Functions – JEE Main MathematicsPractice Questions & PYQs

The total number of onto functions from a set of $5$ elements to a set of $4$ elements is determined by the formula $4! \times S_2(5,4)$, resulting in $240$. To satisfy the condition $f(a) \neq 1$, we subtract the onto functions where $f(a) = 1$ from the total count. Fixing $f(a) = 1$ forces the remaining elements to map onto the target set $S$ in a way that remains onto, which contributes $60$ functions. Subtracting this from the total count yields $240 - 60 = 180$.

The domain of the function $f(x)=\frac{1}{\sqrt{[x]^{2-}3[x]-10}}$ requires the expression under the square root in the denominator to be strictly positive. Solving the inequality $[x]^{2-}3[x]-10 > 0$ yields $([x]-5)([x]+2) > 0$, implying $[x] < -2$ or $[x] > 5$. Converting these into constraints for $x$ gives $x < -2$ for the first case and $x \ge 6$ for the second, leading to the domain $(-\infty , -2) \cup [6, \infty )$.

The multiplicative condition $f(m imes n)=f(m) imes f(n)$ requires $f(1)=1$ and governs the mapping of values in $A=\{1,2,3,5,8,9\}$. Setting $n=3$, the relation $f(9)=f(3)^2$ restricts the potential values for $f(3)$ to those where $f(9)$ remains in $A$, meaning only $f(3)=1$ and $f(3)=3$ are valid. Because the remaining elements $\{2,5,8\}$ lack similar multiplicative constraints within $A$, each can map to any of the 6 elements of $A$, resulting in $6^3 \times 2 \times 1 = 432$ total functions.

The domain of $f(x)$ is determined by the constraints on the logarithm and the inverse sine function, resulting in $x \in [3^{-\frac{23}{6}}, \frac{1}{27})$. Because the fractional part function $g(x)=x-[x]$ on this interval simply outputs $x$ (since $0 < x < 1$ for all values in this domain, making $[x]=0$), the range $(\alpha, \beta)$ corresponds to the domain interval itself, where $\alpha=3^{-\frac{23}{6}}$ and $\beta=\frac{1}{27}$. Calculating $\alpha^{2+}\frac{5}{\beta}$ yields $3^{-\frac{23}{3}}+135$, a value just greater than $135$.

Since the quadratic $x^{2-}x+1$ is always positive with a minimum of $3/4$ occurring at $x=1/2$, the absolute value function $|x^{2-}x+1|$ is simply $x^{2-}x+1$. At this minimum, the value is $3/4$, so the greatest integer part $[x^{2-}x+1]$ evaluates to $[3/4]=0$. Thus, the absolute minimum of the function is $3/4+0 = 3/4$.

The range of the function $f(x)=4\sin^{-1}\left(\frac{x^2}{x^{2+}1}\right)$ depends on the interval of the inner expression $\frac{x^2}{x^{2+}1}$. Since $x^2 \ge 0$, the expression is always non-negative, and because the denominator is always greater than the numerator, the value is strictly less than $1$. Thus, the inner term varies within the interval $[0, 1)$. Applying the inverse sine function maps this interval to $[0, \pi/2)$, and multiplying by $4$ gives the final range of $[0, 2\pi)$.

The domain of the function must satisfy both logarithmic and rational constraints, specifically that the logarithmic argument must be positive and the base must be positive and not equal to one. These conditions for $\log_{x+1}(x-2)$ imply $x-2 > 0$, $x+1 > 0$, and $x+1 \neq 1$, which simplify to $x > 2$. Additionally, the denominator $e^{2\ln x} - (2x+3)$ cannot be zero, which means $x^2 - 2x - 3 \neq 0$, or $(x-3)(x+1) \neq 0$. Excluding $x=3$ and $x=-1$ while maintaining $x > 2$ results in a final domain of $(2, \infty ) - \{3\}$.

