Mathematics: Unit-wise Subjective Question Bank

1. Build the table for all four cases of p,q:

The columns for p\Rightarrow q and \sim p\vee q are identical in all four rows, so p\Rightarrow q\equiv\sim p\vee q.

2. A\cup B=\{1,2,3,4,5,6\}, so (A\cup B)'=\{7,8\}. Next A'=\{5,6,7,8\} and B'=\{1,2,7,8\}, so A'\cap B'=\{7,8\}. Both sides equal \{7,8\}; the law is verified.

3. Let F be the football set and C the cricket set. At least one game: n(F\cup C)=60-10=50.

(i) Both: n(F\cap C)=n(F)+n(C)-n(F\cup C)=35+30-50=15. (ii) Football only: 35-15=20. (iii) Cricket only: 30-15=15. Check: 20+15+15+10=60. Venn diagram: two overlapping circles inside a rectangle, with 20 in the football-only part, 15 in the overlap, 15 in the cricket-only part and 10 outside both circles.

1. |2x-3|\le 5\iff-5\le 2x-3\le 5. Add 3: -2\le 2x\le 8. Divide by 2: -1\le x\le 4. Solution set =[-1,4]; the end points are included because the inequality is non-strict.

2. |x+1|>3 means x+1>3 or x+1<-3, so x>2 or x<-4. Solution set =(-\infty,-4)\cup(2,\infty). On the number line draw open circles at -4 and 2, shading to the left of -4 and to the right of 2; the segment between them is not shaded.

1. Let y=\frac{2x+1}{x-3}. Then y(x-3)=2x+1\Rightarrow xy-3y=2x+1\Rightarrow x(y-2)=3y+1\Rightarrow x=\frac{3y+1}{y-2}. So f^{-1}(x)=\frac{3x+1}{x-2}. Domain of f is \mathbb{R}-\{3\} (denominator zero at 3); range is \mathbb{R}-\{2\} (the value that f^{-1} cannot take).

2. (f\circ g)(x)=f(g(x))=2(x^2+3)-1=2x^2+5. (g\circ f)(x)=g(2x-1)=(2x-1)^2+3=4x^2-4x+4. At x=2: (f\circ g)(2)=8+5=13 and (g\circ f)(2)=16-8+4=12. Since 13\ne12, f\circ g\ne g\circ f.

3. We need 9-x^2\ge 0, so x^2\le 9 and -3\le x\le 3: domain =[-3,3]. Inside the root the value runs from 0 (at x=\pm3) to 9 (at x=0), so f runs from 0 to 3: range =[0,3].

1. f(-x)=(-x)^3+\sin(-x)=-x^3-\sin x=-f(x): odd. g(-x)=x^2\cos(-x)=x^2\cos x=g(x): even. h(-x)=x^2-x, which is neither h(x) nor -h(x): neither.

2. Write y=(x-2)^2-1. Vertex (2,-1) (the minimum point); axis of symmetry x=2. x-intercepts: (x-1)(x-3)=0 gives x=1,3. y-intercept: y=3. The parabola opens upward, passes through (0,3),(1,0),(2,-1),(3,0),(4,3) and is symmetric about x=2.

3. Amplitude =2, period =2\pi. Key points:

Join them with a smooth wave: it rises to a crest of height 2 at \frac{\pi}{2}, crosses the axis at \pi, reaches a trough of -2 at \frac{3\pi}{2} and returns to 0 at 2\pi.

1. t_n=S_n-S_{n-1}=(3n^2+2n)-[3(n-1)^2+2(n-1)]=6n-1. So a=t_1=5, t_2=11, d=6, t_{10}=60-1=59.

2. Take the numbers \frac{a}{r},a,ar. Product: a^3=64\Rightarrow a=4. Sum: \frac{4}{r}+4+4r=14\Rightarrow4+4r^2=10r\Rightarrow2r^2-5r+2=0\Rightarrow(2r-1)(r-2)=0, so r=2 or \frac{1}{2}. The numbers are 2,4,8 (or 8,4,2).

