Q66 / 180 · COMEDK 2026 (9 May Shift 2) · Mathematics
If $\lim\limits_{x \to 0}\left(\dfrac{p\sin 2x + 1 - \cos 2x}{x + \tan x}\right) = 1$ then the value of '$p$' is
- A$2$
- B$-1$
- C$\dfrac{1}{2}$
- Correct answerD$1$
Solution
The given limit is $\lim\limits_{x \to 0}\left(\dfrac{p\sin 2x + 1 - \cos 2x}{x + \tan x}\right) = 1$.
Substituting $x = 0$, we get the form $\dfrac{0}{0}$.
Applying L'Hospital's rule by differentiating the numerator and the denominator with respect to $x$:
$\lim\limits_{x \to 0}\left(\dfrac{2p\cos 2x + 2\sin 2x}{1 + \sec^2 x}\right) = 1$
Substituting $x = 0$ into the expression:
$\dfrac{2p\cos(0) + 2\sin(0)}{1 + \sec^2(0)} = 1$
$\dfrac{2p(1) + 0}{1 + 1} = 1$
$\dfrac{2p}{2} = 1$
$p = 1$
Answer: $1$
Substituting $x = 0$, we get the form $\dfrac{0}{0}$.
Applying L'Hospital's rule by differentiating the numerator and the denominator with respect to $x$:
$\lim\limits_{x \to 0}\left(\dfrac{2p\cos 2x + 2\sin 2x}{1 + \sec^2 x}\right) = 1$
Substituting $x = 0$ into the expression:
$\dfrac{2p\cos(0) + 2\sin(0)}{1 + \sec^2(0)} = 1$
$\dfrac{2p(1) + 0}{1 + 1} = 1$
$\dfrac{2p}{2} = 1$
$p = 1$
Answer: $1$
Hence, the correct answer is option: 'D'.