Functions – JEE Main MathematicsPractice Questions & PYQs

The inverse of a function is determined by isolating the independent variable, yielding $f^{-1}(x) = \frac{3x-2}{x-1}$ and $g^{-1}(x) = \frac{x+3}{2}$. Substituting these into the equation $\frac{3x-2}{x-1} + \frac{x+3}{2} = \frac{13}{2}$ leads to the quadratic equation $x^2 - 5x + 6 = 0$ after algebraic simplification. The roots of this quadratic are $x=2$ and $x=3$, and their sum is $5$.

The greatest integer function [x] is piecewise constant, so we examine the function f(x) on the intervals (1, 2) and [2, 3) where [x] takes the values 1 and 2, respectively. For x in (1, 2), f(x) = x/(1+x^2), which decreases from a limit of 1/2 to 2/5. For x in $[2, 3), f(x) = 2x/(1+x^2), which decreases from 4/5 to a limit of 3/5. Combining the ranges from these intervals, the full range of the function is (2/5, 1/2) union (3/5, 4/5]$.

To find the inverse, set y equal to the expression f(x) and simplify it as y = (8^{4x}-1)/(8^{4x}+1). Rearranging this equation to solve for 8^{4x} yields 8^{4x} = (1+y)/(1-y). Taking the natural logarithm on both sides and solving for x gives 4x * ln(8) = ln((1+y)/(1-y)), which simplifies to x = (1/4) * log_8(e) * ln((1+y)/(1-y)). Substituting x for y, the inverse function is (1/4)(log_8 e)ln((1+x)/(1-x)).

Properties of determinants allow for row operations to be applied without changing the value of the determinant. By applying the transformation $R_1 \rightarrow R_1 + R_3 - 2R_2$, the first row becomes $(a+c-2b, 0, 0)$, and because $a-2b+c=1$, this row simplifies to $(1, 0, 0)$. Expanding the determinant along this row reduces it to a $2 \times 2$ minor, $((x+3)^2 - (x+2)(x+4))$, which simplifies to $x^2 + 6x + 9 - (x^2 + 6x + 8) = 1$. Consequently, the function $f(x)$ is a constant 1 regardless of $x$, meaning $f(50) = 1$.

The Cauchy additive functional equation $f(x+y) = f(x) + f(y)$ characterizes linear functions of the form $f(x) = c \cdot x$, where $c$ is determined by $f(1) = 2$. Substituting this into the sum definition $g(n) = \sum_{k=1}^{n-1} 2k$, we apply the arithmetic series formula to reach $g(n) = 2 \cdot \frac{(n-1)n}{2} = n(n-1)$. Setting this expression equal to 20 leads to the quadratic equation $n^2 - n - 20 = 0$, which factors as $(n-5)(n+4) = 0$, yielding the valid solution $n=5$.

The Cauchy functional equation $f(x+y) = f(x)f(y)$ implies that the function takes the exponential form $f(x) = [f(1)]^x$. The given infinite sum $\sum_{x=1}^{\infty } f(x) = 2$ represents a geometric series $f(1) + f(1)^2 + f(1)^3 + \dots = 2$, which converges to the formula $\frac{f(1)}{1-f(1)} = 2$. Solving for $f(1)$ yields $f(1) = \frac{2}{3}$, and since the required ratio $\frac{f(4)}{f(2)}$ simplifies to $\frac{f(1)^4}{f(1)^2} = f(1)^2$, substituting the calculated value provides $(\frac{2}{3})^2 = \frac{4}{9}$.

The composition g(f(x)) is defined as (f(x))^2 + f(x) - 1, which must equal 4x^2 - 10x + 5. By completing the square for the expression 4x^2 - 10x + 5, we can rewrite it as (2x - 2.5)^2 - 1.25. Setting (f(x) + 0.5)^2 - 0.25 equal to (2x - 2.5)^2 - 1.25 reveals that f(x) = 2 - 2x. Evaluating this at x = 5/4 gives f(5/4) = 2 - 2(5/4) = -0.5, or -1/2.

