Q.If $f$ is a function satisfying $f(x+y) = f(x)\,f(y)$ for all $x, y \in \mathbb{N}$ such that $f(1) = 3$ and $\displaystyle\sum_{x=1}^{n} f(x) = 120$, find the value of $n$.

Showing the 12 most recent of 17 on this concept.

1. AP EAPCET 2026Set eng-2026-05-12-AN1 markMCQ
   Q.If $f$ satisfies the relation $f(x+y) + f(x-y) = 2f(x)f(y)$ for all $x, y \in R$ and $f(0) \neq 0$, then $f(10) - f(-10) =$ (A) $3$ (B) $2$ (C) $1$ (D) $0$

   ›Reveal solutionSolution

   The functional equation $f(x+y)+f(x-y)=2f(x)f(y)$ is the cosine-type equation; setting $x=0$ immediately shows $f$ is even, so $f(10)=f(-10)$ and their difference is 0. Answer: (D).

   Concept and Intuition

   This functional equation is the classical characterization of $\cos(kx)$-type functions (D'Alembert's equation). Rather than solving for the explicit form of $f$, we only need its symmetry — plugging in $x=0$ directly relates $f(y)$ and $f(-y)$, revealing $f$ is an even function, which is all this particular question needs.

   Step-by-Step Solution

   1. Substitute $x=y=0$ into $f(x+y)+f(x-y)=2f(x)f(y)$: $f(0)+f(0) = 2f(0)f(0)$, i.e. $2f(0) = 2f(0)^2$.
   2. This gives $f(0)^2 - f(0) = 0 \Rightarrow f(0)(f(0)-1)=0$. Since it's given $f(0)\ne 0$, we conclude $f(0)=1$.
   3. Substitute $x=0$ (general $y$): $f(0+y)+f(0-y) = 2f(0)f(y)$, i.e. $f(y)+f(-y) = 2(1)f(y) = 2f(y)$.
   4. Rearranging: $f(-y) = 2f(y)-f(y) = f(y)$. So $f$ is an even function: $f(-y)=f(y)$ for all $y$.
   5. Set $y=10$: $f(-10)=f(10)$, hence $f(10)-f(-10) = f(10)-f(10) = 0$.

   Common Mistakes

   • Trying to explicitly solve for $f(x)=\cos(kx)$ and computing numerically — unnecessary work; the evenness alone answers the question.
   • Sign errors when substituting $x=0$ (mixing up which term becomes $f(-y)$ vs $f(y)$).

   ✓Final answer

   The correct option is (D) — $0$.

   ANSWER: D

2. AP EAPCET 2025Set eng-2025-05-22-FN1 markMCQ
   Q.Let $f: N \to N$ be a function such that $f(x+y) = f(x) + f(y) + xy$ for every $x, y \in N$. If $f(1) = 2$, then $\sum_{k=0}^{10} f(k) =$ (A) 1650 (B) 275 (C) 550 (D) 1025

   ›Reveal solutionSolution

   A Cauchy-type functional equation is converted to a recurrence, solved in closed form, then summed — the answer is $275$.

   Concept and Intuition

   Setting $y=1$ in the functional equation turns it into a first-order recurrence in $n$, which telescopes into a closed-form polynomial for $f(n)$. Once you have that closed form, the required sum is just a sum of squares and a sum of integers — both standard formulas.

   Step-by-Step Solution

   1. Put $x=n-1,\ y=1$: $f(n) = f(n-1) + f(1) + (n-1)(1) = f(n-1) + 2 + (n-1) = f(n-1) + (n+1)$.
   2. Also put $x=y=0$: $f(0) = f(0)+f(0)+0 \Rightarrow f(0) = 0$ (needed since the sum starts at $k=0$; $N$ here is taken to include $0$, consistent with the given sum's lower limit).
   3. Telescoping from $f(0)=0$: $f(n) = \sum_{k=1}^{n}(k+1) = \left(\dfrac{n(n+1)}{2}\right) + n = \dfrac{n(n+3)}{2}$.
   4. Check: $f(1)=\frac{1\cdot4}{2}=2$ ✓, $f(2)=\frac{2\cdot5}{2}=5$, $f(3)=\frac{3\cdot6}{2}=9$ — consistent with direct computation from the recurrence.
   5. Now compute $\displaystyle\sum_{k=0}^{10} f(k) = \sum_{k=0}^{10}\frac{k(k+3)}{2} = \frac12\left(\sum_{k=0}^{10}k^2 + 3\sum_{k=0}^{10}k\right)$.
   6. $\sum_{k=0}^{10}k^2 = 385$, $\sum_{k=0}^{10}k = 55$, so the bracket is $385 + 3(55) = 385+165 = 550$.
   7. Dividing by $2$: $\dfrac{550}{2} = 275$.

