Questions are transcribed from NTA's official question paper. Diagrams are described in words; open the official PDF to see them. Every answer was solved twice independently and is shown only when both agree. Where they did not, we say so.
The number of elements in the set $S = \left\{(r, k) : k \in \mathbf{Z} \text{ and } ^{36}C_{r+1} = \frac{6\binom{35}{r}}{\left(k^2-3\right)}\right\}$, is :
(1) 2
(2)
(3)
(4)
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Answer under review. Our two independent solutions did not agree, so we are not showing an answer yet. Check it against NTA's final answer key.
Let the smallest value of $k \in \mathbb{N}$, for which the coefficient of $x^3$ in $(1+x)^3 + (1+x)^4 + (1+x)^5 + ... + (1+x)^{99} + (1+kx)^{100}$, $x \neq 0$, is $\left(43n + \frac{101}{4}\right)\binom{100}{C_3}$ for some $n \in \mathbb{N}$, be $p$. Then the value of $p + n$ is:
(1) 10
(2) 11
(3) 12
(4) 13
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Answer: Option (2)
Key idea: Binomial Theorem, Hockey Stick Identity, Modular Arithmetic
Find coefficient of $x^3$ in $(1+x)^3 + ... + (1+x)^{99}$ using $\sum_{r=3}^{99} \binom{r}{3} = \binom{100}{4}$ (Hockey Stick Identity)
Find coefficient of $x^3$ in $(1+kx)^{100}$ which is $\binom{100}{3}k^3$
Total coefficient: $\binom{100}{4} + \binom{100}{3}k^3 = \left(43n + \frac{101}{4}\right)\binom{100}{3}$
Use $\binom{100}{4} = \frac{97}{4}\binom{100}{3}$ to get: $k^3 = 43n + 1$
Find smallest $k \in \mathbb{N}$: Test $k = 6$ gives $6^3 = 216 = 43(5) + 1$, so $p = 6, n = 5$
If the sum of the coefficients of $x^7$ and $x^{14}$ in the expansion of $\left(\frac{1}{x^3} - x^4\right)^n$, $x \neq 0$, is zero, then the value of $n$ is______.
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Answer: 21
Key idea: Binomial Theorem - General Term and Coefficient Extraction
Write the general term: $T_{r+1} = \binom{n}{r}(x^{-3})^{n-r}(-x^4)^r = \binom{n}{r}(-1)^r x^{-3n+7r}$
For coefficient of $x^7$: Set $-3n+7r=7$, giving $r = \frac{7+3n}{7}$. For integer $r$, need $n \equiv 0 \pmod{7}$, so $n=7k$
When $n=7k$: coefficient of $x^7$ is $\binom{7k}{1+3k}(-1)^{1+3k}$ and coefficient of $x^{14}$ is $\binom{7k}{2+3k}(-1)^{2+3k}$
Sum equals zero: $\binom{7k}{1+3k}(-1)^{1+3k} + \binom{7k}{2+3k}(-1)^{2+3k} = 0$
This requires $\binom{7k}{1+3k} = \binom{7k}{2+3k}$, which gives $(1+3k)+(2+3k)=7k$
If the coefficients of the middle terms in the binomial expansions of $(1 + \alpha x)^{26}$ and $(1 - \alpha x)^{28}$, $\alpha \neq 0$, are equal, then the value of $\alpha$ is:
(1) 1
(2) $\frac{14}{13}$
(3) $\frac{27}{7}$
(4)
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Answer under review. Our two independent solutions did not agree, so we are not showing an answer yet. Check it against NTA's final answer key.
If $26\left(\frac{2^3}{3}\binom{12}{C_2}+\frac{2^5}{5}\binom{12}{C_4}+\frac{2^7}{7}\binom{12}{C_6}+\ldots+\frac{2^{13}}{13}\binom{12}{C_{12}}\right)=3^{13}-\alpha$, then $\alpha$ is equal to:
(1) 45
(2) 48
(3) 51
(4) 54
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Answer: Option (3)
Key idea: Binomial Theorem with Integration
Recognize the general term as $\frac{2^{2k+1}}{2k+1}\binom{12}{2k}$ for $k=1,2,...,6$ and use $\frac{2^{2k+1}}{2k+1} = \int_0^2 t^{2k} dt$
Convert the sum to integral: $\sum_{k=1}^{6} \frac{2^{2k+1}}{2k+1}\binom{12}{2k} = \int_0^2 \sum_{k=1}^{6} \binom{12}{2k} t^{2k} dt$
Use binomial theorem: $(1+t)^{12} + (1-t)^{12} = 2\sum_{k=0}^{6} \binom{12}{2k} t^{2k}$, so $\sum_{k=1}^{6} \binom{12}{2k} t^{2k} = \frac{(1+t)^{12} + (1-t)^{12}}{2} - 1$
