JEE Main Practice — Mathematics Set 3

Explanation: Put $t=1+x^2$: $\int_1^2\frac{dt}{2t}=\frac12\ln2$.

Explanation: They meet at $x=0,1$, with $x\ge x^2$ in between: $\int_0^1(x-x^2)\,dx=\frac12-\frac13=\frac16$.

Explanation: Using $\int_0^af(x)dx=\int_0^af(a-x)dx$, $I=\int_0^{\pi/2}\frac{\cos x}{\cos x+\sin x}dx$. Adding: $2I=\int_0^{\pi/2}1\,dx=\frac\pi2$, so $I=\frac\pi4$.

Explanation: With $x\to\pi-x$: $I=\int_0^\pi\frac{\pi-x}{1+\sin x}dx$, so $2I=\pi\int_0^\pi\frac{dx}{1+\sin x}$. Now $\int\frac{dx}{1+\sin x}=\int\frac{1-\sin x}{\cos^2x}dx=\tan x-\sec x=-\frac{\cos x}{1+\sin x}$; from 0 to $\pi$ this gives $1-(-1)=2$. Hence $2I=2\pi$, $I=\pi$.

Explanation: The curve meets the $x$-axis at $x=0$ and $x=6$ and is non-negative between them. Area $=\int_0^6(6x-x^2)\,dx=108-72=36$.

Explanation: $x^3\cos x$ is odd, so its integral over $[-2,2]$ is 0. $\int_{-2}^23x^2dx=2\times8=16$ and $\int_{-2}^2|x|dx=2\times2=4$. Total $=20$.

Explanation: $\frac{\sin5x}{x}=5\cdot\frac{\sin5x}{5x}\to5\times1=5$.

Explanation: $\left(1+\frac2x\right)^x=\left[\left(1+\frac2x\right)^{x/2}\right]^2\to e^2$.

Explanation: $1-\cos4x=2\sin^22x$, so the limit is $2\cdot\left(\frac{\sin2x}{x}\right)^2\to2\times4=8$.

Explanation: For $x\ne3$, $f(x)=x+3\to6$ as $x\to3$. Continuity needs $k=6$.

Explanation: $f'(0)=\lim_{h\to0}\frac{h|h|}{h}=\lim_{h\to0}|h|=0$ from both sides, so $f$ is differentiable (hence continuous) at 0 with $f'(0)=0$.

Explanation: For $x>1$, $f(x)=\frac1x$ with $f(1^+)=1$ and right derivative $-\frac{1}{x^2}\big|_{1}=-1$. Continuity: $a+b=1$. Equal derivatives: $2a=-1\Rightarrow a=-\frac12$, so $b=\frac32$.

Explanation: $1-\cos2x\approx2x^2$, $3+\cos x\to4$, $x\tan4x\approx4x^2$. Limit $=\frac{2x^2\cdot4}{4x^2}=2$.

Explanation: Each modulus of a function with a simple (sign-changing) zero creates a corner there: $|x-1|$ at 1, $|x-3|$ at 3, $|x^2-4|$ at $\pm2$. The four points are distinct, and at each only one term has a corner, so it cannot be cancelled. Hence 4 points.

Explanation: For an $n\times n$ matrix, $|\mathrm{adj}\,A|=|A|^{n-1}=4^2=16$.

Explanation: $A^2=\begin{pmatrix}1\cdot1+2\cdot3&1\cdot2+2\cdot4\\3\cdot1+4\cdot3&3\cdot2+4\cdot4\end{pmatrix}=\begin{pmatrix}7&10\\15&22\end{pmatrix}$.

Explanation: $|kA|=k^n|A|$ for an $n\times n$ matrix: $2^3\times5=40$.

Explanation: Subtracting the first two: $2y=4\Rightarrow y=2$. Then $x+z=4$ and $2x-z=-1\Rightarrow3x=3$, $x=1$, $z=3$. Check in all equations: $6$, $2$, $2+2-3=1$.

Explanation: $|A|=6-5=1$. $A^{-1}=\frac{1}{|A|}\begin{pmatrix}3&-1\\-5&2\end{pmatrix}$. Check: $\begin{pmatrix}2&1\\5&3\end{pmatrix}\begin{pmatrix}3&-1\\-5&2\end{pmatrix}=\begin{pmatrix}1&0\\0&1\end{pmatrix}$.

Explanation: $\Delta=\begin{vmatrix}1&1&1\\1&2&3\\1&3&k\end{vmatrix}=(2k-9)-(k-3)+(3-2)=k-5$; infinitely many solutions need $\Delta=0\Rightarrow k=5$. Then (eq2 − eq1): $y+2z=1$ and (eq3 − eq2): $y+2z=m-2$; consistency needs $m-2=1\Rightarrow m=3$. $k+m=8$.

Explanation: $|A|=1(50-48)-2(40-42)+3(32-35)=2+4-9=-3$. For $n=3$: $|\mathrm{adj}(\mathrm{adj}A)|=|A|^{(n-1)^2}=(-3)^4=81$.

Explanation: By the Cayley–Hamilton theorem, $A^2-(\mathrm{tr}A)A+|A|I=O$ with $\mathrm{tr}A=5$ and $|A|=6+1=7$. So $\alpha=5$, $\beta=7$, $\alpha+\beta=12$. (Check: $A^2=\begin{pmatrix}3&-5\\5&8\end{pmatrix}$, $5A-7I=\begin{pmatrix}3&-5\\5&8\end{pmatrix}$.)

Explanation: 5 letters with L twice and E twice: $\frac{5!}{2!\,2!}=30$.

Explanation: $\binom73=\frac{7\cdot6\cdot5}{6}=35$.

Explanation: First digit: 9 choices (1–9); then $9,8,7$ choices for the remaining places: $9\times9\times8\times7=4536$.