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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 115

Daily Question 115 $\textbf{Q115.}$ Let $P_1=\begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1\\ \end{bmatrix},$ $P_2=\begin{bmatrix} 1 & 0 & 0\\ 0 & 0 & 1\\ 0 & 1 & 0\\ \end{bmatrix},$ $P_3=\begin{bmatrix} 0 & 1 & 0\\ 1 & 0 & 0\\ 0 & 0 & 1\\ \end{bmatrix},$ $P_4=\begin{bmatrix} 0 & 1 & 0\\ 0 & 0 & 1\\ 1 & 0 & 0\\ \end{bmatrix},$ $P_5=\begin{bmatrix} 0 & 0 & 1\\ 1 & 0 & 0\\ 0 & 1 & 0\\ \end{bmatrix},$ $P_6=\begin{bmatrix} 0 & 0 & 1\\ 0 & 1 & 0\\ 1 & 0 & 0\\ \end{bmatrix}$ and $X=\displaystyle\sum_{k=1}^{6}P_k\begin{bmatrix} 2 & 1 & 3\\ 1 & 0 & 2\\ 3 & 2 & 1\\ \end{bmatrix}P^{T}_{k}$ Where $P^{T}_{k}$ denotes the transpose of matrix $P_k.$ Then which of the following options is/are correct? (A) $X$ is a symmetric matrix. (B) The sum of diagonal entries of $X$ is $18.$ (C) If $X\begin{bmatrix} 1\\ 1\\ 1...
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Daily Question 114

Daily Question 114 $\textbf{Q114.}$ The value of $\displaystyle\sum_{k=0}^{7}\left[\dfrac{\displaystyle{7 \choose k}}{\displaystyle{14 \choose k}}\sum_{r=k}^{14}{r \choose k}{14 \choose r}\right]$ where $\displaystyle{n \choose x}$ denotes ${^{n}}C_{r}$, is (A) $6^7$ (B) $8^7$ (C) greater than $7^6$ (D) greater than $7^{8}$ $\textbf{Ans.} (A), (C)$ $\textbf{Sol.}$ Given $\displaystyle\sum_{k=0}^{7}\left[\dfrac{\displaystyle{7 \choose k}}{\displaystyle{14 \choose k}}\sum_{r=k}^{14}{r \choose k}{14 \choose r}\right]$ $=\displaystyle\sum_{k=0}^{7}\left[\dfrac{{^{7}}C_k}{{^{14}C_k}}\sum_{r=k}^{14}{^{r}}C_k\cdot{^{14}}C_{r}\right]$ $=\displaystyle\sum_{k=0}^{7}\left[{^{7}}C_k\dfrac{k!(14-k)!}{14!}\sum_{r=k}^{14}\dfrac{r!}{k!(r-k!)}\cdot\dfrac{14!}{r!(14-r)!}\right]$ $=\displaystyle\sum_{k=0}^{7}\left[{^{7}}C_k\sum_{r=k}^{14}{^{14-k}}C_{r-k}\right]$ $=\displaystyle\sum_{k=0}^{7}{^{7}}C_k\cdot 2^{14-k}$ $=2^{14}\displaystyle\sum_{k=0}^{7}{^{7}}C_k\left(\dfr...
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Daily Question 113

Daily Question 113 $\textbf{Q113.}$ If $f(x)=\dfrac{x^2}{4+(\ln x)(\ln x)\ldots\infty}$ $\forall$ $x\in[1,\infty)$, then (A) $f(x)=\dfrac{x^2}{5}, x\in[1,e)$ (B) $f(x)=\dfrac{x^2}{4}, x\in \{e\}$ (C) $f(x)=\dfrac{x^2}{4}, x > e$ (D) $\displaystyle\int\limits_{1}^{2}f(x)\mathrm{d}x=\dfrac{7}{12}$ $\textbf{Ans.} (D)$ $\textbf{Sol.}$ Given $f(x)=\dfrac{x^2}{4+(\ln x)(\ln x)\ldots\infty}$ $\forall$ $x\in[1,\infty)$ Let $h(x)=4+(\ln x)(\ln x)\ldots\infty$ $h(x)=\begin{cases} 4, & 1 \le x < e\\ 5, & x=e\\ \infty & x > e\\ \end{cases}$ For $x = e,$ $f(x)=\dfrac{x^2}{5}$ For $1 < x < e,$ $f(x)=\dfrac{x^2}{4}$ Now, $\displaystyle\int\limits_{1}^{2}f(x)\mathrm{d}x=\int\limits_{1}^{2}\dfrac{x^2}{4}\mathrm{d}x$ $\implies\displaystyle\int\limits_{1}^{2}f(x)\mathrm{d}x=\left[\dfrac{x^3}{12}\right]_{1}^{2}$ $\implies\displaystyle\int\limits_{1}^{2}f(x)\mathrm{d}x=\dfrac{7}{12}$ Daily Question 112 Compiled Dail...
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Daily Question 112

