Mathematics: Unit-wise Objective Question Bank

Group A of the NEB Grade 11 Mathematics paper is eleven one-mark multiple-choice questions, taken from across the twenty-one chapters of the syllabus. Eleven marks, and none of them needs more than a minute if you have the right fact ready. This chapter drills every chapter in syllabus order: each section gives five to eight practice MCQs, then an answer key in which every answer carries a one-line reason, and names the misconception where a wrong option is a real trap.

11.1 Logic and Set

1. The negation of the statement "All students are present" is [1]aNo student is presentbSome student is absentcSome student is presentdAll students are absent
2. The implication p \Rightarrow q is false only when [1]ap true, q truebp true, q falsecp false, q truedp false, q false
3. The contrapositive of p \Rightarrow q is [1]aq \Rightarrow pb\sim p \Rightarrow \sim qc\sim q \Rightarrow \sim pd\sim q \Rightarrow p
4. If n(A)=20, n(B)=15 and n(A\cap B)=5, then n(A\cup B) is [1]a40b35c30d25
5. By De Morgan's law, (A\cup B)' equals [1]aA'\cup B'bA'\cap B'cA\cap BdA\cup B'
6. The compound statement p \wedge \sim p is always [1]aa tautologybtrueca contradictionda contingency

Answers to 11.1

1. (b) — the negation of "all" is "at least one is not"; option (a) is the common over-negation.

2. (b) — a true hypothesis with a false conclusion is the only way to break an implication.

3. (c) — contrapositive negates and reverses; (a) is the converse and (b) is the inverse.

4. (c) — n(A\cup B)=n(A)+n(B)-n(A\cap B)=20+15-5=30; (b) forgets to subtract the overlap.

5. (b) — the complement of a union is the intersection of the complements.

6. (c) — a statement and its negation cannot both be true, so the conjunction is always false.

11.2 Real Numbers

1. Which of the following is an irrational number? [1]a0.\overline{3}b\sqrt{16}c\sqrt{7}d\frac{22}{7}
2. The solution set of |x-3|<2 is [1]a(1,5)b[1,5]c(-5,-1)d(-\infty,1)\cup(5,\infty)
3. The solutions of |2x-1|=3 are [1]ax=2 onlybx=-1 onlycx=2 and x=-1dx=1 and x=-2
4. [-2,3]\cap(0,5) equals [1]a(0,3)b[0,3]c(0,3]d[-2,5)
5. The solution set of |x|\ge 4 is [1]a[-4,4]b(-4,4)c(-\infty,-4]\cup[4,\infty)d[4,\infty)

Answers to 11.2

1. (c) — 7 is not a perfect square; 0.\overline{3}=1/3 and 22/7 are rational, \sqrt{16}=4.

2. (a) — |x-3|<2 means -2<x-3<2, so 1<x<5; the strict inequality gives open ends.

3. (c) — 2x-1=3\Rightarrow x=2 and 2x-1=-3\Rightarrow x=-1; (a) forgets the negative case.

4. (c) — the common part starts after 0 (open in the second set) and ends at 3 (closed in the first).

5. (c) — "\ge" gives two outer rays; (a) is the solution of |x|\le 4.

11.3 Functions

1. If f(x)=2x+3, then f^{-1}(x) is [1]a\frac{x+3}{2}b\frac{x-3}{2}c2x-3d\frac{2}{x-3}
2. The domain of f(x)=\sqrt{x-4} is [1]a(4,\infty)b[4,\infty)c(-\infty,4]d\mathbb{R}
3. If f(x)=x^2 and g(x)=x+1, then (f\circ g)(2) is [1]a5b9c7d25
4. The range of f(x)=x^2+1 is [1]a[0,\infty)b(1,\infty)c[1,\infty)d\mathbb{R}
5. The domain of f(x)=\frac{1}{x^2-9} is [1]a\mathbb{R}-\{3\}b\mathbb{R}-\{-3\}c\mathbb{R}-\{3,-3\}d\mathbb{R}
6. If f(x)=\log_2 x, then f(8) equals [1]a2b3c4d16

Answers to 11.3

1. (b) — put y=2x+3 and solve for x: x=(y-3)/2.

