Functions – JEE Main MathematicsPractice Questions & PYQs

The function $\cos^{-1}(u)$ requires $-1 \leq \frac{4x+5}{3x-7} \leq 1$, which leads to the solution $x \in [-12, 2/7]$ by considering the boundaries of the rational inequality. The function $g(x)$ requires $2-6\log_{27}(2x+5) > 0$ and $2x+5 > 0$, resulting in $x < -1$ and $x > -2.5$, defining the interval $(-2.5, -1)$. Using $\alpha=-12, \beta=2/7, \gamma=-2.5,$ and $\delta=-1$, the calculation $|7(\alpha+\beta)+4(\gamma+\delta)|$ equals $|7(-82/7) + 4(-3.5)| = |-82 - 14| = 96$.

Functional equations can be solved by generating a system of linear equations through variable substitution. Replacing $x$ with $\frac{24}{x}$ produces a second equation in the system, which allows for the isolation of $f(x)$ by algebraic manipulation. This results in the expression $f(x) = \frac{36}{x} - \frac{x}{2}$, and substituting the required values leads to $f(3) + f(8) = 10.5 + 0.5 = 11$.

The domain of the nested log function $f(x)$ is determined by $x^{2+}4x+3 < 0$, which yields $x \in (-3, -1)$, giving $\alpha=-3$ and $\beta=-1$. For $g(x)$, the condition $|\frac{7x+10}{x-2}| \leq 1$ restricts $x$ to the interval $[-2, -1]$, giving $\gamma=-2$ and $\delta=-1$. The sum of the squares $(-3)^{2+}(-1)^{2+}(-2)^{2+}(-1)^2$ is $9+1+4+1 = 15$.

A function is one-one if each input maps to a unique output, and onto if the codomain matches the range. By rewriting the function as f(x)=1-rac{2}{2^{2x}+1}, it is clear that $f(x)$ is strictly increasing because $2^{2x}$ is strictly increasing, thus confirming the function is one-one. However, since the term rac{2}{2^{2x}+1} only approaches values between 0 and 2 as $x$ ranges from $(-\infty , \infty )$, the output values of $f(x)$ remain strictly within $(-1, 1)$, meaning the range is not equal to the codomain $(-\infty , 1)$ and the function is not onto.

The total number of functions from set $A$ to set $B$ is given by $|B|^{|A|}$, while the number of one-one functions is the number of permutations $P(|B|, |A|)$. With $|A|=4$ and $|B|=4$, there are $4^4 = 256$ total functions and $4! = 24$ one-one functions, resulting in $256 - 24 = 232$ many-one functions. To isolate those where $1 \in f(A)$, we subtract the cases where $1 \notin f(A)$, which is equivalent to finding all functions from $A$ to the remaining 3 elements of $B$, totaling $3^4 = 81$. Subtracting these from the many-one count yields $232 - 81 = 151$.

Simplifying the function $f(x) = \frac{2^{x+2}+16}{2^{2x+1}+2^{x+4}+32}$ leads to $f(x) = \frac{2}{2^x+4}$. Using the property $f(x) + f(4-x) = 1/2$, the sum $f(\frac{1}{15}) + f(\frac{2}{15}) + \dots + f(\frac{59}{15})$ pairs 29 terms into $1/2$ sums, plus the central term $f(\frac{30}{15}) = f(2) = 1/4$. The total sum equals $29(1/2) + 1/4 = 59/4$, and multiplying by 8 results in 118.

The domain of a composite function defined by sum is the intersection of the domains of the individual terms. For the natural logarithm, the ratio must be positive, and for the inverse sine function, the argument must lie within the interval $[-1, 1]$. Solving these respective inequalities yields $x \in (-\infty , -1.25) \cup (1.5, \infty )$ and $x \in [-3, 2)$, which intersect to form the domain $[-3, -1.25)$. Using $\alpha = -3$ and $\beta = -1.25$ in the expression $\alpha^{2+}4\beta$ gives $9-5 = 4$.

The function $f(x) = 2x^3 - 15x^2 + 36x + 7$ defined on $[0, 3]$ has a derivative $f'(x) = 6x^2 - 30x + 36$, which factors to $6(x-2)(x-3)$. Evaluating the function at the critical points and endpoints, $f(0)=7$, $f(2)=35$, and $f(3)=34$, determines the range $A = [7, 35]$. The function $g(x) = \frac{x^{2025}}{x^{2025}+1}$ for $x \ge 0$ is strictly increasing from $0$ to $1$, yielding the range $B = [0, 1)$. The set $S = \{x \in \mathbb{Z} : x \in [0, 1) \cup [7, 35]\}$ includes $0$ and the integers from $7$ to $35$. Counting these yields $1 + (35-7+1) = 30$ elements.

Logarithmic domains require the argument of the logarithm to be positive, while the base conditions must also be satisfied. The condition $1-\log_4(x^{2-}9x+18) > 0$ restricts the argument of the base-7 logarithm, and the inner quadratic must also be positive. Solving these inequalities simultaneously yields the domain $(2,3) \cup (6,7)$, and the sum of the boundaries $2+3+6+7$ is 18.

Given $f(x) = \frac{2^x}{2^x+\sqrt{2}}$, observe that $f(x) + f(1-x) = 1$ because $f(1-x) = \frac{2^{1-x}}{2^{1-x}+\sqrt{2}} = \frac{2}{2+2^x\sqrt{2}} = \frac{\sqrt{2}}{2^x+\sqrt{2}}$. The sum $\sum_{k=1}^{81} f(\frac{k}{82})$ consists of 40 pairs $(f(\frac{k}{82}) + f(\frac{82-k}{82}))$ each summing to 1, plus the middle term $f(\frac{41}{82}) = f(1/2)$. Since $f(1/2) = \frac{\sqrt{2}}{\sqrt{2}+\sqrt{2}} = 1/2$, the total sum is $40 \times 1 + 0.5 = 40.5$, which is $81/2$.

