Functions – JEE Main MathematicsPractice Questions & PYQs

For a real-valued function involving a square root, the radicand must be non-negative and the denominator must be non-zero. Solving the inequality $\frac{|\lfloor x \rfloor| - 2}{|\lfloor x \rfloor| - 3} \geq 0$ requires evaluating cases where the absolute value of the floor of $x$ is either $\leq 2$ or $> 3$, which leads to the intervals $(-\infty , -3) \cup [-2, 3) \cup [4, \infty )$. Identifying $a = -3$, $b = -2$, and $c = 3$ from the given form $(-\infty , a) \cup [b, c) \cup [4, \infty )$, the sum $a+b+c$ equals $-2$.

The range of the function $f(x) = \log_{\sqrt{5}}(3 + \cos(\frac{3\pi}{4} + x) + \cos(\frac{\pi}{4} + x) + \cos(\frac{\pi}{4} - x) - \cos(\frac{3\pi}{4} - x))$ is determined by the interval of its logarithmic argument. By applying trigonometric identities to the terms within the logarithm, the expression simplifies to $3 - \sqrt{2}\sin x + \sqrt{2}\cos x$. Because the oscillating part $-\sqrt{2}\sin x + \sqrt{2}\cos x$ covers all values between $-2$ and $2$, adding $3$ shifts this interval to $[1, 5]$. Evaluating the logarithm to the base $\sqrt{5}$ at these boundaries gives $\log_{\sqrt{5}}(1) = 0$ and $\log_{\sqrt{5}}(5) = 2$, identifying the final output range as $[0, 2]$.

The number of one-one functions from set $A$ to set $B$ is determined by the permutation formula ${}^n P_r$, where $n$ is the number of elements in the codomain and $r$ is the size of the domain. Calculating $x$ gives ${}^5 P_3 = 60$, while for the set $A \times B$, which contains $3 \times 5 = 15$ elements, the number of one-one functions is $y = {}^{15} P_3 = 2730$. Comparing these values reveals that $2730$ is exactly $\frac{91}{2}$ times $60$, leading to the relation $2y = 91x$.

The requirement $f(1)+f(2)+f(3)=3$ using distinct values from the set $A=\{0, 1, 2, 3, 4, 5, 6, 7\}$ forces the images of $\{1, 2, 3\}$ to be the set $\{0, 1, 2\}$. These three domain elements can be mapped to these three target values in $3!$ permutations to satisfy the bijection. The remaining 5 elements of the domain must map to the remaining 5 values in the codomain in $5!$ ways, leading to a total of $3! \times 5! = 720$ possible bijective functions.

For a multiplicative function $f: S \rightarrow S$, the condition $f(mn) = f(m)f(n)$ implies $f(1) = 1$. When $f(2)=1$, then $f(4)=1$, and $f(3), f(5), f(6), f(7)$ can be any values (with $f(6)=f(2)f(3)=f(3)$), leading to $7 \times 7 \times 7 = 343$ total functions. When $f(2)=2$, then $f(4)=4$, and $f(3)$ must be such that $f(6)=f(2)f(3) \in S$, which restricts $f(3)$ to the values $\{1, 2, 3\}$, resulting in $3 \times 7 \times 7 = 147$ functions for this case. Combining these disjoint possibilities, we obtain a total of $343 + 147 = 490$ functions.

The additive property $f(m+n) = f(m)+f(n)$ implies linear growth $f(n) = n f(1)$ for any natural number $n$. Given $f(6) = 18$, we find $f(3) = 9$ and subsequently $f(1) = 3$. With $f(1)=3$, it follows that $f(2) = 6$, and calculating the product $f(2) \cdot f(3)$ yields $6 \cdot 9 = 54$.

Composing these functions involves calculating $f(g(x)) = 2(\frac{x-1/2}{x-1}) - 1$, which simplifies algebraically to $1 + \frac{1}{x-1}$. Analyzing this resulting expression reveals that the function is strictly monotonic across its domain, which satisfies the one-one condition. Because the horizontal asymptote at $y=1$ is never reached for any $x$ in the domain, the function fails to cover the entire real codomain, meaning it is not onto.

The critical points exist where the derivative $f'(x)$ is zero, so we begin by simplifying the expression using $2 \cot(x/2) \sin^2(x/2) = \sin x$. The derivative is $f'(x) = (4a-3) + (a-7) \cos x$, which equals zero when $\cos x = \frac{4a-3}{7-a}$. Given that the range of $\cos x$ is $[-1, 1]$, we solve $-1 \leq \frac{4a-3}{7-a} \leq 1$. Solving these inequalities, we find $a \in [-4/3, 7)$ and $a \in (-\infty , 2] \cup (7, \infty )$, which intersect to give $a \in [-4/3, 2]$.

The domain of combined functions is generally the intersection of the individual domains, with further restrictions imposed by division. Both $f(x) = \sqrt{x}$ and $g(x) = \sqrt{1-x}$ are defined on $[0, 1]$, which constitutes the initial domain for addition and subtraction. For division, the denominators $\sqrt{1-x}$ and $\sqrt{x}$ cannot be zero, which excludes $x=1$ and $x=0$ respectively, resulting in the common domain $(0, 1)$.

