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### MATHEMATICS 9709

#### Cambridge International AS and A Level

 Name of student Date Adm. number Year/grade Stream Subject Pure Mathematics 1 (P1) Variant(s) P11, P12, P13 Start time Duration Stop time

Qtn No. 1 2 3 4 5 6 7 8 9 10 11 12 Total
Marks 4 4 5 5 6 4 7 5 6 9 11 7 73
Score

Get Mathematics 9709 Topical Questions (2010-2021) for $14.5 per Subject. Attempt all the 12 questions Question 1 Code: 9709/11/M/J/20/1, Topic: Series The sum of the first nine terms of an arithmetic progression is 117. The sum of the next four terms is 91. Find the first term and the common difference of the progression.$[4]$Question 2 Code: 9709/13/M/J/20/1, Topic: Quadratics Find the set of values of$m$for which the line with equation$y=m x+1$and the curve with equation$y=3 x^{2}+2 x+4$intersect at two distinct points.$[4]$Question 3 Code: 9709/11/M/J/10/2, Topic: Series$\text{(i)}$Find the first 3 terms in the expansion of$\displaystyle \left(2 x-\frac{3}{x}\right)^{5}$in descending powers of$x$.$[3]\text{(ii)}$Hence find the coefficient of$x$in the expansion of$\displaystyle \left(1+\frac{2}{x^{2}}\right)\left(2 x-\frac{3}{x}\right)^{5}[2]$Question 4 Code: 9709/12/M/J/13/2, Topic: Series Find the coefficient of$x^{2}$in the expansion of$\text{(i)}\displaystyle\left(2 x-\frac{1}{2 x}\right)^{6}$,$[2]\text{(ii)}\displaystyle\left(1+x^{2}\right)\left(2 x-\frac{1}{2 x}\right)^{6}$.$[3]$Question 5 Code: 9709/12/M/J/17/3, Topic: Trigonometry$\text{(i)}$Prove the identity$\displaystyle\left(\frac{1}{\cos \theta}-\tan \theta\right)^{2} \equiv \frac{1-\sin \theta}{1+\sin \theta}$.$[3]\text{(ii)}$Hence solve the equation$\displaystyle\left(\frac{1}{\cos \theta}-\tan \theta\right)^{2}=\frac{1}{2}$, for$0^{\circ} \leqslant \theta \leqslant 360^{\circ}$.$[3]$Question 6 Code: 9709/11/M/J/19/3, Topic: Circular measure A sector of a circle of radius$r \mathrm{~cm}$has an area of$A \mathrm{~cm}^{2}$. Express the perimeter of the sector in terms of$r$and$A$.$[4]$Question 7 Code: 9709/12/M/J/13/4, Topic: Circular measure The diagram shows a square$A B C D$of side$10 \mathrm{~cm}$. The mid-point of$A D$is$O$and$B X C$is an arc of a circle with centre$O$.$\text{(i)}$Show that angle$B O C$is$0.9273$radians, correct to 4 decimal places.$[2]\text{(ii)}$Find the perimeter of the shaded region.$[3]\text{(iii)}$Find the area of the shaded region.$[2]$Question 8 Code: 9709/12/M/J/15/5, Topic: Trigonometry$\text{(i)}$Prove the identity$\displaystyle\frac{\sin \theta-\cos \theta}{\sin \theta+\cos \theta} \equiv \frac{\tan \theta-1}{\tan \theta+1}$.$[1]\text{(ii)}$Hence solve the equation$\displaystyle\frac{\sin \theta-\cos \theta}{\sin \theta+\cos \theta}=\frac{\tan \theta}{6}$, for$0^{\circ} \leqslant \theta \leqslant 180^{\circ}$.$[4]$Question 9 Code: 9709/11/M/J/20/5, Topic: Quadratics The equation of a line is$y=m x+c$, where$m$and$c$are constants, and the equation of a curve is$x y=16\text{(a)}$Given that the line is a tangent to the curve, express$m$in terms of$c$.$[3]\text{(b)}$Given instead that$m=-4$, find the set of values of$c$for which the line intersects the curve at two distinct points.$[3]$Question 10 Code: 9709/11/M/J/11/9, Topic: Circular measure In the diagram,$O A B$is an isosceles triangle with$O A=O B$and angle$A O B=2 \theta$radians. Arc$P S T$has centre$O$and radius$r$, and the line$A S B$is a tangent to the$\operatorname{arc} P S T$at$S$.$\text{(i)}$Find the total area of the shaded regions in terms of$r$and$\theta$.$[4]\text{(ii)}$In the case where$\theta=\frac{1}{3} \pi$and$r=6$, find the total perimeter of the shaded regions, leaving your answer in terms of$\sqrt{3}$and$\pi$.$[5]$Question 11 Code: 9709/12/M/J/12/9, Topic: Coordinate geometry, Integration The diagram shows part of the curve$y=-x^{2}+8 x-10$which passes through the points$A$and$B$. The curve has a maximum point at$A$and the gradient of the line$B A$is 2.$\text{(i)}$Find the coordinates of$A$and$B$.$[7]\text{(ii)}$Find$\displaystyle\int y \mathrm{~d} x$and hence evaluate the area of the shaded region.$[4]$Question 12 Code: 9709/11/M/J/14/9, Topic: Trigonometry$\text{(i)}$Prove the identity$\displaystyle\frac{\sin \theta}{1-\cos \theta}-\frac{1}{\sin \theta} \equiv \frac{1}{\tan \theta}$.$[4]\text{(ii)}$Hence solve the equation$\displaystyle\frac{\sin \theta}{1-\cos \theta}-\frac{1}{\sin \theta}=4 \tan \theta$for$0^{\circ} < \theta < 180^{\circ}$.$[3]\$

Worked solutions: P1, P3 & P6 (S1)

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