3. A=\frac{2+8}{2}=5; G=\sqrt{2\times8}=4; H=\frac{2\times2\times8}{2+8}=\frac{32}{10}=3.2. Then G^2=16 and A\times H=5\times3.2=16, so G^2=AH.

4. (i) Five terms in all: 64=4r^4\Rightarrow r^4=16\Rightarrow r=2. The means are 8,16,32. (ii) S_{10}=\frac{a(r^{10}-1)}{r-1}=\frac{2(1024-1)}{1}=2046. (iii) A=\frac{4+16}{2}=10 and G=\sqrt{64}=8; 10>8, so A>G. (Equality holds only when the two numbers are equal.)

1. |A|=2\times3-1\times5=1\ne0. The adjoint is found by swapping the diagonal entries and changing the sign of the others: first row (3,\ -1), second row (-5,\ 2). So A^{-1}=\frac{1}{1}\,\text{adj}A has rows (3,-1) and (-5,2). Check: row 1 of A times the columns: 2(3)+1(-5)=1 and 2(-1)+1(2)=0; row 2: 5(3)+3(-5)=0 and 5(-1)+3(2)=1. So AA^{-1}=I.

2. Expand along the first row: 1(5\cdot10-6\cdot8)-2(4\cdot10-6\cdot7)+3(4\cdot8-5\cdot7)=1(2)-2(-2)+3(-3)=2+4-9=-3.

3. AB: row 1 =(1\cdot0+2\cdot1,\ 1\cdot1+2\cdot0)=(2,1); row 2 =(3\cdot0+4\cdot1,\ 3\cdot1+4\cdot0)=(4,3). So (AB)^T has rows (2,4) and (1,3). Now B^T=B (rows (0,1),(1,0)) and A^T has rows (1,3),(2,4). B^TA^T: row 1 =(0\cdot1+1\cdot2,\ 0\cdot3+1\cdot4)=(2,4); row 2 =(1\cdot1+0\cdot2,\ 1\cdot3+0\cdot4)=(1,3). The two results agree.

4. The coefficient determinant has rows (1,1,1),(1,-1,1),(1,1,-1): D=1[(-1)(-1)-1\cdot1]-1[1(-1)-1\cdot1]+1[1\cdot1-(-1)\cdot1]=0+2+2=4. D_x (first column replaced by 6,2,0): 6(1-1)-1(-2-0)+1(2-0)=0+2+2=4. D_y (second column replaced): rows (1,6,1),(1,2,1),(1,0,-1): 1(-2-0)-6(-1-1)+1(0-2)=-2+12-2=8. D_z (third column replaced): rows (1,1,6),(1,-1,2),(1,1,0): 1(0-2)-1(0-2)+6(1+1)=-2+2+12=12. x=\frac{4}{4}=1, y=\frac{8}{4}=2, z=\frac{12}{4}=3. Check in the first equation: 1+2+3=6. ✓

1. Let the roots be \alpha and 2\alpha. Sum: 3\alpha=7\Rightarrow\alpha=\frac{7}{3}. Product: 2\alpha^2=k\Rightarrow k=2\times\frac{49}{9}=\frac{98}{9}.

2. \alpha+\beta=6, \alpha\beta=4. \alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta=36-8=28. \frac{1}{\alpha}+\frac{1}{\beta}=\frac{\alpha+\beta}{\alpha\beta}=\frac{6}{4}=\frac{3}{2}.

3. Equal roots need D=0: [2(k+1)]^2-4(1)(k^2)=4(k^2+2k+1)-4k^2=8k+4=0, so k=-\frac{1}{2}.

4. (i) \alpha+\beta=-\frac{b}{a}, \alpha\beta=\frac{c}{a}. For the new roots: sum =\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\alpha+\beta}{\alpha\beta}=-\frac{b}{c}, product =\frac{1}{\alpha\beta}=\frac{a}{c}. The equation is x^2+\frac{b}{c}x+\frac{a}{c}=0, that is cx^2+bx+a=0. (ii) The roots of x^2-5x+6=0 are 2 and 3, so the new roots are 2+3=5 and 2\times3=6. Sum =11, product =30: x^2-11x+30=0. (iii) The common root must be 2 or 3. Put x=2: 4-6+k=0\Rightarrow k=2. Put x=3: 9-9+k=0\Rightarrow k=0. So k=2 or k=0.

