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 31 on this concept.

1. TG EAPCET 2026Set eng-2026-05-11-FN1 markMCQ
   Q.If $\sum_{n=1}^{10} \left(i^n + i^{n+1} + i^{n+2}\right) = x + iy$, then $2x + 3y =$ (A) $1$ (B) $-5$ (C) $5i$ (D) $-i$

   ›Reveal solutionSolution

   The sum cycles through powers of $i$; grouping terms reveals cancellation, giving $x = -1$ and $y = 0$, so $2x + 3y = -2$.

   The key here is to notice that $i^n$, $i^{n+1}$, and $i^{n+2}$ are three consecutive powers of $i$. Since $i^4 = 1$, the powers repeat every 4. So the sum over $n = 1$ to $10$ can be handled by looking at the pattern of these triplets.

   Let’s first recall the cycle:

   $i^1 = i$, $i^2 = -1$, $i^3 = -i$, $i^4 = 1$, and then it repeats.

   For any $n$, the sum $i^n + i^{n+1} + i^{n+2}$ is just three consecutive terms from this cycle. If we add all such triplets for $n = 1$ to $10$, we are effectively adding every power from $i^1$ up to $i^{12}$, but each power appears in exactly three triplets? Let’s check carefully.

   1. Write out the sum explicitly:

   $S = \sum_{n=1}^{10} \left(i^n + i^{n+1} + i^{n+2}\right)$

   This expands to:

   $S = (i^1 + i^2 + i^3) + (i^2 + i^3 + i^4) + (i^3 + i^4 + i^5) + \cdots + (i^{10} + i^{11} + i^{12})$

   2. Count how many times each power appears.

      $i^1$ appears only in the first triplet.

      $i^2$ appears in the first and second triplets — twice.

      $i^3$ appears in the first, second, and third — three times.

      $i^4$ appears in the second, third, and fourth — three times.

      In general, for $k$ from 3 to 10, $i^k$ appears three times.

      $i^{11}$ appears in the last two triplets — twice.

      $i^{12}$ appears only in the last triplet — once.

      So the sum is:

   $S = i^1 + 2i^2 + 3(i^3 + i^4 + i^5 + i^6 + i^7 + i^8 + i^9 + i^{10}) + 2i^{11} + i^{12}$

   3. Now use the cycle: $i^1 = i$, $i^2 = -1$, $i^3 = -i$, $i^4 = 1$, $i^5 = i$, $i^6 = -1$, $i^7 = -i$, $i^8 = 1$, $i^9 = i$, $i^{10} = -1$, $i^{11} = -i$, $i^{12} = 1$.

      The block $i^3$ to $i^{10}$ covers two full cycles (since $i^3$ to $i^6$ is one cycle, $i^7$ to $i^{10}$ is the next). Each full cycle of four consecutive powers sums to $i + (-1) + (-i) + 1 = 0$. So $i^3 + i^4 + i^5 + i^6 = 0$ and $i^7 + i^8 + i^9 + i^{10} = 0$. Therefore the entire block of eight terms sums to 0.

      So:

   $S = i^1 + 2i^2 + 2i^{11} + i^{12}$

   Substitute values:

   $S = i + 2(-1) + 2(-i) + 1 = i - 2 - 2i + 1 = -1 - i$

   4. Hence $x = -1$ and $y = -1$. Then:

   $2x + 3y = 2(-1) + 3(-1) = -2 - 3 = -5$

   Watch out

   A common mistake is to forget that $i^{11} = -i$ and $i^{12} = 1$, or to miscount how many times each power appears. Double-check the endpoints.

   Tip

   A faster way: notice that $i^n + i^{n+1} + i^{n+2} = i^n(1 + i + i^2) = i^n(1 + i - 1) = i^n \cdot i = i^{n+1}$. So the whole sum is $\sum_{n=1}^{10} i^{n+1} = \sum_{k=2}^{11} i^k$. That’s a geometric series with first term $i^2 = -1$, ratio $i$, and 10 terms. Sum = $-1 \cdot \frac{i^{10} - 1}{i - 1} = -1 \cdot \frac{(-1) - 1}{i - 1} = -1 \cdot \frac{-2}{i-1} = \frac{2}{i-1}$. Rationalize: $\frac{2}{i-1} \cdot \frac{-i-1}{-i-1} = \frac{2(-i-1)}{1+1} = \frac{2(-i-1)}{2} = -1 - i$. Same result.

   ✓Final answer

   The value is $\boxed{-5}$, which corresponds to option (B).

