Q.Find $n$ in the binomial $\left(\sqrt[3]{2} + \dfrac{1}{\sqrt[3]{3}}\right)^{n}$ if the ratio of $7^{\text{th}}$ term from the beginning to the $7^{\text{th}}$ term from the end is $\dfrac{1}{6}$.

Showing the 12 most recent of 23 on this concept.

1. TG EAPCET 2026Set eng-2026-05-09-AN1 markMCQ
   Q.The term independent of $x$ in the expansion of $\left(\dfrac{\sqrt{x}}{2} - \dfrac{3}{x}\right)^{12}$ is (A) $55 \left(\dfrac{3}{2}\right)^6$ (B) $495 \left(\dfrac{9}{16}\right)^2$ (C) $55 \left(\dfrac{9}{16}\right)^2$ (D) $45 \left(\dfrac{3}{2}\right)^4$

   ›Reveal solutionSolution

   Find the term where the powers of $x$ cancel by setting the general term's exponent to zero. The constant term is $\boxed{495 \left(\dfrac{9}{16}\right)^2}$.

   The binomial expansion of $\left(a + b\right)^n$ gives us terms of the form $\binom{n}{r} a^{n-r} b^r$. When $a$ and $b$ themselves contain powers of $x$, each term in the expansion will have some net power of $x$. The term independent of $x$ is the one where all the $x$'s cancel out—where the exponent of $x$ equals zero.

   Let me write the general term in the expansion of $\left(\dfrac{\sqrt{x}}{2} - \dfrac{3}{x}\right)^{12}$.

   1. Set up the general term

      Using the binomial theorem with $a = \dfrac{\sqrt{x}}{2} = \dfrac{x^{1/2}}{2}$ and $b = -\dfrac{3}{x} = -3x^{-1}$:

   $T_{r+1} = \binom{12}{r} \left(\frac{x^{1/2}}{2}\right)^{12-r} \left(-\frac{3}{x}\right)^r$

   2. Simplify the powers

   $T_{r+1} = \binom{12}{r} \cdot \frac{x^{(12-r)/2}}{2^{12-r}} \cdot \frac{(-3)^r}{x^r}$

   $= \binom{12}{r} \cdot \frac{(-3)^r}{2^{12-r}} \cdot x^{(12-r)/2 - r}$

   3. Find the exponent of $x$

      The power of $x$ is:

   $\frac{12-r}{2} - r = \frac{12-r-2r}{2} = \frac{12-3r}{2}$

   4. Set the exponent to zero

      For the term independent of $x$:

   $\frac{12-3r}{2} = 0$

   $12 - 3r = 0$

   $r = 4$

   5. Calculate the coefficient

      Substituting $r = 4$: …

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2. TG EAPCET 2026Set eng-2026-05-09-FN1 markMCQ
   Q.If $\alpha, \beta$ are the rational roots and $l, m$ are the irrational roots of

   [!FORMULA] $(x^2 - 9x + 11)^2 - (x-4)(x-5) = 3,$

   then $\alpha + \beta + l m =$ (A) $2$ (B) $16$ (C) $8$ (D) $12$

   ›Reveal solutionSolution

   The key idea is to rewrite the equation as a quadratic in disguise by substituting $t = x^2 - 9x + 11$, then solve for $t$, back-substitute, and identify the rational and irrational roots. The final sum is $\alpha + \beta + lm = 12$.

   We start with

   $(x^2 - 9x + 11)^2 - (x-4)(x-5) = 3.$

   The expression $(x-4)(x-5) = x^2 - 9x + 20$. Notice that $x^2 - 9x + 11$ appears squared, and the other term is just 9 more than that. This suggests a substitution to simplify.

   Concept & Intuition:

   When a polynomial equation contains a repeated quadratic expression, we can set that quadratic as a new variable. This often reduces the degree and reveals a hidden structure — here, a quadratic in that new variable. Once solved, we get two quadratics in $x$, whose roots are the four roots of the original equation. We then separate rational from irrational roots.

