Functions – JEE Main MathematicsPractice Questions & PYQs

Given the functional property $f(x+y)=f(x)f(y)$ and $f(1)=2$, the function must be $f(x)=2^x$. Substituting this into the sum $\sum_{k=1}^{10} f(a+k)$ gives $\sum_{k=1}^{10} 2^{a+k}$, which equals $2^a(2^1 + 2^2 + \dots + 2^{10})$. Using the geometric series formula, this simplifies to $2^a \cdot 2(2^{10}-1) = 2^{a+1}(2^{10}-1)$. Equating this result to $16(2^{10}-1)$, or $2^4(2^{10}-1)$, leads to the exponent relationship $a+1=4$, confirming that $a=3$.

The range of $f(x) = \frac{x}{1+x^2}$ can be determined by analyzing the function for positive and negative values of $x$. For $x > 0$, rewriting the function as $\frac{1}{x+1/x}$ and applying the arithmetic mean-geometric mean inequality shows that $x+1/x \geq 2$, restricting the range to $(0, 1/2]$. For $x=0$, $f(0)=0$, and because the function is odd, the range extends to $[-1/2, 0)$ for negative $x$, resulting in a total range of $[-1/2, 1/2]$.

A function is injective if distinct domain elements map to distinct values and surjective if its range covers the entire codomain. The expression $f(x) = \frac{2x}{x-1}$ can be rewritten as $f(x) = 2 + \frac{2}{x-1}$, and its derivative $f'(x) = -\frac{2}{(x-1)^2}$ is strictly negative for all $x$ in the domain, confirming it is one-one. However, the value $y = 2$ cannot be achieved because $\frac{2x}{x-1} = 2$ implies $2x = 2x - 2$, which is impossible, so the function is not surjective onto the real numbers.

The function $f(x)=x^2$ maps the interval $S=[0,4]$ to $f(S)=[0,16]$. The preimage set $g(S)$ is the set of all $x$ such that $x^2 \in [0,4]$, which simplifies to $[-2,2]$. Applying $f$ to this result gives $f(g(S))=[0,4]$, so the statement $f(g(S))=S$ is true. However, the composition $g(f(S))$ evaluates to the set of all $x$ such that $x^2 \in [0,16]$, which is $[-4,4]$. Since this set does not equal $g(S)=[-2,2]$, the statement $g(f(S))=g(S)$ is not true.

Using trigonometric power reduction formulas, $\sin^4 x + \cos^4 x$ simplifies to $1 - 2 \sin^2 x \cos^2 x$, which equals $1 - \frac{1}{2} \sin^2 2x$, while $\sin^6 x + \cos^6 x$ simplifies to $1 - 3 \sin^2 x \cos^2 x$, or $1 - \frac{3}{4} \sin^2 2x$. Substituting these into the expressions for $f_4(x)$ and $f_6(x)$ yields $f_4(x) = \frac{1}{4} - \frac{1}{8} \sin^2 2x$ and $f_6(x) = \frac{1}{6} - \frac{1}{8} \sin^2 2x$. Subtracting these values cancels the sine terms, leaving $\frac{1}{4} - \frac{1}{6}$, which results in $\frac{1}{12}$.

Invertibility of a function $f: A \to B$ requires it to be a bijection, which means every element in the codomain has a unique pre-image in the domain. By setting $y = \frac{x-1}{x-2}$ and solving for $x$ in terms of $y$, we perform the algebraic manipulation $y(x-2) = x-1$, resulting in $xy - 2y = x-1$, which simplifies to $x(y-1) = 2y-1$. Isolating $x$ yields $f^{-1}(y) = \frac{2y-1}{y-1}$, demonstrating that a unique pre-image exists for every $y$ in the codomain.

A function is surjective if its range covers the entire codomain and injective if every output maps to a unique input. Analyzing $f(x) = \frac{x}{1+x^2}$ reveals that the function reaches all values in the interval $[-\frac{1}{2}, \frac{1}{2}]$, confirming surjectivity. However, since the function satisfies $f(x) = f(\frac{1}{x})$, distinct inputs can produce the same output, meaning the function fails the injectivity test.

The composition $(f \circ g)(x) = x$ requires substituting the expression for $g(x)$ into $f(x)$ and solving for $x$. By substituting $f(g(x)) = 2^{10}(3^{10}x - 1) + 1 = x$, we isolate $x$ as $\frac{2^{10}-1}{6^{10}-1}$. Dividing the numerator and denominator by $2^{10}$ transforms this fraction into the form $\frac{1-2^{-10}}{3^{10}-2^{-10}}$, which matches the provided result.

The function $f(x) = x - 5[\frac{x}{5}]$ calculates the remainder when a natural number is divided by 5. Because $f(5) = 5 - 5(1) = 0$ and $f(10) = 10 - 5(2) = 0$, the function maps distinct inputs to the same output, confirming it is not one-one. Furthermore, the range of the function is restricted to the set $\{0, 1, 2, 3, 4\}$, which does not encompass all natural numbers in the codomain, meaning it is not onto.