The Cauchy functional equation $f(x+y) = f(x)f(y)$ with $f(1) = 3$ identifies $f(n)$ as an exponential function, specifically $f(n) = 3^n$. The problem requires solving the geometric series sum $3^1 + 3^2 + \dots + 3^n = 363$. Applying the sum formula $\frac{3(3^n - 1)}{3 - 1} = 363$ leads to $3^n - 1 = 242$, which simplifies to $3^n = 243$, confirming that $n = 5$.

To find the number of functions $f: A \rightarrow B$ that include 2 in the range and are not one-one, we subtract the number of injective functions that include 2 in the range from the total number of functions that include 2 in the range. The total number of functions mapping to 2 is calculated by subtracting functions that avoid 2 ($3^3 = 27$) from the total functions ($4^3 = 64$), resulting in 37. To count the injective functions that must include 2, we choose one element from $A$ to map to 2 in $\binom{3}{1} = 3$ ways, and map the remaining two elements to the remaining three elements of $B$ in $3 \times 2 = 6$ ways, totaling 18 injective functions. Subtracting these 18 from 37 gives 19.

The composition of a function with itself, defined as $(f \circ f)(x) = f(f(x))$, involves substituting the original function expression into its own variable. For $f(x) = \frac{a-x}{a+x}$, computing the composition yields $f(f(x)) = \frac{a - \frac{a-x}{a+x}}{a + \frac{a-x}{a+x}}$, which simplifies to $\frac{x(a+1) + a(a-1)}{x(a-1) + a(a+1)}$. Requiring this to equal $x$ for all valid $x$ forces the coefficients of $x$ to align such that $a=1$, simplifying the function to $f(x) = \frac{1-x}{1+x}$. Substituting the input $-\frac{1}{2}$ into this result gives $f(-\frac{1}{2}) = \frac{1 - (-0.5)}{1 + (-0.5)} = \frac{1.5}{0.5} = 3$.

The function $f(x)$ is defined when the logarithm argument is positive and the denominator is non-zero. The condition $x^{3-}x > 0$ simplifies to $x(x-1)(x+1) > 0$, which restricts $x$ to the intervals $(-1,0) \cup (1, \infty )$. Additionally, the denominator $4-x^2 \neq 0$ requires $x \neq 2$ and $x \neq -2$. Combining these conditions results in the domain $(-1,0) \cup (1,2) \cup (2, \infty )$.

The set of inputs consists of numbers $1$ through $20$, with multiples of $4$ being $\{4, 8, 12, 16, 20\}$ and multiples of $3$ being $\{3, 6, 9, 12, 15, 18\}$. The $5$ multiples of $4$ must map to the $6$ multiples of $3$ in an injective manner; we arrange these $5$ inputs into the $6$ available outputs in $6 \times 5 \times 4 \times 3 \times 2$ ways, which is $6!$ ways. The remaining $15$ elements from the domain can then be mapped to the remaining $15$ elements of the codomain in $15!$ ways, leading to a total of $6! \times 15!$ functions.

A function is injective if distinct inputs always yield distinct outputs. Here, for $f(x) = |1 - 1/x|$, evaluating the function at $x = 2/3$ gives $f(2/3) = |1-1.5| = 0.5$, and at $x = 2$ gives $f(2) = |1-0.5| = 0.5$; since $f(2/3) = f(2)$ but $2/3 \neq 2$, the function is not injective. Regarding surjectivity, the codomain is given as $(0, \infty )$, but the function outputs $0$ at $x = 1$, and since $0$ is not in the codomain, the function cannot be surjective, making it neither injective nor surjective.

The composition phi(x) = (h o f o g)(x) is found by substituting g(x) = tan x into f(x) = sqrt(x) and then into h(x) = (1-x^2)/(1+x^2), resulting in (1-tan x)/(1+tan x). This expression simplifies to tan(pi/4 - x) using the tangent subtraction formula. Evaluating at x = pi/3 gives tan(pi/4 - pi/3) = tan(-pi/12), which is equivalent to tan(11pi/12) because tan(pi - theta) = -tan theta.

