac{1}{3}$
- Correct answerC$\dfrac{1}{3}$
- D$1$
Solution
For $f(x)$ to be continuous at $x=3$, the left-hand limit and right-hand limit at $x=3$ must be equal.
$\lim_{x \to 3^-} f(x) = 3 - 1 = 2$
$\lim_{x \to 3^+} f(x) = p(3)^2 + q(3) + 2 = 9p + 3q + 2$
Equating them, we get:
$9p + 3q + 2 = 2 \Rightarrow 9p + 3q = 0 \Rightarrow q = -3p$
The derivative of the function for $x \in (1, 3)$ is $f'(x) = 1$, and for $x > 3$ is $f'(x) = 2px + q$.
Since $f'(x)$ is continuous at $x=3$, the left-hand derivative and right-hand derivative at $x=3$ must be equal.
$\lim_{x \to 3^-} f'(x) = 1$
$\lim_{x \to 3^+} f'(x) = 2p(3) + q = 6p + q$
Equating them, we get:
$6p + q = 1$
Substituting $q = -3p$ into the above equation:
$6p - 3p = 1 \Rightarrow 3p = 1 \Rightarrow p = \dfrac{1}{3}$
Answer: $\dfrac{1}{3}$
$\lim_{x \to 3^-} f(x) = 3 - 1 = 2$
$\lim_{x \to 3^+} f(x) = p(3)^2 + q(3) + 2 = 9p + 3q + 2$
Equating them, we get:
$9p + 3q + 2 = 2 \Rightarrow 9p + 3q = 0 \Rightarrow q = -3p$
The derivative of the function for $x \in (1, 3)$ is $f'(x) = 1$, and for $x > 3$ is $f'(x) = 2px + q$.
Since $f'(x)$ is continuous at $x=3$, the left-hand derivative and right-hand derivative at $x=3$ must be equal.
$\lim_{x \to 3^-} f'(x) = 1$
$\lim_{x \to 3^+} f'(x) = 2p(3) + q = 6p + q$
Equating them, we get:
$6p + q = 1$
Substituting $q = -3p$ into the above equation:
$6p - 3p = 1 \Rightarrow 3p = 1 \Rightarrow p = \dfrac{1}{3}$
Answer: $\dfrac{1}{3}$
Hence, the correct answer is option: 'C'.