Q59 / 150 · MHT CET 2026 (15 April Shift 1) · Mathematics
If the function f is continuous at $x = 1$, where $f(x) = \dfrac{1 + \cos(\pi x)}{\pi(1-x)^2}$, for $x \neq 1$, then the value of $f(1)$ is....
- Correct answerA$\dfrac{\pi}{2}$
- B$\dfrac{\pi}{4}$
- C$\dfrac{\pi}{6}$
- D$\dfrac{\pi}{9}$
Solution
Since $f(x)$ is continuous at $x = 1$, $f(1) = \lim_{x \rightarrow 1} f(x)$.
$f(1) = \lim_{x \rightarrow 1} \dfrac{1 + \cos(\pi x)}{\pi(1-x)^2}$
This is a $\dfrac{0}{0}$ form. Applying L'Hospital's rule:
$f(1) = \lim_{x \rightarrow 1} \dfrac{-\pi \sin(\pi x)}{-2\pi(1-x)}$
$f(1) = \lim_{x \rightarrow 1} \dfrac{\sin(\pi x)}{2(1-x)}$
This is again a $\dfrac{0}{0}$ form. Applying L'Hospital's rule once more:
$f(1) = \lim_{x \rightarrow 1} \dfrac{\pi \cos(\pi x)}{-2}$
Substituting $x = 1$:
$f(1) = \dfrac{\pi \cos(\pi)}{-2} = \dfrac{\pi(-1)}{-2} = \dfrac{\pi}{2}$
Answer: $\dfrac{\pi}{2}$
$f(1) = \lim_{x \rightarrow 1} \dfrac{1 + \cos(\pi x)}{\pi(1-x)^2}$
This is a $\dfrac{0}{0}$ form. Applying L'Hospital's rule:
$f(1) = \lim_{x \rightarrow 1} \dfrac{-\pi \sin(\pi x)}{-2\pi(1-x)}$
$f(1) = \lim_{x \rightarrow 1} \dfrac{\sin(\pi x)}{2(1-x)}$
This is again a $\dfrac{0}{0}$ form. Applying L'Hospital's rule once more:
$f(1) = \lim_{x \rightarrow 1} \dfrac{\pi \cos(\pi x)}{-2}$
Substituting $x = 1$:
$f(1) = \dfrac{\pi \cos(\pi)}{-2} = \dfrac{\pi(-1)}{-2} = \dfrac{\pi}{2}$
Answer: $\dfrac{\pi}{2}$
Hence, the correct answer is option: 'A'.