Limits, Continuity & Differentiability
Differential Calculus-1
star_batch_jee_advanced_2025
Grade 12
Question:
In which of the following cases the given equations has atleast one root in the indicated interval?
$x - \cos x = 0$ in $(0, \pi/2)$
$x + \sin x = 1$ in $(0, \pi/6)$
$\frac{b}{x-1} + \frac{b}{x-3} = 0$, $a, b > 0$ in $(1, 3)$
$f(x) - g(x) = 0$ in $(a, b)$ where $f$ and $g$ are continuous on $[a, b]$ and $f(a) > g(a)$ and $f(b) < g(b)$
Step-by-Step Solution
Key Concept: The Intermediate Value Theorem guarantees a root exists when a continuous function changes sign over an interval.
For option (A), $f(x) = x - \cos x$ gives $f(0) 0$, so a root exists by IVT. For option (B), $f(x) = x + \sin x - 1$ gives $f(0) = -1 0$. For option (C), $f(x) = a(x-3) + b(x-1)$ on $[1,3]$ has $f(1) = -2a 0$, guaranteeing a root. For option (D), $h(x) = f(x) - g(x)$ satisfies $h(a) > 0$ and $h(b) < 0$, so a root exists. All four options have at least one root in the indicated interval.
Correct Answer: 1,2,3,4