1. Multiply by the conjugate: \frac{(1+2i)(2+i)}{(2-i)(2+i)}=\frac{2+i+4i+2i^2}{4+1}=\frac{0+5i}{5}=i. So a=0, b=1 and |i|=1.

2. Let \sqrt{5+12i}=a+ib. Then a^2-b^2=5 and 2ab=12. Also a^2+b^2=|5+12i|=13. Adding: 2a^2=18\Rightarrow a=\pm3; subtracting: 2b^2=8\Rightarrow b=\pm2. Since ab>0 the signs agree: the roots are \pm(3+2i). Check: (3+2i)^2=9-4+12i. ✓

3. |z|=\sqrt{9+16}=5. \bar z=3+4i. z^{-1}=\frac{\bar z}{|z|^2}=\frac{3+4i}{25}.

4. (i) Let z_1=a+ib, z_2=c+id. Then z_1z_2=(ac-bd)+i(ad+bc) and |z_1z_2|^2=(ac-bd)^2+(ad+bc)^2=a^2c^2+b^2d^2+a^2d^2+b^2c^2=(a^2+b^2)(c^2+d^2). Taking square roots, |z_1z_2|=|z_1||z_2|. (ii) x+iy=\frac{-4+7i}{2+3i}=\frac{(-4+7i)(2-3i)}{13}=\frac{-8+12i+14i+21}{13}=\frac{13+26i}{13}=1+2i. So x=1, y=2. (iii) The point is (3,4) in the first quadrant. |z|=5 and \arg z=\tan^{-1}\frac{4}{3}\approx53.13^\circ.

1. Use \tan^{-1}a+\tan^{-1}b=\tan^{-1}\frac{a+b}{1-ab} (valid as ab<1): \tan^{-1}\frac{\frac{1}{2}+\frac{1}{3}}{1-\frac{1}{6}}=\tan^{-1}\frac{5/6}{5/6}=\tan^{-1}1=\frac{\pi}{4}.

2. Let \theta=\cos^{-1}\frac{3}{5}, so \cos\theta=\frac{3}{5} and \sin\theta=\frac{4}{5}. \sin2\theta=2\sin\theta\cos\theta=2\cdot\frac{4}{5}\cdot\frac{3}{5}=\frac{24}{25}.

3. (i) 2\tan^{-1}\frac{1}{3}=\tan^{-1}\frac{2/3}{1-1/9}=\tan^{-1}\frac{3}{4}. Then \tan^{-1}\frac{3}{4}+\tan^{-1}\frac{1}{7}=\tan^{-1}\frac{\frac{3}{4}+\frac{1}{7}}{1-\frac{3}{28}}=\tan^{-1}\frac{25/28}{25/28}=\tan^{-1}1=\frac{\pi}{4}. (ii) LHS =\tan^{-1}\frac{2x}{1-(x^2-1)}=\tan^{-1}\frac{2x}{2-x^2}. So \frac{2x}{2-x^2}=\frac{8}{31}\Rightarrow62x=16-8x^2\Rightarrow4x^2+31x-8=0\Rightarrow x=\frac{-31\pm33}{8}, giving x=\frac{1}{4} or x=-8. For x=-8, (x+1)(x-1)=63>1 and the addition formula does not hold, so reject it. Hence x=\frac{1}{4}. (iii) \frac{5\pi}{6} is outside [-\frac{\pi}{2},\frac{\pi}{2}]; \sin\frac{5\pi}{6}=\frac{1}{2}, so the value is \sin^{-1}\frac{1}{2}=\frac{\pi}{6}.

1. Factorise: (2\sin x-1)(\sin x+1)=0. So \sin x=\frac{1}{2} giving x=\frac{\pi}{6},\frac{5\pi}{6}, or \sin x=-1 giving x=\frac{3\pi}{2}. Solutions: \frac{\pi}{6},\ \frac{5\pi}{6},\ \frac{3\pi}{2}.