   Common Mistakes

   • Forgetting to derive/verify $f(0)=0$ and instead assuming the sum should start effectively from $f(1)$ only.
   • Arithmetic slips in $\sum k^2$ over $0$–$10$ (it is $385$, the same as $1$–$10$, since $0^2=0$ contributes nothing).

   ✓Final answer

   The correct option is (B) — 275.

   ANSWER: B

3. AP EAPCET 2025Set eng-2025-05-24-FN1 markMCQ
   Q.$f(x)$ is an $n^{th}$ degree polynomial satisfying $f(x) = \dfrac{1}{2}\begin{vmatrix} f(x) & f(1/x)-f(x) \\ 1 & f(1/x) \end{vmatrix}$. If $f(2)=33$, then the value of $f(3)$ is (A) 126 (B) 214 (C) 244 (D) -124

   ›Reveal solutionSolution

   This tests recognizing that a functional-equation constraint forces the polynomial's form ($x^n+1$), then using the given value to pin down $n$ and evaluate at $x=3$.

   Concept and Intuition

   The determinant equation looks intimidating, but expanding it turns the condition into a clean relation between $f(x)$ and $f(1/x)$. Once you have that relation, you look for a simple polynomial family satisfying it — here $f(x)=x^n+1$ works beautifully because $f(1/x) = x^{-n}+1$ interacts nicely with $f(x)-1=x^n$.

   Step-by-Step Solution

   1. Expand the given determinant: $\begin{vmatrix} f(x) & f(1/x)-f(x) \\ 1 & f(1/x)\end{vmatrix} = f(x)f(1/x) - \big(f(1/x)-f(x)\big) = f(x)f(1/x) - f(1/x) + f(x)$.
   2. The given equation becomes $f(x) = \tfrac12\big[f(x)f(1/x) - f(1/x) + f(x)\big]$, so $2f(x) = f(x)f(1/x) - f(1/x) + f(x)$.
   3. Simplify: $f(x) = f(x)f(1/x) - f(1/x) = f(1/x)\big[f(x)-1\big]$.
   4. Try $f(x) = x^n + 1$: then $f(1/x) = x^{-n}+1$, and $f(x)-1 = x^n$. So $f(1/x)[f(x)-1] = (x^{-n}+1)x^n = 1+x^n = f(x)$. ✓ This works for any $n$, so $f(x)=x^n+1$ is the required $n$-th degree polynomial family.
   5. Use $f(2)=33$: $2^n+1=33 \Rightarrow 2^n=32=2^5 \Rightarrow n=5$.
   6. So $f(x)=x^5+1$, and $f(3) = 3^5+1 = 243+1 = 244$.

   Common Mistakes

   • Forgetting the minus sign when expanding the $2\times2$ determinant.
   • Not testing a concrete polynomial family and instead trying to solve the functional equation from scratch.

   ✓Final answer

   The correct option is (C) — 244.

   ANSWER: C

4. AP EAPCET 2024Set eng-2024-05-20-FN1 markMCQ
   Q.Let $f(x) = 3+2x$ and $g_n(x) = (f \circ f \circ f \circ \ldots \text{ n times})(x)$. $\forall n \in N$ if all the lines $y = g_n(x)$ pass through a fixed point $(\alpha, \beta)$, then $\alpha+\beta =$ (A) $-5$ (B) $-4$ (C) $-3$ (D) $-6$

   ›Reveal solutionSolution

   Every line $y=g_n(x)$ passes through the one point where the "$x+3$" factor vanishes, namely $(-3,-3)$, giving $\alpha+\beta=-6$.