A function h(x) is increasing where its derivative h'(x) >= 0. Applying the chain rule to the composition f(g(x)), the derivative is h'(x) = f'(g(x))g'(x) = (e^{x^{2-}x}-1)(2x-1). This product is non-negative if both factors are non-negative, occurring when x >= 0.5 and x^{2-}x >= 0, or if both are non-positive, occurring when x <= 0.5 and x^{2-}x <= 0. Solving these inequalities yields the combined interval [0, 0.5] union [1, infinity).

Composite functions follow the structure $(f_2 \circ J \circ f_1)(x) = f_2(J(f_1(x))) = f_3(x)$, where each function is applied sequentially. Substituting the given expressions, we have $1 - J(\frac{1}{x}) = \frac{1}{1-x}$, which rearranges to $J(\frac{1}{x}) = 1 - \frac{1}{1-x} = \frac{-x}{1-x} = \frac{x}{x-1}$. By substituting $t = \frac{1}{x}$, which implies $x = \frac{1}{t}$, we find $J(t) = \frac{1/t}{1/t - 1} = \frac{1}{1-t}$. This identifies $J(x)$ as $\frac{1}{1-x}$, which is exactly $f_3(x)$.

Functional composition $(f \circ g)(n)$ requires determining the output of the inner function first, then applying the outer function to that result. Given $g(n) = n - (-1)^n$, if $n$ is even, $g(n) = n-1$ (which is odd), and if $n$ is odd, $g(n) = n+1$ (which is even). Applying the definition of $f(n)$ to these cases leads to $f(g(n)) = \frac{(n-1)+1}{2} = \frac{n}{2}$ for even $n$, and $f(g(n)) = \frac{(n+1)}{2}$ for odd $n$. Since $f(g(1)) = 1$ and $f(g(2)) = 1$, the function is many-one, but because every natural number in the codomain can be reached by some input $n$, it is onto.

To determine the range $A$, we set $y = \frac{x^2}{1-x^2}$ and express $x^2$ in terms of $y$, yielding $x^2(1+y) = y$ or $x^2 = \frac{y}{1+y}$. Since $x^2$ must be non-negative for all real $x$ in the domain, we require $\frac{y}{1+y} \geq 0$. This inequality holds for $y \in (-\infty , -1) \cup [0, \infty )$, meaning the range of the function is all real numbers except those in the interval $[-1, 0)$.

The function $f(x)=a^x$ is expressed as the sum of its even part $f_1(x) = \frac{a^x+a^{-x}}{2}$ and odd part $f_2(x) = \frac{a^x-a^{-x}}{2}$. Expanding the expression $f_1(x+y) + f_1(x-y)$ using these definitions results in $\frac{1}{2}(a^{x+y} + a^{-x-y} + a^{x-y} + a^{-x+y})$. This expression factors cleanly into $\frac{1}{2}(a^x+a^{-x})(a^y+a^{-y})$, which simplifies directly to $2f_1(x)f_1(y)$.

The logarithmic property $\log(A/B) = \log A - \log B$ applies to the function $f(x) = \log_e(1-x) - \log_e(1+x)$. Substituting $x$ with $\frac{2x}{1+x^2}$ yields $f\left(\frac{2x}{1+x^2}\right) = \log_e\left(\frac{1-\frac{2x}{1+x^2}}{1+\frac{2x}{1+x^2}}\right)$, which simplifies to $\log_e\left(\frac{1+x^{2-}2x}{1+x^{2+}2x}\right)$ or $\log_e\left(\left(\frac{1-x}{1+x}\right)^2\right)$. Using the power rule for logarithms, this simplifies directly to $2 \log_e\left(\frac{1-x}{1+x}\right)$, which is $2f(x)$.