Mathematics: Unit-wise Subjective Question Bank

1. (a) The paper did not say whether digits may repeat, so both cases are answered. (i) Without repetition: the hundreds place can be filled in 6 ways, the tens in 5, the units in 4:

(ii) With repetition each of the three places can be filled in 6 ways:

(b)

Cancel n(n-1), which is non-zero because n\geq4:

So n=7 (reject n=-2). Check: P(7,4)=840 and 20\,P(7,2)=20\times42=840. ✓

2. (a) Boys can be chosen in C(a,2) ways and girls in C(6,3)=20 ways, so

(b) Tie the two particular persons together as one unit. There are then 6 units to seat round the table, which can be done in (6-1)! = 120 ways. The two persons can swap places inside their unit in 2! ways:

3. (a) Write both terms with factorials:

Take out the common factor \dfrac{n!}{r!\,(n-r+1)!}, using (n-r+1)! = (n-r+1)(n-r)! and r! = r\,(r-1)!:

(b) MATHEMATICS has 11 letters, in which M, A and T each occur twice:

4. (a) First seat the 5 boys in a row: 5! = 120 ways. They leave 6 gaps (one before, one after, four between). Put the three girls in different gaps, in order: P(6,3)=120 ways.

(b) A polygon of n sides has C(n,2)-n = \dfrac{n(n-3)}{2} diagonals.

1. (a) INTERVAL has 8 different letters: vowels I, E, A (3 of them) and consonants N, T, R, V, L (5 of them). (i) Treat the three vowels as one block. Arrange 6 objects (block + 5 consonants) in 6! ways, and the vowels inside the block in 3! ways:

(ii) If the positions of vowels and consonants are not to change, the vowels may only be rearranged among the three vowel positions and the consonants among the five consonant positions:

(iii) There are 4 odd positions (1,3,5,7). Choose and arrange the three vowels in them in P(4,3)=24 ways; the five consonants fill the other five places in 5!=120 ways:

(b) "At least 2 women" is easiest by subtraction from all committees. All committees: C(10,5)=252. Committees with no woman: C(6,5)=6. Committees with exactly one woman: C(4,1)\times C(6,4)=4\times15=60.

(c) BANANA has 6 letters with A three times and N twice:

1. (a) The binomial expansion is the formula that expresses a power (a+b)^{n} of a binomial as a sum of terms of the form C(n,r)a^{n-r}b^{r}, for r=0,1,\dots,n.

(b) n=30 is even, so there is one middle term, T_{16}:

with C(30,15)=155\,117\,520.

(c) Put x=1: (1+1)^{n} = C(n,0)+C(n,1)+\cdots+C(n,n), so the sum is 2^{n}.

(d) If z=x+iy=r(\cos\theta+i\sin\theta), then \theta=\tan^{-1}(y/x), measured from the positive x-axis to the line joining the origin to the point (x,y), is the argument of z, written \arg z.

2. (a) n=8, so the middle term is T_{5}:

(b) The coefficient of x^{n} in (1+x)^{2n} is C(2n,n), and in (1+x)^{2n-1} it is C(2n-1,n). Dividing one by the other:

So C(2n,n)=2\,C(2n-1,n). ∎

3. (a) The coefficient of x^{r} is C(11,r) and of x^{r+1} is C(11,r+1). Since C(n,p)=C(n,q) requires p=q or p+q=n, and r\neq r+1,

Check: C(11,5)=C(11,6)=462. ✓

(b) For any index, (1+x)^{n}=1+nx+\dfrac{n(n-1)}{2!}x^{2}+\dfrac{n(n-1)(n-2)}{3!}x^{3}+\cdots with n=-2:

Put x=0.02:

(The exact value is 0.961169.)

4. (a)

(b) T_{r+1}=C(9,r)(2x^{2})^{9-r}\Big(-\dfrac1x\Big)^{r} = C(9,r)\,2^{9-r}(-1)^{r}\,x^{18-2r-r}. The power of x is 18-3r, which is 0 when r=6:

1. (a) For a positive integer n,

and the general term, the (r+1)th, is T_{r+1}=C(n,r)\,a^{n-r}b^{r}.