2. \cos2x=\cos x\Rightarrow2x=2n\pi\pm x. Taking +: x=2n\pi; taking -: 3x=2n\pi, x=\frac{2n\pi}{3}. In [0,2\pi]: x=0,\ \frac{2\pi}{3},\ \frac{4\pi}{3},\ 2\pi.

3. \sin3x=\sin x\Rightarrow3x=n\pi+(-1)^nx. If n=2m: 3x=2m\pi+x\Rightarrow x=m\pi. If n=2m+1: 3x=(2m+1)\pi-x\Rightarrow x=\frac{(2m+1)\pi}{4}. General solution: x=m\pi or x=\frac{(2m+1)\pi}{4}, m\in\mathbb{Z}.

4. Write \sqrt{3}\cos x+\sin x=R\cos(x-\alpha) with R\cos\alpha=\sqrt{3}, R\sin\alpha=1. Then R=\sqrt{3+1}=2 and \tan\alpha=\frac{1}{\sqrt{3}}, so \alpha=\frac{\pi}{6}. The equation becomes 2\cos\left(x-\frac{\pi}{6}\right)=\sqrt{2}, i.e. \cos\left(x-\frac{\pi}{6}\right)=\frac{1}{\sqrt{2}}=\cos\frac{\pi}{4}. Hence x-\frac{\pi}{6}=2n\pi\pm\frac{\pi}{4}, so x=2n\pi+\frac{\pi}{6}\pm\frac{\pi}{4}, that is x=2n\pi+\frac{5\pi}{12} or x=2n\pi-\frac{\pi}{12}, n\in\mathbb{Z}.

1. d=\frac{|3(4)-4(1)-18|}{\sqrt{3^2+4^2}}=\frac{|12-4-18|}{5}=\frac{10}{5}=2 units.

2. Take the point (0,-\frac{5}{4}) on the first line (put x=0: 4y=-5). Its distance from 3x+4y-10=0 is \frac{|0-5-10|}{5}=\frac{15}{5}=3 units. (Equivalently, \frac{|c_1-c_2|}{\sqrt{a^2+b^2}}=\frac{15}{5}.)

3. A point on a bisector is equidistant from both lines: \frac{|4x+3y-7|}{5}=\frac{|3x+4y-7|}{5}. Taking +: 4x+3y-7=3x+4y-7\Rightarrow x-y=0. Taking -: 4x+3y-7=-3x-4y+7\Rightarrow7x+7y-14=0\Rightarrow x+y-2=0. The bisectors are x-y=0 and x+y-2=0.

4. (i) BC has intercepts 4 and 3: \frac{x}{4}+\frac{y}{3}=1\Rightarrow3x+4y-12=0. (ii) Perpendicular from A(0,0): \frac{|-12|}{\sqrt{9+16}}=\frac{12}{5}=2.4 units. (iii) BC=\sqrt{16+9}=5. Area =\frac{1}{2}\times5\times\frac{12}{5}=6. Check: \frac{1}{2}\times AB\times AC=\frac{1}{2}\times4\times3=6. ✓

1. Factorise: 2x^2+5xy+2y^2=(2x+y)(x+2y)=0, so the lines are 2x+y=0 and x+2y=0. Here a=b=2, h=\frac{5}{2}. \tan\theta=\frac{2\sqrt{h^2-ab}}{|a+b|}=\frac{2\sqrt{\frac{25}{4}-4}}{4}=\frac{2\times\frac{3}{2}}{4}=\frac{3}{4}, so \theta=\tan^{-1}\frac{3}{4}\approx36.87^\circ.

2. a=6, b=-6, so a+b=0: the lines are perpendicular. (Indeed 6x^2-5xy-6y^2=(3x+2y)(2x-3y); slopes -\frac{3}{2} and \frac{2}{3} have product -1.) With h=-\frac{5}{2} the bisectors satisfy \frac{x^2-y^2}{a-b}=\frac{xy}{h}: \frac{x^2-y^2}{12}=\frac{xy}{-5/2}=-\frac{2xy}{5}, so 5(x^2-y^2)=-24xy, i.e. 5x^2+24xy-5y^2=0.

