Q.Expand using Binomial Theorem $\left(1 + \dfrac{x}{2} - \dfrac{2}{x}\right)^4$, $x \neq 0$.

Showing the 12 most recent of 60 on this concept.

1. AP EAPCET 2026Set eng-2026-05-12-FN1 markMCQ
   Q.If $(1+x)^n = C_o + C_1 x + C_2 x^2 + \cdots + C_n x^n$, then $C_o - C_2 + C_4 - C_6 + \cdots =$ (A) $2^{\frac{n}{2}}\cos\dfrac{n\pi}{4}$ (B) $2^{n}\cos\dfrac{n\pi}{4}$ (C) $2^{\frac{n}{2}}\sin\dfrac{n\pi}{4}$ (D) $2^{n}\sin\dfrac{n\pi}{4}$

   ›Reveal solutionSolution

   The alternating even-index binomial sum $C_0-C_2+C_4-\cdots$ is the real part of $(1+i)^n$; converting to polar form gives $2^{n/2}\cos(n\pi/4)$.

   Concept and Intuition

   Substituting $x=i$ into $(1+x)^n=\sum C_k x^k$ makes even powers of $i$ real (alternating $+1,-1,+1,-1,\dots$ for $k=0,2,4,6,\dots$) and odd powers purely imaginary. So the real part of $(1+i)^n$ is exactly $C_0-C_2+C_4-C_6+\cdots$, and the imaginary part is $C_1-C_3+C_5-\cdots$. Converting $1+i$ to polar form makes extracting the real part easy.

   Step-by-Step Solution

   1. $(1+i)^n=\displaystyle\sum_{k=0}^n C_k\,i^k = (C_0 - C_2 + C_4 - \cdots) + i(C_1-C_3+C_5-\cdots)$, using $i^0=1,i^1=i,i^2=-1,i^3=-i,i^4=1,\dots$
   2. So $C_0-C_2+C_4-\cdots = \operatorname{Re}\left[(1+i)^n\right]$.
   3. Write $1+i$ in polar form: modulus $=\sqrt{1^2+1^2}=\sqrt2$, argument $=\tan^{-1}(1/1)=\pi/4$. So $1+i=\sqrt2\left(\cos\frac{\pi}{4}+i\sin\frac{\pi}{4}\right)$. …

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2. AP EAPCET 2026Set eng-2026-05-12-FN1 markMCQ
   Q.If in the expansion of $(1+px)^q, q \in N$, the coefficients of $x$ and $x^2$ are 21 and 189 respectively, then $p$ and $q$ are respectively (A) $7, 3$ (B) $3, 7$ (C) $-7, 3$ (D) $-3, 7$

   ›Reveal solutionSolution

   This tests the general binomial coefficient formula $\binom{q}{r}p^r$ for the term $x^r$. Solving the two coefficient equations simultaneously gives $p=3,\ q=7$.

   Concept and Intuition

   The binomial expansion $(1+px)^q = \sum_{r=0}^{q}\binom{q}{r}(px)^r$ has the coefficient of $x^r$ equal to $\binom{q}{r}p^r$. Two given coefficients (for $x^1$ and $x^2$) give two equations in the two unknowns $p$ and $q$, which can be solved together.

   Step-by-Step Solution

   1. Coefficient of $x$: $\binom{q}{1}p = qp = 21$.
   2. Coefficient of $x^2$: $\binom{q}{2}p^2 = \dfrac{q(q-1)}{2}p^2 = 189$.
   3. From step 1, $p = \dfrac{21}{q}$. Substitute into step 2: $\dfrac{q(q-1)}{2}\cdot\dfrac{441}{q^2} = 189 \implies \dfrac{441(q-1)}{2q}=189$ …

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3. AP EAPCET 2026Set eng-2026-05-12-FN1 markMCQ
   Q.If the term independent of $x$ in the expansion of $\left(x^{1/3} - x^{-1/2}\right)^{10}$ is $\alpha$, then the number of proper divisors of $\alpha$ is (A) $16$ (B) $14$ (C) $15$ (D) $17$

   ›Reveal solutionSolution

   The $x$-free term is $^{10}C_4 = 210 = 2\cdot3\cdot5\cdot7$, which has $16$ divisors and hence $14$ proper divisors — option (B).