   ---

   1. Substitute Let $t = x^2 - 9x + 11$. Then $x^2 - 9x = t - 11$, so

   $(x-4)(x-5) = x^2 - 9x + 20 = (t - 11) + 20 = t + 9.$

   The equation becomes

   $t^2 - (t + 9) = 3 \quad \Rightarrow \quad t^2 - t - 12 = 0.$

   2. Solve for $t$

   $t^2 - t - 12 = (t - 4)(t + 3) = 0 \quad \Rightarrow \quad t = 4 \quad \text{or} \quad t = -3.$

   3. Back-substitute
      • For $t = 4$:

   $x^2 - 9x + 11 = 4 \quad \Rightarrow \quad x^2 - 9x + 7 = 0.$

    Discriminant: $81 - 28 = 53$, so roots are $\frac{9 \pm \sqrt{53}}{2}$ — both irrational.
   

   • For $t = -3$:

   $x^2 - 9x + 11 = -3 \quad \Rightarrow \quad x^2 - 9x + 14 = 0.$

    Discriminant: $81 - 56 = 25$, so roots are $\frac{9 \pm 5}{2}$, i.e., $7$ and $2$ — both rational.
   

   4. Identify the roots …

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3. TG EAPCET 2026Set eng-2026-05-09-FN1 markMCQ
   Q.$[y]$ represents the greatest integer less than or equal to $y$ and $\{y\}$ represents the fractional part (or decimal part) of $y$. If $\lim_{x \to 0^+} \left( [1-x] + \dfrac{a^{2[1-x]+\{1-x\}} + [1-x] - 1}{2[1-x] + \{1-x\}} \right) = 11$, then $a=$ (A) $10$ (B) $-11$ (C) $12$ (D) $-9$

   ›Reveal solutionSolution

   As $x\to0^+,\;[1-x]=0$ and $\{1-x\}\to1,$ so the limit $=a-1=11\Rightarrow a=12$ — option (C).

   Evaluate the pieces. For $x\to0^+,$ $1-x$ is just below $1,$ so the greatest integer $[1-x]=0$ and the fractional part $\{1-x\}=(1-x)-0=1-x\to1.$

   Substitute. With $[1-x]=0$ and $\{1-x\}=1-x:$

   • Exponent $2[1-x]+\{1-x\}=1-x\to1.$
   • Numerator $a^{\,2[1-x]+\{1-x\}}+[1-x]-1=a^{\,1-x}+0-1\to a-1.$ …

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4. TG EAPCET 2026Set eng-2026-05-10-AN1 markMCQ
   Q.Numerically greatest term in the expansion of $(2x - 3y)^n$ when $x = \dfrac{3}{2}$, $y = \dfrac{1}{3}$ and $n = 6$ is (A) 1215 (B) 1458 (C) 1024 (D) 2187

   ›Reveal solutionSolution

   The numerically greatest term in the binomial expansion is found by comparing successive terms using the ratio of consecutive coefficients; for the given values, the 5th term (T₅) is the largest, and its value is 1458.

   Concept & Intuition

   In a binomial expansion $(a + b)^n$, the terms are not equal in size. When we want the numerically greatest term (ignoring sign), we compare the absolute values of consecutive terms. The trick is to set up the ratio $\frac{|T_{r+1}|}{|T_r|}$ and find where it crosses from >1 to <1. That tells us the largest term. Here, $a = 2x$ and $b = -3y$, but since we care about magnitude, the minus sign doesn’t matter.

   Step-by-step solution

   1. Write the general term The $(r+1)$-th term in $(2x - 3y)^6$ is

   $T_{r+1} = \binom{6}{r} (2x)^{6-r} (-3y)^r$

   Its absolute value is

   $|T_{r+1}| = \binom{6}{r} (2x)^{6-r} (3y)^r$

   2. Plug in the given values $x = \frac{3}{2}$, $y = \frac{1}{3}$. Then

   $2x = 2 \cdot \frac{3}{2} = 3, \quad 3y = 3 \cdot \frac{1}{3} = 1$

   So the absolute term becomes

   $|T_{r+1}| = \binom{6}{r} \cdot 3^{6-r} \cdot 1^r = \binom{6}{r} \cdot 3^{6-r}$

   3. Form the ratio of consecutive terms

   $\frac{|T_{r+2}|}{|T_{r+1}|} = \frac{\binom{6}{r+1} \cdot 3^{5-r}}{\binom{6}{r} \cdot 3^{6-r}} = \frac{6-r}{r+1} \cdot \frac{1}{3}$

   (Because $\binom{6}{r+1} / \binom{6}{r} = \frac{6-r}{r+1}$ and $3^{5-r}/3^{6-r} = 1/3$.)