3. The lines coincide when h^2=ab. Here a=1, 2h=\lambda, b=4: \left(\frac{\lambda}{2}\right)^2=4\Rightarrow\lambda^2=16\Rightarrow\lambda=\pm4.

4. (i) a=3, h=\frac{7}{2}, b=2, g=\frac{5}{2}, f=\frac{5}{2}, c=2. The condition is abc+2fgh-af^2-bg^2-ch^2=0. Compute: abc=12; 2fgh=2\cdot\frac{5}{2}\cdot\frac{5}{2}\cdot\frac{7}{2}=\frac{175}{4}; af^2=\frac{75}{4}; bg^2=\frac{50}{4}; ch^2=\frac{98}{4}. Sum: 12+\frac{175-75-50-98}{4}=12-12=0. ✓ (ii) 3x^2+7xy+2y^2=(3x+y)(x+2y). Write (3x+y+p)(x+2y+q): matching x: 3q+p=5; y: q+2p=5; constant pq=2. Solving, q=1, p=2, and pq=2 ✓. The lines are 3x+y+2=0 and x+2y+1=0. (iii) \tan\theta=\frac{2\sqrt{h^2-ab}}{|a+b|}=\frac{2\sqrt{\frac{49}{4}-6}}{5}=\frac{2\times\frac{5}{2}}{5}=1, so \theta=45^\circ.

1. AB^2=1+1+16=18; BC^2=9+9+0=18; CA^2=16+4+16=36. AB=BC so the triangle is isosceles, and AB^2+BC^2=36=CA^2, so by the converse of Pythagoras the angle at B is a right angle.

2. Direction ratios: (-1-2,\ 1-3,\ 5+1)=(-3,-2,6). Magnitude \sqrt{9+4+36}=7. Direction cosines: -\frac{3}{7},\ -\frac{2}{7},\ \frac{6}{7} (or all signs reversed).

3. Let the point be P(0,y,0). PA^2=1+(y-2)^2+9 and PB^2=1+(y-4)^2+25. Equating: (y-2)^2+10=(y-4)^2+26\Rightarrow y^2-4y+14=y^2-8y+42\Rightarrow4y=28\Rightarrow y=7. The point is (0,7,0).

4. (i) Let the line pass through the origin and let P on it be at distance r. Its coordinates are (lr,mr,nr), where l=\cos\alpha, m=\cos\beta, n=\cos\gamma. Then OP^2=r^2=l^2r^2+m^2r^2+n^2r^2, so l^2+m^2+n^2=1. (ii) \sqrt{1+4+4}=3, so the direction cosines are \frac{1}{3},\ -\frac{2}{3},\ \frac{2}{3} (or all signs reversed). (iii) OP=\sqrt{9+16+25}=\sqrt{50}=5\sqrt{2}.

1. Name the points A,B,C in order. \overrightarrow{AB}=(2\mathbf{a}+3\mathbf{b}-4\mathbf{c})-(\mathbf{a}-2\mathbf{b}+3\mathbf{c})=\mathbf{a}+5\mathbf{b}-7\mathbf{c}. \overrightarrow{BC}=(-7\mathbf{b}+10\mathbf{c})-(2\mathbf{a}+3\mathbf{b}-4\mathbf{c})=-2\mathbf{a}-10\mathbf{b}+14\mathbf{c}=-2(\mathbf{a}+5\mathbf{b}-7\mathbf{c}). So \overrightarrow{BC}=-2\overrightarrow{AB}: the vectors are parallel and share the point B, hence A,B,C are collinear.

2. Let 5\mathbf{i}+3\mathbf{j}=x(\mathbf{i}+\mathbf{j})+y(\mathbf{i}-\mathbf{j})=(x+y)\mathbf{i}+(x-y)\mathbf{j}. Comparing, x+y=5 and x-y=3, so x=4, y=1. Hence 5\mathbf{i}+3\mathbf{j}=4(\mathbf{i}+\mathbf{j})+1(\mathbf{i}-\mathbf{j}).