(b) The 5th, 6th, 7th coefficients are C(n,4), C(n,5), C(n,6). In an A.P. the middle term is the mean of the other two:

Divide by C(n,5) using \dfrac{C(n,4)}{C(n,5)}=\dfrac{5}{n-4} and \dfrac{C(n,6)}{C(n,5)}=\dfrac{n-5}{6}:

So n=7 or n=14. Check for n=7: C(7,4)=35, C(7,5)=21, C(7,6)=7; 35,21,7 is an A.P. with common difference -14. ✓

(c) T_{r+1}=C(8,r)\,x^{8-r}\Big(\dfrac{2}{x}\Big)^{r} = C(8,r)\,2^{r}\,x^{8-2r}. For x^{4}: 8-2r=4, r=2:

so the coefficient is 112. Next, \sqrt{1.04}=(1+0.04)^{1/2}:

(The exact value is 1.0198039\ldots)

1. (i) De Moivre's theorem. (ii) [r(\cos\theta+i\sin\theta)]^{-n}=r^{-n}[\cos n\theta-i\sin n\theta]. (iii) r is the modulus of the complex number, r=|z|=\sqrt{x^{2}+y^{2}}, its distance from the origin on the Argand plane. (iv) \theta is the argument (amplitude) of the complex number, the angle its radius vector makes with the positive real axis. (v) Finding powers and roots of complex numbers, for instance the nth roots of unity; also proving trigonometric identities for multiple angles.

2. (a) The complex number is the square of either root:

(b) r=\sqrt{2^{2}+(2\sqrt{3})^{2}}=\sqrt{16}=4 and \tan\theta=\dfrac{2\sqrt{3}}{2}=\sqrt{3}. The point (2,2\sqrt{3}) is in the first quadrant, so \theta=\dfrac{\pi}{3}:

Check: square roots 2\big(\cos\frac{\pi}{6}+i\sin\frac{\pi}{6}\big)=\sqrt{3}+i. ✓

3. Let z=-2-2\sqrt{3}\,i. Then r=\sqrt{4+12}=4 and \tan\theta=\sqrt{3}. The point (-2,-2\sqrt{3}) is in the third quadrant, so \theta=\pi+\dfrac{\pi}{3}=\dfrac{4\pi}{3}:

By De Moivre's theorem for roots,

For k=0: 2\Big(\cos\dfrac{2\pi}{3}+i\sin\dfrac{2\pi}{3}\Big)=2\Big(-\dfrac12+\dfrac{\sqrt{3}}{2}i\Big)=-1+\sqrt{3}\,i. For k=1: 2\Big(\cos\dfrac{5\pi}{3}+i\sin\dfrac{5\pi}{3}\Big)=2\Big(\dfrac12-\dfrac{\sqrt{3}}{2}i\Big)=1-\sqrt{3}\,i. The square roots are \pm(1-\sqrt{3}\,i). Check: (1-\sqrt{3} i)^{2}=1-2\sqrt{3} i-3=-2-2\sqrt{3} i. ✓

4. (a) The cube roots of unity satisfy x^{3}-1=0, that is (x-1)(x^{2}+x+1)=0. The non-real roots \omega,\omega^{2} satisfy x^{2}+x+1=0, so \omega^{2}+\omega+1=0. ∎ (b) From (a), 1+\omega=-\omega^{2} and 1+\omega^{2}=-\omega. Hence 1-\omega+\omega^{2}=(1+\omega^{2})-\omega=-2\omega and 1+\omega-\omega^{2}=(1+\omega)-\omega^{2}=-2\omega^{2}:

∎

1. (a) r=\sqrt{1+3}=2 and \tan\theta=\sqrt{3} with the point (1,\sqrt{3}) in the first quadrant, so \theta=\dfrac{\pi}{3}:

(b) 8i=8\Big(\cos\dfrac{\pi}{2}+i\sin\dfrac{\pi}{2}\Big), so

k=0: 2(\cos30^{\circ}+i\sin30^{\circ})=\sqrt{3}+i. k=1: 2(\cos150^{\circ}+i\sin150^{\circ})=-\sqrt{3}+i. k=2: 2(\cos270^{\circ}+i\sin270^{\circ})=-2i. Check: (-2i)^{3}=-8i^{3}=8i. ✓

(c) The cube roots of unity are the roots of x^{3}=1. Factorising, x^{3}-1=(x-1)(x^{2}+x+1)=0. The root x=1 comes from the first factor; the other two roots, \omega=\dfrac{-1+i\sqrt{3}}{2} and \omega^{2}=\dfrac{-1-i\sqrt{3}}{2}, are the roots of x^{2}+x+1=0. Direct check: sum of roots =\omega+\omega^{2}=-1 and product =\omega^{3}=1, matching x^{2}+x+1. ∎

1. (a) Let P(n):\ 1^{2}+2^{2}+\cdots+n^{2}=\dfrac{n(n+1)(2n+1)}{6}. Step 1. For n=1: LHS =1, RHS =\dfrac{1\cdot2\cdot3}{6}=1. So P(1) is true. Step 2. Assume P(k) is true: 1^{2}+\cdots+k^{2}=\dfrac{k(k+1)(2k+1)}{6}. Add (k+1)^{2} to both sides:

which is P(k+1). Step 3. P(1) is true and P(k)\Rightarrow P(k+1), so P(n) is true for every natural number n. ∎ (b)

2. (a) Let P(n):\ 1+7+13+\cdots+(6n-5)=n(3n-2). For n=1: LHS =6(1)-5=1, RHS =1(3-2)=1. True. Assume P(k). Add the (k+1)th term, 6(k+1)-5=6k+1:

which is P(k+1). By the principle of mathematical induction the result holds for all n. ∎ (b) The rth term is r(r+1)=r^{2}+r, so

Check: the general formula \dfrac{n(n+1)(n+2)}{3} gives \dfrac{10\cdot11\cdot12}{3}=440. ✓

3. Let P(n):\ n!\geq2^{n}, for n\geq4. Step 1. For n=4: 4!=24 and 2^{4}=16; 24\geq16, so P(4) is true. Step 2. Assume P(k): k!\geq2^{k} for some k\geq4. Then

Since k\geq4, k+1\geq5>2, so (k+1)\,2^{k}>2\cdot2^{k}=2^{k+1}. Hence (k+1)!\geq2^{k+1}, which is P(k+1). Step 3. By induction, n!\geq2^{n} for all n\geq4. ∎ Note that the base case is n=4, not n=1: the inequality is false for n=3 (6<8).

4. (a) The sum of the first n natural numbers is S=\dfrac{n(n+1)}{2}, so S^{2}=\dfrac{n^{2}(n+1)^{2}}{4}. Prove \sum k^{3}=S^{2} by induction. For n=1: 1=\Big(\dfrac{1\cdot2}{2}\Big)^{2}=1. Assume it for k and add (k+1)^{3}:

which is the formula for n=k+1. ∎ So 1^{3}+2^{3}+\cdots+n^{3}=(1+2+\cdots+n)^{2}. (b) n=15: \Big(\dfrac{15\times16}{2}\Big)^{2}=120^{2}=14\,400.