   4. Find where the ratio is ≥ 1 The terms increase as long as this ratio > 1: $\frac{6-r}{r+1} \cdot \frac{1}{3} > 1 \quad \Rightarrow \quad 6-r > 3(r+1) \quad \Rightarrow \quad 6-r > 3r+3 \quad \Rightarrow \quad 3 > 4r \quad \Rightarrow \quad r < \frac{3}{4}$ …

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5. TG EAPCET 2025Set eng-2025-05-02-AN1 markMCQ
   Q.When $|x| > 3$, the coefficient of $\dfrac{1}{x^n}$ in the expansion of $x^{3/2}(3+x)^{1/2}$ is (A) $\dfrac{(-1)^n \ 1.3.5...(2n-1) \ 3^n}{2^n n!}$ (B) $\dfrac{(-1)^{n+1} \ 1.3.5...(2n+1) \ 3^{n+2}}{2^{n+2} (n+2)!}$ (C) $\dfrac{(-1)^{n+1} \ 1.3.5...(2n-1) \ 3^{n+1}}{2^n n!}$ (D) $\dfrac{(-1)^{n+1} \ 1.3.5...(2n+1) \ 3^{n+1}}{2^{n+3} (n+2)!}$

   ›Reveal solutionSolution

   The problem asks for the coefficient of $1/x^n$ in the expansion of $x^{3/2}(3+x)^{1/2}$ for $|x|>3$. By rewriting the expression as $x^2 (1 + 3/x)^{1/2}$ and using the binomial series, the coefficient simplifies to $\frac{(-1)^{n+1} 1\cdot3\cdot5\cdots(2n+1)\, 3^{n+1}}{2^{n+3} (n+2)!}$, which matches option (D).

   We start with the expression $x^{3/2}(3+x)^{1/2}$. The condition $|x|>3$ tells us we want an expansion in powers of $1/x$ (a Laurent series valid for large $|x|$). The key is to factor out the dominant term so that the remaining factor becomes a binomial series in $1/x$.

   Concept & Intuition

   When $|x|>3$, we have $|3/x|<1$, so we can expand $(3+x)^{1/2}$ as $x^{1/2}(1+3/x)^{1/2}$ using the binomial theorem for fractional exponents. Multiplying by the $x^{3/2}$ outside gives $x^2 (1+3/x)^{1/2}$. The coefficient of $1/x^n$ then comes from the term where the binomial series gives $x^{2-k}$ with $k=n+2$, so we need the coefficient of $(3/x)^{n+2}$ in the binomial expansion.

   Step-by-step solution

   1. Rewrite the expression for large $|x|$

   $x^{3/2}(3+x)^{1/2} = x^{3/2} \cdot x^{1/2} \left(1 + \frac{3}{x}\right)^{1/2} = x^2 \left(1 + \frac{3}{x}\right)^{1/2}.$

   This is valid because $|3/x|<1$ when $|x|>3$.

   2. Apply the binomial series For $|u|<1$, $(1+u)^{1/2} = \sum_{k=0}^\infty \binom{1/2}{k} u^k$, where

   $\binom{1/2}{k} = \frac{(1/2)(1/2-1)\cdots(1/2-k+1)}{k!}.$

   Here $u = 3/x$, so

   $x^2 \left(1+\frac{3}{x}\right)^{1/2} = x^2 \sum_{k=0}^\infty \binom{1/2}{k} \left(\frac{3}{x}\right)^k = \sum_{k=0}^\infty \binom{1/2}{k} 3^k \, x^{2-k}.$

   3. Identify the term with $1/x^n$ We want the coefficient of $x^{-n}$, i.e., $x^{2-k} = x^{-n}$ implies $2-k = -n$, so $k = n+2$. Thus the coefficient is

   $\binom{1/2}{n+2} \, 3^{n+2}.$

   4. Simplify the binomial coefficient

   $\binom{1/2}{n+2} = \frac{(1/2)(-1/2)(-3/2)\cdots(1/2 - (n+2) + 1)}{(n+2)!}.$

   The product runs from $1/2$ down to $1/2 - (n+1) = -(2n+1)/2$. There are $n+2$ factors.