2. (b) — we need x-4\ge 0; \sqrt{0} is defined, so 4 is included.

3. (b) — g(2)=3, then f(3)=9; (a) wrongly computes (g\circ f)(2)=5.

4. (c) — x^2\ge 0, so x^2+1\ge 1 and the value 1 is attained at x=0.

5. (c) — the denominator vanishes at both x=3 and x=-3.

6. (b) — 2^3=8.

11.4 Curve Sketching

1. Which of the following is an even function? [1]ax^3b\sin xcx^2+\cos xdx+\frac{1}{x}
2. The period of \sin 2x is [1]a2\pib\pic\frac{\pi}{2}d4\pi
3. The graph of y=x^3 is symmetric about [1]athe x-axisbthe y-axiscthe origindthe line y=x
4. The amplitude of y=3\sin 2x is [1]a2b3c6d\pi
5. The graph of y=\frac{1}{x+2} has a vertical asymptote at [1]ax=2bx=-2cy=-2dx=0
6. The graph of y=\ln x passes through the point [1]a(0,1)b(1,0)c(e,0)d(0,0)

Answers to 11.4

1. (c) — both x^2 and \cos x are even, so their sum is even; x^3, \sin x and x+1/x are odd.

2. (b) — the period of \sin bx is 2\pi/b=\pi; (a) forgets to divide by b.

3. (c) — (-x)^3=-x^3, so the function is odd and symmetric about the origin.

4. (b) — amplitude is |a| in a\sin bx; the 2 changes the period, not the height.

5. (b) — the denominator is zero at x=-2.

6. (b) — \ln 1=0; the graph never touches x=0.

11.5 Sequence and Series

1. The 10th term of the A.P. 3, 7, 11, \dots is [1]a39b43c40d36
2. The sum of the first 20 natural numbers is [1]a200b210c420d190
3. The 5th term of the G.P. 2, 6, 18, \dots is [1]a54b162c486d90
4. The sum to infinity of 1+\frac12+\frac14+\cdots is [1]a1b\frac32c2d\infty
5. The G.M. of 4 and 16 is [1]a10b8c6.4d12
6. If a, b, c are in H.P., then \frac1a,\frac1b,\frac1c are in [1]aG.P.bH.P.cA.P.dnone of these
7. The sum of the first 15 odd natural numbers is [1]a225b240c120d255

Answers to 11.5

1. (a) — t_{10}=a+9d=3+9\times4=39; (b) uses 10d by mistake.

2. (b) — \frac{n(n+1)}{2}=\frac{20\times21}{2}=210.

3. (b) — r=3, so t_5=2\times3^4=162; (c) is t_6.

4. (c) — S_\infty=\frac{a}{1-r}=\frac{1}{1-1/2}=2.

5. (b) — \sqrt{4\times16}=8; (a) is the A.M. and (c) is the H.M.

6. (c) — that is the definition of a harmonic progression.

7. (a) — the sum of the first n odd numbers is n^2=225.

11.6 Matrices and Determinants

1. For two matrices A and B for which the product is defined, (AB)^T equals [1]aA^TB^TbB^TA^TcABdBA
2. The determinant of the matrix with rows (2,3) and (1,4) is [1]a11b5c-5d8
3. If A is a 2\times2 matrix with |A|=5, then |2A| is [1]a10b20c25d40
4. For a square matrix A, A\cdot\text{adj}A equals [1]aA^2b|A|\,IcIdO
5. The inverse of a square matrix A exists if and only if [1]a|A|=0b|A|\ne 0cA is symmetricdA=A^T
6. If A is a 3\times3 matrix with |A|=4, then |\text{adj}A| is [1]a4b8c16d64

Answers to 11.6

1. (b) — the transpose of a product reverses the order; (a) is the commonest slip.

2. (b) — 2\times4-3\times1=5.

3. (b) — for an n\times n matrix |kA|=k^n|A|, so 2^2\times5=20; (a) multiplies once only.

4. (b) — A\,\text{adj}A=|A|I, the identity behind the inverse formula.

5. (b) — a singular matrix (|A|=0) has no inverse.

6. (c) — |\text{adj}A|=|A|^{n-1}=4^2=16.