3. Let ABCD be a parallelogram with position vectors \mathbf{a},\mathbf{b},\mathbf{c},\mathbf{d}. Since \overrightarrow{AB}=\overrightarrow{DC}, \mathbf{b}-\mathbf{a}=\mathbf{c}-\mathbf{d}, so \mathbf{a}+\mathbf{c}=\mathbf{b}+\mathbf{d}. The midpoint of AC is \frac{\mathbf{a}+\mathbf{c}}{2} and the midpoint of BD is \frac{\mathbf{b}+\mathbf{d}}{2}. These are equal, so the diagonals bisect each other.

4. (i) Take A as origin, with B=\mathbf{b} and C=\mathbf{c}. The midpoints of AB and AC are D=\frac{\mathbf{b}}{2} and E=\frac{\mathbf{c}}{2}. \overrightarrow{DE}=\frac{\mathbf{c}}{2}-\frac{\mathbf{b}}{2}=\frac{1}{2}(\mathbf{c}-\mathbf{b})=\frac{1}{2}\overrightarrow{BC}. So DE\parallel BC and DE=\frac{1}{2}BC. (ii) Let D be the midpoint of BC: \frac{\mathbf{b}+\mathbf{c}}{2}. The point dividing AD in the ratio 2:1 is \frac{1\cdot\mathbf{a}+2\cdot\frac{\mathbf{b}+\mathbf{c}}{2}}{3}=\frac{\mathbf{a}+\mathbf{b}+\mathbf{c}}{3}=G. The expression is symmetric in \mathbf{a},\mathbf{b},\mathbf{c}, so the same point divides the other two medians 2:1 as well; the three medians concur at G.

1. n=8, mean =\frac{40}{8}=5. Deviations: -3,-1,-1,-1,0,0,2,4; squares 9,1,1,1,0,0,4,16, sum =32. Variance =\frac{32}{8}=4, so \sigma=2.

2. C.V. of A =\frac{10}{50}\times100=20\%. C.V. of B =\frac{12}{80}\times100=15\%. The smaller C.V. means greater consistency, so series B is more consistent.

3. Mode =3\times\text{median}-2\times\text{mean}=174-120=54. S_k=\frac{\text{Mean}-\text{Mode}}{\sigma}=\frac{60-54}{8}=0.75. Positive, so the distribution is positively skewed (longer tail to the right).

4. Mid-values: 5,15,25,35,45. Take A=25, u=\frac{x-25}{10}: -2,-1,0,1,2. N=2+8+30+8+2=50. (i) \sum fu=2(-2)+8(-1)+0+8(1)+2(2)=0, so \bar x=25+10\times0=25. (ii) \sum fu^2=2(4)+8(1)+0+8(1)+2(4)=32. \sigma^2=100\left[\frac{32}{50}-0^2\right]=64, so \sigma=8. (iii) C.V. =\frac{8}{25}\times100=32\%.

1. (i) P=\frac{5}{8}\times\frac{4}{7}=\frac{20}{56}=\frac{5}{14}. (ii) With replacement the draws are independent: P=\frac{5}{8}\times\frac{5}{8}=\frac{25}{64}.

2. P(A\cap B)=P(A)+P(B)-P(A\cup B)=0.5+0.4-0.7=0.2. Since P(A)P(B)=0.5\times0.4=0.2=P(A\cap B), the events are independent.

3. Total outcomes =36. Sum \ge 10: (4,6),(5,5),(6,4),(5,6),(6,5),(6,6), so 6 outcomes. P=\frac{6}{36}=\frac{1}{6}.

4. (i) There are 4 kings and 13 hearts, and the king of hearts is counted in both, so P=\frac{4}{52}+\frac{13}{52}-\frac{1}{52}=\frac{16}{52}=\frac{4}{13}. (ii) 1-\frac{4}{13}=\frac{9}{13}. (iii) The sample space has 2^3=8 outcomes. Exactly two heads: HHT, HTH, THH, so P=\frac{3}{8}.

1. The direct substitution gives \frac{0}{0}, so factorise: \frac{(x-3)(x^2+3x+9)}{(x-3)(x+3)}=\frac{x^2+3x+9}{x+3}\to\frac{27}{6}=\frac{9}{2}.

