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Inverse and Periodic Functions — JEE Main Maths PYQ MCQs with Solutions
Free JEE Main Maths PYQ Inverse and Periodic Functions MCQs with step-by-step solutions (8 questions). Part of Functions. Practise online on Prepizo — no login needed.
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Questions with solutions
Q1 — Inverse and Periodic Functions · hard · theory
$X$ and $Y$ are two non-empty sets where $f: X \to Y$ is a function defined such that $f(C) = \{f(x): x \in C\}$ for $C \subseteq X$ and $f^{-1}(D) = \{x: f(x) \in D\}$ for $D \subseteq Y$. For any $A \subseteq X$ and $B \subseteq Y$, then
A. $f^{-1}\{f(A)\} = A$
B. $f^{-1}\{f(A)\} = A$, only if $f(X) = Y$
C. $f\{f^{-1}(B)\} = B$, only if $B \subseteq f(X)$ ✓ Correct
D. $f\{f^{-1}(B)\} = B$
Solution: By the given definitions, $f\{f^{-1}(B)\} = B$ holds only when $B \subseteq f(X)$ (the range of $f$); in general $f\{f^{-1}(B)\} \subseteq B$. So option (c) is the correct statement.
Q2 — Inverse and Periodic Functions · medium · theory
If $f(x) = \sin x + \cos x$, $g(x) = x^2 - 1$, then $g\{f(x)\}$ is invertible in the domain
A. $\left[0, \dfrac{\pi}{2}\right]$
B. $\left[-\dfrac{\pi}{4}, \dfrac{\pi}{4}\right]$ ✓ Correct
C. $\left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$
D. $[0, \pi]$
Solution: $g(f(x)) = (\sin x + \cos x)^2 - 1 = \sin 2x$, which is bijective when $2x \in \left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$, i.e. $x \in \left[-\dfrac{\pi}{4}, \dfrac{\pi}{4}\right]$.
Q3 — Inverse and Periodic Functions · medium · theory
Suppose $f(x) = (x-1)^2$ for $x \ge 1$. If $g(x)$ is the function whose graph is the reflection of the graph of $f(x)$ with respect to the line $y = x$, then $g(x)$ equals
A. $\sqrt{x} + 1,\ x \ge 0$ ✓ Correct
B. $\dfrac{1}{(x-1)^2},\ x \ne 1$
C. $\sqrt{x-1},\ x \ge 1$
D. $\sqrt{x} - 1,\ x \ge 0$
Solution: The reflection of $f$ about $y = x$ is $f^{-1}$. Let $y = (x-1)^2, x \ge 1 \Rightarrow \sqrt{y} = x - 1 \Rightarrow x = \sqrt{y} + 1$. So $f^{-1}(x) = g(x) = \sqrt{x} + 1$, $x \ge 0$.
Q4 — Inverse and Periodic Functions · medium · theory
If $f: [1, \infty) \to [2, \infty)$ is given by $f(x) = x + \dfrac{1}{x}$, then $f^{-1}(x)$ equals
A. $\dfrac{x + \sqrt{x^2 - 4}}{2}$ ✓ Correct
B. $\dfrac{x}{1 + x^2}$
C. $\dfrac{x - \sqrt{x^2 - 4}}{2}$
D. $1 + \sqrt{x^2 - 4}$
Solution: Let $y = x + \dfrac{1}{x} \Rightarrow x^2 - xy + 1 = 0 \Rightarrow x = \dfrac{y + \sqrt{y^2 - 4}}{2}$ (taking the root giving range $[1,\infty)$). So $f^{-1}(x) = \dfrac{x + \sqrt{x^2 - 4}}{2}$.
Q5 — Inverse and Periodic Functions · hard · theory
If the function $f: [1, \infty) \to [1, \infty)$ is defined by $f(x) = 2^{x(x-1)}$, then $f^{-1}(x)$ is
A. $\left(\dfrac{1}{2}\right)^{x(x-1)}$
B. $\dfrac{1}{2}\left(1 + \sqrt{1 + 4\log_2 x}\right)$ ✓ Correct
C. $\dfrac{1}{2}\left(1 - \sqrt{1 + 4\log_2 x}\right)$
D. not defined
Solution: Let $y = 2^{x(x-1)}$. Taking $\log_2$: $\log_2 y = x^2 - x \Rightarrow x^2 - x - \log_2 y = 0 \Rightarrow x = \dfrac{1 + \sqrt{1 + 4\log_2 y}}{2}$ (rejecting the negative-sign root since $x \ge 1$). So $f^{-1}(x) = \dfrac{1}{2}\left(1 + \sqrt{1 + 4\log_2 x}\right)$.
Q6 — Inverse and Periodic Functions · easy · theory
If $f(x) = 3x - 5$, then $f^{-1}(x)$
A. is given by $\dfrac{1}{3x - 5}$
B. is given by $\dfrac{x + 5}{3}$ ✓ Correct
C. does not exist because $f$ is not one-one
D. does not exist because $f$ is not onto
Solution: Let $y = 3x - 5 \Rightarrow x = \dfrac{y + 5}{3}$. So $f^{-1}(x) = \dfrac{x + 5}{3}$.
Q7 — Inverse and Periodic Functions · easy · theory
Which of the following functions is periodic?
A. $f(x) = x - [x]$, where $[x]$ denotes the greatest integer less than or equal to the real number $x$ ✓ Correct
B. $f(x) = \sin\left(\dfrac{1}{x}\right)$ for $x \ne 0$, $f(0) = 0$
C. $f(x) = x \cos x$
D. None of the above
Solution: $f(x) = x - [x] = \{x\}$, the fractional part function, satisfies $f(x+1) = f(x)$ for all $x$, so it is periodic with period $1$. The functions $\sin(1/x)$ and $x\cos x$ are not periodic.
Q8 — Inverse and Periodic Functions · hard · theory
Let $f: \left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right) \to R$ be given by $f(x) = [\log(\sec x + \tan x)]^3$. Then,
A. $f(x)$ is an odd function ✓ Correct
B. $f(x)$ is a one-one function ✓ Correct
C. $f(x)$ is an onto function ✓ Correct
D. $f(x)$ is an even function
Solution: $f(-x) = [\log(\sec x - \tan x)]^3 = [-\log(\sec x + \tan x)]^3 = -f(x)$, so $f$ is odd. Also $f'(x) = 3\sec x\,[\log(\sec x + \tan x)]^2 > 0$ on $\left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)$, so $f$ is strictly increasing, hence one-one. As $x \to \pm\dfrac{\pi}{2}$, $\sec x + \tan x \to \infty$ or $0^+$, so $\log(\sec x+\tan x) \to \pm\infty$ and $f(x) \to \pm\infty$, giving range $R$, so $f$ is onto. It is not even.