JEE Main Practice — Mathematics Set 1

Explanation: $f'(x)=3x^2-6x-9=3(x-3)(x+1)$. $f''(x)=6x-6$ is negative at $x=-1$, so the local maximum is $f(-1)=-1-3+9+5=10$. ($-22=f(3)$ is the local minimum.)

Explanation: $\frac{dy}{dx}=2x-4=-2$ at $x=1$. Slope of normal $=-\frac{1}{-2}=\frac12$.

Explanation: $f'(x)=e^{-x}(1-x)>0\iff x<1$. So $f$ increases on $(-\infty,1)$ and decreases on $(1,\infty)$.

Explanation: $V=\frac43\pi r^3\Rightarrow\frac{dV}{dt}=4\pi r^2\frac{dr}{dt}=4\pi(25)(0.1)=10\pi\ \mathrm{cm^3/s}$.

Explanation: Vertices $(\pm5\cos\theta,\pm5\sin\theta)$: area $=4\cdot25\sin\theta\cos\theta=50\sin2\theta\le50$, attained at $\theta=\frac\pi4$ (a square of side $5\sqrt2$).

Explanation: Point on the parabola: $(t^2,2t)$. $D^2=(t^2-3)^2+4t^2=t^4-2t^2+9$. $\frac{d}{dt}D^2=4t^3-4t=0\Rightarrow t=0,\pm1$. $D^2(0)=9$, $D^2(\pm1)=8$ (minimum). $D_{min}=2\sqrt2$ at $(1,\pm2)$.

Explanation: $f'(x)=6(x-2)(x-3)$. Candidates: $f(1)=24$, $f(2)=29$, $f(3)=28$, $f(5)=250-375+180+1=56$. Maximum $=56$ (at the endpoint $x=5$).

Explanation: $y=\frac8x\Rightarrow y'=-\frac{8}{x^2}=-2$ at $x=2$. Tangent: $y-4=-2(x-2)\Rightarrow2x+y=8$, intercepts 4 and 8. Area $=\frac12\times4\times8=16$ (for $xy=c^2$ the area is always $2c^2$).

Explanation: $T_{r+1}=\binom5r(2x)^r$. For $r=3$: $\binom53\cdot8=10\times8=80$.

Explanation: There are 11 terms; the middle one is $T_6=\binom{10}{5}x^5\left(\frac1x\right)^5=252$.

Explanation: $T_{r+1}=\binom6r(x^2)^{6-r}\left(-\frac2x\right)^r=\binom6r(-2)^rx^{12-3r}$. $12-3r=0\Rightarrow r=4$: $\binom64(16)=240$.

Explanation: Sum of coefficients $=(3-1)^n=2^n=256\Rightarrow n=8$. Middle term $T_5=\binom84(3x)^4(-1)^4$, coefficient $=70\times81=5670$.

Explanation: $2^3=8\equiv1\pmod7$. $2^{100}=(2^3)^{33}\cdot2\equiv2\pmod7$. (Binomially: $8^{33}=(7+1)^{33}=7k+1$.)

Explanation: $\sum r\binom nr=n2^{n-1}$ and $\sum r(r-1)\binom nr=n(n-1)2^{n-2}$. So $\sum r^2\binom nr=n(n-1)2^{n-2}+n2^{n-1}=n(n+1)2^{n-2}=10\times11\times256=28160$.

Explanation: Coefficient of $x^r$: $\binom nr2^{n-r}3^{-r}$. Equating $r=2$ and $r=3$: $\binom n2\frac{2^{n-2}}{9}=\binom n3\frac{2^{n-3}}{27}\Rightarrow\frac{\binom n3}{\binom n2}=\frac{2\times27}{9}=6\Rightarrow\frac{n-2}{3}=6\Rightarrow n=20$.

Explanation: $T_{r+1}=\binom{12}{r}2^{(12-r)/2}3^{r/3}$ is rational iff $12-r$ is even and $r$ is a multiple of 3, i.e. $r$ is a multiple of 6: $r=0,6,12$. Hence 3 rational terms.

Explanation: Centre $(-g,-f)=(3,-4)$; radius $=\sqrt{g^2+f^2-c}=\sqrt{9+16+11}=6$.

Explanation: $(x-1)(x-5)+(y-2)(y-6)=0\Rightarrow x^2-6x+5+y^2-8y+12=0\Rightarrow x^2+y^2-6x-8y+17=0$.

Explanation: Length $=\sqrt{S_1}=\sqrt{25+9-20-6-4}=\sqrt4=2$.

Explanation: Condition for $y=mx+c$ to touch $x^2+y^2=r^2$: $c^2=r^2(1+m^2)=8\times2=16\Rightarrow c=\pm4$.

Explanation: $C_1=(0,0)$, $r_1=3$; $C_2=(4,0)$, $r_2=\sqrt{16-12}=2$. $d=4$ satisfies $|r_1-r_2|=1<d<r_1+r_2=5$, so the circles cut at two points: 2 common tangents.

Explanation: Family: $(x^2+y^2-4)+\lambda(x^2+y^2-2x-4y+4)=0$. Through $(0,0)$: $-4+4\lambda=0\Rightarrow\lambda=1$: $2x^2+2y^2-2x-4y=0$, i.e. $x^2+y^2-x-2y=0$. Radius $=\sqrt{\frac14+1}=\frac{\sqrt5}{2}$.

Explanation: The angle at the origin between the two axes is $90^\circ$, so the segment joining $(6,0)$ and $(0,8)$ is a diameter: length $\sqrt{36+64}=10$, radius 5.

Explanation: Centre $(1,2)$, radius $\sqrt{1+4-1}=2$. Tangency: $\frac{|3+8-k|}{5}=2\Rightarrow|11-k|=10\Rightarrow k=1$ or $21$. Sum $=22$.

Explanation: $(1+i)^2=2i$, so $(1+i)^8=(2i)^4=16i^4=16$.