Functions – JEE Main MathematicsPractice Questions & PYQs

The comparison $e^\pi$ vs $\pi^e$ follows from the property of the function $h(x) = \frac{\ln x}{x}$, which attains its absolute maximum at $x=e$. Since $h(e) = 1/e$ is the maximum value, for any other $x > 0$ including $x = \pi$, we have $\frac{\ln \pi}{\pi} < \frac{1}{e}$. This inequality simplifies to $e \ln \pi < \pi$, which is equivalent to $\ln \pi^e < \ln e^\pi$, confirming that $\pi^e < e^\pi$.

The range of the composite function $f(g(x))$ is determined by the output values of $g(x)$ serving as the input for $f(x)$. Analyzing $g(x)$, we see its output spans the interval $[0, 3]$ for the domain $[-3, 1]$. Substituting this into the definition of $f(x)$ for the interval $[0, 3]$, which is $1 - \frac{x}{3}$, the values vary from $1$ down to $0$, establishing the range as $[0, 1]$.

The domain of a composite function $f(g(x))$ is the set of all $x$ in the domain of $g$ such that $g(x)$ lies within the domain of $f$. Since the domain of $f(x)$ excludes $x = -1/2$ and the domain of $g(x)$ excludes $x = -5/2$, we check if $g(x) = -1/2$ has any solutions. Solving $\frac{|x|+1}{2x+5} = -1/2$ yields no valid solutions for $x$, meaning the condition $g(x) \ne -1/2$ is always satisfied for the domain of $g$. Thus, the domain remains restricted only by the original definition of $g(x)$, which is $R \setminus \{-5/2\}$.

The functional equation $f(m+n) = f(m)+f(n)$ with $f(1)=1$ characterizes the identity function $f(x)=x$ for all $x \in N$. Substituting this into the sum, we get $\sum_{k=1}^{2022} (\lambda + k) = 2022\lambda + \frac{2022 \times 2023}{2} \leq (2022)^2$. Simplifying this inequality results in $\lambda + 1011.5 \leq 2022$, or $\lambda \leq 1010.5$. The largest natural number $\lambda$ satisfying this condition is 1010.

Utilizing the identity $a = [a] + \{a\}$, the equation becomes $|2a-1| = 3[a] + 2(a - [a])$, which simplifies to $|2a-1| = [a] + 2a$. Considering the case a < rac{1}{2}, the absolute value expression negates to $1 - 2a = [a] + 2a$, leading to a = rac{1 - [a]}{4}. Given the floor function property $[a] = n$, we solve the inequality n \le rac{1 - n}{4} < n + 1 to find $n = 0$, yielding a = rac{1}{4}. Since this is the only solution, the sum is rac{1}{4}, which equals 18 when multiplied by 72.

The expression $\sin 3x + \cos 3x$ fluctuates between $-\sqrt{2}$ and $\sqrt{2}$, forcing the denominator $2 + \sin 3x + \cos 3x$ into the interval $[2-\sqrt{2}, 2+\sqrt{2}]$. Taking the reciprocal of these bounds, the range of the function is $[a, b] = [\frac{1}{2+\sqrt{2}}, \frac{1}{2-\sqrt{2}}]$. Calculating the arithmetic mean $\alpha = \frac{a+b}{2}$ results in 1, while the geometric mean $\beta = \sqrt{ab}$ simplifies to $\sqrt{\frac{1}{4-2}} = \frac{1}{\sqrt{2}}$. The final ratio $\frac{\alpha}{\beta}$ is therefore $\sqrt{2}$.

Determining the sets $A$ and $B$ begins with finding integer pairs $(x, y)$ that satisfy $2x + 3y = 23$, which yields the four pairs $(1, 7), (4, 5), (7, 3),$ and $(10, 1)$. Set $A$ consists of these four pairs, and set $B$ consists of the corresponding $x$-coordinates, which are the four values $1, 4, 7,$ and $10$. Since both sets contain exactly 4 elements, the number of one-one functions between them is defined by the total number of permutations of 4 elements, which is $4! = 24$.

The function $g(x)$ is defined piecewise by evaluating $f(|x|)$ and $|f(x)|$ over the domain $[-a, a]$. For $x \in [0, a]$, $g(x)$ simplifies to $0$, whereas for $x \in [-a, 0]$, it takes the form $-x/2$. Because $g(x)$ maps the entire sub-interval $[0, a]$ to a constant value, it fails to be one-one, and since the output range is limited to $[0, a/2]$ rather than the entire codomain $[-a, a]$, it is also not onto.

The composite function $f(g(x))$ is obtained by substituting $g(x)$ into $f(x) = |x-1|$. For $x \ge 0$, $f(g(x)) = |e^x - 1| = e^x - 1$, and for $x \le 0$, $f(g(x)) = |(x+1) - 1| = |x| = -x$. Since $e^x - 1$ and $-x$ can yield identical positive values for distinct inputs, the function is not one-one. Additionally, because the outputs are always non-negative, the range $[0, \infty )$ does not span the entire codomain $\mathbb{R}$, meaning it is not onto.

The composition $f(f(x))$ involves substituting the function into itself, yielding $f(f(x))=\frac{4x}{\sqrt{1+9x^2(1+2^2)}}$. Repeating this process $n$ times results in the general expression $f_n(x)=\frac{2^n x}{\sqrt{1+9x^2(1+2^{2+}\dots+2^{2n-2})}}$, where the denominator series is a geometric progression with sum $\frac{4^n-1}{3}$. Setting $n=10$, we identify $\alpha = \frac{4^{10}-1}{3}$, which simplifies the expression $\sqrt{3\alpha+1}$ to $\sqrt{4^{10}}$, resulting in $2^{10}=1024$.