   Concept and Intuition

   Two separate ideas are being tested.

   (i) The term independent of $x$ in a binomial expansion is found by writing the general term, collecting the exponent of $x$, and setting it to zero. No expansion is ever needed.

   (ii) Counting divisors. If $N = p_1^{a_1}p_2^{a_2}\cdots$, the number of divisors is $(a_1+1)(a_2+1)\cdots$ — because a divisor is built by independently choosing how many copies of each prime to take. A proper divisor (standard convention in this syllabus) is a divisor other than $1$ and the number itself, so you subtract $2$.

   Step-by-Step Solution

   1. General term:

   $T_{r+1} = {}^{10}C_r\left(x^{1/3}\right)^{10-r}\left(-x^{-1/2}\right)^{r} = (-1)^r\,{}^{10}C_r\,x^{\frac{10-r}{3}-\frac r2}.$

   2. Set the exponent to $0$:

   $\frac{10-r}{3}-\frac r2 = 0 \Rightarrow 2(10-r) = 3r \Rightarrow 20 = 5r \Rightarrow r = 4.$

   3. That term is $T_5 = (-1)^4\,{}^{10}C_4 = 210$, so $\alpha = 210$.
   4. Factorise: $210 = 2\times 3\times 5\times 7$ (each exponent $=1$). …

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4. AP EAPCET 2026Set eng-2026-05-12-AN1 markMCQ
   Q.The number of integral terms in the expansion of $\left(5^{\frac{1}{2}} + 7^{\frac{1}{8}}\right)^{1024}$ is (A) $128$ (B) $129$ (C) $130$ (D) $131$

   ›Reveal solutionSolution

   A term in this binomial expansion is an integer exactly when the exponent of $7^{1/8}$ is a whole number, i.e. $r$ is a multiple of 8; counting such $r$ from $0$ to $1024$ gives $129$ integral terms.

   Concept and Intuition

   In $(a^{p}+b^{q})^n$, the general term $\binom{n}{r}a^{p(n-r)}b^{qr}$ is an integer (given $a,b$ are integers) exactly when both exponents $p(n-r)$ and $qr$ are non-negative integers. Here we need both $(1024-r)/2$ and $r/8$ to be whole numbers.

   Step-by-Step Solution

   1. General term: $T_{r+1}=\binom{1024}{r}\left(5^{1/2}\right)^{1024-r}\left(7^{1/8}\right)^{r}=\binom{1024}{r}5^{(1024-r)/2}7^{r/8}$, for $r=0,1,\dots,1024$.
   2. For $7^{r/8}$ to be an integer power of 7, need $8\mid r$.
   3. If $8\mid r$, then $r$ is automatically even, so $1024-r$ is even too (since 1024 is even), making $5^{(1024-r)/2}$ automatically an integer power as well. …

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5. AP EAPCET 2026Set eng-2026-05-13-FN1 markMCQ
   Q.In the binomial expansion of $(x + a)^{15}$, if the eleventh term is the geometric mean of the eighth and twelfth terms, then the numerically greatest term in its expansion is (A) $8$ (B) $9$ (C) $10$ (D) $11$

   ›Reveal solutionSolution

   This tests using a geometric-mean condition on three binomial-expansion terms to pin down the ratio $a/x$, then applying the standard "numerically greatest term" test. The greatest term turns out to be $T_8$.

   Concept and Intuition

   In the expansion of $(x+a)^{15}$, each term is $T_{r+1}=\binom{15}{r}x^{15-r}a^r$. A geometric-mean relation between three specific terms fixes the ratio $a/x$ numerically. Once that ratio is known, the standard technique for the numerically greatest term is to look at the ratio of consecutive terms $T_{r+1}/T_r$ and find where it transitions from $\ge 1$ (still increasing) to $<1$ (starts decreasing) — the last term before that transition is the maximum.