11.7 Quadratic Equation

1. The roots of x^2-5x+6=0 are [1]a1, 6b2, 3c-2,-3d-1,-6
2. The nature of the roots of 2x^2-4x+2=0 is [1]areal and distinctbreal and equalcimaginarydirrational
3. The sum and product of the roots of 2x^2-7x+3=0 are [1]a\frac72,\frac32b-\frac72,\frac32c\frac72,-\frac32d7,3
4. The quadratic equation whose roots are 2 and -3 is [1]ax^2-x-6=0bx^2+x-6=0cx^2+x+6=0dx^2-5x-6=0
5. If the roots of x^2+kx+9=0 are equal, then k is [1]a6 onlyb-6 onlyc\pm6d\pm3
6. If \alpha,\beta are the roots of x^2-4x+1=0, then \alpha^2+\beta^2 is [1]a16b14c18d12

Answers to 11.7

1. (b) — (x-2)(x-3)=0; (c) has the wrong signs.

2. (b) — D=16-16=0, so the roots are real and equal.

3. (a) — sum =-b/a=7/2, product =c/a=3/2; (b) forgets the minus sign in -b/a.

4. (b) — sum =-1, product =-6, so x^2-(\text{sum})x+\text{product}=x^2+x-6.

5. (c) — D=k^2-36=0 gives k=\pm6; both signs work.

6. (b) — \alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta=16-2=14.

11.8 Complex Numbers

1. The value of i^{46} is [1]a1b-1cid-i
2. (2+3i)(2-3i) equals [1]a-5b13c4-9id7
3. The modulus of 3+4i is [1]a7b25c5d1
4. The square roots of -3+4i are [1]a\pm(1+2i)b\pm(2+i)c1-2i onlyd\pm(2-i)
5. The conjugate of \frac{1}{1+i} is [1]a\frac{1-i}{2}b\frac{1+i}{2}c1-id\frac{1}{1-i} only
6. The argument of the complex number -1 is [1]a0b\frac{\pi}{2}c\pid\frac{3\pi}{2}

Answers to 11.8

1. (b) — 46=4\times11+2, so i^{46}=i^2=-1.

2. (b) — z\bar z=a^2+b^2=4+9=13; (c) forgets i^2=-1.

3. (c) — \sqrt{9+16}=5; (b) is |z|^2.

4. (a) — (1+2i)^2=1-4+4i=-3+4i; both \pm roots are needed, and (c) leaves out the other one.

5. (b) — \frac{1}{1+i}=\frac{1-i}{2}, whose conjugate is \frac{1+i}{2}; (a) stops one step short.

6. (c) — the point -1 lies on the negative real axis, angle \pi.

11.9 Inverse Circular Functions

1. \sin^{-1}\left(\frac12\right) is [1]a\frac{\pi}{3}b\frac{\pi}{6}c\frac{\pi}{4}d\frac{5\pi}{6}
2. The range (principal values) of \cos^{-1}x is [1]a[-\frac{\pi}{2},\frac{\pi}{2}]b[0,\pi]c(0,\pi)d[0,2\pi]
3. \sin^{-1}x+\cos^{-1}x equals, for -1\le x\le 1, [1]a0b\pic\frac{\pi}{2}d\frac{\pi}{4}
4. \tan^{-1}\frac12+\tan^{-1}\frac13 equals [1]a\frac{\pi}{6}b\frac{\pi}{3}c\frac{\pi}{4}d\frac{\pi}{2}
5. \cos^{-1}\left(-\frac12\right) is [1]a\frac{\pi}{3}b-\frac{\pi}{3}c\frac{2\pi}{3}d\frac{4\pi}{3}
6. \sin\left(\cos^{-1}\frac35\right) equals [1]a\frac35b\frac45c\frac34d\frac53

Answers to 11.9

1. (b) — principal value in [-\pi/2,\pi/2]; 5\pi/6 also has sine \frac12 but is outside the range.

2. (b) — \cos^{-1} takes values in [0,\pi]; (a) is the range of \sin^{-1}.

3. (c) — the standard complementary identity.