   Write them as:

   $\frac{1}{2},\; -\frac{1}{2},\; -\frac{3}{2},\; \dots,\; -\frac{2n+1}{2}.$

   The first factor is positive; the remaining $n+1$ factors are negative. So the product is

   $\frac{1}{2} \cdot (-1)^{n+1} \cdot \frac{1\cdot3\cdot5\cdots(2n+1)}{2^{n+1}}.$

   Hence

   $\binom{1/2}{n+2} = \frac{(-1)^{n+1} \, 1\cdot3\cdot5\cdots(2n+1)}{2^{n+2} \, (n+2)!}.$

   5. Multiply by $3^{n+2}$ The coefficient becomes

   $\frac{(-1)^{n+1} \, 1\cdot3\cdot5\cdots(2n+1) \, 3^{n+2}}{2^{n+2} \, (n+2)!}.$

   Notice $3^{n+2} = 3^{n+1} \cdot 3$, so we can also write

   $\frac{(-1)^{n+1} \, 1\cdot3\cdot5\cdots(2n+1) \, 3^{n+1} \cdot 3}{2^{n+2} \, (n+2)!}.$

   But the given options have $3^{n+1}$ in the numerator and an extra factor of $2$ in the denominator. Check option (D):

   $\frac{(-1)^{n+1} \, 1\cdot3\cdot5\cdots(2n+1) \, 3^{n+1}}{2^{n+3} \, (n+2)!}.$

   Our result has $3^{n+2}$ and $2^{n+2}$; factor one $3$ into the denominator? No—let's compare directly:

   Our coefficient = $\frac{(-1)^{n+1} (2n+1)!! \, 3^{n+2}}{2^{n+2} (n+2)!}$.

   Option (D) = $\frac{(-1)^{n+1} (2n+1)!! \, 3^{n+1}}{2^{n+3} (n+2)!}$.

   Multiply numerator and denominator of (D) by $3$: gives $\frac{(-1)^{n+1} (2n+1)!! \, 3^{n+2}}{3 \cdot 2^{n+3} (n+2)!} = \frac{(-1)^{n+1} (2n+1)!! \, 3^{n+2}}{2^{n+3} \cdot 3 (n+2)!}$, which is not the same. Wait—check arithmetic:

   Actually, $2^{n+3} = 2 \cdot 2^{n+2}$, so (D) = $\frac{(-1)^{n+1} (2n+1)!! \, 3^{n+1}}{2 \cdot 2^{n+2} (n+2)!} = \frac{1}{2} \cdot \frac{(-1)^{n+1} (2n+1)!! \, 3^{n+1}}{2^{n+2} (n+2)!}$.

   Our result = $\frac{(-1)^{n+1} (2n+1)!! \, 3^{n+2}}{2^{n+2} (n+2)!} = 3 \cdot \frac{(-1)^{n+1} (2n+1)!! \, 3^{n+1}}{2^{n+2} (n+2)!}$.

   So our result is $3$ times larger than (D)? That can't be—we must have mis-indexed. Let's re-check step 3 carefully.

   Watch out

   A common mistake: forgetting that the exponent $k$ in the binomial series starts at $0$, so the term with $1/x^n$ corresponds to $k = n+2$, not $k=n$. Always match the power of $x$ exactly.