1. (a) Let P(n) be the statement. For n=1: LHS =1\cdot3=3; RHS =\dfrac{1\cdot2\cdot9}{6}=3. True. Assume P(k) and add the next term (k+1)(k+3):

which is P(k+1). By induction the result holds for all n. ∎

(b) The series on the right is the expansion of e^{y}-1, so x=e^{y}-1, that is 1+x=e^{y}. Taking natural logarithms, y=\log_e(1+x), and expanding,

∎

(c) P(n): n(n+1)(n+2) is a multiple of 6. For n=1: 1\cdot2\cdot3=6. True. Assume k(k+1)(k+2)=6m for an integer m. Then

Of two consecutive integers k+1 and k+2, one is even, so (k+1)(k+2)=2q and the expression equals 6m+6q=6(m+q), a multiple of 6. This is P(k+1). ∎

1. Expand each determinant along its first row (the four determinants are drawn above):

Since D\neq0 the solution is unique:

Check: 2(1)+2=4 ✓, 3(2)-3=3 ✓, 2(1)+3(3)=11 ✓.

2. Write the augmented matrix and use row operations; the figure shows the stages. Stage 1. Interchange R_1\leftrightarrow R_3 so that the pivot is 1. Stage 2. R_2\to R_2-3R_1 and R_3\to R_3-2R_1 to clear the first column. Stage 3. R_2\to R_2-3R_3 to clear the second column of the last row.

In the stage-3 matrix (rows reordered so that the pivots descend), the last row says -10z=-30, so z=3. The middle row says -5y+7z=11, so -5y=11-21=-10 and y=2. The first row says x+4y-z=6, so x=6-8+3=1.

Check in the original: 2(1)+3(2)+5(3)=23 ✓; 3(1)-3(2)+8(3)=21 ✓; 1+8-3=6 ✓.

3. The equations in full are 7x-2y+0z=18, 3x+0y+7z=33, x+y+z=12. Interchange R_1\leftrightarrow R_3 first, then R_2\to R_2-3R_1 and R_3\to R_3-7R_1:

Now R_3\to R_3-3R_2 gives -19z=-57, so z=3. Then -3y+12=-3 gives y=5, and x=12-5-3=4.

Check: 7(4)-2(5)=18 ✓; 3(4)+7(3)=33 ✓; 4+5+3=12 ✓.

4. (a) Subtract 2\times the first equation from the second: R_2\to R_2-2R_1 gives 0x+0y+0z=0, a zero row with zero constant. The system reduces to two equations in three unknowns:

Subtracting, 2y=2 so y=1, and x+z=2. Putting x=t gives z=2-t: the solutions are (t,\,1,\,2-t) for every real t. The system is consistent with infinitely many solutions; this is the situation "0\times z=0" of the 2079 Group A item. (b) Subtracting the equations gives 0=1, which is impossible: the row-reduced augmented matrix has a row of zeros in the coefficient columns but the constant 1 in the last column, a zero row with a non-zero constant. The system is inconsistent — no solution: two parallel lines never meet.

1. (a) Let x, y, z kg of A, B, C be used. Counting each nutrient:

(b) In matrix form AX=B, with A, X and B as in the first figure. Then X=A^{-1}B provided |A|\neq0. Determinant, expanding along the first row:

Cofactors, C_{ij}=(-1)^{i+j}M_{ij}:

The adjoint is the transpose of the cofactor matrix (second figure) and A^{-1}=\dfrac{1}{|A|}\,\text{adj}\,A, so X=A^{-1}B=-\dfrac{1}{24}\,(\text{adj}\,A)\,B. Multiply the adjoint into B row by row:

Dividing each by -24:

So the mixture uses 5 kg of A, 4 kg of B and 2 kg of C. Check: P: 10+12+8=30 ✓; Q: 10+20+6=36 ✓; R: 20+0+10=30 ✓.

(c) Total cost =40x+60y+80z (the product of the price row and the solution column):

1. (a) In triangle PQR with sides p,q,r opposite P,Q,R, the half-angle formulae are

Substituting,

Since (s-p)+(s-r)=2s-p-r=q, the left-hand side is \dfrac{s-q}{q}\cdot q=s-q. Given that it equals \dfrac{q}{2}:

so p,\,q,\,r are in A.P. ∎ (b) By the sine law a=2R\sin A, b=2R\sin B, c=2R\sin C. Then