4. (c) — \tan^{-1}\frac{\frac12+\frac13}{1-\frac16}=\tan^{-1}1=\frac{\pi}{4}.

5. (c) — \cos^{-1}(-x)=\pi-\cos^{-1}x=\pi-\frac{\pi}{3}; (b) lies outside [0,\pi].

6. (b) — a 3-4-5 triangle: if the cosine is 3/5, the sine is 4/5.

11.10 Trigonometric Equations and General Values

1. The general solution of \sin x=\frac12 is [1]a2n\pi\pm\frac{\pi}{6}bn\pi+(-1)^n\frac{\pi}{6}cn\pi\pm\frac{\pi}{6}dn\pi+\frac{\pi}{6}
2. The general solution of \cos x=0 is [1]an\pib2n\pic(2n+1)\frac{\pi}{2}dn\pi+\frac{\pi}{4}
3. The general solution of \tan x=\sqrt{3} is [1]an\pi+\frac{\pi}{3}b2n\pi+\frac{\pi}{3}cn\pi\pm\frac{\pi}{3}dn\pi+\frac{\pi}{6}
4. The number of solutions of \sin x=\frac12 in [0,2\pi] is [1]a1b2c3d4
5. If \cos x=\cos\alpha, then the general value of x is [1]an\pi+\alphabn\pi+(-1)^n\alphac2n\pi\pm\alphad2n\pi+\alpha
6. The smallest positive solution of \tan x=-1 is [1]a\frac{\pi}{4}b\frac{3\pi}{4}c\frac{5\pi}{4}d\frac{7\pi}{4}

Answers to 11.10

1. (b) — for \sin x=\sin\alpha the general value is n\pi+(-1)^n\alpha with \alpha=\pi/6.

2. (c) — cosine vanishes at odd multiples of \pi/2; (a) is the answer for \sin x=0.

3. (a) — tangent has period \pi, so n\pi+\alpha with \alpha=\pi/3.

4. (b) — x=\frac{\pi}{6} and x=\frac{5\pi}{6}.

5. (c) — cosine repeats every 2\pi and is even, hence \pm\alpha.

6. (b) — tangent is negative in the second quadrant; the reference angle is \pi/4, giving \pi-\frac{\pi}{4}=\frac{3\pi}{4}.

11.11 Straight Line

1. The perpendicular distance of the point (2,1) from the line 3x+4y+5=0 is [1]a3b5c\frac{15}{25}d15
2. The distance of the origin from the line 5x+12y-26=0 is [1]a26b13c2d1
3. The equations of the bisectors of the angles between the coordinate axes are [1]ay=0 and x=0by=\pm xcy=\pm 2xdx+y=1
4. The distance of the point (-1,3) from the line x+y=0 is [1]a2b\sqrt{2}c2\sqrt{2}d1
5. The two bisectors of the angles between two intersecting straight lines are [1]aparallelbperpendicular to each otherccoincidentdinclined at 45^\circ always

Answers to 11.11

1. (a) — \frac{|3(2)+4(1)+5|}{\sqrt{9+16}}=\frac{15}{5}=3; (c) forgets the square root.

2. (c) — \frac{|-26|}{\sqrt{25+144}}=\frac{26}{13}=2.

3. (b) — the axes are perpendicular, so the bisectors make 45^\circ with each: y=\pm x.

4. (b) — \frac{|-1+3|}{\sqrt{2}}=\sqrt{2}; (a) forgets to divide by \sqrt{2}.

5. (b) — the internal and external bisectors always meet at a right angle.

11.12 Pair of Straight Lines

1. The equation x^2-y^2=0 represents [1]aa circlebthe lines y=x and y=-xcone linedparallel lines
2. The angle between the lines x^2-5xy+6y^2=0 has \tan\theta equal to [1]a\frac17b\frac15c1d\frac{1}{6}
3. The lines given by ax^2+2hxy+by^2=0 are perpendicular if [1]ah=0ba=bca+b=0dab=h^2
4. The lines given by ax^2+2hxy+by^2=0 coincide if [1]ah^2=abbh=0ca+b=0da=b
5. A homogeneous equation of the second degree in x and y represents a pair of lines [1]aparallel to the axesbpassing through the origincnot intersectingdof equal slope
6. The equation of the bisectors of ax^2+2hxy+by^2=0 is [1]a\frac{x^2-y^2}{a-b}=\frac{xy}{h}b\frac{x^2+y^2}{a+b}=\frac{xy}{h}c\frac{x^2-y^2}{a+b}=\frac{xy}{h}d\frac{x^2-y^2}{a-b}=\frac{xy}{a}

Answers to 11.12

1. (b) — (x-y)(x+y)=0.