   Concept and Intuition

   A family of lines $y=m_n x + c_n$ (here indexed by $n$, with $m_n=2^n$) shares a common point for all $n$ exactly when the equation can be rewritten in the form $y = m_n(x-\alpha)+\beta$ — because then plugging $x=\alpha$ gives $y=\beta$ regardless of $m_n$'s value.

   Step-by-Step Solution

   1. $f(x)=2x+3$. Compute the composition explicitly: $g_1(x)=2x+3$; $g_2(x)=f(f(x))=2(2x+3)+3=4x+9$; $g_3(x)=f(g_2(x))=2(4x+9)+3=8x+21$.
   2. The pattern is $g_n(x)=2^n x + 3(2^n-1)$ (each step doubles the coefficient of $x$ and adds $3$ to the constant via the geometric series $3(2^{n-1}+2^{n-2}+\cdots+1)=3(2^n-1)$).
   3. Rewrite: $g_n(x) = 2^n x + 3\cdot 2^n - 3 = 2^n(x+3) - 3$.
   4. For this to equal a fixed value of $y$ regardless of $n$, the term $2^n(x+3)$ must vanish for all $n$, which forces $x+3=0$, i.e. $x=-3$. Then $y = 0 - 3 = -3$ for every $n$.
   5. So the common fixed point is $(\alpha,\beta)=(-3,-3)$, and $\alpha+\beta = -3+(-3) = -6$.

   Common Mistakes

   • Trying to solve for the fixed point using only $g_1$ and $g_2$ (two lines determine an intersection point, but you must verify it's genuinely common to all $n$, which the algebraic rewriting guarantees here).
   • Sign errors in expanding the geometric series for the constant term.

   ✓Final answer

   The correct option is (D) — $-6$.

   ANSWER: D

5. AP EAPCET 2024Set eng-2024-05-20-AN1 markMCQ
   Q.$f: \mathbb{R} \to \mathbb{R}$ is defined by $f(x+y) = f(x) + 12y$, $\forall x, y \in \mathbb{R}$. If $f(1) = 6$, then $\sum_{r=1}^{n} f(r) =$ (A) $n^2$ (B) $5n^2$ (C) $6n^2$ (D) $\dfrac{3n(n+1)}{2}$

   ›Reveal solutionSolution

   The functional equation forces $f$ to be linear in its argument; using $f(1)=6$ pins the constant, and summing the resulting arithmetic sequence gives $6n^2$.

   Concept and Intuition

   The equation $f(x+y)=f(x)+12y$ says shifting the input by $y$ always changes the output by exactly $12y$, regardless of $x$ — that is the defining property of an affine (linear-plus-constant) function $f(t)=12t+c$. Once we know $c$ (via $f(1)=6$), $f$ is completely determined and the sum becomes a routine arithmetic-progression sum.

   Step-by-Step Solution

   1. Put $x=0$: $f(y)=f(0)+12y$ for all $y\in\mathbb{R}$, so $f$ is of the form $f(t)=12t+f(0)$.
   2. Use $f(1)=6$: $12(1)+f(0)=6\Rightarrow f(0)=-6$.
   3. So $f(r)=12r-6$.
   4. $\displaystyle\sum_{r=1}^n f(r)=\sum_{r=1}^n(12r-6)=12\sum_{r=1}^n r-6n=12\cdot\frac{n(n+1)}{2}-6n=6n(n+1)-6n=6n^2+6n-6n=6n^2$.

   Common Mistakes

   • Assuming $f(x+y)=f(x)+f(y)$ (Cauchy's equation) instead of correctly using the given asymmetric form $f(x+y)=f(x)+12y$.
   • Forgetting to solve for $f(0)$ and instead assuming $f(0)=0$.

   ✓Final answer

   The correct option is (C) — $6n^2$.

   ANSWER: C

6. AP EAPCET 2024Set eng-2024-05-21-AN1 markMCQ
   Q.If $1.3.5 + 3.5.7 + 5.7.9 + \cdots n \text{ terms} = n(n+1)f(n) - 3n$, then $f(1) =$ (A) $9$ (B) $11$ (C) $12$ (D) $8$

   ›Reveal solutionSolution

   Summing the series $\sum (2r-1)(2r+1)(2r+3)$ using the standard power-sum formulas gives $S(n)=n(n+1)(2n^2+6n+1)-3n$, so $f(n)=2n^2+6n+1$ and $f(1)=9$.

