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JEE MAIN 2021 March Attempt Mathematics Q1

Question 1: The middle term in the expansion of $\left(x^2+\dfrac{1}{x^2}+2\right)^{n}$ is (a) $\dfrac{n!}{\left(\dfrac{n}{2}!\right)^2}$ (b) $\dfrac{(2n)!}{\left(\dfrac{n}{2}!\right)^2}$ (c) $\dfrac{(2n)!}{\left(n!\right)^2}$ (d) $\dfrac{1.3.5\ldots(2n+1)}{n!}2^{n}$ Solution :
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Daily Question 108

Daily Question 108 $\textbf{Q108.}$ If $f(x)$ is a periodic function having period $7$ and $g(x)$ is periodic function having period $11$ then the period of $D(x)=\begin{vmatrix} f(x) & f\left(\dfrac{x}{3}\right)\\ g(x) & g\left(\dfrac{x}{5}\right)\\ \end{vmatrix}$ is (A) $231$ (B) $385$ (C) $1155$ (D) $77$ $\textbf{Ans.} (C)$ $\textbf{Sol.}$ Given $D(x)=\begin{vmatrix} f(x) & f\left(\dfrac{x}{3}\right)\\ g(x) & g\left(\dfrac{x}{5}\right)\\ \end{vmatrix}$ $\implies D(x)=f(x)g\left(\dfrac{x}{5}\right)-g(x)f\left(\dfrac{x}{3}\right)$ Period of $f(x)g\left(\dfrac{x}{5}\right)$ is $7\times 55 = 385$ Period of $g(x)f\left(\dfrac{x}{3}\right)$ is $11\times 21 = 231$ Hence the period of $D(x) = \text{LCM}$ of $(385, 231)=1155$ Compiled Daily Questions 1-100
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