2. (a) — a=1,\ h=-\frac52,\ b=6; \tan\theta=\frac{2\sqrt{h^2-ab}}{|a+b|}=\frac{2\times\frac12}{7}=\frac17.

3. (c) — the angle formula has a+b in the denominator; perpendicular means that denominator is zero.

4. (a) — h^2-ab=0 makes the numerator of \tan\theta zero, so the angle is zero.

5. (b) — with no linear or constant term the origin satisfies the equation.

6. (a) — the standard result; (b) and (c) mix up the signs.

11.13 Coordinates in Space

1. The distance between (1,2,3) and (4,6,15) is [1]a11b12c13d14
2. The direction cosines of the x-axis are [1]a(0,1,0)b(1,0,0)c(0,0,1)d(1,1,1)
3. The distance of the point (3,-4,5) from the xy-plane is [1]a3b4c5d\sqrt{50}
4. The direction cosines of the line joining (1,2,3) and (3,5,9) are [1]a\frac27,\frac37,\frac67b\frac13,\frac13,\frac13c2,3,6d\frac{2}{49},\frac{3}{49},\frac{6}{49}
5. Which of the following can be the direction cosines of a line? [1]a\frac12,\frac12,\frac12b\frac13,\frac23,\frac23c\frac12,\frac13,\frac14d1,1,1
6. The distance of the point (2,-3,6) from the origin is [1]a5b6c7d11

Answers to 11.13

1. (c) — \sqrt{3^2+4^2+12^2}=\sqrt{169}=13.

2. (b) — the x-axis makes 0^\circ with itself and 90^\circ with the others.

3. (c) — the distance from the xy-plane is |z|.

4. (a) — direction ratios 2,3,6 with \sqrt{4+9+36}=7; (c) leaves the ratios unnormalised.

5. (b) — only (b) satisfies l^2+m^2+n^2=\frac19+\frac49+\frac49=1.

6. (c) — \sqrt{4+9+36}=7.

11.14 Vectors

1. Which pair of vectors is collinear? [1]a2\mathbf{i}-4\mathbf{j} and -\mathbf{i}+2\mathbf{j}b\mathbf{i}+\mathbf{j} and \mathbf{i}-\mathbf{j}c\mathbf{i} and \mathbf{j}d2\mathbf{i}+\mathbf{j} and \mathbf{i}+2\mathbf{j}
2. The value of m for which 2\mathbf{i}+3\mathbf{j} and 4\mathbf{i}+m\mathbf{j} are collinear is [1]a4b5c6d8
3. Two non-collinear vectors \mathbf{a},\mathbf{b} are linearly independent; if x\mathbf{a}+y\mathbf{b}=\mathbf{0} then [1]ax=y=0bx=y=1cx=-ydx=0 only
4. The position vector of the midpoint of the points with position vectors \mathbf{a} and \mathbf{b} is [1]a\mathbf{a}+\mathbf{b}b\frac{\mathbf{a}-\mathbf{b}}{2}c\frac{\mathbf{a}+\mathbf{b}}{2}d\frac{2\mathbf{a}+\mathbf{b}}{3}
5. The magnitude of 3\mathbf{i}+4\mathbf{j} is [1]a7b5c25d1
6. Three vectors are coplanar if [1]athey are all unit vectorsbone of them is a linear combination of the other twocthey are mutually perpendiculardthey have equal magnitudes

Answers to 11.14

1. (a) — the second is -\frac12 times the first; the others have unequal component ratios.

2. (c) — \frac24=\frac3m gives m=6.

3. (a) — that is the definition of linear independence.

4. (c) — the midpoint formula is the average of the position vectors.

