JEE Main Practice — Mathematics Set 2

Explanation: $\alpha+\beta=5$, $\alpha\beta=6$. $\alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta=25-12=13$.

Explanation: $|z|=\frac{2}{\sqrt2}=\sqrt2$. $\arg z=\arg(1+i\sqrt3)-\arg(1-i)=\frac\pi3-\left(-\frac\pi4\right)=\frac{7\pi}{12}$, which lies in $(-\pi,\pi]$.

Explanation: $(x-2)^2+y^2=x^2+(y+2)^2\Rightarrow-4x+4=4y+4\Rightarrow x+y=0$: the perpendicular bisector of the points $2$ and $-2i$.

Explanation: Discriminant $k^2-36>0\iff|k|>6$ (i.e. $k<-6$ or $k>6$). $|k|=6$ gives equal roots.

Explanation: $1+\omega+\omega^2=0\Rightarrow1+\omega=-\omega^2$. So $1+\omega-\omega^2=-2\omega^2$ and $(-2\omega^2)^7=-128\,\omega^{14}=-128\,\omega^{12}\omega^2=-128\omega^2$.

Explanation: Let $z=x+iy$: $x^2-y^2+\sqrt{x^2+y^2}=0$ and $2xy=0$. If $y=0$: $x^2+|x|=0\Rightarrow x=0$. If $x=0$: $-y^2+|y|=0\Rightarrow y=0$ or $|y|=1$. Solutions: $0, i, -i$ — three in all.

Explanation: $|3+4i|^2=25$ and $|1+i|^4=(\sqrt2)^4=4$. Product $=100$.

Explanation: $y^2=4ax$ with $4a=12\Rightarrow a=3$; focus $(a,0)=(3,0)$.

Explanation: $e=\sqrt{1-\frac{b^2}{a^2}}=\sqrt{1-\frac{16}{25}}=\frac35$.

Explanation: $c^2=a^2+b^2=25$, $c=5$; foci $(\pm5,0)$, distance 10. ($2\sqrt7$ would wrongly use $a^2-b^2$.)

Explanation: For $y^2=4ax$ the tangent of slope $m$ is $y=mx+\frac am$. Here $a=2$: $y=2x+1$. (Check: substituting gives $(2x+1)^2=8x\Rightarrow(2x-1)^2=0$, a double root.)

Explanation: $a=5$, $ae=4$, $b^2=a^2-a^2e^2=25-16=9$. Major axis along the $x$-axis: $\frac{x^2}{25}+\frac{y^2}{9}=1$.

Explanation: $a=2$; $(8,8)=(at^2,2at)$ with $t=2$. The other end has $t_2=-\frac1{t}=-\frac12$: $\left(\frac12,-2\right)$. Length $=\sqrt{(8-\frac12)^2+(8+2)^2}=\sqrt{56.25+100}=12.5=\frac{25}{2}$ (formula $a\left(t+\frac1t\right)^2=2\times\frac{25}{4}$).

Explanation: Latus rectum $=\frac{2b^2}{a}=\frac{2\times9}{2}=9$.

Explanation: $\frac{2b^2}{a}=\frac{2b}{2}=b\Rightarrow\frac ba=\frac12$. $e^2=1-\frac{b^2}{a^2}=\frac34$, so $4e^2=3$.

Explanation: Highest derivative is $\frac{d^2y}{dx^2}$ (order 2); it appears to power 3 in a polynomial equation in derivatives (degree 3).

Explanation: Separate: $\frac{dy}{y}=\frac{dx}{x}\Rightarrow\ln|y|=\ln|x|+k\Rightarrow y=cx$.

Explanation: Linear with integrating factor $e^{\int\frac1xdx}=x$: $\frac{d}{dx}(xy)=x^3\Rightarrow xy=\frac{x^4}{4}+C$.

Explanation: $\int\frac{dy}{1+y^2}=\int\frac{dx}{1+x^2}\Rightarrow\tan^{-1}y-\tan^{-1}x=k\Rightarrow\frac{y-x}{1+xy}=\tan k=C$, i.e. $y-x=C(1+xy)$.

Explanation: $\frac{dy}{dx}-\frac yx=1$; integrating factor $\frac1x$: $\frac{d}{dx}\left(\frac yx\right)=\frac1x\Rightarrow\frac yx=\ln x+C$, so $y=x\ln x+Cx$. (Equivalently homogeneous with $y=vx$: $x\frac{dv}{dx}=1$.)

Explanation: The left side is $\frac{d}{dx}\left[(1+x^2)y\right]$. So $(1+x^2)y=\frac{4x^3}{3}+C$; $y(0)=0\Rightarrow C=0$. $y=\frac{4x^3}{3(1+x^2)}$, $y(1)=\frac{4}{6}=\frac23$.

Explanation: Integrating factor $\frac1x$: $\frac{d}{dx}\left(\frac yx\right)=1\Rightarrow\frac yx=x+C$. $y(1)=2\Rightarrow C=1$. $y=x^2+x$, $y(3)=12$.

Explanation: $e^{2x}$ and $e^{3x}$ solve $y''+ay'+by=0$ iff $2$ and $3$ are roots of $m^2+am+b=0$: $a=-(2+3)=-5$, $b=6$. (Direct elimination: $y'=2c_1e^{2x}+3c_2e^{3x}$, $y''=4c_1e^{2x}+9c_2e^{3x}$ gives $y''-5y'+6y=0$.) $b-a=11$.

Explanation: $\sin^2x=\frac{1-\cos2x}{2}$: $\int_0^{\pi/2}=\frac12\left[x-\frac{\sin2x}{2}\right]_0^{\pi/2}=\frac\pi4$.

Explanation: By parts: $\int xe^x\,dx=xe^x-\int e^x\,dx=e^x(x-1)+C$. (Check: $\frac{d}{dx}[e^x(x-1)]=e^x(x-1)+e^x=xe^x$.)