2. TG EAPCET 2026Set eng-2026-05-11-AN1 markMCQ
   Q.The sum of all the integers in the domain of the real valued function

   [!FORMULA] $f(x) = \sqrt{\log_{1/3} \left( \frac{9x-30}{x-26} \right)}$

   is (A) 6 (B) 9 (C) 12 (D) 10

   ›Reveal solutionSolution

   The domain requires the log argument positive and the log value non-negative. Because the base $\tfrac13<1$, this forces $0<\tfrac{9x-30}{x-26}\le 1$, giving $\tfrac12\le x<\tfrac{10}{3}$. The integers are $1,2,3$, summing to $6$ — option (A).

   Conditions for the domain

   For $f(x)=\sqrt{\log_{1/3}\!\left(\dfrac{9x-30}{x-26}\right)}$ we need:

   1. the logarithm to be defined: $\dfrac{9x-30}{x-26} > 0$;
   2. the radicand non-negative: $\log_{1/3}\!\left(\dfrac{9x-30}{x-26}\right) \ge 0$.

   Since the base $\tfrac13<1$, the logarithm is a decreasing function, so $\log_{1/3}(u)\ge 0 \iff 0<u\le 1$. Both conditions combine into

   $0 < \frac{9x-30}{x-26} \le 1.$

   Step 1 — Positivity

   $\dfrac{9x-30}{x-26}>0$: numerator zero at $x=\tfrac{10}{3}$, denominator zero at $x=26$. Sign analysis gives

   $x < \tfrac{10}{3} \quad\text{or}\quad x > 26.$

   Step 2 — Upper bound $\le 1$

   $\frac{9x-30}{x-26} - 1 = \frac{9x-30-(x-26)}{x-26} = \frac{8x-4}{x-26} = \frac{4(2x-1)}{x-26} \le 0.$

   Critical points $x=\tfrac12$ and $x=26$; the expression is $\le 0$ for

   $\tfrac12 \le x < 26.$

   Step 3 — Intersection

   Combining $\left(x<\tfrac{10}{3}\ \text{or}\ x>26\right)$ with $\tfrac12\le x<26$:

   $\frac12 \le x < \frac{10}{3}.$

   Step 4 — Integers in the domain

   With $\tfrac{10}{3}\approx 3.33$, the integers satisfying $\tfrac12\le x<\tfrac{10}{3}$ are $1,\,2,\,3$.

   Check $x=3$: $\dfrac{27-30}{3-26}=\dfrac{-3}{-23}=\dfrac{3}{23}\in(0,1]$, valid. Check $x=4$: $\dfrac{36-30}{4-26}=\dfrac{6}{-22}<0$, log undefined — correctly excluded.

   Sum $=1+2+3=6$.

   ✓Final answer

   The sum of all integers in the domain is $6$ — option (A).

   ANSWER: A

3. 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

4. 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

5. 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

6. KCET 2024Set A-11 markMCQ
   Q.In the expansion of $(1 + x)^n$ $\dfrac{C_1}{C_0} + \dfrac{C_2}{C_1} + \dfrac{3C_3}{2} + \ldots + n \dfrac{C_n}{C_{n-1}}$ is equal to (A) $\dfrac{n(n+1)}{2}$ (B) $\dfrac{n}{2}$ (C) $\dfrac{n+1}{2}$ (D) $3n(n+1)$

   ›Reveal solutionSolution

   Use $\dfrac{C_k}{C_{k-1}} = \dfrac{n-k+1}{k}$ so the $k$-th term $k\cdot\frac{C_k}{C_{k-1}}$ simplifies to $(n-k+1)$; summing gives the sum of the first $n$ natural numbers.

   Note on the printed stem. The series is the standard one

   $\frac{C_1}{C_0} + \frac{2C_2}{C_1} + \frac{3C_3}{C_2} + \cdots + \frac{n\,C_n}{C_{n-1}},$

   whose last term is printed correctly as $n\dfrac{C_n}{C_{n-1}}$. (The third term appears in the paper as "$\tfrac{3C_3}{2}$", a typographical mangling of $\tfrac{3C_3}{C_2}$ — the general term stated by the final printed term is unambiguous, so the intended series is clear.)

   Step 1 — The concept: the ratio of consecutive binomial coefficients.