   Step-by-Step Solution

   1. Write the relevant terms: $T_{11}=\binom{15}{10}x^{5}a^{10}$, $T_8=\binom{15}{7}x^{8}a^{7}$, $T_{12}=\binom{15}{11}x^{4}a^{11}$.
   2. Apply the GM condition $T_{11}^2 = T_8\cdot T_{12}$:

   $\binom{15}{10}^2 x^{10}a^{20} = \binom{15}{7}\binom{15}{11}\,x^{12}a^{18}$

   $\Rightarrow \binom{15}{10}^2 a^2 = \binom{15}{7}\binom{15}{11}\,x^2 \Rightarrow \left(\frac{a}{x}\right)^2 = \frac{\binom{15}{7}\binom{15}{11}}{\binom{15}{10}^2}.$

   3. Compute the coefficients: $\binom{15}{7}=6435$, $\binom{15}{10}=\binom{15}{5}=3003$, $\binom{15}{11}=\binom{15}{4}=1365$.

   $\left(\frac{a}{x}\right)^2 = \frac{6435\times1365}{3003^2}=\frac{8{,}783{,}775}{9{,}018{,}009} = \frac{75}{77}\ \ (\approx 0.974).$

   4. So $\dfrac{a}{x}=\sqrt{\dfrac{75}{77}}\approx0.9869$ (taking the positive root, as is standard for this test).
   5. For the numerically greatest term, examine $\left|\dfrac{T_{r+1}}{T_r}\right| = \dfrac{(16-r)}{r}\cdot\dfrac{a}{x} \ge 1$. Substituting the value of $a/x$: …

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6. AP EAPCET 2026Set eng-2026-05-14-FN1 markMCQ
   Q.The coefficient of $x^{12}$ in the expansion of $(3 + 2x)^{-5}$ is (A) $^{17}C_5\, 3^{12} 2^5$ (B) $^{16}C_4\, \dfrac{2^{12}}{3^{17}}$ (C) $^{16}C_{12}\, \dfrac{2^{17}}{3^{12}}$ (D) $^{17}C_5\, \dfrac{3^{12}}{2^{17}}$

   ›Reveal solutionSolution

   This tests the binomial expansion for a negative index, $(1+u)^{-n}=\sum(-1)^k\,^{n+k-1}C_k\,u^k$. The answer is (B).

   Concept and Intuition

   When the exponent is negative (or fractional), the binomial theorem no longer terminates — it becomes an infinite series, valid only for $|u|<1$. To use it we must first factor out a constant so the series is written in the standard form $(1+u)^{-n}$, and then pick out the general term $T_{k+1}=(-1)^k\,^{n+k-1}C_k\,u^k$.

   Step-by-Step Solution

   1. Write $(3+2x)^{-5} = 3^{-5}\left(1+\dfrac{2x}{3}\right)^{-5}$, pulling the constant $3$ out so the bracket has the form $(1+u)^{-n}$ with $u = \dfrac{2x}{3}$, $n=5$.
   2. The general term of $(1+u)^{-n}$ is $T_{k+1} = (-1)^k\,^{n+k-1}C_k\, u^k$.
   3. Here $n=5$, so $T_{k+1} = 3^{-5}(-1)^k\,^{k+4}C_k\left(\dfrac{2x}{3}\right)^k = 3^{-5}(-1)^k\,^{k+4}C_k\left(\dfrac{2}{3}\right)^k x^k$.
   4. We need the coefficient of $x^{12}$, so set $k=12$: $^{k+4}C_k = {}^{16}C_{12} = {}^{16}C_4 = 1820$. …

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7. AP EAPCET 2026Set eng-2026-05-14-FN1 markMCQ
   Q.If $x$ is so small that the values of $x^n, n \geq 2$ are negligible, then the approximate value of $\dfrac{\sqrt{2 - 3x}}{(3 + 2x)}(x + 1)$ is (A) $\dfrac{\sqrt{2}}{3}\left(1 - \dfrac{5x}{12}\right)$ (B) $\dfrac{\sqrt{2}}{3}(1 + 5x)$ (C) $\dfrac{\sqrt{2}}{3}(1 - 7x)$ (D) $\dfrac{\sqrt{2}}{3}(1 + 6x)$

   ›Reveal solutionSolution

   This tests the small-$x$ binomial approximation $(1+t)^n \approx 1+nt$ applied to several factors at once, keeping only terms linear in $x$. The answer is (A).

