Differential Calculus
Differential Calculus
star_batch_jee_advanced_2025
Grade None
Question:
The least value of $'u'$ for which the equation, $\frac{4}{\sin x} + \frac{1}{1 - \sin x} = u$ has atleast one solution on the interval $(0, \pi/2)$ is:
Step-by-Step Solution
Key Concept: Use the definition of derivative to handle absolute value functions by examining the limit from both sides.
We compute $f'(x) = \lim_{h\to 0}\frac{f(x+h)-f(x)}{h} = \lim_{h\to 0}\frac{t(h)+|x|h+h\delta^2}{h}$ where $t(h)$ is the change from the absolute value term. Since $f(0) = 0$ and $f'(0) = \lim_{h\to 0}\left(\frac{t(h)-t(0)}{h}+|x|\right)$, we find $f'(x) = f'(0) + |x| = |x|$ (assuming $f'(0)=0$).
Correct Answer: 2