   With $C_k = \binom{n}{k} = \dfrac{n!}{k!\,(n-k)!}$,

   $\frac{C_k}{C_{k-1}} = \frac{\dfrac{n!}{k!\,(n-k)!}}{\dfrac{n!}{(k-1)!\,(n-k+1)!}} = \frac{(k-1)!\,(n-k+1)!}{k!\,(n-k)!}.$

   Now simplify using $k! = k\,(k-1)!$ and $(n-k+1)! = (n-k+1)(n-k)!$:

   $\frac{C_k}{C_{k-1}} = \frac{(k-1)!}{k\,(k-1)!}\cdot\frac{(n-k+1)(n-k)!}{(n-k)!} = \frac{n-k+1}{k}.$

   This is the key identity:

   $\boxed{\frac{C_k}{C_{k-1}} = \frac{n-k+1}{k}}$

   Step 2 — Simplify the general term of the series.

   The $k$-th term of the given sum is $T_k = k\cdot\dfrac{C_k}{C_{k-1}}$. Substituting the identity:

   $T_k = k \cdot \frac{n-k+1}{k} = n - k + 1.$

   The $k$ cancels beautifully — this is why the identity is the right tool: it turns a messy ratio of factorials into a simple linear term.

   Step 3 — Verify the first and last terms.

   • $k = 1$: $T_1 = n - 1 + 1 = n$. Directly, $\dfrac{C_1}{C_0} = \dfrac{n}{1} = n$ ✓
   • $k = n$: $T_n = n - n + 1 = 1$. Directly, $n\dfrac{C_n}{C_{n-1}} = n\cdot\dfrac{1}{n} = 1$ ✓

   Step 4 — Sum from $k=1$ to $n$.

   $\sum_{k=1}^{n} T_k = \sum_{k=1}^{n} (n-k+1) = n + (n-1) + (n-2) + \cdots + 2 + 1.$

   This is just the sum of the first $n$ natural numbers written backwards:

   $\sum_{k=1}^{n}(n-k+1) = \sum_{j=1}^{n} j = \frac{n(n+1)}{2}.$

   Step 5 — Spot-check with a small $n$.

   Take $n = 3$, so $C_0=1, C_1=3, C_2=3, C_3=1$:

   $\frac{C_1}{C_0} + \frac{2C_2}{C_1} + \frac{3C_3}{C_2} = \frac{3}{1} + \frac{2(3)}{3} + \frac{3(1)}{3} = 3 + 2 + 1 = 6.$

   Formula: $\dfrac{n(n+1)}{2} = \dfrac{3\cdot 4}{2} = 6$ ✓ — matches, and rules out (B) $\tfrac{n}{2}=1.5$, (C) $\tfrac{n+1}{2}=2$, and (D) $3n(n+1)=36$.

   ✓Final answer

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

   ANSWER: A

7. TG EAPCET 2024Set eng-2024-05-09-FN1 markMCQ
   Q.If $^nC_r = C_r$ and $2 \dfrac{C_1}{C_0} + 4 \dfrac{C_2}{C_1} + 6 \dfrac{C_3}{C_2} + \ldots + 2n \dfrac{C_n}{C_{n-1}} = 650$, then $^nC_2 =$ (A) $25$ (B) $300$ (C) $225$ (D) $625$

   ›Reveal solutionSolution

   The key is to simplify each term $2k \frac{C_k}{C_{k-1}}$ using the binomial coefficient ratio formula, which reduces to $2(n-k+1)$. Summing this arithmetic series gives $n(n+1)=650$, so $n=25$, and then $^nC_2 = 300$.

   We are given that $^nC_r$ is written as $C_r$ for brevity. The sum

   $2 \frac{C_1}{C_0} + 4 \frac{C_2}{C_1} + 6 \frac{C_3}{C_2} + \dots + 2n \frac{C_n}{C_{n-1}} = 650$

   and we need $^nC_2$.

   Concept and intuition:

   Each term looks like $2k \cdot \frac{C_k}{C_{k-1}}$. The ratio $\frac{C_k}{C_{k-1}}$ for binomial coefficients has a simple closed form:

   $\frac{\binom{n}{k}}{\binom{n}{k-1}} = \frac{n-k+1}{k}.$

   Multiplying by $2k$ cancels the denominator, leaving a linear expression in $k$ and $n$. The sum then becomes an arithmetic series, easy to evaluate.

   Step-by-step:

   1. Write the general term.

      For $k = 1, 2, \dots, n$, the $k$-th term is $2k \cdot \frac{C_k}{C_{k-1}}$.