   Concept and Intuition

   The $r$-th term of this series is a product of three consecutive odd-type linear factors in $r$; expanding it into a cubic polynomial in $r$ lets us sum term-by-term using $\sum r$, $\sum r^2$, $\sum r^3$, then factor out $n(n+1)$ to match the given closed form and read off $f(n)$.

   Step-by-Step Solution

   1. General term: $T_r = (2r-1)(2r+1)(2r+3)$.
   2. Expand: $(2r-1)(2r+1) = 4r^2-1$. So $T_r = (4r^2-1)(2r+3) = 8r^3+12r^2-2r-3$.
   3. Sum from $r=1$ to $n$: $S(n) = 8\sum r^3 + 12\sum r^2 - 2\sum r - 3n$.
   4. Use $\sum r^3 = \left[\dfrac{n(n+1)}{2}\right]^2$, $\sum r^2=\dfrac{n(n+1)(2n+1)}{6}$, $\sum r=\dfrac{n(n+1)}{2}$:

   $S(n) = 8\cdot\frac{n^2(n+1)^2}{4} + 12\cdot\frac{n(n+1)(2n+1)}{6} - 2\cdot\frac{n(n+1)}{2} - 3n$

   $= 2n^2(n+1)^2 + 2n(n+1)(2n+1) - n(n+1) - 3n$

   5. Factor $n(n+1)$ out of the first three terms:

   $S(n) = n(n+1)\big[2n(n+1) + 2(2n+1) - 1\big] - 3n = n(n+1)\big[2n^2+2n+4n+2-1\big]-3n$

   $= n(n+1)(2n^2+6n+1) - 3n$

   6. Comparing with $n(n+1)f(n)-3n$: $f(n) = 2n^2+6n+1$. So $f(1) = 2(1)+6(1)+1 = 9$.
   7. Cross-check directly at $n=1$: the sum of "1 term" is just $1\cdot3\cdot5=15$, and the RHS formula gives $1\cdot2\cdot f(1)-3\cdot1 = 2f(1)-3$; setting $15=2f(1)-3$ gives $f(1)=9$ — consistent.

   Common Mistakes

   • Arithmetic slips while combining the $\sum r^3, \sum r^2, \sum r$ terms before factoring $n(n+1)$ — always double check by plugging in a small $n$ (like $n=1$) directly into both the raw sum and the derived closed form.

   ✓Final answer

   The correct option is (A) — $9$.

   ANSWER: A

7. AP EAPCET 2023Set eng-2023-05-15-AN1 markMCQ
   Q.If $f(a) = \log\left|\dfrac{1-a}{1+a}\right|$ for $a \neq \{-1,1\}$, then the set of values of all 'a', for which $f\left(\dfrac{2a}{1+a^2}\right) > 0$ is (A) $(0,\infty) - \{1\}$ (B) $(-\infty,0) - \{-1\}$ (C) $(-\infty,\infty) - \{-1,1\}$ (D) $(-1,1)$

   ›Reveal solutionSolution

   The substitution $x=\frac{2a}{1+a^2}$ into $f(x)=\log\left|\frac{1-x}{1+x}\right|$

   turns out to double the function: $f\!\left(\frac{2a}{1+a^2}\right)=2f(a)$. So the

   given inequality reduces to simply $f(a)>0$, solved to give $a\in(-\infty,0)-\{-1\}$.

   Concept and Intuition

   This looks intimidating only because of the nested substitution. The trick is to

   notice that $\frac{2a}{1+a^2}$ is exactly the tangent-half-angle-style substitution

   that turns $\log\left|\frac{1-x}{1+x}\right|$ into a clean multiple of $f(a)$ itself —

   so instead of manipulating the messy compound expression directly, first simplify

   $f\!\left(\frac{2a}{1+a^2}\right)$ symbolically in terms of $f(a)$, then solve the

   resulting simple inequality in $a$.

