Answers For Cambridge AS & A Level P1 (Pure) 2014 (9709/11)

*These answers are only for reference purposes and are not the official marking scheme for the relevant paper. Download Paper

1. $(2+ax)^7=\displaystyle\sum_{r=0}^{7} {}^7C_r 2^{7-r} (ax)^r$
   The terms of $x$ and $x^2$ can get by $r=1$ and $r=2$ respectively.
   $T_1={}^7C_1 2^{7-1} (ax)$ and $T_2={}^7C_2 2^{7-2} (ax)^2$
   $\therefore {}^7C_1 2^{\not{6}} \not{a}={}^7C_2 \not{2^5} a^{\not{2}}$
   $\mathbf{a=\cfrac{2}{3}}$
    
2. Assume $\theta=\tan^{-1}(3)$, Then $t=\tan{\theta}=3$
   $\sin^{-1}(x-1)=\tan^{-1}3$
   $(x-1)=\sin{\theta}$
   $(x-1)^2=\sin^2{\theta}$
   $=\cfrac{1-\cos 2\theta}{2}$
   $=\cfrac{1-\cfrac{1-t^2}{1+t^2}}{2}$
   $=\cfrac{1-\cfrac{1-9}{1+9}}{2}$
   $=\cfrac{18}{20}$
   $\therefore x-1=\pm \cfrac{3}{\sqrt{10}}$
   $\mathbf{x=1\pm\cfrac{3}{\sqrt{10}}}$
   *The answer can be easily taken if you are using table or calculators.
    
3. $\cfrac{13\sin^2\theta}{2+\cos\theta}+\cos\theta=2$
   $13\sin^2\theta=(2-\cos\theta)(2+\cos\theta)$
   $13\sin^2\theta=4-\cos^2\theta$
   $12\sin^2\theta+\sin^2\theta+\cos^2\theta=4$
   $12\sin^2\theta=4-1$
   $\sin\theta=\pm \cfrac{1}{2}$
   $\mathbf{\theta=30^0}$ or $\mathbf{\theta=150^0}$
    
4.   
1. As $4x+ky=20$ passes through $A(8,-4)$ and $B(b,2b)$,
   $4\times 8+k\times(-4)=20$
   $\mathbf{k=3}$
   $4\times b+3\times (2b)=20$
   $\mathbf{b=2}$
    
2. If the mid point is $M$,
   $M=\left(\cfrac{8+2}{2},\cfrac{-4+4}{2}\right)$
   $\mathbf{M=(5,0)}$
5. If the line $y_1=2x-k$ meets the curve $y_2=x^2+kx-2$ at two distinct points, then the quadratic equation $y_1=y_2$ must have two distinct roots. For that, the discriminant $\Delta _x$ must be positive.
   $2x-k=x^2+kx-2$
   $x^2+(k-2)x+k-2=0$
   $\therefore \Delta _x=(k-2)^2-4\times1\times(k-2)$
   But $\Delta _x> 0$
   $\therefore (k-2)^2-4(k-2)> 0$
   $(k-2)(k-2-4)>0$
   $(k-2)(k-6)>0$
   $\therefore$ either $k>2$ and $k>6$ or $k<2$ and $k<6$
   So the set of values of $k$ is $\mathbf{k>6}$ or $\mathbf{k<2}$
   
   
    
6.  
1. Here I show two methods
   Method 01
   $\overrightarrow{OA}=3i+2j-k$ and $\overrightarrow{OB}=7i-3j+k$
   $\therefore \overrightarrow{AB}=\overrightarrow{AO}+\overrightarrow{OB}$
   $=-3i-2j+k+7i-3j+k=4i-5j+2k$
   If $OAB\measuredangle=90^0$ then the scalar product between $\overrightarrow{OA}$ and $\overrightarrow{AB}$ must be zero.
   $\overrightarrow{OA}\cdot\overrightarrow{AB}=(3i+2j-k)\cdot(4i-5j+2k)$
   $=12-10-2=\mathbf{0}$
   
   Method 02
   If $\hat{OAB}$ is a right angle, then $OAB$ triangle must be a right angle triangle. So we must be use Pythagorean theorem for the triangle.
   $|\overrightarrow{OA}|^2=3^2+2^2+(-1)^2=14$
   $|\overrightarrow{OB}|^2=7^2+(-3)^2+1^2=59$
   $|\overrightarrow{AB}|^2=4^2+(-5)^2+2^2=45$
   $\therefore$ We can see that,
   $|\overrightarrow{OB}|^2=|\overrightarrow{OA}|^2+|\overrightarrow{AB}|^2$
   Hence, according to the converse of the Pythagorean theorem we can say $OAB$ is a right angle triangle.
    