The monotonicity of a function determines its one-to-one nature, which can be analyzed using the derivative of the given function $f(x) = \frac{x^{2+}2x+1}{x^{2+}1}$. Rewriting the function as $f(x) = 1 + \frac{2x}{x^{2+}1}$ allows us to calculate the derivative $f'(x) = \frac{2(1-x^2)}{(x^{2+}1)^2}$. For $x \in [1, \infty )$, the derivative $f'(x) \le 0$, indicating the function is monotonic and thus one-to-one in this interval, though it fails to maintain this property over the entire real domain because the derivative changes sign.

The composite function $fog(x)=f(g(x))$ can be analyzed by letting $u=g(x)=\sqrt{x}+1$. Since $\sqrt{x}=u-1$ and $x=(u-1)^2$, we substitute these into the expression $x+3-\sqrt{x}$ to find $f(u)=(u-1)^{2+}3-(u-1)$, which simplifies to $f(u)=u^{2-}3u+5$. Although the real-valued domain of $g(x)$ limits $u$ to values greater than or equal to $1$, the algebraic form $f(u)=u^{2-}3u+5$ allows for the evaluation at $u=0$, yielding $f(0)=5$.

Setting the rational function equal to $y$ and rearranging it into a quadratic equation in terms of $x$ allows the use of the discriminant condition $D \ge 0$ to identify reachable values. For $x$ to be real, the discriminant of $x^2(y-1) - x(8y+2) + (12y-1) = 0$ must satisfy $16y^{2+}84y \ge 0$, which yields $y \in (-\infty , -21/4] \cup [0, \infty )$. As the value $y=1$ is also achievable by linear simplification when the quadratic coefficient vanishes, this entire set constitutes the range of the function.

A function satisfying the Cauchy functional equation $f(x+y) = f(x) + f(y)$ takes the form $f(x) = cx$, and given $f(1) = \frac{1}{5}$, we have $f(n) = \frac{n}{5}$. Substituting this into the sum results in $\frac{1}{5} \sum_{n=1}^{m} \frac{n}{n(n+1)(n+2)} = \frac{1}{5} \sum_{n=1}^{m} (\frac{1}{n+1} - \frac{1}{n+2})$. This telescoping series simplifies to $\frac{1}{5} (\frac{1}{2} - \frac{1}{m+2}) = \frac{m}{10(m+2)}$, and setting this equal to $\frac{1}{12}$ yields $12m = 10m + 20$, which simplifies to $m = 10$.

Determining the number of one-one functions $f: S \rightarrow P(S)$ such that $f(n) \subset f(m)$ for $n < m$ is equivalent to counting the number of strictly increasing chains of 6 subsets within the power set $P(S)$ of a six-element set $S$. Each element of $S$ must be included in one of the transition steps between successive subsets in the chain to maintain strict inclusion. Accounting for all possible ways to distribute these elements across the six required set transitions leads to the total count of 3240.

Given $f(1)+f(2)=f(4)-1$, we can rewrite this as $f(4)=f(1)+f(2)+1$. Since the range of $f$ is limited to integers up to $6$, the condition $f(1)+f(2) \le 5$ must hold. Counting the possible pairs $(f(1), f(2))$ results in $10$ valid combinations ($4$ with $f(1)=1$, $3$ with $f(1)=2$, $2$ with $f(1)=3$, and $1$ with $f(1)=4$). Since $f(4)$ is fixed for each pair and $f(3)$ and $f(5)$ can each take $6$ different values, the total number of functions is $10 \times 6 \times 6 = 360$.

The range of the function $\sin x - \cos x$ is $[-\sqrt{2}, \sqrt{2}]$, which implies that $\sqrt{2}(\sin x - \cos x)$ varies between $-2$ and $2$. Consequently, the argument of the logarithm, given by $\sqrt{2}(\sin x - \cos x) + m - 2$, ranges from $[m-4, m]$. Since the range of the logarithmic function $f(x)$ is $[0, 2]$, the argument must span $[(\sqrt{m})^0, (\sqrt{m})^2]$, meaning $[1, m]$. Equating the lower bounds, $m - 4 = 1$ leads to $m = 5$.