   6. Re-verify the exponent matching We have $x^2 \sum_{k} \binom{1/2}{k} 3^k x^{-k} = \sum_k \binom{1/2}{k} 3^k x^{2-k}$. We want the coefficient of $x^{-n}$. So $2-k = -n \implies k = n+2$. That is correct. So the coefficient is $\binom{1/2}{n+2} 3^{n+2}$. …

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6. TG EAPCET 2025Set eng-2025-05-04-AN1 markMCQ
   Q.Numerically greatest term in the expansion of $(3x - 4y)^{23}$ when $x = \dfrac{1}{6}$ and $y = \dfrac{1}{8}$ is (A) $\dfrac{{}^{23}C_{11}}{6^{23}}$ (B) ${}^{23}C_{11} \left( \dfrac{8}{6} \right)^{23}$ (C) ${}^{23}C_{11} \left( \dfrac{6}{8} \right)^{23}$ (D) ${}^{23}C_{11} \left( \dfrac{1}{2} \right)^{23}$

   ›Reveal solutionSolution

   The numerically greatest term in a binomial expansion is found by comparing consecutive terms using the ratio $\frac{T_{r+1}}{T_r}$; for $(3x-4y)^{23}$ with $x=1/6$, $y=1/8$, the greatest term corresponds to $r=11$ and simplifies to ${}^{23}C_{11}\left(\frac12\right)^{23}$, which is option (D).

   The key idea is that in a binomial expansion $(a+b)^n$, the terms are $T_{r+1} = \binom{n}{r} a^{n-r} b^r$. To find which term has the largest numerical value (ignoring sign), we compare the absolute values of consecutive terms. The ratio $\left|\frac{T_{r+1}}{T_r}\right|$ tells us when the terms stop increasing and start decreasing. We set this ratio $\ge 1$ to find the range of $r$ where terms are still growing, then pick the integer $r$ giving the maximum.

   Here, $a = 3x$ and $b = -4y$, so we must be careful with signs — but since we care about numerical (absolute) value, we take absolute values.

   ---

   1. Write the general term The expansion is $(3x - 4y)^{23}$. The $(r+1)$-th term is

   $T_{r+1} = \binom{23}{r} (3x)^{23-r} (-4y)^r.$

   Its absolute value is

   $|T_{r+1}| = \binom{23}{r} (3|x|)^{23-r} (4|y|)^r.$

   2. Substitute the given values $x = \frac16$, $y = \frac18$, so

   $3|x| = 3 \cdot \frac16 = \frac12, \quad 4|y| = 4 \cdot \frac18 = \frac12.$

   Hence

   $|T_{r+1}| = \binom{23}{r} \left(\frac12\right)^{23-r} \left(\frac12\right)^r = \binom{23}{r} \left(\frac12\right)^{23}.$

   Notice that the powers of $\frac12$ combine to a constant factor $(\frac12)^{23}$ independent of $r$! So every term has the same factor $(\frac12)^{23}$, and the only variation comes from the binomial coefficient $\binom{23}{r}$.

   3. Find the largest binomial coefficient For a fixed $n=23$, the binomial coefficients $\binom{23}{r}$ are largest when $r$ is as close as possible to $n/2 = 11.5$. So the maximum occurs at $r=11$ and $r=12$ (they are equal). …

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7. TG EAPCET 2025Set eng-2025-05-04-FN1 markMCQ
   Q.Numerically greatest term in the expansion of $(3x-4y)^{23}$ when $x = \dfrac{1}{6}$ and $y = \dfrac{1}{8}$ is (A) ${}^{23}C_{11} \left(\dfrac{1}{2}\right)^{23}$ (B) ${}^{23}C_{11} \left(\dfrac{6}{8}\right)^{23}$ (C) ${}^{23}C_{11} \left(\dfrac{8}{6}\right)^{23}$ (D) $\dfrac{{}^{23}C_{11}}{6^{23}}$

   ›Reveal solutionSolution

   With $x=\tfrac16,y=\tfrac18$, both $3x$ and $4y$ equal $\tfrac12$, so every term's magnitude is $\binom{23}{r}\left(\tfrac12\right)^{23}$, and the numerically greatest term occurs where $\binom{23}{r}$ is largest ($r=11$ or $12$). Its value is ${}^{23}C_{11}\left(\tfrac12\right)^{23}$ — option (A).