∎

2. (a) By the half-angle formula, \sin^{2}\dfrac{C}{2}=\dfrac{(s-a)(s-b)}{ab} and \sin^{2}\dfrac{A}{2}=\dfrac{(s-b)(s-c)}{bc}. So

because (s-a)+(s-c)=b. Hence

and a+c-b<a+b+c (the perimeter) because b>0. ∎ (In fact it is smaller by exactly 2b.) (b) From the cosine law, \cos A=\dfrac{b^{2}+c^{2}-a^{2}}{2bc}, and 2\cos^{2}\dfrac{A}{2}=1+\cos A:

With b+c+a=2s and b+c-a=2(s-a) this is \dfrac{4s(s-a)}{2bc}, so \cos^{2}\dfrac{A}{2}=\dfrac{s(s-a)}{bc}. ∎

3. (a) "Cosines proportional to the opposite sides" means \dfrac{\cos A}{a}=\dfrac{\cos B}{b}. By the sine law a=2R\sin A and b=2R\sin B, so

since A,B lie between 0^{\circ} and 180^{\circ}, where cotangent is one-to-one. Equal angles A=B mean equal opposite sides a=b: the triangle is isosceles. ∎ (b) By the sine law a=2R\sin A, b=2R\sin B:

Since A+B=180^{\circ}-C, \cot\dfrac{A+B}{2}=\tan\dfrac{C}{2}. Therefore \dfrac{a-b}{a+b}=\tan\dfrac{C}{2}\tan\dfrac{A-B}{2}, that is

∎

4. (a) By the sine law, a\cos A=b\cos B becomes \sin A\cos A=\sin B\cos B, that is \sin2A=\sin2B. Two angles have equal sines when they are equal or supplementary:

In the first case the triangle is isosceles (a=b); in the second C=90^{\circ}, so it is right-angled. ∎ (b) Let a=k\sin A, b=k\sin B, c=k\sin C (k=2R). Then the expression is

because the six terms cancel in pairs. ∎

1. (a) Place vertex A at the origin and AB along the positive x-axis, so B=(c,0). Vertex C has coordinates (b\cos A,\ b\sin A), since AC=b makes angle A with AB. By the distance formula,

using \cos^{2}A+\sin^{2}A=1. ∎ (b)

(c) By the cosine law, \cos C=\dfrac{a^{2}+b^{2}-c^{2}}{2ab} and \cos B=\dfrac{a^{2}+c^{2}-b^{2}}{2ac}. So

Multiplying by a gives a(b\cos C-c\cos B)=b^{2}-c^{2}. ∎

1. Tangent law:

so \dfrac{A-B}{2}=45^{\circ}, that is A-B=90^{\circ}. Also A+B=180^{\circ}-60^{\circ}=120^{\circ}. Adding and subtracting,

Third side by the cosine law:

Check by the sine law: \dfrac{c}{\sin C}=\dfrac{\sqrt{6}}{\sin60^{\circ}}=2.828 and \dfrac{a}{\sin105^{\circ}}=\dfrac{2.732}{0.9659}=2.828. ✓

2. All three sides are given, so use the cosine law:

Check: \cos C=\dfrac{a^{2}+b^{2}-c^{2}}{2ab}=\dfrac{32}{112}=\dfrac{2}{7}, and \cos^{-1}(2/7)=73.40^{\circ}. ✓

3. C=180^{\circ}-60^{\circ}-45^{\circ}=75^{\circ}. By the sine law \dfrac{a}{\sin A}=\dfrac{12}{\sin60^{\circ}}=\dfrac{12}{0.8660}=13.856, so

4. By the sine law,

Because a<b and a>b\sin A=5, this is the ambiguous case: B=38.68^{\circ} or B=180^{\circ}-38.68^{\circ}=141.32^{\circ}. Both satisfy A+B<180^{\circ} (30^{\circ}+141.32^{\circ}=171.32^{\circ}), so there are two triangles. Triangle 1: B=38.68^{\circ}, C=111.32^{\circ}, c=\dfrac{a\sin C}{\sin A}=\dfrac{8\times0.9317}{0.5}=14.91. Triangle 2: B=141.32^{\circ}, C=8.68^{\circ}, c=\dfrac{8\times0.1510}{0.5}=2.42.