Functional composition involves substituting one function into another, and the range on an interval is determined by the behavior of the resulting function. After substitution, the composite function simplifies to h(x) = rac{9x-7}{17x-12}. Since its derivative is positive, the function is strictly increasing on $[2,4]$, and the range is obtained by evaluating the function at the endpoints to get $[1/2, 29/56]$. The final calculation yields rac{1}{29/56 - 1/2} = 56.

For the rational function $y = \frac{5-x}{x^{2-}3x+2}$ to have a real domain $x$, the quadratic equation $yx^2 + x(1-3y) + (2y-5) = 0$ must possess a discriminant $D = (1-3y)^2 - 4y(2y-5) \geq 0$. This condition simplifies to $y^{2+}14y+1 \geq 0$, with roots given by the quadratic formula as $-7 \pm 4\sqrt{3}$. The resulting range is $(-\infty , -7-4\sqrt{3}] \cup [-7+4\sqrt{3}, \infty )$, where $\alpha^{2+}\beta^2$ calculates to $(-7-4\sqrt{3})^2 + (-7+4\sqrt{3})^2 = 194$.

Radical functions in denominators require their radicands to be strictly positive to ensure the function is defined and non-zero. The first term forces $x$ into the interval $(-2, 5)$ based on the roots of the quadratic, while the second term requires $x > 0$ due to the properties of the absolute value. The intersection of these intervals is $(0, 5)$, leading to $a=0$ and $b=5$. Substituting these values into $(1+a)^{2+}b^2$ results in $1+25 = 26$.

For a function $f$ mapping from set $A$ to its power set $P(A)$, the constraint $a \in f(a)$ requires that the subset chosen for each element must contain that specific element. Since the set $A$ contains $7$ elements, for any given element $a$, there are $2^{7-1} = 2^6$ possible subsets of $A$ that include $a$. Applying the fundamental principle of counting for all $7$ elements, the total number of such functions is $(2^6)^7 = 2^{42}$. This gives $m=2$ and $n=42$, resulting in a sum $m+n = 44$.

To satisfy the condition $f(1) + f(3) = 14$, choose pairs of distinct elements from set $B$ that sum to 14, which are $(2, 12)$, $(12, 2)$, $(4, 10)$, and $(10, 4)$. There are 4 such ordered pairs. For each of these assignments, the remaining 3 elements of $A$ must be mapped to the remaining 5 elements of $B$ in a one-one fashion, which can be done in $P(5, 3) = 5 \times 4 \times 3 = 60$ ways. Multiplying these counts gives $4 \times 60 = 240$ possible mappings.

The composition of a function with itself can be evaluated by substituting the algebraic definition of $f(x)$ directly into the expression. Computing $f(f(x)) = \frac{4(\frac{4x+3}{6x-4})+3}{6(\frac{4x+3}{6x-4})-4}$ simplifies to $x$, which means $g(x) = x$ is an identity function. Therefore, the triple composition $(g \circ g \circ g)(4)$ simplifies directly to $g(4) = 4$.

The domain of the inverse sine function $\sin^{-1}(u)$ is restricted to $u \in [-1, 1]$, which implies the inequality $\left|\frac{x-1}{2x+3}\right| \leq 1$. Squaring both sides yields $(x-1)^2 \leq (2x+3)^2$, which simplifies to $3x^2 + 14x + 8 \geq 0$. Factoring the quadratic expression gives $(3x+2)(x+4) \geq 0$, identifying the domain intervals as $(-\infty , -4] \cup [-2/3, \infty )$ while excluding $x = -3/2$ where the denominator vanishes. With the domain given as $R-(\alpha, \beta)$, we identify $\alpha = -4$ and $\beta = -2/3$, resulting in $12\alpha\beta = 12(-4)(-2/3) = 32$.

To determine injectivity, observe that $f(x)$ takes the value 0 at both $x = -5$ and $x = 3$. Since two distinct values in the domain map to the same output, the function fails the one-one condition. Analyzing the range of $f$ by solving y = rac{x^{2+}2x-15}{x^{2-}4x+9} for $x$ leads to a quadratic inequality involving $y$, demonstrating that the range does not encompass all real numbers, thus the function is not onto.

The domain is determined by requiring the square root argument $x^{2-}25 \ge 0$, the logarithmic argument $x^{2+}2x-15 > 0$, and the non-zero denominator $4-x^2 \ne 0$. Solving $x^2 \ge 25$ gives $x \in (-\infty , -5] \cup [5, \infty )$, while the logarithm requires $x \in (-\infty , -5) \cup (3, \infty )$. Combining these conditions with $x \ne \pm 2$ leads to the intersection $(-\infty , -5) \cup [5, \infty )$, identifying $\alpha = -5$ and $\beta = 5$. Thus, $\alpha^2 + \beta^3 = 25 + 125 = 150$.

The domain is defined by the intersection of the conditions for the logarithmic argument $\frac{2x+3}{4x^{2+}x-3} > 0$ and the $\cos^{-1}$ argument $-1 \le \frac{2x-1}{x+2} \le 1$. Solving the logarithmic inequality restricts $x$ such that the function is valid only for $x > 0.75$, while the trigonometric constraint restricts $x$ to the interval $[-1/3, 3]$. The intersection of these intervals is $(0.75, 3]$, identifying $\alpha = 0.75$ and $\beta = 3$. Evaluating $5\beta - 4\alpha$ results in $5(3) - 4(0.75) = 15 - 3 = 12$.