JEE Main Practice — Mathematics Set 5

Explanation: All functions: $2^4=16$. Exclude the 2 constant functions (not onto): $16-2=14$.

Explanation: Not reflexive: $\left(\frac12,\frac12\right)\notin R$ since $\frac12>\frac14$. Not symmetric: $(1,2)\in R$ ($1\le4$) but $(2,1)\notin R$ ($2>1$). Not transitive: $(3,2)\in R$ ($3\le4$) and $(2,1.5)\in R$ ($2\le2.25$) but $(3,1.5)\notin R$ ($3>2.25$).

Explanation: $n(A\times B)=6$, total subsets $2^6=64$. Remove subsets with 0, 1 or 2 elements: $1+6+15=22$. Answer $64-22=42$.

Explanation: $n(M\cup P)=60+50-30=80$. Neither $=100-80=20$.

Explanation: $m=\frac{11-3}{6-2}=\frac84=2$.

Explanation: $d=\frac{|3(3)+4(4)-5|}{\sqrt{9+16}}=\frac{20}{5}=4$.

Explanation: Slopes $\sqrt3$ and $\frac1{\sqrt3}$ correspond to inclinations $60^\circ$ and $30^\circ$; the angle between the lines is $30^\circ$. (Formula: $\tan\theta=\left|\frac{\sqrt3-1/\sqrt3}{1+1}\right|=\frac1{\sqrt3}$.)

Explanation: The perpendicular through $(1,2)$ has direction $(1,1)$: points $(1+t,2+t)$. On the line: $3+2t=7\Rightarrow t=2$, giving $(3,4)$.

Explanation: The first two meet where $3x=6$: $(2,3)$. Substituting into the third: $2k+3=7\Rightarrow k=2$.

Explanation: Altitude from $C$ is perpendicular to $AB$ (the $x$-axis): $x=1$. $BC$ has slope $\frac{3-0}{1-4}=-1$, so the altitude from $A$ has slope 1: $y=x$. Intersection: $(1,1)$. Check with the altitude from $B$ (slope $-\frac13$ since $AC$ has slope 3): $y=-\frac{x-4}{3}$ gives $y=1$ at $x=1$. ($\left(\frac53,1\right)$ is the centroid.)

Explanation: Intercepts: $x=8$, $y=6$. Area $=\frac12\times8\times6=24$.

Explanation: Write the second line as $3x+4y+8=0$ and the first as $3x+4y-7=0$. Distance $=\frac{|-7-8|}{\sqrt{9+16}}=\frac{15}{5}=3$.

Explanation: Divide by $\sqrt{4+1+4}=3$: $\left(\frac23,-\frac13,\frac23\right)$.

Explanation: $\sqrt{3^2+4^2+12^2}=\sqrt{169}=13$.

Explanation: $\cos\theta=\frac{|1\cdot2+1\cdot(-1)+2\cdot1|}{\sqrt6\sqrt6}=\frac36=\frac12\Rightarrow\theta=60^\circ$.

Explanation: $\frac{|2+1+2-14|}{\sqrt{4+1+4}}=\frac93=3$.

Explanation: General point $(1+2t,2+3t,3+4t)$. Sum $=6+9t=15\Rightarrow t=1$: $(3,5,7)$.

Explanation: $\vec b_1\times\vec b_2=\begin{vmatrix}\hat i&\hat j&\hat k\\2&-1&1\\3&-5&2\end{vmatrix}=3\hat i-\hat j-7\hat k$, magnitude $\sqrt{59}$. $\vec a_2-\vec a_1=\hat i-\hat k$. SD $=\frac{|(1)(3)+0+(-1)(-7)|}{\sqrt{59}}=\frac{10}{\sqrt{59}}$.

Explanation: Direction from the first to the second point: $(1,2,2)$. From the first to the third: $(3,k-2,6)$, which must be $3(1,2,2)$. Hence $k-2=6$, $k=8$.

Explanation: Second plane: $2x-y+2z+15=0$; first: $2x-y+2z-3=0$. Distance $=\frac{|-3-15|}{\sqrt{4+1+4}}=\frac{18}{3}=6$.

Explanation: $\sin(45^\circ+30^\circ)=\sin45^\circ\cos30^\circ+\cos45^\circ\sin30^\circ=\frac{1}{\sqrt2}\cdot\frac{\sqrt3}{2}+\frac1{\sqrt2}\cdot\frac12=\frac{\sqrt6+\sqrt2}{4}$.

Explanation: The principal range of $\cos^{-1}$ is $[0,\pi]$, and $\cos\frac{2\pi}{3}=-\frac12$.

Explanation: $\sin x(2\cos x-1)=0$. $\sin x=0$: $x=0,\pi,2\pi$. $\cos x=\frac12$: $x=\frac\pi3,\frac{5\pi}3$. Total 5 distinct solutions.

Explanation: $3\sin x+4\cos x=5\sin(x+\alpha)$ with $\tan\alpha=\frac43$, so its maximum is $\sqrt{9+16}=5$. Maximum of the expression $=5+5=10$.

Explanation: Since $\frac12\cdot\frac13<1$: $\tan^{-1}\frac12+\tan^{-1}\frac13=\tan^{-1}\frac{\frac12+\frac13}{1-\frac16}=\tan^{-1}1=\frac\pi4$.