Expressing $f(\theta)$ as a function of $t = \cos^2\theta$, we write $f(t) = \frac{(1-t)^2 + 3t}{(1-t)^2 + t} = 1 + \frac{2t}{t^{2-}t+1}$. The range of the function $h(t) = \frac{2t}{t^{2-}t+1}$ for $t \in [0, 1]$ is $[0, 2]$, which makes the range of $f(\theta)$ equal to $[1, 3]$, implying $\alpha=1$ and $\beta=3$. The sum of the infinite geometric series with first term 64 and common ratio $1/3$ is $S = \frac{64}{1-1/3} = 96$.

The sine function oscillates between -1 and 1, which dictates the bounds for its argument in the denominator. Subtracting $\sin 5x$ from 7 restricts the denominator's values to the closed interval $[6, 8]$. Taking the reciprocal of these values inverts the boundaries, resulting in a range of [rac{1}{8}, rac{1}{6}].

The composition $g(f(x))$ is piecewise defined by analyzing the range of $f(x)$ for different input intervals. For $x \le 0$, $f(x) = e^{-x} > 0$, so $g(f(x)) = f(x) = e^{-x}$, while for $x > 0$, we observe that $f(x)$ takes both positive and negative values, leading to overlapping range values for $g(f(x))$. Since the function covers the same output values for different inputs, it fails the one-one test, and its range is a subset of the codomain, meaning it is not onto. Consequently, the function is neither one-one nor onto.

A function is one-to-one if distinct inputs yield distinct outputs and onto if every element in the codomain is reached by at least one input. Here, $f(2) = 2$ and $f(4) = 2$, which immediately shows the function is not one-to-one because two different values map to the same prime. Additionally, since the output is always a prime number, no input can produce a composite number like $4$ as an image, indicating the function is not onto.

The domain of this function depends on the conditions for the inverse cosine and the logarithmic term. For $\cos^{-1}(u)$, the input $u$ must lie in $[-1, 1]$, which yields $-1 \le \frac{2-|x|}{4} \le 1$, simplifying to $x \in [-6, 6]$. Additionally, the argument of the logarithm must be positive ($3-x > 0$) and the expression $\log_e(3-x)$ cannot be zero, implying $3-x \ne 1$. Combining these restrictions, we find the interval $[-6, 3)$ excluding the value $2$, meaning $\alpha=6$, $\beta=3$, and $\gamma=2$, leading to a sum of $11$.

Defining the constants $A = \tan 1^{\circ}$, $B = \ln 123$, and $C = \ln 1234$ simplifies the function to $f(x) = \frac{Ax + B}{xC - A}$. Calculating the composition $f(f(x))$ results in the identity function $x$, which indicates the function is its own inverse. Consequently, the expression simplifies to $x + \frac{4}{x}$, and applying the Arithmetic Mean-Geometric Mean inequality $x + \frac{4}{x} \geq 2\sqrt{x \cdot \frac{4}{x}}$ yields a least value of $4$.

Given the inverse function property where $y = fog(x)$ implies $x = (fog)^{-1}(y)$, we can rearrange the expression to find $fog(x) = 2x^3 + 7$. Comparing this to the composite form $a(x^b + c) - 3 = ax^b + ac - 3$, we identify the constants as $a = 2$, $b = 3$, and $ac = 5 + 5 = 10$, implying $c = 5$. Evaluating the required components, $fog(ac) = fog(10) = 2(10^3) + 7 = 2007$, and $gof(b) = gof(3) = (2(3)-3)^3 + 5 = 32$. The final sum is $2007 + 32 = 2039$.

Substituting $x = 2$, $x = -1$, and $x = \frac{1}{2}$ into the functional equation $f(x) + f\left(\frac{1}{1-x}\right) = 1+x$ generates a system of linear equations: $f(2) + f(-1) = 3$, $f(-1) + f(1/2) = 0$, and $f(1/2) + f(2) = 3/2$. By adding the first and third equations and subtracting the second, the variables $f(-1)$ and $f(1/2)$ are eliminated. This leads to $2f(2) = 3 + \frac{3}{2} = \frac{9}{2}$, resulting in $f(2) = \frac{9}{4}$.

For a function of the form $f(x) = \frac{a^x}{a^x + \sqrt{a}}$, the property $f(x) + f(1-x) = 1$ holds because $f(1-x)$ simplifies to $\frac{\sqrt{a}}{a^x + \sqrt{a}}$. Substituting $a = 4$ into the given expression $f(x) = \frac{4^x}{4^x+2}$, we confirm that $f(x) + f(1-x) = 1$. The sum $f(\frac{1}{2023}) + f(\frac{2}{2023}) + \dots + f(\frac{2022}{2023})$ contains 2022 terms, which pair up to form 1011 instances of the sum $f(x) + f(1-x) = 1$, resulting in a total value of 1011.

The functional equation $f(x+y) = f(x)f(y)$ characterizes an exponential function of the form $f(x) = a^x$. Given $f(1) = 3$, it follows that $f(x) = 3^x$, turning the summation into a geometric series $3^1 + 3^2 + \dots + 3^n = 3279$. Applying the sum formula $\frac{a(r^n - 1)}{r - 1}$, we get $\frac{3(3^n - 1)}{3 - 1} = 3279$, which simplifies to $3^n - 1 = 2186$. Since $3^n = 2187$ and $3^7 = 2187$, the value of $n$ is 7.