Functions – JEE Main MathematicsPractice Questions & PYQs

By applying the definition of $f(n)$, we determine the outputs for all values from 1 to 10. Using the composition $f(g(n))$, we isolate $g(n)$ for each $n$ by finding the corresponding preimage under $f$. Evaluating the sum of $g(1)$ through $g(5)$ and multiplying this result by $g(10)$ leads to the final value of 190.

Since the derivative $f'(x) = \frac{2x}{x^{2+}1} + e^{-x} > 0$ and $g'(x) = -e^{-x} - 2e^x < 0$, the function $f$ is strictly increasing while $g$ is strictly decreasing. The inequality $f(g(A)) > f(g(B))$ implies $g(A) > g(B)$ because $f$ is increasing, and therefore $A < B$ because $g$ is decreasing. This requires solving the inequality $\frac{(\alpha-1)^2}{3} < \alpha - \frac{5}{3}$, which simplifies to $\alpha^2 - 5\alpha + 6 < 0$. The roots of this quadratic are 2 and 3, meaning the condition holds for $\alpha \in (2,3)$.

A quadratic function $f(x) = ax^2 + bx + c$ can be determined by solving the system of linear equations derived from the given functional values. Using $f(1) = a + b + c = 3$, $f(3) = 9a + 3b + c = 4$, and the condition $f(0) + f(1) + f(-2) + f(3) = 14$, which simplifies to $4a - 2b + 2c = 7$, we solve for the constants. Substituting these back, we find $a = 1/6$, $b = -1/6$, and $c = 3$. Evaluating $f(-2) = 4a - 2b + c$ with these coefficients yields $4(1/6) - 2(-1/6) + 3$, which equals $4/6 + 2/6 + 3 = 4$.

The domain has 50 elements and the range has 50 elements. We must select 17 distinct values from the range to assign to the 17 domain elements involved in the restriction $f(3) \geq f(9) \geq \ldots \geq f(99)$, which can be done in $\binom{50}{17}$ ways. Once these 17 values are chosen, there is exactly one way to arrange them in non-increasing order to satisfy the condition. The remaining $50-17=33$ elements in the domain can be mapped to the remaining 33 elements in the range in $33!$ ways. Multiplying these counts gives $\binom{50}{17} \times 33! = \frac{50!}{17!33!} \times 33! = \frac{50!}{17!} = {}^{50}P_{33}$.

The number of such functions is determined by the Fundamental Counting Principle, where we identify the number of valid choices for $f(x)$ for every $x$ in the domain $A$. Solving $x^2 - 10x + 9 \le 0$ identifies $A$ as the set of integers from 1 to 9, and the codomain $B$ as the set of perfect squares. For each $x$, we count how many $n^2 \in B$ satisfy $n^2 \le (x-3)^2 + 1$. Calculating these gives 2 choices for $x=1$, 1 for $x=2$, 1 for $x=3$, 1 for $x=4$, 2 for $x=5$, 3 for $x=6$, 4 for $x=7$, 5 for $x=8$, and 6 for $x=9$. Multiplying these choices results in $2 \times 1 \times 1 \times 1 \times 2 \times 3 \times 4 \times 5 \times 6 = 1440$.

The function $f(x) = \alpha x^5 + \beta x^3 + \gamma x$ is an odd function, satisfying $f(-x) = -f(x)$. Since the arithmetic progression $a_1, a_2, \dots, a_n$ has a mean of zero, the terms are symmetrically distributed about zero, which causes the sum $\sum_{i=1}^n f(a_i)$ to vanish because $f(a_i) + f(-a_i) = 0$ for every pair. Given that $g$ is the inverse of $f$ defined by $g(f(x)) = x$, applying $f$ to $g$ at the point zero returns the original input, so $f(g(0)) = 0$. Consequently, the value of the entire expression simplifies to zero.

The condition $f(x+y) = 2f(x)f(y)$ with $f(1)=2$ identifies $f(n) = 2^{2n-1}$ as the functional form. The sum $\sum_{k=1}^{10} f(\alpha+k) = f(\alpha) \sum_{k=1}^{10} 4^k$ forms a geometric series evaluating to $f(\alpha) \cdot \frac{4(4^{10}-1)}{3} = f(\alpha) \cdot \frac{4(2^{20}-1)}{3}$. Equating this to $\frac{512}{3}(2^{20}-1)$ leads to $4f(\alpha) = 512$, so $f(\alpha) = 128$. Solving $2^{2\alpha-1} = 2^7$ gives $2\alpha-1=7$, which results in $\alpha=4$.

