Functions – JEE Main MathematicsPractice Questions & PYQs

To determine the nature of the function, we analyze its range by setting $y = \frac{2x^{2-}3x+2}{3x^{2+}x+3}$. Rearranging this leads to the quadratic equation in $x$, $(3y-2)x^2 + (y+3)x + (3y-2) = 0$. Since $x$ must be real, the discriminant $D = (y+3)^2 - 4(3y-2)^2$ must be non-negative, which simplifies to the inequality $(7y-1)(5y-7) \le 0$, resulting in $y \in [\frac{1}{7}, \frac{7}{5}]$. Because the range is a restricted interval and not the entire set of real numbers, the function is not onto. Furthermore, testing values reveals the function does not maintain a one-to-one mapping, making it neither one-one nor onto.

Starting with $x=0$, the functional equation simplifies to $f(y) = 2y^2 + y - 1$. Substituting $x=1$ and $y=1$ into the original expression gives $f(2) = f(1) + 2(1)^2 + 1 + \alpha(1)(1)$, and since $f(1) = 2(1)^{2+}1-1=2$, we have $f(2) = 2 + 3 + \alpha = 5 + \alpha$. Using $f(2) = 2(2)^{2+}2-1 = 9$, we find $9 = 5 + \alpha$, so $\alpha = 4$, and the required sum becomes $\sum_{n=1}^{5}(4 + 2n^2 + n - 1) = \sum_{n=1}^{5}(2n^2 + n + 3) = 110 + 15 + 15 = 140$.

The logarithmic term requires the inner expression $7-\log_2(x^{2-}10x+85)$ to be greater than $1$, which simplifies to $x^{2-}10x+85 < 2^6$, narrowing $x$ to $(3, 7)$. Simultaneously, the inverse sine term requires the absolute value of the inner ratio to be at most $1$, meaning $|\frac{3x-7}{17-x}| \leq 1$, which simplifies to $x \in [-5, 6]$. The intersection of these domains is $(3, 6]$, where $\alpha=3$ and $\beta=6$, leading to $\alpha+\beta=9$.

First, evaluate $g(2) = 3(2)^2 + 2(2) - 3 = 13$, so we must determine $f(13)$. Using the relation $4g(f(x)) = 3x^2 - 32x + 72$, we find $g(f(x)) = 0.75x^2 - 8x + 18$. Assuming $f(x) = ax + b$ and noting $f(0) = -3$, we have $b = -3$, so $f(x) = ax - 3$. Substituting this into $g(x)$ yields $g(f(x)) = 3(ax-3)^2 + 2(ax-3) - 3 = 3a^2x^2 - 16ax + 18$. Comparing coefficients with $0.75x^2 - 8x + 18$, we solve $3a^2 = 0.75$ and $-16a = -8$, confirming $a = 0.5$. Calculating $f(13) = 0.5(13) - 3$ gives $6.5 - 3 = 3.5$, or $\frac{7}{2}$.

The domain of $\cos^{-1}(x)$ is restricted to $[-1, 1]$, so the expression inside requires $-1 \le \frac{4x+2[x]}{3} \le 1$, simplifying to $-3 \le 4x+2[x] \le 3$. By testing the intervals of $[x]$, when $x \in [-1/4, 0)$, the inequality becomes $-3 \le 4x-2 \le 3$, leading to $x \in [-1/4, 0)$, and when $x \in [0, 3/4]$, it becomes $-3 \le 4x \le 3$, leading to $x \in [0, 3/4]$. Combining these, the domain is $[-1/4, 3/4]$, so $12(\alpha+\beta) = 12(-1/4 + 3/4) = 12(1/2) = 6$.

The domain of the function is defined by the requirement that the inner argument of $\sin^{-1}$ must lie within $[-1, 1]$ and the denominator $\log_e(10-x)$ must be non-zero while its argument $10-x$ must be positive. Solving $-1 \leq \frac{5-x}{3+2x} \leq 1$ yields $x \in (-\infty , -8] \cup [2/3, \infty )$, and the conditions $10-x > 0$ and $10-x \neq 1$ restrict $x$ to $x < 10$ and $x \neq 9$. Combining these constraints results in the interval $(-\infty , -8] \cup [2/3, 9) \cup (9, 10)$, identifying $\alpha=-8, \beta=2/3, \gamma=10,$ and $\delta=9$, which gives a final value of $6(-8 + 2/3 + 10 + 9) = 70$.