2. 1-\cos x=2\sin^2\frac{x}{2}, so \frac{1-\cos x}{x^2}=\frac{1}{2}\left(\frac{\sin(x/2)}{x/2}\right)^2\to\frac{1}{2}(1)^2=\frac{1}{2}.

3. Left limit at 3: 3k+1. Right limit at 3: 3(3)-5=4. Also f(3)=3k+1. Continuity needs left limit = right limit: 3k+1=4\Rightarrow k=1.

4. (i) Multiply and divide by the conjugate: \frac{(x^2+3x)-x^2}{\sqrt{x^2+3x}+x}=\frac{3x}{\sqrt{x^2+3x}+x}=\frac{3}{\sqrt{1+\frac{3}{x}}+1}\to\frac{3}{1+1}=\frac{3}{2}. (ii) \frac{\sin5x}{\tan3x}=\frac{\sin5x}{5x}\cdot\frac{3x}{\tan3x}\cdot\frac{5}{3}\to1\cdot1\cdot\frac{5}{3}=\frac{5}{3}. (iii) f(0)=0. Left limit: for x<0, f(x)=-x\to0. Right limit: for x>0, f(x)=x\to0. Left limit = right limit =f(0), so f is continuous at 0 (though not differentiable there).

1. f'(x)=\lim_{h\to0}\frac{\sqrt{x+h}-\sqrt{x}}{h}=\lim_{h\to0}\frac{(x+h)-x}{h(\sqrt{x+h}+\sqrt{x})}=\lim_{h\to0}\frac{1}{\sqrt{x+h}+\sqrt{x}}=\frac{1}{2\sqrt{x}}.

2. Quotient rule: y'=\frac{(x-1)(2x)-(x^2+1)(1)}{(x-1)^2}=\frac{2x^2-2x-x^2-1}{(x-1)^2}=\frac{x^2-2x-1}{(x-1)^2}.

3. Differentiate implicitly: 3x^2+3y^2\frac{dy}{dx}=3y+3x\frac{dy}{dx}. Collect: \frac{dy}{dx}(y^2-x)=y-x^2, so \frac{dy}{dx}=\frac{y-x^2}{y^2-x}.

4. \frac{dy}{dx}=2x-3, which is 3 at x=3. Tangent: y-2=3(x-3)\Rightarrow y=3x-7, i.e. 3x-y-7=0.

5. \frac{dx}{dt}=a(1-\cos t) and \frac{dy}{dt}=a\sin t. \frac{dy}{dx}=\frac{\sin t}{1-\cos t}=\frac{2\sin\frac{t}{2}\cos\frac{t}{2}}{2\sin^2\frac{t}{2}}=\cot\frac{t}{2}.

6. f'(x)=3x^2-12x+9=3(x-1)(x-3), zero at x=1,3. f''(x)=6x-12. At x=1: f''=-6<0, a maximum, f(1)=1-6+9+1=5. At x=3: f''=6>0, a minimum, f(3)=27-54+27+1=1. Sign of f': positive for x<1, negative for 1<x<3, positive for x>3. So f is increasing on (-\infty,1) and (3,\infty) and decreasing on (1,3). (The point of inflection is at x=2, where f''=0 and changes sign; f(2)=3.)

7. Let the parts be x and 40-x; P=x(40-x)=40x-x^2. P'=40-2x=0\Rightarrow x=20. P''=-2<0, so x=20 gives a maximum: the parts are 20 and 20 and the greatest product is 400. For y=x^3, y''=6x is zero at x=0 and changes sign from negative to positive there, so (0,0) is the point of inflection.

1. \frac{(1+x)^2}{x}=\frac{1}{x}+2+x. So the integral is \ln|x|+2x+\frac{x^2}{2}+C.