5. (b) — \sqrt{9+16}=5; (a) adds the components.

6. (b) — that is the condition for coplanarity.

11.15 Measures of Dispersion

1. The variance of the data 1,2,3,4,5 is [1]a1b2c2.5d3
2. The coefficient of variation is defined as [1]a\frac{\bar x}{\sigma}\times100b\frac{\sigma}{\bar x}\times100c\sigma^2\times100d\frac{\sigma}{\sigma^2}
3. If the mean is 50 and the standard deviation is 5, the coefficient of variation is [1]a10%b5%c25%d50%
4. Karl Pearson's coefficient of skewness is [1]a\frac{\text{Mean}-\text{Mode}}{\text{S.D.}}b\frac{\text{Mean}+\text{Mode}}{\text{S.D.}}c\frac{\text{S.D.}}{\text{Mean}-\text{Mode}}d\text{Mean}-\text{Median}
5. If the mean is 40, the mode 30 and the S.D. 10, the coefficient of skewness is [1]a-1b0c1d3
6. The skewness of a perfectly symmetrical distribution is [1]apositivebnegativeczerodinfinite

Answers to 11.15

1. (b) — mean 3, squared deviations 4+1+0+1+4=10, divide by n=5 to get 2; (c) wrongly divides by n-1.

2. (b) — C.V. =\frac{\sigma}{\bar x}\times100\%; (a) is inverted.

3. (a) — \frac{5}{50}\times100=10\%.

4. (a) — the mean minus the mode, divided by the standard deviation.

5. (c) — \frac{40-30}{10}=1, positive skew.

6. (c) — mean, median and mode coincide, so the numerator is zero.

11.16 Probability

1. A fair die is thrown. The probability of getting a prime number is [1]a\frac13b\frac12c\frac23d\frac16
2. For two events, P(A\cup B) equals [1]aP(A)+P(B)bP(A)P(B)cP(A)+P(B)-P(A\cap B)dP(A)-P(B)
3. If A and B are independent with P(A)=0.3 and P(B)=0.4, then P(A\cap B) is [1]a0.7b0.1c0.12d0.58
4. Two coins are tossed. The probability of at least one head is [1]a\frac14b\frac12c\frac34d1
5. The probability of drawing a king or a queen from a pack of 52 cards is [1]a\frac{1}{13}b\frac{2}{13}c\frac{1}{26}d\frac{4}{13}
6. Two dice are thrown. The probability that the sum is 7 is [1]a\frac{1}{12}b\frac{1}{9}c\frac16d\frac{7}{36}

Answers to 11.16

1. (b) — primes 2,3,5: \frac36=\frac12; the trap is counting 1 as prime.

2. (c) — the addition law; (a) double-counts the overlap.

3. (c) — independence gives 0.3\times0.4=0.12; (a) adds instead.

4. (c) — 1-P(\text{no head})=1-\frac14.

5. (b) — \frac{4+4}{52}=\frac{2}{13} (mutually exclusive events add).

6. (c) — six favourable outcomes out of 36.

11.17 Limits and Continuity

1. \lim_{x\to2}\frac{x^2-4}{x-2} equals [1]a0b2c4ddoes not exist
2. \lim_{x\to0}\frac{\sin3x}{x} equals [1]a1b3c\frac13d0
3. \lim_{x\to\infty}\frac{3x+1}{x+5} equals [1]a0b1c3d\infty
4. \lim_{x\to0}\frac{e^x-1}{x} equals [1]a0b1ced\infty
5. \lim_{x\to\infty}\left(1+\frac1x\right)^x equals [1]a1b0ced\infty
6. The function f(x)=\frac{x^2-1}{x-1} for x\ne 1, f(1)=k, is continuous at x=1 when k is [1]a0b1c2d-2

Answers to 11.17

1. (c) — factorise: \frac{(x-2)(x+2)}{x-2}=x+2\to4; it is the \frac00 form, not "does not exist".

2. (b) — \frac{\sin3x}{x}=3\cdot\frac{\sin3x}{3x}\to3; (a) forgets the factor 3.

3. (c) — divide by x: the ratio of leading coefficients 3/1.

4. (b) — a standard limit.

