Q59 / 150 · MHT CET 2026 (16 April Shift 2) · Mathematics

If the function f is continuous at $x = \pi$, where $f(x) = \dfrac{1 - \cos[7(x - \pi)]}{5(x - \pi)^2}$, for $x \neq \pi$, then $f(\pi) = $

  • A
    $\dfrac{49}{4}$
  • B
    $\dfrac{4}{49}$
  • Correct answer
    C
    $\dfrac{49}{10}$
  • D
    $\dfrac{10}{49}$

Solution

Since the function $f(x)$ is continuous at $x = \pi$, $f(\pi) = \lim_{x \rightarrow \pi} f(x)$.

Substituting $t = x - \pi$, as $x \rightarrow \pi$, $t \rightarrow 0$.

$f(\pi) = \lim_{t \rightarrow 0} \dfrac{1 - \cos(7t)}{5t^2}$

Using the identity $1 - \cos \theta = 2 \sin^2\left(\dfrac{\theta}{2}\right)$:

$f(\pi) = \lim_{t \rightarrow 0} \dfrac{2 \sin^2\left(\dfrac{7t}{2}\right)}{5t^2}$

$f(\pi) = \dfrac{2}{5} \lim_{t \rightarrow 0} \left( \dfrac{\sin\left(\dfrac{7t}{2}\right)}{t} \right)^2$

$f(\pi) = \dfrac{2}{5} \times \left(\dfrac{7}{2}\right)^2 \lim_{t \rightarrow 0} \left( \dfrac{\sin\left(\dfrac{7t}{2}\right)}{\dfrac{7t}{2}} \right)^2$

Since $\lim_{\theta \rightarrow 0} \dfrac{\sin \theta}{\theta} = 1$:

$f(\pi) = \dfrac{2}{5} \times \dfrac{49}{4} \times 1 = \dfrac{49}{10}$

Answer: $\dfrac{49}{10}$

Hence, the correct answer is option: 'C'.

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