2. Put u=x^2, du=2x\,dx: \int xe^{x^2}dx=\frac{1}{2}\int e^u\,du=\frac{1}{2}e^{x^2}+C.

3. By parts with u=x, dv=\sin x\,dx: \int x\sin x\,dx=-x\cos x+\int\cos x\,dx=-x\cos x+\sin x+C.

4. Put u=1+\sin x, du=\cos x\,dx. Limits: x=0\to u=1; x=\frac{\pi}{2}\to u=2. \int_1^2\frac{du}{u}=\ln2-\ln1=\ln2.

5. (i) Intersections: x^2=2x\Rightarrow x=0,2. On [0,2] the line is above the parabola: area =\int_0^2(2x-x^2)dx=\left[x^2-\frac{x^3}{3}\right]_0^2=4-\frac{8}{3}=\frac{4}{3} square units. (ii) Area =\int_0^\pi\sin x\,dx=[-\cos x]_0^\pi=1+1=2 square units.

6. By parts with u=\ln x, dv=dx: \int\ln x\,dx=x\ln x-\int x\cdot\frac{1}{x}\,dx=x\ln x-x+C. Then \int_1^e\ln x\,dx=[x\ln x-x]_1^e=(e-e)-(0-1)=1.

1. f(1)=-1<0, f(2)=5>0, so a root lies in [1,2].

The root lies in (1.3125,1.375), so x\approx1.34 (the true root is 1.3247).

2. f(x)=x^2-5, f'(x)=2x, x_{n+1}=x_n-\frac{x_n^2-5}{2x_n}. x_1=2-\frac{4-5}{4}=2.25. x_2=2.25-\frac{5.0625-5}{4.5}=2.25-0.01389=2.23611. So \sqrt{5}\approx2.2361.

3. h=1; y values at x=0,1,2,3,4: 0,1,4,9,16. I\approx\frac{h}{2}[y_0+2(y_1+y_2+y_3)+y_4]=\frac{1}{2}[0+2(14)+16]=22. Exact: \frac{4^3}{3}=21.33. The trapezoidal rule overestimates here because x^2 is convex.

4. h=0.25; y=\frac{1}{1+x^2} at 0,0.25,0.5,0.75,1: 1,\ 0.94118,\ 0.8,\ 0.64,\ 0.5. Simpson: \frac{h}{3}[y_0+4(y_1+y_3)+2y_2+y_4]=\frac{0.25}{3}[1+4(1.58118)+1.6+0.5]=0.083333\times9.42471=0.78539. (Exact value \frac{\pi}{4}=0.78540.)

5. f(x)=x^3-2x-5, f'(x)=3x^2-2. At x_0=2: f=8-4-5=-1 and f'=10, so x_1=2-\frac{-1}{10}=2.1. At x_1=2.1: f=9.261-4.2-5=0.061 and f'=13.23-2=11.23, so x_2=2.1-\frac{0.061}{11.23}=2.1-0.00543=2.09457. The root is \approx2.0946.

1. R=\sqrt{5^2+12^2}=13 N. The angle with the 5 N force: \tan\alpha=\frac{12}{5}, so \alpha=\tan^{-1}2.4\approx67.4^\circ.

2. R^2=P^2+Q^2+2PQ\cos\theta=9+25+2(3)(5)\cos60^\circ=34+15=49, so R=7 N.

3. v=u+at: 0=10-2t\Rightarrow t=5 s. Distance: s=\frac{u^2}{2|a|}=\frac{100}{4}=25 m (check: ut+\frac{1}{2}at^2=50-25=25).

4. Maximum height h=\frac{u^2}{2g}=\frac{900}{20}=45 m. Time to the top =\frac{u}{g}=3 s, so total time of flight =6 s.

5. a=g\sin30^\circ=10\times\frac{1}{2}=5\ \text{m s}^{-2}. After 4 s: v=at=20\ \text{m s}^{-1} and s=\frac{1}{2}at^2=\frac{1}{2}\times5\times16=40 m.

6. (i) v=u+at=4+2\times5=14\ \text{m s}^{-1}. (ii) s=ut+\frac{1}{2}at^2=4(5)+\frac{1}{2}(2)(25)=20+25=45 m. (iii) Distance in the 5th second =s_5-s_4. s_4=4(4)+\frac{1}{2}(2)(16)=32 m, so the distance is 45-32=13 m. (Formula check: u+\frac{a}{2}(2n-1)=4+1\times9=13.)