   Step-by-Step Solution

   1. Let $x = \dfrac{2a}{1+a^2}$. Compute $1-x$ and $1+x$:

   $1-x = \frac{1+a^2-2a}{1+a^2} = \frac{(1-a)^2}{1+a^2}, \qquad 1+x = \frac{1+a^2+2a}{1+a^2} = \frac{(1+a)^2}{1+a^2}.$

   2. So $\dfrac{1-x}{1+x} = \dfrac{(1-a)^2}{(1+a)^2}$, and

   $f(x) = \log\left|\frac{(1-a)^2}{(1+a)^2}\right| = \log\left(\frac{1-a}{1+a}\right)^2 = 2\log\left|\frac{1-a}{1+a}\right| = 2f(a).$

   3. The condition becomes $2f(a) > 0 \iff f(a) > 0 \iff \log\left|\frac{1-a}{1+a}\right|>0 \iff \left|\frac{1-a}{1+a}\right| > 1$.
   4. Since both sides are non-negative, square: $(1-a)^2 > (1+a)^2$. Expand: $1 - 2a + a^2 > 1 + 2a + a^2 \Rightarrow -4a > 0 \Rightarrow a < 0$.
   5. Domain of $f$ excludes $a=\pm 1$. Check which of these lands in $a<0$: only $a=-1$ (since $a=1$ is already outside $a<0$). Also verify the inner domain: $\frac{2a}{1+a^2}=1 \Rightarrow a=1$; $\frac{2a}{1+a^2}=-1 \Rightarrow a=-1$ — the same two exclusions, consistent.
   6. Final answer: $a \in (-\infty, 0) - \{-1\}$.

   Common Mistakes

   • Trying to directly simplify $\frac{2a}{1+a^2}$ inside the log without first spotting the "doubling" identity — leads to a much messier, error-prone algebra path.
   • Forgetting to exclude $a=-1$ from the final answer (the domain restriction on $f$ survives through the substitution).
   • Sign error when squaring the absolute-value inequality (forgetting both sides are already non-negative, so squaring is valid and direction-preserving here).

   ✓Final answer

   The correct option is (B) — $(-\infty,0)-\{-1\}$.

   ANSWER: B

8. AP EAPCET 2023Set eng-2023-05-16-AN1 markMCQ
   Q.If $f: \mathbb{R} \to \mathbb{R}$ is defined by $f(x+y) = f(x) + f(y)$ $\forall x,y \in \mathbb{R}$ and $f(1) = 7$, then $\sum_{r=1}^{n} f(r) =$ (A) $\dfrac{3n(n+2)}{4}$ (B) $\dfrac{n(n-1)}{2}$ (C) $\dfrac{7n(n+1)}{2}$ (D) $\dfrac{(n+1)(n+2)}{4}$

   ›Reveal solutionSolution

   The additive functional equation forces $f(r)=7r$, so the sum telescopes to $\frac{7n(n+1)}{2}$.

   Concept and Intuition

   $f(x+y)=f(x)+f(y)$ for all reals means $f$ is additive. Setting $x=y=0$ gives $f(0)=0$; setting $y=x$ repeatedly (or by induction) gives $f(nx)=nf(x)$ for integers $n$. In particular $f(n)=nf(1)$ for every positive integer $n$ — this is the key fact used here (we don't need the full pathological-solutions theory since we only evaluate at integers).

   Step-by-Step Solution

   1. By induction, $f(n)=n\,f(1)$ for every positive integer $n$: $f(2)=f(1)+f(1)=2f(1)$, $f(3)=f(2)+f(1)=3f(1)$, and so on.
   2. Given $f(1)=7$, we get $f(r)=7r$ for every positive integer $r$.
   3. $\displaystyle\sum_{r=1}^n f(r)=\sum_{r=1}^n 7r=7\cdot\frac{n(n+1)}{2}=\frac{7n(n+1)}{2}$.

   Common Mistakes

   • Trying to "solve" the Cauchy equation in full generality (needs continuity/regularity for non-integer arguments) when the problem only needs the simple integer induction.
   • Arithmetic slip turning $\frac{n(n+1)}{2}$ into $\frac{n(n-1)}{2}$.

   ✓Final answer

   The correct option is (C) — $\dfrac{7n(n+1)}{2}$.

   ANSWER: C

9. AP EAPCET 2023Set eng-2023-05-17-AN1 markMCQ
   Q.If $f(0)=0$, $f(1)=1$, $f(2)=2$ and $f(x)=f(x-2)+f(x-3)$ for $x=3,4,5,\ldots$, then $f(10)=$ (A) 13 (B) 9 (C) 11 (D) 10

   ›Reveal solutionSolution

   Tests direct application of a linear recurrence relation. Answer: $f(10)=13$ (option A).

