Q.Find the value of $\left(a^2 + \sqrt{a^2 - 1}\right)^4 + \left(a^2 - \sqrt{a^2 - 1}\right)^4$.

Showing the 12 most recent of 31 on this concept.

1. KEAM 2026Set eng-2026-04174 marksMCQ
   Q.Let $(2-x)^9=a_0+a_1x+a_2x^2+\ldots\ldots+a_9x^9$. Then the value of $a_1+a_2+a_3+\ldots\ldots+a_8$ is equal to (A) -511 (B) 510 (C) -512 (D) 512 (E) -510

   ›Reveal solutionSolution

   Sum all coefficients ($=1$), then remove $a_0=512$ and $a_9=-1$ to get $-510$.

   Substituting $x=1$ into $(2-x)^9=\sum a_kx^k$:

   $a_0+a_1+\cdots+a_9=(2-1)^9=1.$

   The endpoint coefficients are $a_0=2^9=512$ (constant term) and $a_9=(-1)^9=-1$ (coefficient of $x^9$). …

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2. KEAM 2026Set eng-2026-04174 marksMCQ
   Q.If the coefficient of $x^3$ in the binomial expansion of $(2+x)^n$ is 160, then the coefficient of $x^6$ in the binomial expansion of $(2-x^2)^n$ is (A) 160 (B) 320 (C) -160 (D) -320 (E) -960

   ›Reveal solutionSolution

   The condition gives $n=6$; the $x^6$ term of $(2-x^2)^6$ is $\binom63 2^3(-1)^3=-160$.

   The coefficient of $x^3$ in $(2+x)^n$ is $\binom{n}{3}2^{n-3}$. Setting it to $160$ and testing, $n=6$ works: $\binom63 2^3=20\cdot8=160$. …

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3. KEAM 2026Set eng-2026-04184 marksMCQ
   Q.If the coefficient of $x^{4}$ in the binomial expansion of $(4x+a)^{7}$ is $-1120$, then the value of $a$ is equal to (A) $\dfrac{1}{2}$ (B) $\dfrac{1}{4}$ (C) $\dfrac{-1}{2}$ (D) $\dfrac{1}{8}$ (E) $\dfrac{-1}{8}$

   ›Reveal solutionSolution

   The $x^4$ term gives $8960a^3=-1120$, so $a=-\tfrac12$.

   The general term of $(4x+a)^7$ is ${}^{7}C_{k}(4x)^{k}a^{7-k}$. For $x^4$ take $k=4$:

   ${}^{7}C_{4}(4)^{4}a^{3}=35\cdot256\cdot a^{3}=8960a^{3}.$

   Setting this equal to $-1120$: …

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4. KEAM 2026Set eng-2026-04194 marksMCQ
   Q.The coefficient of $\frac{1}{x^2}$ in the binomial expansion of $\left(3x - \frac{1}{3x}\right)^4$, is (A) $\frac{4}{7}$ (B) $\frac{3}{8}$ (C) $\frac{2}{9}$ (D) $\frac{4}{9}$ (E) $\frac{-4}{9}$

   ›Reveal solutionSolution

   Write the general term, set the exponent of $x$ equal to $-2$ to find $k=3$, then evaluate the coefficient: $-\dfrac{4}{9}$.

   Concept. For $(a+b)^n$ the general (,$(k+1)$-th,) term is $T_{k+1}=\binom{n}{k}a^{n-k}b^{k}$. Here $a=3x$, $b=-\dfrac{1}{3x}$, $n=4$.

   Step 1 — general term.

   $T_{k+1}=\binom{4}{k}(3x)^{4-k}\left(-\frac{1}{3x}\right)^{k}=\binom{4}{k}(-1)^k\,3^{4-k}\,3^{-k}\,x^{4-k}x^{-k}.$

   Combining powers,

   $T_{k+1}=\binom{4}{k}(-1)^k\,3^{4-2k}\,x^{4-2k}.$ …

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5. KEAM 2026Set eng-2026-04204 marksMCQ
   Q.The coefficient of $x^8$ in the expansion of $\left(x^2 + \sqrt{1 - x^2}\right)^5 + \left(x^2 - \sqrt{1 - x^2}\right)^5$, is (A) $-20$ (B) $-10$ (C) $-30$ (D) $-40$ (E) $20$

   ›Reveal solutionSolution

   Adding $(u+v)^5+(u-v)^5$ with $u=x^2,\ v=\sqrt{1-x^2}$ leaves only even powers of $v$: $2[u^5+10u^3v^2+5uv^4]$. Substituting $v^2=1-x^2$ and collecting, the $x^8$ term has coefficient $-20$.

   Let $u=x^2$ and $v=\sqrt{1-x^2}$, so $v^2=1-x^2$. Then

   $(u+v)^5+(u-v)^5=2\left[\binom{5}{0}u^5+\binom{5}{2}u^3v^2+\binom{5}{4}uv^4\right]=2[u^5+10u^3v^2+5uv^4].$

   Substitute $v^2=1-u$ (in $u=x^2$):

   $u^5=x^{10},$

   $10u^3(1-u)=10x^6-10x^8,$ …

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6. KEAM 2026Set eng-2026-04214 marksMCQ
   Q.If the 17th and 18th term in the expansion of $(2 + x)^{50}$ are equal, then the value of $x$ is equal to (A) 1 (B) 2 (C) 4 (D) 6 (E) 8

   ›Reveal solutionSolution

   Equating the 17th and 18th terms gives $x=2\cdot\binom{50}{16}/\binom{50}{17}=1$.