The value $m = \sum_{i=1}^{9} i^2 = \frac{9(10)(19)}{6} = 285$, so the expression simplifies to $3f(x) + 2f(15/x) = 5x$. Substituting $x$ with $15/x$ yields $3f(15/x) + 2f(x) = 5(15/x) = 75/x$. Solving this system for $f(x)$ by eliminating $f(15/x)$ leads to $f(x) = 3x - 30/x$. Evaluating the terms, $f(5) = 15 - 6 = 9$ and $f(2) = 6 - 15 = -9$. The difference $f(5) - f(2)$ is $9 - (-9) = 18$.

For statement (I), points satisfying $f(x) = f^{-1}(x)$ are equivalent to solving $f(x) = x$ because the function is strictly increasing. Solving $(x-1)^4 + 1 = x$ yields (x-1)\[(x-1)^3 - 1]$ = 0$, which gives the solutions$x=1$and$x=2$, confirming the set contains exactly two elements. For statement (II), the equation$f(x) = f^{-1}(x+1)$is equivalent to$f(f(x)) = x+1$. Analysis shows this equation possesses at least one real root, meaning the set$S$ is not empty, which contradicts the assertion in statement (II). Therefore, only statement (I) is true.

A function is not onto if its range is a proper subset of the codomain, meaning it misses at least one element. The total number of functions from a set of size $4$ to a set of size $3$ is $3^4 = 81$, while the number of onto functions is calculated via the inclusion-exclusion principle as $3^4 - \binom{3}{1}2^4 + \binom{3}{2}1^4 = 81 - 3(16) + 3(1) = 36$. Subtracting the onto functions from the total gives $81 - 36 = 45$ functions that are not onto.

For the square root to be defined, the logarithmic argument must be positive, and for $\log_{0.6}(u) \ge 0$ with a base less than 1, we require $u \le 1$. This leads to the inequality $|\frac{2x-5}{x^{2-}4}| \le 1$, which solves to $(x-1)^2(x^{2+}2x-9) \ge 0$ after excluding points where the denominator is zero, specifically $x \neq 2, -2, 2.5$. The solution set $(-\infty , -1-\sqrt{10}] \cup \{1\} \cup [-1+\sqrt{10}, 2.5) \cup (2.5, \infty )$ matches the given interval structure, where $a = -1-\sqrt{10}$, $b = 1$, $c = -1+\sqrt{10}$, $d = 2.5$, and $e = 2.5$, summing to $a+b+c+d+e = 4$.

For the function $f(x) = \sin^{-1}\left(\frac{x+[x]}{3}\right)$ to be defined, its argument must satisfy $-1 \le \frac{x+[x]}{3} \le 1$, or $-3 \le x + [x] \le 3$. Letting $x = [x] + \{x\}$ where $0 \le \{x\} < 1$, the inequality becomes $-3 \le 2[x] + \{x\} \le 3$. Testing integer values for $[x]$, we find valid solutions for $[x] = -1$ where $x \in [-1, 0)$, $[x] = 0$ where $x \in [0, 1)$, and $[x] = 1$ where $x \in [1, 2)$. The union of these intervals gives the domain $[-1, 2)$, so $\alpha = -1$ and $\beta = 2$. Consequently, $\alpha^2 + \beta^2 = (-1)^2 + 2^2 = 5$.

The function simplifies using the identity $[x+3] = [x]+3$, which leads to $f(x) = [x]^2 - [x] - 6$, factoring into $([x]-3)([x]+2)$. The condition $f(x) < 0$ implies $-2 < [x] < 3$, meaning $[x]$ must be one of the integers $\{-1, 0, 1, 2\}$. These values of $[x]$ correspond precisely to the interval $x \in [-1, 3)$.

The expression inside the logarithm involves an infinite geometric series with first term $a=2$ and common ratio $r=1/3$, which sums to $\frac{2}{1-1/3} = 3$. Substituting this value, the equation becomes $\log_2(f(x)) = \log_2(3) \cdot \log_3 \left( 1 + \frac{f(x)}{f(1/x)} \right)$, which simplifies to $\log_2(f(x)) = \log_2 \left( 1 + \frac{f(x)}{f(1/x)} \right)$ using the change of base property. This implies $f(x) = 1 + \frac{f(x)}{f(1/x)}$, rearranging to $f(x)f(1/x) = f(x) + f(1/x)$. This functional equation is satisfied by polynomials of the form $f(x) = x^n + 1$. Given $f(6) = 37$, we have $6^n + 1 = 37$, so $6^n = 36$, yielding $n=2$. Finally, evaluating the summation $\sum_{n=1}^{10} (n^2 + 1) = \frac{10(11)(21)}{6} + 10 = 385 + 10 = 395$.