   The general term in the expansion of $(3x-4y)^{23}$ is

   $T_{r+1} = {}^{23}C_r (3x)^{23-r}(-4y)^r.$

   1. Substitute the given values. $3x = 3\cdot\tfrac16=\tfrac12$ and $4y=4\cdot\tfrac18=\tfrac12$. So with $a=3x=\tfrac12$ and $b=-4y=-\tfrac12$,

   $T_{r+1} = {}^{23}C_r \left(\tfrac12\right)^{23-r}\left(-\tfrac12\right)^r = {}^{23}C_r\left(\tfrac12\right)^{23}(-1)^r,$

   so every term's numerical value is $|T_{r+1}| = {}^{23}C_r \left(\tfrac12\right)^{23}$ — the power of $\tfrac12$ is the same ($23$) for every $r$, so the magnitude is governed entirely by how $\binom{23}{r}$ varies with $r$.

   2. Find which $r$ maximises $\binom{23}{r}$. Use the ratio test: since $|b/a|=1$, $\frac{|T_{r+1}|}{|T_r|} = \frac{23-r+1}{r}\cdot\left|\frac{b}{a}\right| = \frac{24-r}{r}.$ …

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8. TG EAPCET 2025Set eng-2025-05-04-FN1 markMCQ
   Q.If $2^{4n+3} + 3^{3n+1}$ is divisible by P for all natural numbers $n$, then P is (A) an integer less than 9 (B) an odd integer, not a prime (C) an even integer (D) an odd prime integer

   ›Reveal solutionSolution

   The expression $2^{4n+3} + 3^{3n+1}$ is always divisible by 11, which is an odd prime integer — so the correct option is (D).

   The key is to notice that this is a divisibility problem for all natural numbers $n$. When a statement must hold for every $n \in \mathbb{N}$, we are looking for a common divisor that never changes — a fixed number that divides the expression regardless of $n$. The natural tool here is modular arithmetic (or, equivalently, the principle of mathematical induction). We want to find the largest possible $P$ that always works, and then see which option describes it.

   Let’s simplify the expression by rewriting the bases in a more convenient form. Notice that $2^{4n+3} = 2^3 \cdot (2^4)^n = 8 \cdot 16^n$, and $3^{3n+1} = 3^1 \cdot (3^3)^n = 3 \cdot 27^n$. So the expression becomes:

   $8 \cdot 16^n + 3 \cdot 27^n.$

   Now, we want a number that divides this for every $n$. A classic trick: if two numbers $a$ and $b$ have the same remainder modulo some $m$, then $a^n$ and $b^n$ also have the same remainder modulo $m$, and the expression simplifies. Here, $16$ and $27$ are close: $27 - 16 = 11$. That suggests checking divisibility by 11.

   1. Check modulo 11.

      Compute $16 \mod 11$: $16 \equiv 5 \pmod{11}$.

      Compute $27 \mod 11$: $27 \equiv 5 \pmod{11}$ as well (since $27 = 22 + 5$).

      So $16 \equiv 27 \equiv 5 \pmod{11}$.

      Therefore $16^n \equiv 5^n \pmod{11}$ and $27^n \equiv 5^n \pmod{11}$.

   2. Substitute into the expression.

   $8 \cdot 16^n + 3 \cdot 27^n \equiv 8 \cdot 5^n + 3 \cdot 5^n \pmod{11}.$

   Factor $5^n$:

   $\equiv (8 + 3) \cdot 5^n = 11 \cdot 5^n \equiv 0 \pmod{11}.$

   So the expression is divisible by 11 for every natural number $n$.

   3. Is 11 the only such divisor? …

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9. TG EAPCET 2024Set eng-2024-05-09-AN1 markMCQ
   Q.If the coefficients of 3 consecutive terms in the expansion of $(1+x)^{23}$ are in arithmetic progression, then those terms are (A) $T_{10}, T_{11}, T_{12}$ (B) $T_{8}, T_{9}, T_{10}$ (C) $T_{13}, T_{14}, T_{15}$ (D) $T_{14}, T_{15}, T_{16}$

   ›Reveal solutionSolution

   For $(1+x)^{23}$, the AP condition on three consecutive coefficients gives $(23-2r)^2=25$, so the middle index is $r=9$ or $r=14$; the terms $T_{14},T_{15},T_{16}$ are one such triple.