   In $(2+x)^{50}$, $T_{k+1}=\binom{50}{k}2^{50-k}x^{k}$. The 17th term is $k=16$, the 18th is $k=17$:

   $\binom{50}{16}2^{34}x^{16}=\binom{50}{17}2^{33}x^{17}.$ …

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7. KEAM 2026Set eng-2026-04224 marksMCQ
   Q.If the coefficient of $y^3$ in the binomial expansion of $\left(2\alpha - \frac{y}{2}\right)^8$ is $-7$, then the value of $\alpha$ is equal to (A) $\frac{1}{2}$ (B) $2$ (C) $3$ (D) $6$ (E) $8$

   ›Reveal solutionSolution

   Pick the general term with $y^3$, set its coefficient to $-7$, solve for $\alpha$.

   General term of $\left(2\alpha-\frac{y}{2}\right)^8$: $\binom{8}{r}(2\alpha)^{8-r}\left(-\frac{y}{2}\right)^{r}$. …

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8. KEAM 2026Set eng-2026-04224 marksMCQ
   Q.Let $(1 + ax)(1 - 2x)^3 = \sum_{n=0}^{4} a_n x^n$, where $a$ is a constant. If $a_2 = 0$, then the value of $a$ is (A) $1$ (B) $9$ (C) $3$ (D) $5$ (E) $2$

   ›Reveal solutionSolution

   Expand $(1-2x)^3$, collect the $x^2$ term, set it to zero.

   $(1-2x)^3=1-6x+12x^2-8x^3$.

   Multiplying by $(1+ax)$, the $x^2$ coefficient comes from $1\cdot12x^2$ and $ax\cdot(-6x)$:

   $a_2=12-6a$. …

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9. KEAM 2025Set eng-2025-04234 marksMCQ
   Q.In the binomial expansion of $(2x + \alpha)^8$, the co-efficients of $x^2$ and $x^3$ are equal. Then the value of $\alpha$ is equal to (A) $2$ (B) $\dfrac{1}{4}$ (C) $4$ (D) $\dfrac{1}{2}$ (E) $3$

   ›Reveal solutionSolution

   Equating the x^2 and x^3 coefficients of (2x+alpha)^8 gives alpha = 4.

   Concept and Intuition

   In (2x + alpha)^8 the general term is C(8,k) (2x)^k alpha^(8-k), so the coefficient of x^k is C(8,k) 2^k alpha^(8-k). Setting the k=2 and k=3 coefficients equal isolates alpha.

   Step-by-Step Solution

   1. Coefficient of x^2: C(8,2) 2^2 alpha^6 = 284alpha^6 = 112 alpha^6.
   2. Coefficient of x^3: C(8,3) 2^3 alpha^5 = 568alpha^5 = 448 alpha^5. …

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10. KEAM 2025Set eng-2025-04254 marksMCQ
    Q.The constant term in the binomial expansion of $\left(2x - \dfrac{5}{x^2}\right)^6$ is (A) 4800 (B) 3200 (C) 5600 (D) 5400 (E) 6000

    ›Reveal solutionSolution

    Set the exponent of $x$ to zero and evaluate that term.

    The general term of $\left(2x - \tfrac{5}{x^2}\right)^6$ is

    $\binom{6}{k}(2x)^{6-k}\left(-\tfrac{5}{x^2}\right)^k = \binom{6}{k} 2^{6-k}(-5)^k\, x^{6-3k}.$ …

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11. KEAM 2025Set eng-2025-04264 marksMCQ
    Q.The coefficient of $x^{10}$ in $(1 - x^2)(1 - x^3)^9$ is (A) ${}^9C_4$ (B) $-{}^9C_6$ (C) $-{}^9C_4$ (D) ${}^9C_6$ (E) $0$

    ›Reveal solutionSolution

    $(1-x^{3})^{9}=\sum\binom{9}{k}(-1)^{k}x^{3k}$; to make $x^{10}$ need $3k=10$ or $3k=8$ — impossible — so coefficient $=0$.

    Expand. $(1-x^{3})^{9}=\sum_{k=0}^{9}\binom{9}{k}(-1)^{k}x^{3k}$, giving exponents $0,3,6,9,\dots$ …

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12. KEAM 2025Set eng-2025-04264 marksMCQ
    Q.The constant term in $\left(\frac{\sqrt{x}}{2} + \frac{1}{3x^2}\right)^{10}$ is (A) $\frac{5}{128}$ (B) $\frac{9}{128}$ (C) $\frac{5}{256}$ (D) $\frac{9}{256}$ (E) $0$

    ›Reveal solutionSolution

    Set the exponent of $x$ to zero: $\frac{10-r}{2}-2r=0\Rightarrow r=2$; evaluate the term to get $\frac{5}{256}$.

    General term. $T_{r+1}=\binom{10}{r}\left(\frac{\sqrt{x}}{2}\right)^{10-r}\left(\frac{1}{3x^{2}}\right)^{r}=\binom{10}{r}\frac{1}{2^{10-r}}\frac{1}{3^{r}}\,x^{\frac{10-r}{2}-2r}$. …

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