For a quadratic polynomial $f(x)=(x-\alpha)(x-\beta)$ with $f(0)=p$ and $f(1)=1/3$, we have $\alpha\beta=p$ and $(1-\alpha)(1-\beta)=1/3$. The condition that $f(x)=0$ and its iteration have a common root implies $f(p)=\alpha$, leading to the relation $(\alpha\beta-\alpha)(\alpha\beta-\beta)=\alpha$, which simplifies to $\beta(\beta-1)(\alpha-1)=1$ given $\alpha \neq 0$. Using the second initial condition $(\alpha-1)(\beta-1)=1/3$, we solve for $\beta=3$ and consequently $\alpha=7/6$. Thus, $f(x)=(x-7/6)(x-3)$, and evaluating this at $x=-3$ yields $(-3-7/6)(-3-3) = (-25/6)(-6) = 25$.

A function is injective (one-one) if every unique input maps to a unique output and surjective (onto) if every element in the codomain is reached by at least one input. For the sum $f(a)+g(a)$, we calculate $(f+g)(4) = f(4)+g(4) = 2+5 = 7$ and $(f+g)(5) = f(5)+g(5) = 1+6 = 7$, demonstrating that the function is not one-one because distinct inputs map to the same value. Furthermore, since the minimum value of $f(a)$ is 1 and the minimum of $g(a)$ is 3, the minimum value of $(f+g)(a)$ is 4, implying that values like 1, 2, and 3 are never attained in the codomain, so the function is not onto.

The equation $2f(a) - f(b) + 3f(c) + f(d) = 0$ is rearranged to $f(b) = 2f(a) + 3f(c) + f(d)$, where $f(a), f(b), f(c), f(d)$ must be distinct values from the set $\{0, 1, \ldots, 10\}$ because the function is one-one. By methodically selecting possible values for $f(c)$ and $f(a)$ that ensure $f(b) \leq 10$ and maintain distinctness, one determines the number of valid outcomes for each case. Summing these counts (17 for $f(c)=0$, 9 for $f(c)=1$, 4 for $f(c)=2$, and 1 for $f(c)=3$) provides a total of 31 possible functions.

Setting $x=y=0$ in the functional equation shows $f(0)=0$, confirming $k=0$ in $f(x)=(c+1)x^{2+}(1-c^2)x$. By applying the functional equation $f(x+1)-f(x)=f(1)-x$, we compare coefficients to determine $f(x)=-\frac{1}{2}x^{2-}\frac{5}{4}x$. We calculate the sum $2 \sum_{i=1}^{20} f(i) = 2 \left( -\frac{1}{2} \sum_{i=1}^{20} i^2 - \frac{5}{4} \sum_{i=1}^{20} i \right)$, which simplifies to $| -2870 - 525 | = 3395$.

Iterating the function $f(x)$ reveals that f^2(x) = -rac{1}{x}, f^3(x) = rac{-(x+1)}{x-1}, and $f^4(x) = x$, establishing a cycle of length 4. Using this periodicity, $f^6(6)$ simplifies to f^2(6) = -rac{1}{6} and $f^7(7)$ simplifies to f^3(7) = rac{-(7+1)}{7-1} = -rac{8}{6}. Summing these values gives -rac{3}{2}.

The absolute difference between the roots of $f(x) = (x-p)^2 - q = 0$ is $2\sqrt{q}$. We represent the terms of the arithmetic progression as $p-3d, p-d, p+d, p+3d$ with $d > 0$, so $|f(p \pm d)| = |d^2 - q| = 500$ and $|f(p \pm 3d)| = |9d^2 - q| = 500$. For both to hold, we set $d^2 - q = - (9d^2 - q)$, which simplifies to $10d^2 = 2q$, or $q = 5d^2$. Substituting $q = 5d^2$ into $|d^2 - 5d^2| = 500$ yields $4d^2 = 500$, so $d^2 = 125$ and $q = 625$. The absolute difference $2\sqrt{q}$ is therefore $2\sqrt{625} = 50$.

The recurrence relation $f(1)+2f(2)+3f(3)+\dots+xf(x)=x(x+1)f(x)$ allows us to isolate terms by subtracting the equation for $S_{x-1}$ from $S_x$ for $x \ge 3$. This manipulation yields $x(x+1)f(x) - (x-1)x f(x-1) = xf(x)$, which simplifies to $f(x) = \frac{x-1}{x} f(x-1)$. Given $f(2)=1/4$, this recurrence yields $f(x) = \frac{1}{2x}$ for $x \ge 2$, leading to the final calculation $\frac{1}{f(2022)} + \frac{1}{f(2028)} = 4044 + 4056 = 8100$.

