CUET UG 2025 — Mathematics Calculus

The function $f(x) = \tan x - x$f(x)=tanx−x is defined on the interval $[0, \pi/2)$[0,π/2).

To determine if the function is increasing or decreasing, find the derivative:

$$f'(x) = \sec^2 x - 1$$f′(x)=sec2x−1

---

Using the identity $\sec^2 x = 1 + \tan^2 x$sec2x=1+tan2x:

$$f'(x) = 1 + \tan^2 x - 1$$f′(x)=1+tan2x−1

$$f'(x) = \tan^2 x$$f′(x)=tan2x

---

For any real number, the square is non-negative:

$\tan^2 x \geq 0$tan2x≥0 for all $x$x in $[0, \pi/2)$[0,π/2)

$\tan^2 x = 0$tan2x=0 only when $x = 0$x=0

$\tan^2 x > 0$tan2x>0 for all $x$x in $(0, \pi/2)$(0,π/2)

Therefore $f'(x) \geq 0$f′(x)≥0 on $[0, \pi/2)$[0,π/2).

---

Since $f'(x) > 0$f′(x)>0 for $x \in (0, \pi/2)$x∈(0,π/2), the function $f(x) = \tan x - x$f(x)=tanx−x is increasing on $[0, \pi/2)$[0,π/2).