   2. Simplify the ratio.

   $\frac{C_k}{C_{k-1}} = \frac{\binom{n}{k}}{\binom{n}{k-1}} = \frac{\frac{n!}{k!(n-k)!}}{\frac{n!}{(k-1)!(n-k+1)!}} = \frac{(k-1)!(n-k+1)!}{k!(n-k)!} = \frac{n-k+1}{k}.$

   3. Multiply by $2k$.

   $2k \cdot \frac{C_k}{C_{k-1}} = 2k \cdot \frac{n-k+1}{k} = 2(n-k+1).$

   4. Write the sum. The sum becomes

   $\sum_{k=1}^{n} 2(n-k+1) = 2\sum_{k=1}^{n} (n-k+1).$

   Let $j = n-k+1$. As $k$ runs from $1$ to $n$, $j$ runs from $n$ down to $1$. So

   $\sum_{k=1}^{n} (n-k+1) = \sum_{j=1}^{n} j = \frac{n(n+1)}{2}.$

   5. Set equal to 650.

   $2 \cdot \frac{n(n+1)}{2} = n(n+1) = 650.$

   Solve $n^2 + n - 650 = 0$. Discriminant: $1 + 2600 = 2601 = 51^2$. So

   $n = \frac{-1 \pm 51}{2}.$

   Positive root: $n = \frac{50}{2} = 25$.

   6. Find $^nC_2$.

   $\binom{25}{2} = \frac{25 \cdot 24}{2} = 25 \cdot 12 = 300.$

   Watch out

   A common mistake is to forget that the sum runs from $k=1$ to $n$, not $k=0$. The term for $k=0$ would be undefined because $C_{-1}$ doesn't exist. Always check the index range.

   Tip

   The simplification $2k \cdot \frac{C_k}{C_{k-1}} = 2(n-k+1)$ is very neat: it turns a complicated-looking sum into a simple arithmetic progression. Memorizing the ratio formula $\frac{\binom{n}{k}}{\binom{n}{k-1}} = \frac{n-k+1}{k}$ saves time in many combinatorial sums.

   ✓Final answer

   The correct option is (B).

   ANSWER: B

8. 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

9. 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

10. 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

11. WBJEE 2024Set math-20241 markMCQ
    Q.For every real number $x \neq -1$, let $f(x) = \frac{x}{x + 1}$. Write $f_1(x) = f(x)$ & for $n \geq 2$, $f_n(x) = f(f_{n-1}(x))$. Then $f_1(-2) \cdot f_2(-2) \ldots f_n(-2)$ must be (A) $\frac{2^n}{1 \cdot 3 \cdot 5 \ldots (2n - 1)}$ (B) $1$ (C) $\frac{1}{2}\binom{2n}{n}$ (D) $\binom{2n}{n}$

    ›Reveal solutionSolution

    Find a closed form for the iterate $f_n$, then multiply.

    With $f(x)=\dfrac{x}{x+1}$, compute the iterate. Claim $f_n(x)=\dfrac{x}{1+nx}$:

    $f_1(x)=\frac{x}{1+x},\quad f_2(x)=\frac{f_1}{f_1+1}=\frac{x/(1+x)}{(1+2x)/(1+x)}=\frac{x}{1+2x},$

    and inductively $f_n(x)=\dfrac{x}{1+nx}$.

    At $x=-2$: $f_k(-2)=\dfrac{-2}{1-2k}$. So

    $\prod_{k=1}^{n} f_k(-2)=\prod_{k=1}^{n}\frac{-2}{1-2k}=\frac{(-2)^n}{\prod_{k=1}^{n}(1-2k)}.$

    Here $1-2k=-(2k-1)$, so $\prod_{k=1}^{n}(1-2k)=(-1)^n\,[1\cdot3\cdot5\cdots(2n-1)]$. Thus

    $\prod_{k=1}^{n} f_k(-2)=\frac{(-1)^n 2^n}{(-1)^n\,1\cdot3\cdot5\cdots(2n-1)}=\frac{2^n}{1\cdot3\cdot5\cdots(2n-1)}.$

    ✓Final answer

    $\dfrac{2^n}{1\cdot3\cdot5\cdots(2n-1)}$.

    ANSWER: A

12. TG EAPCET 2023Set eng-2023-05-13-AN1 markMCQ
    Q.The numerically greatest term in the binomial expansion of $(2x-3y)^5$ when $x = \dfrac{3}{2}$ and $y = \dfrac{2}{3}$ is (A) 360 (B) 1080 (C) 720 (D) 2160

    ›Reveal solutionSolution

    The numerically greatest term in the expansion is found by comparing the ratio of consecutive terms; after substituting the given values, the 4th term (index 3) is the largest, and its value is 1080.

    Concept & Intuition

    In a binomial expansion like $(a+b)^n$, the terms are not equal in size. To find which term has the largest numerical value (ignoring sign), we compare consecutive terms using their ratio. If the ratio $T_{r+1}/T_r$ is greater than 1, terms are still growing; when it drops below 1, we’ve passed the maximum. This is similar to finding the peak of a sequence by checking where it stops increasing.