1. Angles (cosine law).

(a) The semi-perimeter is s=\dfrac{13+14+15}{2}=21, so

(b)

Check: \dfrac{a}{\sin A}=\dfrac{13}{0.8}=16.25=2R. ✓

1. (a) The foci and vertices lie on the y-axis, so the hyperbola is \dfrac{y^{2}}{a^{2}}-\dfrac{x^{2}}{b^{2}}=1 with centre at the origin. The vertices give a=4 and the foci give c=6. Then

(b)

2. (a) Complete the square in x:

This is a hyperbola with a^{2}=1 and b^{2}=\dfrac34:

(b) Take the ellipse \dfrac{x^{2}}{a^{2}}+\dfrac{y^{2}}{b^{2}}=1. "Major axis four times the minor axis" means 2a=4\times2b, so a=4b. Then a^{2}=16b^{2}. The point (4,2) lies on it:

(If the major axis were taken along the y-axis the point gives \dfrac{16}{b^{2}}+\dfrac{4}{16b^{2}}=1, so b^{2}=\dfrac{65}{4}, and e is still \dfrac{\sqrt{15}}{4}: the eccentricity depends only on the ratio a:b.)

3. (a) Here a^{2}=100 under y^{2}, which is the larger denominator, so the major axis lies along the y-axis, with a=10 and b=5.

Foci: (0,\,5\sqrt{3}) and (0,\,-5\sqrt{3}). Latus rectum =\dfrac{2b^{2}}{a}=\dfrac{2\times25}{10}=5. (b) Example: the circle x^{2}+y^{2}=25 (a=5) and the line y=\dfrac34x+\dfrac{25}{4} (m=\dfrac34, c=\dfrac{25}{4}).

The condition holds. Verification: the line is 3x-4y+25=0; its distance from the centre is \dfrac{|25|}{\sqrt{9+16}}=5=a, equal to the radius, so the line touches the circle. ✓

4. (a) Divide by 144: \dfrac{x^{2}}{16}-\dfrac{y^{2}}{9}=1. So a=4, b=3, c=\sqrt{16+9}=5, and the centre is the origin. Foci: (\pm5,\,0). (b) Place the vertex at the origin and the axis along the y-axis: x^{2}=4ay. The rim of the reflector is 10 cm from the axis and 5 cm from the vertex, so (10,5) lies on the parabola:

The focus is at (0,5), that is, 5 cm from the vertex — exactly at the level of the rim.

5. (a) The tangent at (x_1,y_1) to y^{2}=4ax with 4a=8, a=2 is yy_1=2a(x+x_1):

Its slope is 1, so the normal has slope -1 and passes through (2,4):

(b) The tangent at (x_1,y_1) to x^{2}+y^{2}=a^{2} is xx_1+yy_1=a^{2}:

The normal to a circle passes through the centre, so it is the line through (0,0) and (3,4): y=\dfrac43x, that is 4x-3y=0.

1. (a) Let the foci be F_1(-c,0) and F_2(c,0) and let P(x,y) satisfy PF_1+PF_2=2a with a>c.

Move one root across and square:

Square again:

Put b^{2}=a^{2}-c^{2} and divide by a^{2}b^{2}:

∎ (b) Divide by 144: \dfrac{x^{2}}{16}+\dfrac{y^{2}}{9}=1, so a=4, b=3, c=\sqrt{16-9}=\sqrt{7}. Major axis 2a=8, minor axis 2b=6. Eccentricity e=\dfrac{c}{a}=\dfrac{\sqrt{7}}{4}. Foci (\pm\sqrt{7},0). Latus rectum \dfrac{2b^{2}}{a}=\dfrac{2\times9}{4}=\dfrac{9}{2}. (c) Write the parabola x^{2}=4Ay and take the point (2At,\,At^{2}) on it. From 2x=4A\,\dfrac{dy}{dx} the slope of the tangent is \dfrac{x}{2A}=t, so the slope of the normal is -\dfrac{1}{t}. The line ax+by+c=0 has slope -\dfrac{a}{b}; for it to be the normal at that point,

The normal point must also lie on the line:

Substitute t=\dfrac{b}{a}:

This is the required condition. ∎ (For a parabola written x^{2}=4ay with the same letter a as the line, rename the parabola's constant first, as above.)