   Concept and Intuition

   When $x$ is so small that $x^2, x^3,\dots$ are negligible, every factor of the form $(1+\text{something}\cdot x)^n$ can be replaced by its first-order approximation $1+n(\text{something})x$. When several such approximate factors are multiplied together, we again drop any product term that is $O(x^2)$ or higher, keeping only the constant term and the term linear in $x$.

   Step-by-Step Solution

   1. $\sqrt{2-3x} = \sqrt{2}\left(1-\dfrac{3x}{2}\right)^{1/2} \approx \sqrt2\left(1 - \dfrac12\cdot\dfrac{3x}{2}\right) = \sqrt2\left(1-\dfrac{3x}{4}\right)$.
   2. $\dfrac{1}{3+2x} = \dfrac13\left(1+\dfrac{2x}{3}\right)^{-1} \approx \dfrac13\left(1-\dfrac{2x}{3}\right)$.
   3. So $\dfrac{\sqrt{2-3x}}{3+2x} \approx \dfrac{\sqrt2}{3}\left(1-\dfrac{3x}{4}\right)\left(1-\dfrac{2x}{3}\right) \approx \dfrac{\sqrt2}{3}\left(1-\dfrac{3x}{4}-\dfrac{2x}{3}\right)$ (dropping the $x^2$ cross term).
   4. Multiply by $(x+1)$: $\dfrac{\sqrt2}{3}\left(1-\dfrac{3x}{4}-\dfrac{2x}{3}\right)(1+x) \approx \dfrac{\sqrt2}{3}\left(1 - \dfrac{3x}{4}-\dfrac{2x}{3}+x\right)$ (again dropping $x^2$ terms). …

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8. AP EAPCET 2026Set eng-2026-05-14-AN1 markMCQ
   Q.If $|x|$ is so small that $x^2$ and higher powers of $x$ may be neglected, then the approximate value of $\left(\dfrac{3x+64}{3x+8}\right)^{2/3}$ is (A) $4 + \dfrac{9x}{8}$ (B) $1 - \dfrac{7x}{32}$ (C) $1 + \dfrac{9x}{32}$ (D) $4 - \dfrac{7x}{8}$

   ›Reveal solutionSolution

   Pull out the constant parts (64 and 8) to write each bracket as $(1+\text{small})$, apply the binomial approximation $(1+u)^n\approx 1+nu$ to each factor separately, multiply, and drop $x^2$ terms. Result: $4-\tfrac{7x}{8}$.

   Concept and Intuition

   When $|x|$ is small enough that $x^2$ and higher powers are negligible, any expression $(1+u)^n$ with $u$ small can be replaced by its first-order Taylor/binomial approximation $1+nu$. The trick here is that the given fraction isn't already in the form $(1+\text{small})$ — we must first factor out the dominant constant from numerator and denominator so that what's left over is genuinely small (proportional to $x$), and only then apply the approximation to each piece independently before recombining.

   Step-by-Step Solution

   1. Factor 64 from the numerator and 8 from the denominator:

   $\frac{3x+64}{3x+8} = \frac{64\left(1+\frac{3x}{64}\right)}{8\left(1+\frac{3x}{8}\right)} = 8\cdot\frac{1+\frac{3x}{64}}{1+\frac{3x}{8}}.$

   2. Raise to the power $2/3$:

   $\left(\frac{3x+64}{3x+8}\right)^{2/3} = 8^{2/3}\left(1+\frac{3x}{64}\right)^{2/3}\left(1+\frac{3x}{8}\right)^{-2/3}.$

   3. $8^{2/3}=(2^3)^{2/3}=2^2=4$.
   4. Apply $(1+u)^n\approx1+nu$ to each bracket (dropping $x^2$ terms):