Daily Question 112 $\textbf{Q112.}$ If $x^{15}-x^{13}+x^{11}-x^9+x^7-x^5+x^3-x=7$, where $x>0$, then (A) $x^{16}$ is equal to 15 (B) $x^{16}$ is less than 15 (C) $x^{16}$ is greater than 15 (D) none of these $\textbf{Ans.} (C)$ $\textbf{Sol.}$ Given $x^{15}-x^{13}+x^{11}-x^9+x^7-x^5+x^3-x=7$ $\implies(x^2-1)(x^4+1)(x^8+1)x=7\qquad\ldots(1)$ Now, $x^{16}-1=(x^{2}-1)(x^{2}+1)(x^{4}+1)(x^8+1)$ Using Eq. $(1)$ $\implies x^{16}-1=\dfrac{7}{x}(x^2+1)$ $\implies x^{16}-1=7\left[\left(\sqrt{x}-\dfrac{1}{\sqrt{x}}\right)^2+2\right]> 14$ $\therefore x^{16}>15$ Daily Question 111 Compiled Daily Questions 1-100
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Daily Question 111

Daily Question 111 $\textbf{Q111.}$ The set of values of $x$ for which $\dfrac{\tan 3x - \tan 2x}{1+\tan 3x\tan2x}=1$ (A) $x=n\pi+\dfrac{\pi}{4}$ (B) $x=n\pi\pm\dfrac{\pi}{4}$ (C) $x=n\pi-\dfrac{\pi}{4}$ (D) none of these $\textbf{Ans.} (D)$ $\textbf{Sol.}$ Given $\dfrac{\tan 3x - \tan 2x}{1+\tan 3x\tan2x}=1$ $\implies \tan({3x-2x})=1$ $\implies \tan x=1=\tan\dfrac{\pi}{4}$ $\implies x=n\pi+\dfrac{\pi}{4}$ But $\tan 2x=\tan\left(2\left(n\pi+\dfrac{\pi}{4}\right)\right)=$ $\tan\left(2n\pi+\dfrac{\pi}{2}\right)=$ $\tan\dfrac{\pi}{2}=\infty$ Which does not satisfy the given equation. So, no solution exits. Daily Question 110 Compiled Daily Questions 1-100
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Daily Question 110

Daily Question 110 $\textbf{Q110.}$ If $I_{1}=\displaystyle\int\limits_{0}^{1}\dfrac{1+x^8}{1+x^{4}}\mathrm{d}x$ and $I_2=\displaystyle\int\limits_{0}^{1}\dfrac{1+x^{9}}{1+x^{3}}\mathrm{d}x$, then (A) $I_1 > 1, I_2 < 1$ (B) $I_1 < 1, I_2 > 1$ (C) $1 < I_1 < I_2$ (D) $I_2 < I_1 < 1$ $\textbf{Ans.} (D)$ $\textbf{Sol.}$ As $x\in(0, 1)$ $\implies 1+x^8 < 1+x^4$ $\therefore I_{1}< 1$ Similarly, $I_2 < 1$ Now, $1+x^8 > 1+x^9$ and $1+x^4< 1+x^3$ Thus $\dfrac{1+x^8}{1+x^{4}}>\dfrac{1+x^{9}}{1+x^{3}}$ $\implies I_1 > I_2$ Hence $I_2 < I_1 < 1$ Compiled Daily Questions 1-100
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Daily Question 109

Daily Question 109 $\textbf{Q109.}$ $\displaystyle\int\limits_{1}^{5}\left(\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}\right)=$ (A) $\dfrac{16}{3}$ (B) $\dfrac{32}{3}$ (C) $\dfrac{34}{3}$ (D) $\dfrac{8}{3}$ $\textbf{Ans.} (C)$ $\textbf{Sol.}$ Let $I=\displaystyle\int\limits_{1}^{5}\left[\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}\right]\mathbb{d}x$ Substitute $\sqrt{x-1}=u\to\mathbb{d}x=2\sqrt{x-1}\mathbb{d}u$ $I=\displaystyle\int\limits_{0}^{2}2u\left[\sqrt{u^2+2u+1}+\sqrt{u^2-2u+1}\right]\mathbb{d}u$ $I=\displaystyle\int\limits_{0}^{2}2u\left[|u+1|+|u-1|\right]\mathbb{d}u$ $I=\displaystyle\int\limits_{0}^{1}2u\left[u+1-u+1|\right]\mathbb{d}u+\int\limits_{1}^{2}2u\left[u+1+u-1|\right]\mathbb{d}u$ $I=\displaystyle\int\limits_{0}^{1}4u\mathbb{d}u+\int\limits_{1}^{2}4u^2\mathbb{d}u$ $I=\left[2u^2\right]_{0}^{1}+\left[\dfrac{4u^3}{3}\right]_{1}^{2}$ $I=\dfrac{34}{3}$ Compiled Daily Questions 1-100
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