Statement I is true because the function $f(x) = \frac{x}{1+|x|}$ is strictly increasing on both $(-\infty , 0)$ and $[0, \infty )$, mapping distinct inputs to distinct outputs across the entire real line. Statement II is also true because the function $f(x) = \frac{x^{2+}4x-30}{x^{2-}8x+18}$ has a derivative $f'(x) = \frac{-12(x^{2-}8x+14)}{(x^{2-}8x+18)^2}$, which vanishes at $x = 4 \pm \sqrt{2}$, indicating that the function is not monotonic and thus many-one.

The range of the function is determined by evaluating the sum of the signs of the trigonometric functions in each of the four quadrants, given that x eq rac{n imes ext{pi}}{2}. In the first quadrant, $\sin x, \cos x, \tan x,$ and $\cot x$ are all positive, yielding $1+1+1+1=4$. In the second quadrant, $\sin x$ is positive while the others are negative, resulting \in $1-1-1-1=-2$. In the third quadrant, $\sin x$ and $\cos x$ are negative while $\tan x$ and $\cot x$ are positive, giving $-1-1+1+1=0$. In the fourth quadrant, only $\cos x$ is positive, producing $-1+1-1-1=-2$. The set of unique values \in the range is $\{4, 0, -2\}$, which sums to $4+0+(-2)=2$.

The property $f(x+y)=f(x)f(y)$ with $f(1)=7$ implies $f(x)=7^x$, while $g(x+y)=g(xy)$ with $g(1)=1$ identifies $g(x)=1$. The summation $\sum_{x=1}^{n} \frac{7^x}{1} = 19607$ is a geometric series with sum $\frac{7(7^n - 1)}{6} = 19607$. Multiplying by 6 and dividing by 7 gives $7^n - 1 = 16806$, meaning $7^n = 16807$. Since $7^5 = 16807$, the value of $n$ is 5.

We identify valid mappings for $(f(2), f(3))$ given the constraint $f(2)+f(3)=5$ with $f(3) \le 4$, noting that $f$ is one-to-one. The possible pairs are $(4, 1), (3, 2), (2, 3), (1, 4)$. For each pair, we determine the number of valid choices for $f(1)$ such that $f(1) \ge 3$ and $f(1)$ is distinct from $f(2)$ and $f(3)$. For cases $(4, 1)$ and $(1, 4)$, $f(1) \in \{3, 5, 6\}$ providing 3 choices, and for cases $(3, 2)$ and $(2, 3)$, $f(1) \in \{4, 5, 6\}$ providing 3 choices. In every scenario, the remaining 3 elements are mapped to the remaining 3 values in $3! = 6$ ways, leading to $4 \times 3 \times 6 = 72$ total functions.

The domain of the logarithmic function $f(x) = \log_b(A)$ is defined where the argument $A > 0$ and the base $b > 0, b \neq 1$. For $f_1(x) = \log_5(18x - x^2 - 77)$, the argument $18x - x^2 - 77$ must be positive, leading to the inequality $x^2 - 18x + 77 < 0$, which factors as $(x-7)(x-11) < 0$, giving the interval $(7, 11)$. Thus $\alpha = 7$ and $\beta = 11$. For $f_2(x)$, the argument $\frac{(2x-1)(x+2)}{(x-4)(x+1)}$ must be positive and the base $x-1$ must be positive and not equal to 1, satisfying $x > 1$ and $x \neq 2$. Analysis of the rational expression shows it is positive for $x > 4$, so $\gamma = 4$. Summing the squares $\alpha^{2+}\beta^{2+}\gamma^2$ results in $49+121+16 = 186$.

The functional equation $f(x+y) = f(x) + f(y) + 1 - \frac{2}{7}xy$ indicates a quadratic form $f(x) = Ax^2 + Bx + C$, where $f(0) = -1$ leads to $C = -1$. Substituting this into the equation reveals $2Axy - C = 1 - \frac{2}{7}xy$, so $A = -1/7$. Comparing with $f(x) = (2+3a)x^2 + \frac{a+2}{a-1}x + b$, we find $A = 2+3a = -1/7$, implying $a = -5/7$, and $B = -3/4$. Evaluating the sum $28 \sum_{i=1}^5 |f(i)|$ where $f(i) = -1/7 i^2 - 3/4 i - 1$ leads to a total value of $675$.

The composition $f(g(x))$ requires the inner function $g(x)$ to be within the domain of the logarithm, which means $g(x) > 0$. The denominator $2x^2 - 2x + 1$ has a discriminant of $D = (-2)^2 - 4(2)(1) = -4$, confirming it is always positive for any real number $x$. The numerator $x^4 - 2x^3 + 3x^2 - 2x + 2$ can be refactored into $x^2(x-1)^2 + (x-1)^2 + 1$, which is also strictly greater than zero for all real $x$. Since $g(x)$ is positive for every real number, the domain of the composite function is the entire set of real numbers.