    Here, $a = 2x$ and $b = -3y$, so the expansion alternates signs. But “numerically greatest” means we consider absolute values. We’ll substitute $x = 3/2$ and $y = 2/3$ first, then find the term with the largest absolute value.

    ---

    Step-by-step solution

    1. Substitute the given values

    $x = \frac{3}{2},\quad y = \frac{2}{3}$

    Then

    $2x = 2 \cdot \frac{3}{2} = 3,\qquad 3y = 3 \cdot \frac{2}{3} = 2$

    So the expression becomes $(3 - 2)^5$. Wait — that’s just $1^5 = 1$. That seems too trivial. But careful: the binomial expansion is of $(2x - 3y)^5$, and after substitution we get $(3 - 2)^5 = 1$. That means the sum of all terms is 1, but individual terms can be much larger in magnitude (they cancel). So we must find the term with the largest absolute value.

    2. General term in the expansion The $(r+1)$-th term (starting with $r=0$) is

    $T_{r+1} = \binom{5}{r} (2x)^{5-r} (-3y)^r$

    Substitute $2x = 3$ and $3y = 2$:

    $T_{r+1} = \binom{5}{r} (3)^{5-r} (-2)^r$

    The absolute value is

    $|T_{r+1}| = \binom{5}{r} \cdot 3^{5-r} \cdot 2^r$

    3. Find where terms stop growing Consider the ratio of consecutive absolute terms:

    $\frac{|T_{r+2}|}{|T_{r+1}|} = \frac{\binom{5}{r+1} \cdot 3^{4-r} \cdot 2^{r+1}}{\binom{5}{r} \cdot 3^{5-r} \cdot 2^r}$

    Simplify:

    $= \frac{5-r}{r+1} \cdot \frac{2}{3}$

    The terms increase as long as this ratio > 1:

    $\frac{5-r}{r+1} \cdot \frac{2}{3} > 1 \quad\Rightarrow\quad \frac{5-r}{r+1} > \frac{3}{2}$

    Cross-multiply (positive denominators):

    $2(5-r) > 3(r+1) \quad\Rightarrow\quad 10 - 2r > 3r + 3 \quad\Rightarrow\quad 7 > 5r \quad\Rightarrow\quad r < \frac{7}{5} = 1.4$

    So for $r = 0$ and $r = 1$, the ratio > 1 (terms increase). For $r = 1$, check: ratio = $\frac{5-1}{1+1} \cdot \frac{2}{3} = \frac{4}{2} \cdot \frac{2}{3} = \frac{4}{3} > 1$. So $T_2 < T_3$.

    For $r = 2$: ratio = $\frac{5-2}{2+1} \cdot \frac{2}{3} = \frac{3}{3} \cdot \frac{2}{3} = \frac{2}{3} < 1$. So $T_4 < T_3$.

    Hence the maximum occurs at $r = 2$ (the 3rd term? Wait: $r=0$ is first term, $r=1$ second, $r=2$ third, $r=3$ fourth). Let’s check:

    • $r=1$ gives $T_2$, $r=2$ gives $T_3$, ratio from $T_2$ to $T_3$ > 1, so $T_3 > T_2$.
    • Ratio from $T_3$ to $T_4$ < 1, so $T_3 > T_4$. So the largest term is $T_3$ (the third term, $r=2$).

    Watch out

    A common mistake is to think the largest term is the middle one ($r=2$ or $3$ for $n=5$), but the ratio test is needed because the base values (3 and 2) are not equal. Here it happens to be the third term, but always verify.

    4. Compute the value of the largest term For $r=2$:

    $T_3 = \binom{5}{2} (3)^{5-2} (-2)^2 = 10 \cdot 3^3 \cdot 4 = 10 \cdot 27 \cdot 4 = 1080$

    The sign is positive (since $(-2)^2 = 4$), so the numerical value is 1080.

    Tip

    You could also compute all terms quickly:

    $T_1 = 3^5 = 243$

    $T_2 = 5 \cdot 3^4 \cdot (-2) = -810$ (abs 810)

    $T_3 = 10 \cdot 3^3 \cdot 4 = 1080$

    $T_4 = 10 \cdot 3^2 \cdot (-8) = -720$ (abs 720)

    $T_5 = 5 \cdot 3 \cdot 16 = 240$

    $T_6 = -32$ (abs 32)

    Clearly 1080 is largest.

    ✓Final answer

    The correct option is (B).

    ANSWER: B