2. Area $=\cfrac{1}{2}\times|OA|\times|AB|=\cfrac{1}{2}\times\sqrt{14}\times\sqrt{45}$
   $\therefore$ Area $=\cfrac{3}{2}\sqrt{70}$
    
7.  
1. According to the given data,
   $S=\cfrac{a\left(1-r^{\infty}\right)}{1-r}$  --------- (1)
   $3S=\cfrac{a\left(1-(2r)^{\infty}\right)}{1-2r}$ ------- (2)
   By dividing (2) by (1),
   $3=\cfrac{1-r}{1-2r}$    $(\because r^{\infty},$ $(2r)^{\infty} \to 0)$
   $3-6r=1-r$
   $\therefore r=\cfrac{2}{5}$
    
2. According to the given data,
   $T_1=7$, $T_n=84$ and $T_{3n}=245$
   If the common difference is $d$, then the general term is,
   $T_n=a+(n-1)d$  where $a$ is the first term.
   $84=7+(n-1)d$
   $(n-1)d=77$ -------- (A)
   $T_{3n}=a+(3n-1)d$
   $245=7+(3n-1)d$
   $(3n-1)d=238$ -------- (B)
   Dividing (B) by (A),
   $\cfrac{3n-1}{n-1}=\cfrac{238}{77}$
   $11(3n-1)=34(n-1)$
   $\therefore \mathbf{n=23}$
    
8.  
1. When we consider the given figure $OA=4 cm$
   By resolving the side,
   $OD=4\cos\alpha$ $cm$
   $\therefore$ The arc length $CD=4\alpha\cos\alpha$   $(\because A=r\theta)$
   $\therefore$ Total perimeter of $ABCD=4\alpha\cos\alpha+4\alpha+2(4-4\cos\alpha)$
   $=4\alpha(1+\cos\alpha)+8(1-\cos\alpha)$ $cm$
    
2. Area of a sector $=\cfrac{1}{2}r^2\theta$
   $\therefore$ The area of the shaded region $=\cfrac{1}{2}4^2\alpha-\cfrac{1}{2}(4\cos\alpha)^2\alpha$
   $=8\cdot\cfrac{\pi}{6}\left(1-\left(\cfrac{\sqrt{3}}{2}\right)^2\right)$
   $=\cfrac{1}{3}\pi$ $cm^2$
    
9.  
1. As $\mathrm{f}'(x)=2x-\cfrac{2}{x^2}$ we can get the gradient of the tangent ($m_T$) at $P(2,6)$,
   $m_T=2(2)-\cfrac{2}{2^2}=\cfrac{7}{2}$
   $\therefore$ the gradient of the normal $m_N$ satisfies,
   $m_T m_N=-1$  ($\because$ they are perpendicular to each other)
   $\therefore m_N=-\cfrac{2}{7}$
   As the normal passes through point $P$,
   $\cfrac{y-6}{x-2}=\cfrac{-2}{7}$
   $\therefore$ the equation of the normal is $\mathbf{7y+2x-46=0}$
    
2. By integrating w.r.t $x$, $\int \mathrm{f}'(x) \: \mathrm{d}x=\mathrm{f}(x)$
   
   $\mathrm{f}(x)=2\cdot\cfrac{x^2}{2}-2\cdot\cfrac{x^{-2+1}}{-2+1}+c$  Where $c$ is a constant
   $\mathrm{f}(x)=x^2+\cfrac{2}{x}+c$
   As $y=\mathrm{f}(x)$ passes through $P$,
   $6=2^2+\cfrac{2}{2}+c$
   $\therefore c=1$
   Then $\mathrm{f}(x)=x^2+\cfrac{2}{x}+1$
    