   Concept and Intuition

   This is a Tribonacci-like recurrence where each term depends on two earlier terms (2 and 3 steps back, not consecutive). The only reliable approach is to compute term by term from the given initial values — there's no shortcut needed since we just need one specific value.

   Step-by-Step Solution

   1. Given: $f(0)=0,\ f(1)=1,\ f(2)=2$.
   2. $f(3)=f(1)+f(0)=1+0=1$
   3. $f(4)=f(2)+f(1)=2+1=3$
   4. $f(5)=f(3)+f(2)=1+2=3$
   5. $f(6)=f(4)+f(3)=3+1=4$
   6. $f(7)=f(5)+f(4)=3+3=6$
   7. $f(8)=f(6)+f(5)=4+3=7$
   8. $f(9)=f(7)+f(6)=6+4=10$
   9. $f(10)=f(8)+f(7)=7+6=13$

   Common Mistakes

   • Misreading the recurrence as $f(x-1)+f(x-2)$ (standard Fibonacci-style) instead of the actual $f(x-2)+f(x-3)$ given.
   • Arithmetic slips while building up the sequence — best to tabulate each value as computed.

   ✓Final answer

   The correct option is (A) — 13.

   ANSWER: A

10. AP EAPCET 2022Set eng-2022-07-04-FN1 markMCQ
    Q.If $f(x) = ax^2 + bx + c$ for some $a, b, c \in \mathbb{R}$ with $a + b + c = 3$ and $f(x+y) = f(x) + f(y) + xy$ $\forall x, y \in \mathbb{R}$. Then $\sum_{n=1}^{10} f(n) =$ (A) $330$ (B) $255$ (C) $165$ (D) $190$

    ›Reveal solutionSolution

    The functional equation pins down $a=1/2$, $b=5/2$, $c=0$; summing $f(n)$ from 1 to 10 then gives 330.

    Concept and Intuition

    A functional equation like $f(x+y)=f(x)+f(y)+xy$ constrains the coefficients of the assumed quadratic form directly — matching same-degree terms on both sides (in $x$, $y$, and the cross term $xy$) determines $a$, $b$, $c$ uniquely (together with the given linear constraint).

    Step-by-Step Solution

    1. Put $x=y=0$: $f(0)=f(0)+f(0)+0 \Rightarrow f(0)=0 \Rightarrow c=0$.
    2. Expand $f(x+y)=a(x+y)^2+b(x+y)+c = ax^2+2axy+ay^2+bx+by+c$.
    3. Expand RHS: $f(x)+f(y)+xy = ax^2+bx+c+ay^2+by+c+xy$.
    4. Equate: $ax^2+2axy+ay^2+bx+by+c = ax^2+ay^2+bx+by+2c+xy$.
    5. Cancel common terms: $2axy+c = 2c+xy \Rightarrow (2a-1)xy = c$ for all $x,y \Rightarrow 2a-1=0$ and $c=0$ (consistent with step 1). So $a=1/2$.
    6. From $a+b+c=3$: $1/2+b+0=3 \Rightarrow b=5/2$.
    7. $f(n) = \dfrac{1}{2}n^2+\dfrac{5}{2}n$.
    8. $\sum_{n=1}^{10}f(n) = \dfrac{1}{2}\sum n^2 + \dfrac{5}{2}\sum n = \dfrac{1}{2}(385)+\dfrac{5}{2}(55) = 192.5+137.5=330$.

    Common Mistakes

    • Forgetting to also use the given constraint $a+b+c=3$ after finding $a$ and $c$ from the functional equation.
    • Using wrong standard sums (e.g. $\sum_{1}^{10}n^2=385$, $\sum_1^{10}n=55$ — mixing these up is a common slip).

    ✓Final answer

    The correct option is (A) — 330.