When dealing with sums of functions like $f(x) = \frac{2 e^{2 x}}{e^{2 x}+e}$ over arithmetic progressions, it is helpful to examine the behavior of $f(x) + f(1-x)$. Calculating $f(1-x)$ yields the expression $\frac{2 e^{2(1-x)}}{e^{2(1-x)}+e} = \frac{2e}{e^{2x}+e}$, and combining this with the original function gives $f(x) + f(1-x) = \frac{2 e^{2 x}}{e^{2 x}+e} + \frac{2 e}{e^{2 x}+e} = \frac{2(e^{2 x}+e)}{e^{2 x}+e} = 2$.

Because this sum is constant, we can pair terms in the sequence $f(\frac{1}{100}) + f(\frac{2}{100}) + \cdots + f(\frac{99}{100})$ by grouping terms from the beginning and the end. There are 99 terms in total, allowing for 49 pairs that each sum to 2, contributing a total of 98. The single remaining middle term is $f(\frac{50}{100})$, which is equivalent to $f(\frac{1}{2})$. Since the identity $f(x) + f(1-x) = 2$ holds for all $x$, setting $x = \frac{1}{2}$ results in $f\left(\frac{1}{2}\right) + f\left(\frac{1}{2}\right) = 2$, which implies that the middle term is 1. Adding this to the sum of the pairs provides the final result of 99.

The condition $|f(n)| \le 800$ corresponds to the inequality $-800 \le 2n^2 - n - 1 \le 800$. Solving $2n^2 - n - 801 \le 0$ provides the range for $n$ as approximately $[-19.7, 20.2]$, meaning $n$ takes integer values from $-19$ to $20$. We calculate the sum $\sum_{n=-19}^{20} (2n^2 - n - 1)$ by splitting it into $2 \sum n^2 - \sum n - \sum 1$. Using standard sum formulas, $2(5340) - 20 - 40$ evaluates to $10680 - 60$, which equals $10620$.

The function $f(x)$ simplifies significantly when evaluating composite values. Calculating $f(1) = (2(1-0.5)(2+1))^{1/50} = 3^{1/50}$. Substituting this result into the function for the next layer yields $f(3^{1/50}) = (2(1 - \frac{\sqrt{3}}{2})(2 + \sqrt{3}))^{1/50} = (4-3)^{1/50} = 1$. Consequently, $f(f(f(1))) = f(1) = 3^{1/50}$, and since $3^{1/50} \approx 1.02$, $g(1) = 3^{1/50} + 1 \approx 2.02$. The greatest integer less than or equal to $g(1)$ is 2.

The function condition $f(1)+f(2)=f(3)$ restricts the possible outputs for $f(1), f(2)$, and $f(3)$, while $f(4)$ can take any of the 6 values in the range. We evaluate the number of integer pairs $(f(1), f(2))$ that sum to $f(3)$ for each possible value of $f(3)$ in the set $\{1, 2, 3, 4, 5, 6\}$. There is 1 pair for $f(3)=2$, 2 pairs for $f(3)=3$, 3 pairs for $f(3)=4$, 4 pairs for $f(3)=5$, and 5 pairs for $f(3)=6$, totaling $1+2+3+4+5=15$ combinations for the first three inputs. Multiplying this by the 6 possible values for $f(4)$ gives $15 \times 6 = 90$ total functions.

The condition $f(a,b)=f(b,a) eq a$ with $f: S imes S ightarrow S$ implies that for any pair $(a,b)$, the image must be greater than or equal to both $a$ and $b$ to maintain symmetry and the constraint. By grouping pairs based on their necessary lower bounds, such as those that must map to 4, 3, or 2, we can determine the valid configurations for an onto function. Accounting for the two cases-where 3 is provided by the set of pairs with a lower bound of 2 and where at least one element with a lower bound of 3 maps to 3-results in a total of 37 possible onto functions.

Composing the two functions yields f ig( g(x) ig) = rac{1}{x^{2-}1}. Since the derivative rac{-2x}{(x^{2-}1)^2} changes sign at $x=0$, the function is not monotonic and fails the one-to-one test. Furthermore, the range of the resulting function is $(- ext{inf}, -1] ext{ \cup } (0, ext{inf})$, which does not cover all real numbers, so the function is not onto.