3. At stationary points $\mathrm{f}'(x)=0$
   $\therefore 2x-\cfrac{2}{x^2}=0$
   $x^3-1=0$
   $(x-1)(x^2+x+1)=0$
   As $x^2+x+1\ne 0$, $x-1=0$
   Therefore there is a stationary point at $\mathbf{x=1}$
   If you substitute two values in positive and negative sides of $x=1$, to the differential $\mathrm{f}'(x)$, you will find that this is a minimum. or
   differentiate $\mathrm{f}'(x)$ w.r.t $x$ again and substitute $x=1$, you'll get a positive value.
10.  
1. $x^2-2x-15=x^2-2x+\left(\cfrac{-2}{2}\right)^2-\left(\cfrac{-2}{2}\right)^2-15$
   $=(x^2-2x+1)-16=(x-1)^2-16$
    
2. $\mathrm{f}(x)=(x-1)^2-16$
   $c$ is the minimum value that can be taken for $\mathrm{f}(x)$
   $\therefore \mathrm{f}_{min}(x)=0-16=\mathbf{-16}\quad(\because 0$ is the minimum value for a quadratic$)$
   $c_{min}=-16$
    
3. If $\mathrm{f}(x)\to c$ then $x\to p$
   $9=(p-1)^2-16$
   $(p-1)^2=25$
   $p-1=\pm 5$
   $p=\mathbf{6}$  as $p$ is a positive constant
   $65=(q-1)^2-16$
   $(q-1)^2=81$
   $q-1=\pm 9$
   $q=\mathbf{10}$ as $q$ is positive.
    
4. $y=\mathrm{f}(x)=(x-1)^2-16$
   $(x-1)^2=y+16$
   $x-1=\pm \sqrt{y+16}$
   $x=1\pm \sqrt{y+16}$
   $0<x$ and $9<y$
   $\therefore \mathrm{f}^{-1}(x)=1+\sqrt{x+16}$
    
11.  


1. By differentiating both curves w.r.t $x$, at the point $Q(2,3)$, we can get the gradient of to tangents.
   $m_1=\left.\cfrac{\mathrm{d}}{\mathrm{d}x}\left(\sqrt{4x+1}\right)\right |_{x=2}=\left.\cfrac{1}{2}(4x+1)^{\frac{1}{2}-1}\cdot 4\right |_{x=2}$
   $m_1=\cfrac{2}{\sqrt{4\times 2+1}}=\cfrac{2}{3}$
   $m_2=\left.\cfrac{\mathrm d}{\mathrm d x}\left(\cfrac{1}{2}x^2+1\right)\right |_{x=2}=\left. \cfrac{1}{2}\cdot 2x\right |_{x=2}$
   $m_2=2$
   
   $\theta=\theta _2-\theta _1$
   $\tan\theta=\tan(\theta_2-\theta_1)$
   $\tan\theta=\cfrac{\tan\theta_2-\tan\theta_1}{1+\tan\theta_2\tan\theta_1}$
   $\tan\theta=\left|\cfrac{m_2-m_1}{1+m_2 m_1}\right|$
   $\theta=\tan^{-1}\left|\cfrac{2-\cfrac{2}{3}}{1+2\cdot\cfrac{2}{3}}\right|$
   $\theta=\tan^{-1}\left(\cfrac{4}{7}\right)=29.745^0$
    
2. Area $=\int\limits_0^2 y_1-y_2 \: \mathrm{d}x$
   $=\int\limits_0^2 \sqrt{4x+1}-\left(\cfrac{1}{2}x^2+1\right)\:\mathrm{d}x$
   $=\left[\cfrac{(4x+1)^{\frac{1}{2}+1}}{\frac{1}{2}+1}\cdot\cfrac{1}{4}-\cfrac{1}{2}\cfrac{x^3}{3}-x\right]_0^2$
   $=\left[\cfrac{(4x+1)^{\frac{3}{2}}}{6}-\cfrac{x^3}{6}-x\right]_0^2$
   $=\left\{\cfrac{(4\times2+1)^\frac{3}{2}}{6}-\cfrac{2^3}{6}-2\right\}-\left\{\cfrac{(0+1)^\frac{3}{2}}{6}-0-0\right\}$
   $=\cfrac{9^{^3/_2}}{6}-\cfrac{8}{6}-2-\cfrac{1}{6}$
   $=1$ unit
   

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