The function $f(x)$ behaves as $\frac{n}{1+x^2}$ on the interval $[n, n+1)$ for integers $n \in \{2, 3, 4, 5\}$. By computing the output range of each segment, the global minimum occurs at $x \to 6$ for $n=5$, resulting in $\frac{5}{1+6^2} = \frac{5}{37}$, and the global maximum occurs at $x=2$ for $n=2$, resulting in $\frac{2}{1+2^2} = \frac{2}{5}$. Because these individual ranges overlap, the total range of the function is simply the union, covering $(\frac{5}{37}, \frac{2}{5}]$.

Squaring the function $f(x)$ yields $f^2(x) = (\sqrt{3-x}+\sqrt{2+x})^2 = 5 + 2\sqrt{(3-x)(2+x)}$. Completing the square for the quadratic inside the radical gives $6+x-x^2 = \frac{25}{4} - (x-1/2)^2$, which ranges from $0$ at the domain boundaries $x=-2, 3$ to $25/4$ at the vertex $x=1/2$. Consequently, $f^2(x)$ ranges from $5+0=5$ to $5+2(2.5)=10$, making the range of the original function $[\sqrt{5}, \sqrt{10}]$.

The domain is determined by the intersection of the conditions for the logarithm, where the argument $\frac{(3x+1)(2x+1)}{2x-1} > 0$, and the inverse cosine, where the argument lies within $[-1, 1]$. Solving these inequalities simultaneously restricts the possible values of $x$ to the interval $\left( -\frac{1}{2}, -\frac{1}{3} \right) \cup \left( \frac{1}{2}, \frac{1}{\sqrt{2}} \right]$. Calculating $18(\alpha^{2+}\beta^{2+}\gamma^{2+}\delta^2)$ with $\alpha = -\frac{1}{2}$, $\beta = -\frac{1}{3}$, $\gamma = \frac{1}{2}$, and $\delta = \frac{1}{\sqrt{2}}$ gives $18 \left( \frac{1}{4} + \frac{1}{9} + \frac{1}{4} + \frac{1}{2} \right)$, which simplifies to $18 + 2 = 20$.

Given $f(x) = 2x^n + \lambda$, we use the values $f(4) = 133$ and $f(5) = 255$ to form the system $2(4^n) + \lambda = 133$ and $2(5^n) + \lambda = 255$. Subtracting these equations gives $2(5^n - 4^n) = 122$, or $5^n - 4^n = 61$, which holds for $n=3$. Solving for $\lambda$ yields $\lambda = 5$, and calculating $f(3) - f(2)$ gives $2(27) + 5 - (2(8) + 5) = 38$. The sum of the positive integer divisors of $38$, which are $1, 2, 19, 38$, equals $60$.

The recurrence relation $f(n+1) = n(1 - f(n))$ allows us to express all values of the function based on $f(2)$, resulting in $f(3) = 2(1 - f(2))$ and $f(4) = 6f(2) - 3$. Applying the constraint $|f(n)| \le 8$ for all $n \in \{1, 2, 3, 4\}$, we require $|f(2)| \le 8$, $|1 - f(2)| \le 8$, $|2 - 2f(2)| \le 8$, and $|6f(2) - 3| \le 8$. Testing integer values for $f(2)$ against these inequalities confirms that only $f(2) = 0$ and $f(2) = 1$ are valid solutions.

Given $f(x)$ is a quadratic and $g(x)$ is linear, we define $f(x)=Ax^{2+}Bx+C$ and $g(x)=ax+b$. By substituting these forms into $f(g(x))=8x^{2-}2x$ and $g(f(x))=4x^{2+}6x+1$ and equating the coefficients of corresponding powers of $x$, we identify $g(x)=2x-1$ and $f(x)=2x^{2+}3x+1$. Calculating the values at $x=2$ gives $f(2)=2(2^2)+3(2)+1=15$ and $g(2)=2(2)-1=3$. The final value is the sum $f(2)+g(2)=15+3=18$.