1. (a)

(b) |\vec{a}|=\sqrt{4+1+4}=3 and |\vec{b}|=\sqrt{9+4+36}=7:

2. (a) \vec{a}\cdot\vec{b}=12+6+8=26 and |\vec{a}|=\sqrt{16+9+4}=\sqrt{29}:

(b) Take unit vectors \vec{u}=(\cos A,\sin A) and \vec{v}=(\cos B,\sin B) in the xy-plane, making angles A and B with the x-axis. The angle between them is A-B and both have magnitude 1, so

By components,

Equating the two values gives \cos(A-B)=\cos A\cos B+\sin A\sin B. ∎

3. (a) The area of a parallelogram with diagonals \vec{d_1}, \vec{d_2} is half the modulus of \vec{d_1}\times\vec{d_2}. With the determinant expansion,

(b) Expand using the distributive law and \vec{a}\times\vec{a}=\vec{b}\times\vec{b}=\vec{0}, \vec{b}\times\vec{a}=-\vec{a}\times\vec{b}:

Half of its modulus is the modulus of \vec{a}\times\vec{b}, which is |\vec{a}||\vec{b}|\sin\theta. ∎ (Geometrically \vec{a}-\vec{b} and \vec{a}+\vec{b} are the diagonals of the parallelogram on \vec{a} and \vec{b}.)

4. (a) Let the sides be the vectors \vec{BC}=\vec{a}, \vec{CA}=\vec{b}, \vec{AB}=\vec{c}, taken round the triangle, so \vec{a}+\vec{b}+\vec{c}=\vec{0}. Cross this with \vec{a} and with \vec{b} in turn:

So \vec{a}\times\vec{b}=\vec{b}\times\vec{c}=\vec{c}\times\vec{a}. Taking moduli, and using that the angle between the head-to-tail vectors \vec{a} and \vec{b} is 180^{\circ}-C (similarly 180^{\circ}-A and 180^{\circ}-B):

Divide throughout by abc:

∎ (b) \vec{a}\cdot\vec{b}=2-2+3=3. The cross product is

so |\vec{a}\times\vec{b}|^{2}=75. Then |\vec{a}\times\vec{b}|^{2}+(\vec{a}\cdot\vec{b})^{2}=75+9=84, and |\vec{a}|^{2}|\vec{b}|^{2}=14\times6=84. ✓ This is the relation between the scalar and the vector product: together they give \tan\theta=\dfrac{|\vec{a}\times\vec{b}|}{\vec{a}\cdot\vec{b}}.

1. (a) A vector perpendicular to both is \vec{a}\times\vec{b}:

Since |\vec{a}|=|\vec{b}|=\sqrt{6},

(The negative of the unit vector is equally correct.) (b) \vec{AB}=2\vec{i}+3\vec{j}+4\vec{k} and \vec{AC}=\vec{i}+2\vec{j}+2\vec{k}:

(c) The displacement is \vec{d}=\vec{AB}=2\vec{i}+3\vec{j}+4\vec{k} metre. Work done is the scalar product:

1. (a) Write the first equation as y=0.22x+81.1175: this is the line of y on x, so b_{yx}=0.22. Write the second as x=1.623y-105.695: the line of x on y, so b_{xy}=1.623. (b) r^{2}=b_{yx}\,b_{xy}=0.22\times1.623=0.3571, and both coefficients are positive, so

(c) Both lines pass through (\bar{x},\bar{y}). Substitute the first into the second:

Mean age \approx40.4 years, mean blood pressure =90 units. (If you assign the lines the other way round, b_{yx}=1/1.623=0.616 and b_{xy}=1/0.22=4.55, whose product is 2.8>1 — impossible. That test confirms the assignment above.)