   $\left(1+\frac{3x}{64}\right)^{2/3}\approx 1+\frac{2}{3}\cdot\frac{3x}{64}=1+\frac{x}{32},\qquad \left(1+\frac{3x}{8}\right)^{-2/3}\approx 1-\frac{2}{3}\cdot\frac{3x}{8}=1-\frac{x}{4}.$

   5. Multiply the two approximations, discarding the $x^2$ cross term: …

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9. AP EAPCET 2026Set eng-2026-05-13-AN1 markMCQ
   Q.The coefficient of $x^5$ in the expansion of $\left(x+\sqrt{x^2-x}\right)^6 + \left(x-\sqrt{x^2-x}\right)^6$ is (A) $96$ (B) $48$ (C) $-48$ (D) $-96$

   ›Reveal solutionSolution

   Only the even-power binomial terms survive in $(a+b)^6+(a-b)^6$; substituting $b^2=x^2-x$ and collecting the $x^5$ terms gives coefficient $-96$.

   Concept and Intuition

   For $(a+b)^n+(a-b)^n$, all odd-power terms in $b$ cancel (they appear with opposite signs in the two expansions), leaving only the even-power terms doubled: $2\left[\binom{n}{0}a^n + \binom{n}{2}a^{n-2}b^2+\binom{n}{4}a^{n-4}b^4+\cdots\right]$. Here $b^2=x^2-x$ is a genuine polynomial in $x$, so after substitution the whole thing becomes an ordinary polynomial in $x$, and we just need to track the $x^5$ coefficient.

   Step-by-Step Solution

   1. With $a=x$, $b=\sqrt{x^2-x}$, $n=6$: $(a+b)^6+(a-b)^6 = 2\left[a^6+\binom{6}{2}a^4b^2+\binom{6}{4}a^2b^4+\binom{6}{6}b^6\right] = 2\left[a^6+15a^4b^2+15a^2b^4+b^6\right]$.
   2. $a^6=x^6$ (no $x^5$ term).
   3. $15a^4b^2 = 15x^4(x^2-x)=15x^6-15x^5$ (contributes $-15$ to $x^5$).
   4. $15a^2b^4 = 15x^2(x^2-x)^2=15x^2(x^4-2x^3+x^2)=15x^6-30x^5+15x^4$ (contributes $-30$ to $x^5$). …

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10. AP EAPCET 2026Set eng-2026-05-13-AN1 markMCQ
    Q.If $f(x)$ is the third term in the expansion of $\dfrac{1}{\sqrt[3]{9-48x+64x^2}}$, when $|x| > \dfrac{3}{8}$, then $f(1) =$ (A) $\dfrac{5}{128}$ (B) $\dfrac{5}{256}$ (C) $-\dfrac{1}{128}$ (D) $-\dfrac{1}{256}$

    ›Reveal solutionSolution

    Recognizing $9-48x+64x^2$ as a perfect square $(3-8x)^2$ turns this into a fractional-power binomial expansion valid for $|x|>3/8$; the third term of that series evaluated at $x=1$ gives $\dfrac{5}{256}$.

    Concept and Intuition

    The condition $|x|>\tfrac38$ is the giveaway that we must expand around $x=\infty$ (i.e. in powers of $1/x$), not around $x=0$: the binomial series $(1-u)^{-2/3}$ only converges for $|u|<1$, so we must factor the dominant $-8x$ term out of $(3-8x)$ so that the remaining small quantity is $\dfrac{3}{8x}$, which is indeed less than 1 in magnitude when $|x|>\tfrac38$.

    Step-by-Step Solution

    1. Factor the quadratic: $9-48x+64x^2 = (8x)^2-2(8x)(3)+3^2 = (8x-3)^2=(3-8x)^2$.
    2. So the given function is $\left[(3-8x)^2\right]^{-1/3} = (3-8x)^{-2/3}$.
    3. Since $|x|>3/8$, write $3-8x = -8x\left(1-\dfrac{3}{8x}\right)$, so $(3-8x)^{-2/3} = (-8x)^{-2/3}\left(1-\dfrac{3}{8x}\right)^{-2/3}$, and now $\left|\dfrac{3}{8x}\right|<1$ so the binomial series converges. …