This problem involves a functional equation that identifies a quadratic relationship $f(x) = ax^{2+}bx+c$. By evaluating $f(x+1) - f(x) = x+3$ and comparing coefficients, we determine the function is $f(x) = \frac{1}{2}x^2 + \frac{5}{2}x$. Calculating the sum $\sum_{n=1}^{10} f(n)$ requires applying standard summation formulas for $n^2$ and $n$, resulting in $\frac{1}{2} \left( \frac{10(11)(21)}{6} + 5 \cdot \frac{10(11)}{2} \right) = 330$.

Repeated composition of $f_0(x) = \frac{1}{1-x}$ reveals a cyclic pattern where $f_2(x) = x$, establishing a period of three for the sequence of functions. Using this periodicity, we simplify the terms $f_{100}(3) = f_1(3) = \frac{2}{3}$, $f_1(\frac{2}{3}) = 1 - \frac{3}{2} = -\frac{1}{2}$, and $f_2(\frac{3}{2}) = \frac{3}{2}$. Summing these results yields $\frac{2}{3} - \frac{1}{2} + \frac{3}{2} = \frac{5}{3}$.

This functional equation can be solved by creating a system of two equations with $f(x)$ and $f(\frac{1}{x})$ as variables, allowing us to isolate $f(x) = \frac{2}{x} - x$. The condition $f(x) = f(-x)$ translates to $\frac{2}{x} - x = -(\frac{2}{x} - x)$, which simplifies to $\frac{2}{x} - x = -\frac{2}{x} + x$, reducing to $x^2 = 2$. Solving this quadratic equation provides two real roots, $\pm\sqrt{2}$, confirming that the set $S$ contains exactly two elements.

The domain of the expression $\frac{1}{\sqrt{|x|-x}}$ depends on the condition that the quantity inside the square root must be strictly positive, meaning $|x|-x > 0$. Because the absolute value of $x$ is always greater than or equal to $x$ for all real numbers, this inequality simplifies to the requirement that $|x| > x$, which is satisfied only when $x < 0$. Consequently, the domain is the set of all negative real numbers, denoted as $(-\infty , 0)$.

A function is one-to-one if its derivative maintains a consistent sign, indicating it is strictly monotonic. Calculating the derivative gives $f'(x) = 3x^2 + 5$, which is strictly positive for all real $x$, confirming the function is strictly increasing and therefore one-to-one. Furthermore, as $x$ ranges from negative to positive infinity, the cubic polynomial outputs all real values, satisfying the requirement for the function to be onto.

The intersection of a function and its inverse occurs where $f(x) = x$ for an increasing function. Solving $(x+1)^2 - 1 = x$ leads to $x^2 + 2x = x$, which simplifies to $x(x+1) = 0$, yielding the solutions $x = 0$ and $x = -1$. While these points confirm the set for the first statement, the second statement is false because the problem does not define the codomain to be equal to the range, failing the condition required for a bijection.

To determine the inverse of a function, one must solve the equation $y = f(x)$ for $x$ in terms of $y$. Beginning with $y = 4x + 3$, subtracting 3 from both sides and then dividing by 4 isolates $x$ to yield $x = \frac{y-3}{4}$. This algebraic rearrangement identifies the inverse function as $g(y) = \frac{y-3}{4}$.

The overall domain is determined by the intersection of the domains of individual components within the interval $(-\frac{\pi}{2}, \frac{\pi}{2})$. The term $\cos^{-1}\left(\frac{x}{2}-1\right)$ is defined when $-1 \le \frac{x}{2}-1 \le 1$, which simplifies to $0 \le x \le 4$. Since $\log(\cos x)$ requires $\cos x > 0$, which holds true for all $x$ in the interval $(-\frac{\pi}{2}, \frac{\pi}{2})$, the final domain is the intersection of these two sets, resulting in $[0, \frac{\pi}{2})$.

The expression f(x) = an^{-1}rac{2x}{1-x^2} simplifies to $2 an^{-1}(x)$ for the interval $x imes (-1,1)$. As $x$ ranges from $-1$ to $1$, $an^{-1}(x)$ varies within (-rac{rac{\pi}{4}}, rac{rac{\pi}{4}}), which scales the range of $f(x)$ to (-rac{rac{\pi}{2}}, rac{rac{\pi}{2}}). For the function to be onto, the codomain $B$ must match this calculated range.

The functional equation $f(x-y)=f(x)f(y)-f(a-x)f(a+y)$ allows for specific substitutions to determine functional properties. By setting $y=x-a$ and utilizing the given $f(0)=1$, one can derive the relationship $f(2a-x) = -f(x)$. This specific form demonstrates how the function behaves under reflection across the constant $a$.

Symmetry about a vertical line $x=k$ implies that the function values at points equidistant from the line are identical, mathematically expressed as $f(k-h) = f(k+h)$. For the line $x=2$, substituting $h=x$ directly yields the identity $f(2-x) = f(2+x)$. This geometric condition ensures the graph maintains identical shape on both sides of the axis of symmetry.