NIMCET 2022 — Mathematics PYQ

1. Simplify the denominator: 

The sum in the denominator is an arithmetic progression: $(na+1)+(na+2)+\cdots +(na+n)$. The sum of an arithmetic progression is given by $\frac{N}{2}(first\text{ term}+last\text{ term})$, where $N$ is the number of terms. In this case, $N=n$, the first term is $na+1$, and the last term is $na+n$. Therefore, the sum is $\frac{n}{2}((na+1)+(na+n))=\frac{n}{2}(2na+n+1)=\frac{n^2}{2}(2a+1+\frac{1}{n})$. 

1. Rewrite the limit expression: 

The given limit can be written as: ${\lim }_{n\to \infty }\frac{{\sum }_{k=1}^n{k}^a}{(n+1{)}^{a-1}\cdot \frac{n^2}{2}(2a+1+\frac{1}{n})}=\frac{1}{60}$. 

1. Apply the integral approximation for the numerator: 

For large $n$, the sum ${\sum }_{k=1}^n{k}^a$ can be approximated by the integral ${\int }_0^n{x}^{a}dx$. ${\int }_0^n{x}^{a}dx={\left[\frac{x^{a+1}}{a+1}\right]}_0^n=\frac{n^{a+1}}{a+1}$, for $a\ne -1$. 

1. Substitute the approximations into the limit: 

The limit becomes: ${\lim }_{n\to \infty }\frac{\frac{n^{a+1}}{a+1}}{(n+1{)}^{a-1}\cdot \frac{n^2}{2}(2a+1+\frac{1}{n})}=\frac{1}{60}$. 

1. Simplify the expression inside the limit: 

${\lim }_{n\to \infty }\frac{n^{a+1}}{(a+1)\cdot n^{a-1}(1+\frac{1}{n}{)}^{a-1}\cdot \frac{n^2}{2}(2a+1+\frac{1}{n})}=\frac{1}{60}$. ${\lim }_{n\to \infty }\frac{n^{a+1}}{(a+1)\cdot n^{a-1}\cdot n^2\cdot (1+\frac{1}{n}{)}^{a-1}\cdot \frac{1}{2}(2a+1+\frac{1}{n})}=\frac{1}{60}$. ${\lim }_{n\to \infty }\frac{n^{a+1}}{(a+1)\cdot n^{a+1}\cdot (1+\frac{1}{n}{)}^{a-1}\cdot \frac{1}{2}(2a+1+\frac{1}{n})}=\frac{1}{60}$. 

1. Evaluate the limit: 

As $n\to \infty$, $(1+\frac{1}{n}{)}^{a-1}\to 1^{a-1}=1$, and $(2a+1+\frac{1}{n})\to (2a+1)$. Therefore, the limit simplifies to: $\frac{1}{(a+1)\cdot 1\cdot \frac{1}{2}(2a+1)}=\frac{1}{60}$. $\frac{2}{(a+1)(2a+1)}=\frac{1}{60}$. 

1. Solve for 'a': 

$(a+1)(2a+1)=120$. $2a^2+a+2a+1=120$. $2a^2+3a+1=120$. $2a^2+3a-119=0$. 

1. Solve the quadratic equation: 

Using the quadratic formula $a=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$: $a=\frac{-3\pm \sqrt{3^2-4(2)(-119)}}{2(2)}$. $a=\frac{-3\pm \sqrt{9+952}}{4}$. $a=\frac{-3\pm \sqrt{961}}{4}$. $a=\frac{-3\pm 31}{4}$. 

1. Find the possible values of 'a': 

$a_1=\frac{-3+31}{4}=\frac{28}{4}=7$. $a_2=\frac{-3-31}{4}=\frac{-34}{4}=-\frac{17}{2}$. 
Final Answer 
One of the values of $a$ is $7$.