5. (c) — the definition of e; (a) is the trap of putting \frac1x\to0 first.

6. (c) — the limit is x+1\to2, so continuity needs f(1)=2.

11.18 Derivatives

1. \frac{d}{dx}(x^5) equals [1]ax^4b5x^4c5x^5d\frac{x^6}{6}
2. \frac{d}{dx}(e^{2x}) equals [1]ae^{2x}b2e^{x}c2e^{2x}d2xe^{2x}
3. The slope of the tangent to y=x^2 at x=3 is [1]a3b6c9d2
4. \frac{d}{dx}(x\sin x) equals [1]a\cos xbx\cos xc\sin x+x\cos xd\sin x-x\cos x
5. If x^2+y^2=25, then \frac{dy}{dx} equals [1]a\frac{x}{y}b-\frac{x}{y}c-\frac{y}{x}d\frac{y}{x}
6. If x=t^2 and y=2t, then \frac{dy}{dx} equals [1]atb\frac1tc2td\frac{1}{2t}
7. The function f(x)=x^2-4x+5 has a minimum at [1]ax=1bx=2cx=4dx=-2
8. For f(x)=x^3, the point x=0 is [1]aa maximumba minimumca point of inflectionda discontinuity

Answers to 11.18

1. (b) — power rule; (d) is the integral.

2. (c) — the chain rule brings down the inner derivative 2.

3. (b) — y'=2x=6; (c) gives the value of y, not the slope.

4. (c) — product rule: 1\cdot\sin x+x\cos x.

5. (b) — 2x+2y\,y'=0.

6. (b) — \frac{dy/dt}{dx/dt}=\frac{2}{2t}=\frac1t.

7. (b) — f'(x)=2x-4=0 at x=2 and f''=2>0.

8. (c) — f''(x)=6x changes sign at 0, so the curve changes concavity there; f'(0)=0 but f keeps increasing, so it is not an extremum.

11.19 Anti-derivatives

1. \int x^3\,dx equals [1]a3x^2+Cb\frac{x^4}{4}+Ccx^4+Cd\frac{x^3}{3}+C
2. \int_0^2 x\,dx equals [1]a1b2c4d0
3. \int \frac{2x}{x^2+1}\,dx equals [1]a\frac{1}{x^2+1}+Cb\ln(x^2+1)+Cc\tan^{-1}x+Cd2\ln x+C
4. \int xe^x\,dx equals [1]ae^x(x-1)+Cbe^x(x+1)+Ccxe^x+Cd\frac{x^2}{2}e^x+C
5. The area under y=x from x=0 to x=4 is [1]a4b8c16d2
6. \int_0^{\pi/2}\sin x\,dx equals [1]a0b1c-1d\frac{\pi}{2}
7. \int \frac1x\,dx equals [1]a\ln|x|+Cb-\frac{1}{x^2}+Cc\frac{1}{x^2}+Cdx+C

Answers to 11.19

1. (b) — raise the power by one and divide by it; (a) is differentiation.

2. (b) — \left[\frac{x^2}{2}\right]_0^2=2.

3. (b) — the numerator is the derivative of the denominator.

4. (a) — by parts: xe^x-\int e^x\,dx=e^x(x-1); (b) has the wrong sign.

5. (b) — \left[\frac{x^2}{2}\right]_0^4=8, which is also the area of the triangle \frac12\cdot4\cdot4.

6. (b) — [-\cos x]_0^{\pi/2}=0-(-1)=1.

7. (a) — the one case the power rule cannot give.

11.20 Computational Methods

1. The bisection method can be applied to f(x)=0 on [a,b] if [1]af(a)fb>0 (b) f(a)f(b)<0cf(a)=f(b)df'(a)=0
2. The Newton–Raphson iteration formula is [1]ax_{n+1}=x_n+\frac{f(x_n)}{f'(x_n)}bx_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}cx_{n+1}=x_n-\frac{f'(x_n)}{f(x_n)}dx_{n+1}=\frac{f(x_n)}{f'(x_n)}
3. The first midpoint in the bisection method for a root in [1,2] is [1]a1b1.25c1.5d2
4. Simpson's rule requires the number of sub-intervals to be [1]aoddbevenca multiple of 3dany number
5. For f(x)=x^2-2 with x_0=2, the first Newton–Raphson approximation x_1 is [1]a1.5b1.4c1.414d1
6. The trapezoidal rule with n=2 for \int_0^2x^2\,dx gives [1]a\frac83b3c4d2

Answers to 11.20

1. (b) — opposite signs at the ends guarantee a root in between for a continuous f.

