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JEE Advanced · Mathematics · Binomial Theorem

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1
Consider f(x)=d10dx10((x2−1)10)f(x)=\dfrac{d^{10}}{dx^{10}}\left((x^2-1)^{10}\right) for all x∈Rx\in\mathbb R. Then which of the following statements is (are) TRUE?
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A
The coefficient of x8x^8 in the polynomial f(x)f(x) is (−10)(18!8!)(-10)\left(\dfrac{18!}{8!}\right)
B
The value of f(1)+f(−1)f(1)+f(-1) is equal to 10! 21110!\,2^{11}
C
The degree of the polynomial f(x)f(x) is 1010
D
The constant term of the polynomial f(x)f(x) is −(10!5!)-\left(\dfrac{10!}{5!}\right)
2
Let a0,a1,…,a23a_0,a_1,\dots,a_{23} be real numbers such that (1+25x)23=∑i=023aixi\left(1+\dfrac25x\right)^{23} = \displaystyle\sum_{i=0}^{23}a_ix^i for every real number xx. Let ara_r be the largest among the numbers aja_j for 0≤j≤230\le j\le23. Then the value of rr is ___________.
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Your answer: —
Correct answer: 6
3
Let aa and bb be two nonzero real numbers. If the coefficient of x5x^5 in the expansion of (ax2+7027bx)4\left(ax^2+\dfrac{70}{27bx}\right)^4 is equal to the coefficient of x−5x^{-5} in the expansion of (ax−1bx2)7\left(ax-\dfrac{1}{bx^2}\right)^7, then the value of 2b2b is
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Your answer: —
Correct answer: 3
4
For nonnegative integers ss and rr, let (sr)={s!r! (s−r)!if r≤s,0if r>s.\dbinom{s}{r} = \begin{cases} \dfrac{s!}{r!\,(s-r)!} & \text{if } r \le s, \\ 0 & \text{if } r>s. \end{cases} For positive integers mm and nn, let g(m,n)=∑p=0m+nf(m,n,p)(n+pp)g(m,n) = \displaystyle\sum_{p=0}^{m+n} \dfrac{f(m,n,p)}{\dbinom{n+p}{p}}, where for any nonnegative integer pp, f(m,n,p)=∑i=0p(mi)(n+ip)(p+np−i)f(m,n,p) = \displaystyle\sum_{i=0}^{p} \dbinom{m}{i}\dbinom{n+i}{p}\dbinom{p+n}{p-i}. Then which of the following statements is/are TRUE?
Not attempted
A
g(m,n)=g(n,m)g(m,n)=g(n,m) for all positive integers m,nm,n
B
g(m,n+1)=g(m+1,n)g(m,n+1)=g(m+1,n) for all positive integers m,nm,n
C
g(2m,2n)=2 g(m,n)g(2m,2n)=2\,g(m,n) for all positive integers m,nm,n
D
g(2m,2n)=(g(m,n))2g(2m,2n)=(g(m,n))^2 for all positive integers m,nm,n
5
Suppose det⁡(∑k=0nk∑k=0nnCk k2∑k=0nnCk k∑k=0nnCk 3k)=0\det\begin{pmatrix}\displaystyle\sum_{k=0}^n k & \displaystyle\sum_{k=0}^n {}^nC_k\,k^2\\[6pt] \displaystyle\sum_{k=0}^n {}^nC_k\,k & \displaystyle\sum_{k=0}^n {}^nC_k\,3^k\end{pmatrix}=0 holds for some positive integer nn. Then ∑k=0nnCkk+1\displaystyle\sum_{k=0}^n \dfrac{{}^nC_k}{k+1} equals ____
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Your answer: —
Correct answer: 6.2
6
Let X=(10C1)2+2(10C2)2+3(10C3)2+⋯+10(10C10)2X=\left({}^{10}C_1\right)^2+2\left({}^{10}C_2\right)^2+3\left({}^{10}C_3\right)^2+\cdots+10\left({}^{10}C_{10}\right)^2, where 10Cr{}^{10}C_r, r∈{1,2,…,10}r\in\{1,2,\ldots,10\} denote binomial coefficients. Then, the value of 11430X\frac{1}{1430}X is _____.
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Your answer: —
Correct answer: 646
7
Let mm be the smallest positive integer such that the coefficient of x2x^2 in the expansion of (1+x)2+(1+x)3+⋯+(1+x)49+(1+mx)50(1+x)^2+(1+x)^3+\cdots+(1+x)^{49}+(1+mx)^{50} is (3n+1) 51C3(3n+1)\,{}^{51}C_3 for some positive integer nn. Then the value of nn is
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Your answer: —
Correct answer: 5
8
Coefficient of x11x^{11} in the expansion of (1+x2)4(1+x3)7(1+x4)12(1+x^2)^4(1+x^3)^7(1+x^4)^{12} is
Not attempted
A
10511051
B
11061106
C
11131113
D
11201120
9
The coefficients of three consecutive terms of (1+x)n+5(1+x)^{n+5} are in the ratio 5:10:145:10:14. Then n=n=
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Your answer: —
Correct answer: 6
Write the three consecutive binomial coefficients as (n+5r−1),(n+5r),(n+5r+1)\binom{n+5}{r-1},\binom{n+5}{r},\binom{n+5}{r+1} and use the standard ratio (Nr)(Nr−1)=N−r+1r\dfrac{\binom{N}{r}}{\binom{N}{r-1}}=\dfrac{N-r+1}{r} twice.
Deconstruction\textbf{Deconstruction}: Three consecutive binomial coefficients of a known expansion are in a given ratio; find the index nn. Strategy\textbf{Strategy}: Use the consecutive-coefficient ratio formula twice (once for each adjacent pair) to get two equations in nn and the position rr, then solve simultaneously. Execution\textbf{Execution}: Let N=n+5N=n+5. (Nr)(Nr−1)=N−r+1r=105=2⇒N−r+1=2r⇒N+1=3r\dfrac{\binom{N}{r}}{\binom{N}{r-1}}=\dfrac{N-r+1}{r}=\dfrac{10}{5}=2 \Rightarrow N-r+1=2r \Rightarrow N+1=3r ... (i) - (Nr+1)(Nr)=N−rr+1=1410=75⇒5(N−r)=7(r+1)⇒5N−5r=7r+7⇒5N−7=12r\dfrac{\binom{N}{r+1}}{\binom{N}{r}}=\dfrac{N-r}{r+1}=\dfrac{14}{10}=\dfrac75 \Rightarrow 5(N-r)=7(r+1) \Rightarrow 5N-5r=7r+7 \Rightarrow 5N-7=12r ... (ii) - From (i): r=N+13r=\dfrac{N+1}{3}. Substituting into (ii): 5N−7=12⋅N+13=4(N+1)=4N+45N-7=12\cdot\dfrac{N+1}{3}=4(N+1)=4N+4. - 5N−7=4N+4⇒N=115N-7=4N+4 \Rightarrow N=11. - N=n+5=11⇒n=6N=n+5=11 \Rightarrow n=6. Verification\textbf{Verification}: With N=11N=11, r=123=4r=\dfrac{12}{3}=4: (113)=165, (114)=330, (115)=462\binom{11}{3}=165,\ \binom{11}{4}=330,\ \binom{11}{5}=462. Ratios: 165:330:462=5:10:14165:330:462 = 5:10:14 (dividing through by 33) — exact match. Final Answer\textbf{Final Answer}: n=6n=6