Tangent Function Graph

How the real exam tests it

Questions from real CSCA papers that test this concept. Work each one out before you open the answer.

1. CSCA Math January 2026 Past Paper
   About the tangent function $y=\tan x$, which statement is correct?
   • A．The range is $[-1,1]$
   • B．The domain is $\mathbf{R}$
   • C．The function is even
   • D．The minimum positive period is $\pi$
   Show answer & solution

   Answer: D

   Answer: D The domain of $y=\tan x$ is not all real numbers; $x=\frac{\pi}{2}+k\pi$ must be excluded. Its range is $\mathbf{R}$, it is odd, and its minimum positive period is $\pi$. Therefore D is correct.
   Take the full CSCA Math January 2026 Past Paper — timed & graded
2. CSCA Math March 2026 Past Paper
   Which description of $f ( x ) = \tan x$ is INCORRECT?
   • A．$f ( x )$ satisfies $f ( - x ) = - f ( x )$ on its domain
   • B．$f ( x )$ is an even function
   • C．$f ( x )$ is monotonically increasing on $( - \pi / 2 , \pi / 2 )$
   • D．$f ( x )$ has exactly one zero on ( $- \pi / 2 , \pi / 2$ )
   Show answer & solution

   Answer: B

   tan $x$ is an odd function satisfying $f ( - x ) = - f ( x )$, not an even function. $C$ and $D$ correctly describe properties of $\tan \mathrm { x }$.
   Take the full CSCA Math March 2026 Past Paper — timed & graded
3. CSCA Math April 2026 Past Paper
   The minimum positive period of $y = 2 \tan \left( 3 x + \frac { \pi } { 4 } \right)$ is
   • A．$\frac { \pi } { 3 }$
   • B．$2 \pi$
   • C．$\frac { \pi } { 4 }$
   • D．$\pi$
   Show answer & solution

   Answer: A

   For $\tan ( \omega x + \varphi )$, minimum positive period is $\frac { \pi } { | \omega | }$. Here $\omega = 3$, so $T = \frac { \pi } { 3 }$.
   Take the full CSCA Math April 2026 Past Paper — timed & graded