The condition $g(f(k)) = f(k)$ implies that $g$ acts as an identity function for any value in the range of $f$. Since $f$ maps even values to themselves and odd values $k$ to $k+1$, the range of $f$ consists of all even integers $\{2, 4, 6, 8, 10\}$. This fixes the image of $g$ for these five even values, while for the remaining five odd values, $g$ can map each to any of the 10 values in the set $A$. Therefore, the total number of such functions is $10^5$.

The invertibility of a composite function $(g \circ f)^{-1}$ requires $g \circ f$ to be a bijection, which necessitates that the function $g \circ f$ is both injective and surjective. If $f$ were not one-one, the composition could not be injective, and if $g$ were not onto, the composition could not be surjective. Therefore, $f$ must be one-one and $g$ must be onto to ensure every input maps to a unique output and the entire target set is reachable.

A function involving $\csc^{-1} x$ requires $|x| \geq 1$ to yield real values. The denominator $\sqrt{x - \lfloor x \rfloor}$ must be strictly positive, which means the fractional part $\{x\}$ must be greater than zero, thereby excluding all integers from the domain. Combining $|x| \geq 1$ with the exclusion of all integers results in the set of all real numbers such that $|x| > 1$, which is equivalent to all non-integers except for the interval $[-1, 1]$.

The polynomial $P(x)$ being divisible by $x^{2+}x+1$ implies that the roots of $x^{2+}x+1=0$, which are the complex cube roots of unity $\omega$ and $\omega^2$, are also roots of $P(x)$. Setting $P(\omega) = 0$ results in $f(\omega^3) + \omega g(\omega^3) = f(1) + \omega g(1) = 0$, where $\omega$ is a complex number. Since this equation involves real coefficients for $f(1)$ and $g(1)$, both $f(1)$ and $g(1)$ must independently be zero, confirming that $P(1) = f(1) + 1 \cdot g(1) = 0$.

By substituting $\frac{1}{x}$ for $x$ in the given linear equation, a system of two equations is formed involving $f(x)$ and $f(\frac{1}{x})$. Adding these equations results in $(a+\alpha)[f(x) + f(\frac{1}{x})] = (b+\beta)(x + \frac{1}{x})$. Dividing both sides by the term $(a+\alpha)(x + \frac{1}{x})$ isolates the expression, yielding $\frac{f(x)+f(1/x)}{x+1/x} = \frac{b+\beta}{a+\alpha}$, which evaluates to $\frac{2}{1} = 2$.

The condition for a function to be its own inverse is that $(f \circ f)(x) = x$ implies $f(x) = f^{-1}(x)$. Determining the inverse by solving $y = \frac{5x+3}{6x-\alpha}$ for $x$ yields $x = \frac{\alpha y + 3}{6y - 5}$, so $f^{-1}(x) = \frac{\alpha x + 3}{6x - 5}$. Comparing the original function $f(x) = \frac{5x+3}{6x-\alpha}$ with its inverse demonstrates that the coefficients must match, directly resulting in $\alpha = 5$.

The functional equation $f(n+1) = f(n) + f(1)$ defines an arithmetic progression, which implies $f(n) = n \cdot f(1)$ for natural numbers. Since this is a linear mapping, $f$ is inherently one-one. However, if $g$ is an onto function that maps non-distinct values to the same output, the composition $f \circ g$ cannot be guaranteed to be one-one, making the claim that $f \circ g$ must be one-one if $g$ is onto generally false.

Given the function $y = 5^{\log x}$, we take the logarithm on both sides to obtain $\log y = \log x \cdot \log 5$. Rearranging this equation to solve for $\log x$ results in $\log x = \frac{\log y}{\log 5}$. Since $\log_y x = \frac{\log x}{\log y}$, we arrive at the relation $\log_y x = \frac{1}{\log 5}$, which implies $x = y^{\frac{1}{\log 5}}$.

The defined function $g$ acts as a cyclic permutation on elements of $N$ in groups of three, such that $g^3(x)=x$ for all $x$. For a function $f$ to satisfy $f \circ g = f$, it must map all elements within each cycle of $g$ to the same value. By defining $f$ to map each triplet $\{3n+1, 3n+2, 3n+3\}$ to a unique natural number $n+1$, we ensure the resulting function is onto and satisfies the required invariance property $f(g(x)) = f(x)$.

The functional equation $f(x+y)+f(x-y)=2f(x)f(y)$ characterizes a cosine function of the form $f(x) = \cos(\lambda x)$. Given $f(1/2) = -1$, we find $\cos(\lambda/2) = -1$, which implies $\lambda = 2\pi$, so $f(x) = \cos(2\pi x)$ and $f(k) = 1$ for any integer $k$. The sum then becomes $\sum_{k=1}^{20} \frac{1}{\sin(k)\sin(k+1)}$. By applying the identity $\frac{\sin((k+1)-k)}{\sin k \sin(k+1)} = \cot k - \cot(k+1)$, the series telescopes to $\csc(1) (\cot(1) - \cot(21))$, which simplifies to $\csc^{2}(1) \csc(21) \sin(20)$.

Evaluating the function $f(x) = \frac{5^x}{5^x+5}$, we find that $f(2-x) = \frac{5}{5^x+5}$, which satisfies $f(x) + f(2-x) = 1$. The series contains 39 terms from $x=1/20$ to $x=39/20$, which form 19 pairs that each sum to 1, with the middle term being $f(20/20) = f(1) = 1/2$. Thus, the total sum of the series is $19 + 1/2 = 39/2$.