    ANSWER: A

11. AP EAPCET 2022Set eng-2022-07-04-FN1 markMCQ
    Q.Let $f: R^+ \to R^+$ be a function satisfying $f(x) - x = \lambda$ (constant), $\forall x \in R^+$ and $f(xf(y)) = f(xy) + x$, $\forall x, y \in R^+$. Then $\lim_{x \to 0} \frac{(f(x))^{1/3} - 1}{(f(x))^{1/2} - 1} =$ (A) $\frac{1}{3}$ (B) $0$ (C) $\frac{2}{3}$ (D) $1$

    ›Reveal solutionSolution

    The functional equation forces $f(x)=x+1$; then the requested limit reduces to a standard $(1+x)^n-1\sim nx$ comparison, giving $2/3$.

    Concept and Intuition

    First pin down the unknown function using the given functional equation, then the limit becomes a routine application of the small-$x$ expansion $(1+x)^n \approx 1+nx$, since both numerator and denominator vanish at $x=0$ (a $0/0$ form whose ratio is governed by the leading-order linear terms).

    Step-by-Step Solution

    1. $f(x)-x=\lambda$ (constant) $\Rightarrow f(x)=x+\lambda$.
    2. Substitute into $f(xf(y))=f(xy)+x$: LHS $=f(x(y+\lambda))=x(y+\lambda)+\lambda=xy+x\lambda+\lambda$. RHS $=xy+\lambda+x$.
    3. Equate: $xy+x\lambda+\lambda=xy+\lambda+x \Rightarrow x\lambda=x$ for all $x>0 \Rightarrow \lambda=1$.
    4. So $f(x)=x+1$.
    5. Limit: $\lim_{x\to0}\dfrac{(x+1)^{1/3}-1}{(x+1)^{1/2}-1}$. As $x\to0$, using $(1+x)^n-1\approx nx$: numerator $\approx\dfrac13x$, denominator $\approx\dfrac12x$.
    6. Ratio $\to \dfrac{1/3}{1/2}=\dfrac23$.

    Common Mistakes

    • Forgetting to first determine $\lambda$ from the functional equation before evaluating the limit.
    • Using the wrong order term in the binomial-type expansion (mixing up which exponent goes with which root).

    ✓Final answer

    The correct option is (C) — $\dfrac{2}{3}$.

    ANSWER: C

12. AP EAPCET 2022Set eng-2022-07-06-FN1 markMCQ
    Q.Let f be a function defined by $f(xy) = \dfrac{f(x)}{y}$ for all positive real numbers x and y. If f(30) = 20, then f(40) = (A) 10 (B) 15 (C) 25 (D) 17

    ›Reveal solutionSolution

    Choosing $y=40/30$ in the functional equation directly relates $f(40)$ to the known value $f(30)=20$, giving $f(40)=15$.

    Concept and Intuition

    The given rule $f(xy)=f(x)/y$ lets us compute $f$ at any target point by picking a convenient factorisation of that point as $x\cdot y$ where $x$ is a value we already know $f$ at. Here we know $f(30)$, so we should express $40$ as $30\times y$ for the right choice of $y$, then read off $f(40)$ directly. (As a cross-check, this functional equation is exactly satisfied by any function of the form $f(t)=k/t$, since $f(xy)=k/(xy)=(k/x)/y=f(x)/y$ — confirming this is an inverse-proportionality relationship.)

    Step-by-Step Solution

    1. We want $f(40)$ and we know $f(30)=20$. Write $40=30\cdot y$, so $y=\dfrac{40}{30}=\dfrac{4}{3}$ (a valid positive real number, as required).
    2. Apply the functional equation with $x=30$: $f(30\cdot y)=\dfrac{f(30)}{y}$, i.e. $f(40)=\dfrac{f(30)}{4/3}=f(30)\cdot\dfrac{3}{4}$.
    3. Substitute $f(30)=20$: $f(40)=20\cdot\dfrac34=15$.
    4. Cross-check with the form $f(t)=k/t$: $f(30)=k/30=20\Rightarrow k=600$; then $f(40)=600/40=15$, matching.

    Common Mistakes

    • Trying to directly scale $f(30)$ by $40/30$ (i.e. multiplying rather than dividing by $y$) — the rule divides by $y$, since $f(xy)=f(x)/y$, not $f(x)\cdot y$.
    • Picking an inconvenient factorisation of 40 that doesn't use the known value $f(30)$.

    ✓Final answer

    The correct option is (B) — 15.

    ANSWER: B