   Let the three consecutive terms be $T_{r},T_{r+1},T_{r+2}$, whose coefficients are $\binom{23}{r-1},\binom{23}{r},\binom{23}{r+1}$. For these to be in arithmetic progression,

   $2\binom{23}{r}=\binom{23}{r-1}+\binom{23}{r+1}.$

   Dividing through by $\binom{23}{r}$ and using $\dfrac{\binom{n}{r-1}}{\binom{n}{r}}=\dfrac{r}{\,n-r+1\,}$, $\dfrac{\binom{n}{r+1}}{\binom{n}{r}}=\dfrac{n-r}{\,r+1\,}$ with $n=23$, this simplifies to the standard result …

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10. TG EAPCET 2024Set eng-2024-05-10-AN1 markMCQ
    Q.If $3^{2n+2} - 8n - 9$ is divisible by $2^P \ \forall \ n \in \mathbb{N}$, then the maximum value of $P$ is (A) $8$ (B) $7$ (C) $6$ (D) $9$

    ›Reveal solutionSolution

    The expression $3^{2n+2} - 8n - 9$ is always divisible by $2^6 = 64$ for all natural $n$, but not by $2^7 = 128$; hence the maximum $P$ is 6.

    We need the highest power of 2 that divides $3^{2n+2} - 8n - 9$ for every natural number $n$. That means we want the largest $P$ such that $2^P$ is a factor of the expression for all $n \in \mathbb{N}$.

    ---

    Concept and intuition

    The expression mixes an exponential term $3^{2n+2}$ with a linear term $-8n - 9$. The exponential part grows fast, but we care about its 2-adic valuation — the exponent of 2 dividing it. A classic trick: rewrite $3^{2n+2}$ as $9^{n+1}$ and expand using the binomial theorem. Since $9 = 8 + 1$, the expansion will produce powers of 8 (which are powers of 2). The linear part cancels the lowest-order terms, leaving a remainder whose 2-adic valuation we can read off.

    ---

    Step-by-step reasoning

    1. Rewrite the exponential base $3^{2n+2} = 3^{2(n+1)} = (3^2)^{n+1} = 9^{n+1}$. So the expression becomes

    $E_n = 9^{n+1} - 8n - 9.$

    2. Expand $9^{n+1}$ using the binomial theorem Write $9 = 8 + 1$. Then

    $9^{n+1} = (1 + 8)^{n+1} = \sum_{k=0}^{n+1} \binom{n+1}{k} 8^k.$

    The first few terms:

    $9^{n+1} = 1 + (n+1)\cdot 8 + \binom{n+1}{2} 8^2 + \binom{n+1}{3} 8^3 + \cdots + 8^{n+1}.$

    3. Substitute into $E_n$

    $E_n = \left[1 + 8(n+1) + \binom{n+1}{2} 8^2 + \binom{n+1}{3} 8^3 + \cdots + 8^{n+1}\right] - 8n - 9.$

    Simplify the constant and linear parts:

    $1 + 8(n+1) - 8n - 9 = 1 + 8n + 8 - 8n - 9 = 0$.

    So the constant and $n$-linear terms cancel exactly.

    Hence

    $E_n = \binom{n+1}{2} 8^2 + \binom{n+1}{3} 8^3 + \cdots + 8^{n+1}.$

    4. Factor out the smallest power of 2 Each term has a factor $8^2 = 2^6$ because $8^2 = 64 = 2^6$. So we can write

    $E_n = 2^6 \left[ \binom{n+1}{2} + \binom{n+1}{3} 8 + \binom{n+1}{4} 8^2 + \cdots + 8^{n-1} \right].$

    The bracket is an integer. Thus $2^6$ always divides $E_n$.

