Sets, Relations & Functions
Functions
star_batch_jee_advanced_2025
Grade 11
Question:
The equation $\sin(\cos x) = x$ has only one root $x_1$ in $(0, \pi/2)$ and the equation $\cos(\sin x) = x$ has also only one root $x_2$ in $(0, \pi/2)$. Then:
$x_1 > x_2$
$x_1 < x_2$
$x_1 = x_2$
$x_1 = 2x_2$
Step-by-Step Solution
Key Concept: Composite trigonometric functions are bounded and their graphs can be compared with linear and simple trigonometric functions.
From the graph, three curves are shown: $y = x$, $y = \cos(\sin x)$, and $y = \sin(\cos x)$. At the intersection point marked $x_1, x_2, 1$, we need to identify which curve corresponds to which equation. The curve $y = x$ is the straight line. By analyzing the behavior near the origin and comparing slopes, $y = \cos(\sin x)$ lies above $y = \sin(\cos x)$ in the visible region, and both are bounded by the line $y = x$.
Correct Answer: 2