2. (b) — subtract the ratio; (a) has the wrong sign and (c) inverts the ratio.

3. (c) — \frac{1+2}{2}=1.5.

4. (b) — Simpson's \frac13 rule pairs the intervals.

5. (a) — 2-\frac{2}{4}=1.5.

6. (b) — h=1: \frac{1}{2}[0+2(1)+4]=3; the exact value \frac83 is what Simpson's rule gives with n=2.

11.21 Mechanics

1. The resultant of two forces P and Q acting at right angles is [1]aP+Qb\sqrt{P^2+Q^2}cP-QdPQ
2. Forces of 3 N and 4 N act at right angles. Their resultant is [1]a7 Nb1 Nc5 Nd12 N
3. The horizontal component of a 10 N force acting at 30^\circ above the horizontal is [1]a5 Nb5\sqrt{3} Nc10 Nd10\sqrt{3} N
4. A particle starts with u=5\ \text{m s}^{-1} and has a=2\ \text{m s}^{-2}. Its velocity after 3 s is [1]a6 m s⁻¹b9 m s⁻¹c11 m s⁻¹d15 m s⁻¹
5. A body falls from rest for 4 s (g=10\ \text{m s}^{-2}). The distance fallen is [1]a40 mb80 mc160 md20 m
6. The acceleration of a body down a smooth plane inclined at 30^\circ (g=10\ \text{m s}^{-2}) is [1]a10 m s⁻²b5\sqrt{3} m s⁻²c5 m s⁻²d20 m s⁻²
7. A ball is thrown vertically up at 20\ \text{m s}^{-1} (g=10\ \text{m s}^{-2}). The maximum height reached is [1]a10 mb20 mc40 md200 m

Answers to 11.21

1. (b) — the parallelogram law with \cos90^\circ=0.

2. (c) — \sqrt{9+16}=5 N; (a) simply adds the magnitudes.

3. (b) — 10\cos30^\circ=5\sqrt{3}; (a) is the vertical component.

4. (c) — v=u+at=5+6=11.

5. (b) — s=\frac12gt^2=\frac12\times10\times16=80 m; (c) forgets the \frac12.

6. (c) — a=g\sin\theta=10\times\frac12=5.

7. (b) — h=\frac{u^2}{2g}=\frac{400}{20}=20 m.

Chapter summary

• Group A is eleven one-mark MCQs; they are spread across algebra, trigonometry, geometry, vectors, statistics, calculus and the computational-or-mechanics unit.
• Logic: the contrapositive of p\Rightarrow q is \sim q\Rightarrow\sim p; n(A\cup B)=n(A)+n(B)-n(A\cap B).
• Algebra: t_n=a+(n-1)d, S_\infty=\frac{a}{1-r}, \alpha+\beta=-\frac ba, \alpha\beta=\frac ca, |kA|=k^n|A|, |\text{adj}A|=|A|^{n-1}, |z|=\sqrt{a^2+b^2}.
• Trigonometry: \sin^{-1}x+\cos^{-1}x=\frac{\pi}{2}; general values n\pi+(-1)^n\alpha, 2n\pi\pm\alpha, n\pi+\alpha.
• Geometry: distance =\frac{|ax_1+by_1+c|}{\sqrt{a^2+b^2}}; a pair of lines is perpendicular when a+b=0; l^2+m^2+n^2=1.
• Statistics and probability: C.V. =\frac{\sigma}{\bar x}\times100; P(A\cup B)=P(A)+P(B)-P(A\cap B).
• Calculus: \int\frac{f'(x)}{f(x)}dx=\ln|f(x)|; standard limits \frac{\sin x}{x}\to1 and \frac{e^x-1}{x}\to1.
• Computational methods: x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}; Simpson's rule needs an even number of intervals.