    5. Check if a higher power of 2 always divides $E_n$ For $P = 7$ (i.e., factor $2^7 = 128$), we need the bracket to be even for all $n$. Look at the bracket modulo 2:

    $\text{Bracket} \equiv \binom{n+1}{2} \pmod{2},$

    because every other term contains a factor 8, which is even, so they vanish mod 2.

    Now $\binom{n+1}{2} = \frac{(n+1)n}{2}$.

    • If $n$ is odd, say $n=1$: $\binom{2}{2}=1$ (odd).
    • If $n$ is even, say $n=2$: $\binom{3}{2}=3$ (odd). In fact, $\binom{n+1}{2}$ is odd for all $n$? Let’s test: …

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11. TG EAPCET 2024Set eng-2024-05-10-FN1 markMCQ
    Q.If $\lim_{x \to a} \dfrac{2x^2 + (3+2a)x + 3a}{x^3 - 2x^2 - 23x + 60} = \dfrac{11}{9}$, then $\lim_{x \to a} \dfrac{x^2 + 9x + 20}{x^2 - x - 20} =$ (A) $-9$ (B) $-4$ (C) $-\dfrac{1}{4}$ (D) $-\dfrac{1}{9}$

    ›Reveal solutionSolution

    The condition fixes $a=-4$, so the required limit is $\dfrac{a+5}{a-5}=\dfrac{1}{-9}=-\dfrac19$.

    Factor both expressions. The first quotient factors as

    $\frac{2x^2+(3+2a)x+3a}{x^3-2x^2-23x+60}=\frac{(2x+3)(x+a)}{(x-3)(x-4)(x+5)},$

    and imposing $\displaystyle\lim_{x\to a}=\frac{11}{9}$ pins the parameter at $a=-4$.

    Evaluate the required limit. Factor numerator and denominator:

    $x^2+9x+20=(x+4)(x+5),\qquad x^2-x-20=(x-5)(x+4).$

    Cancelling the common factor $(x+4)$, …

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12. TG EAPCET 2023Set eng-2023-05-13-AN1 markMCQ
    Q.If the equation $x + y + n = 0$ represents a normal to the hyperbola $\dfrac{x^2}{6} - \dfrac{y^2}{2} = 1$, then $n =$ (A) $\pm \sqrt{3}$ (B) $\pm 4$ (C) $\pm \sqrt{2}$ (D) $\pm 2$

    ›Reveal solutionSolution

    For a line to be normal to a hyperbola, its slope must match the negative reciprocal of the tangent slope at the point of contact. Using the standard normal condition for a hyperbola, we find that $n = \pm 4$, so option (B) is correct.

    ---

    The key idea: A line $x + y + n = 0$ has slope $-1$. For it to be a normal to the hyperbola $\frac{x^2}{6} - \frac{y^2}{2} = 1$, the slope of the tangent at the point of contact must be $1$ (since normal slope $\times$ tangent slope $= -1$). We use the condition for a line to be normal to a hyperbola in standard form.

    ---

    1. Recall the normal condition for a hyperbola For the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, any line $y = mx + c$ is a normal if

    $c = \pm \frac{m(a^2 + b^2)}{\sqrt{a^2 - b^2 m^2}}$

    provided $a^2 > b^2 m^2$. This comes from solving the condition that the line is perpendicular to the tangent at the point of contact.

    2. Identify parameters Here $a^2 = 6$, $b^2 = 2$. The given line is $x + y + n = 0$, which we rewrite as

    $y = -x - n$

    So the slope $m = -1$ and the intercept $c = -n$.

    3. Apply the formula Substitute $m = -1$, $a^2 = 6$, $b^2 = 2$ into the normal condition:

    $c = \pm \frac{m(a^2 + b^2)}{\sqrt{a^2 - b^2 m^2}}$

    First compute numerator: $m(a^2 + b^2) = (-1)(6 + 2) = -8$.

    Denominator: $a^2 - b^2 m^2 = 6 - 2(1) = 4$, so $\sqrt{4} = 2$.

    Thus

    $c = \pm \frac{-8}{2} = \mp 4$

    That is, $c = -4$ or $c = 4$.

    4. Relate to $n$ …

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