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11. AP EAPCET 2026Set eng-2026-05-15-AN1 markMCQ
    Q.The sum of the coefficients of fractional powers of 'x' in the expansion of $\left(1-3x^{1/2}\right)^{2026}$ is (A) $2^{2025} - 2^{2026}$ (B) $2^{2025}(2^{2026}-1)$ (C) $2^{2025}(1-2^{2026})$ (D) $2^{2026} - 2^{2025}$

    ›Reveal solutionSolution

    Substituting $y=x^{1/2}$ turns this into a standard even/odd-coefficient-sum problem; the fractional-power (odd $k$) sum comes out to $2^{2025}(1-2^{2026})$.

    Concept and Intuition

    In $(1-3x^{1/2})^{2026}$, each term is $\binom{2026}{k}(-3)^k x^{k/2}$. The power of $x$, namely $k/2$, is an integer when $k$ is even and a genuine fraction when $k$ is odd. So "coefficients of fractional powers" means exactly the terms with odd $k$. The standard trick to isolate the odd-indexed terms of a polynomial $f(y)=\sum a_k y^k$ is $\dfrac{f(1)-f(-1)}{2}$, since $f(1)=\sum a_k$ and $f(-1)=\sum a_k(-1)^k$ — subtracting cancels all even-$k$ terms and doubles the odd-$k$ ones.

    Step-by-Step Solution

    1. Let $y=x^{1/2}$; then $(1-3x^{1/2})^{2026}=(1-3y)^{2026}=f(y)$, a genuine polynomial of degree 2026 in $y$.
    2. $f(1) = (1-3)^{2026} = (-2)^{2026} = 2^{2026}$ (even exponent kills the sign).
    3. $f(-1) = (1+3)^{2026} = 4^{2026} = 2^{4052}$.
    4. Sum over odd $k$ (the fractional-power terms of $x$) $=\dfrac{f(1)-f(-1)}{2}=\dfrac{2^{2026}-2^{4052}}{2}=2^{2025}-2^{4051}$. …

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12. AP EAPCET 2026Set eng-2026-05-15-AN1 markMCQ
    Q.If $a>0$ and the expansion of $(a-5x)^{5/2}$ is valid for all $x \in \left(-\dfrac{4}{5}, \dfrac{4}{5}\right)$, then the third term in the expansion of $(a-5x)^{5/2}$ is (A) $-\dfrac{625}{32}x^2$ (B) $\dfrac{125}{8}x^2$ (C) $\dfrac{475}{16}x^2$ (D) $\dfrac{375}{4}x^2$

    ›Reveal solutionSolution

    The stated interval of validity pins down $a=4$; expanding the generalized binomial series and picking out the $k=2$ term gives $\dfrac{375}{4}x^2$.

    Concept and Intuition

    For a non-integer exponent $n$, $(1+u)^n$ only has a convergent (valid) binomial series when $|u|<1$. Writing $(a-5x)^{5/2}=a^{5/2}(1-5x/a)^{5/2}$ puts it in that form with $u=-5x/a$, so the series is valid exactly for $|x|<a/5$ — matching that radius to the given interval pins down $a$.

    Step-by-Step Solution

    1. $(a-5x)^{5/2}=a^{5/2}\left(1-\dfrac{5x}{a}\right)^{5/2}$, valid when $\left|\dfrac{5x}{a}\right|<1 \iff |x|<\dfrac{a}{5}$ (using $a>0$).
    2. Given validity on $\left(-\dfrac45,\dfrac45\right)$, so $\dfrac a5=\dfrac45 \Rightarrow a=4$.
    3. Expand: $(4-5x)^{5/2}=4^{5/2}\sum_{k=0}^\infty \binom{5/2}{k}\left(-\dfrac{5x}{4}\right)^k$, where $4^{5/2}=(\sqrt4)^5=2^5=32$.
    4. Third term is $k=2$: $\binom{5/2}{2}=\dfrac{(5/2)(3/2)}{2!}=\dfrac{15/4}{2}=\dfrac{15}{8}$, and $\left(-\dfrac{5x}{4}\right)^2=\dfrac{25x^2}{16}$. …

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