2. \bar{x}=\dfrac{810}{5}=162, \bar{y}=\dfrac{310}{5}=62.

The correlation is positive and moderate: taller persons tend to be somewhat heavier.

3. \bar{X}=\dfrac{348}{6}=58, \bar{Y}=\dfrac{480}{6}=80.

(a)

a very strong positive correlation. (b) To estimate the supply from the price, use the regression of X on Y:

The estimated supply at a price of Rs. 150 is 107 units.

4. \bar{x}=\dfrac{623}{7}=89 and \bar{y}=\dfrac{117}{7}=16.71. Deviations of x: -9,-5,-3,-1,3,7,8, with \sum dx=0 and \sum dx^{2}=81+25+9+1+9+49+64=238. Then

Because \sum dx=0, \sum dx\,dy=\sum dx\cdot y=131. So

The likely price when the supply is 120 is about Rs. 33.8.

5. d=R_1-R_2: -1,\ 1,\ -1,\ 1,\ -1,\ 1, so \sum d^{2}=6 and n=6:

The two judges largely agree: a strong positive rank correlation (perfect agreement would give 1).

1. \bar{x}=6, \bar{y}=8.

(a)

There is a strong positive correlation: students who score more in Mathematics tend to score more in Physics. (b)

Check: b_{yx}\,b_{xy}=0.845=r^{2} ✓. The regression line of y on x:

The regression line of x on y:

(c) When x=12: y=0.65(12)+4.1=11.9. When y=10: x=1.3(10)-4.4=8.6.

1. (a)

(b)

(Only P(A) and P(A\cap B) are used; P(B) is a distractor.)

2. From 2P(A)=5P(B)=\dfrac{5}{18} we get P(A)=\dfrac{5}{36}.

Independence would need P(A\cap B)=P(A)P(B)=\dfrac{5}{36}\times\dfrac{1}{18}=\dfrac{5}{648}, but P(A\cap B)=\dfrac{1}{18}. So A and B are dependent. (Indeed P(A\cap B)=P(B) means B is wholly contained in A.)

3. (a) A leap year has 366=52\times7+2 days: 52 complete weeks and 2 extra days. The extra pair is one of seven equally likely pairs: (Sun, Mon), (Mon, Tue), (Tue, Wed), (Wed, Thu), (Thu, Fri), (Fri, Sat), (Sat, Sun). A year has 53 Fridays or 53 Saturdays if the pair contains Friday or Saturday: (Thu, Fri), (Fri, Sat), (Sat, Sun) — 3 pairs.

(b) P(\text{first ace})=\dfrac{4}{52} and, given that, P(\text{second ace})=\dfrac{3}{51}:

4. (a)

(b) Red then white, or white then red:

(c) "At least one white" is the complement of "both red":

Check: P(RR)+P(RW\text{ or }WR)+P(WW)=\dfrac{20}{56}+\dfrac{30}{56}+\dfrac{6}{56}=1. ✓

1. (a) The conditional probability of B given that A has occurred is P(B/A)=\dfrac{P(A\cap B)}{P(A)}, P(A)>0. The multiplication theorem states P(A\cap B)=P(A)\,P(B/A). (b) From the addition theorem:

Since P(A/B)=0.6=P(A) (equivalently P(A)P(B)=0.3=P(A\cap B)), the events are independent. (c)

Sum: \dfrac27+\dfrac47+\dfrac17=1. ✓

1. (a) Take natural logarithms of y=(\tan x)^{\ln x}: \ln y=\ln x\cdot\ln(\tan x). Differentiate:

(b) At x=2: y=2(8)+4-14-2=4, so the point is (2,4). The slope is

(As recorded, the 2081 question names the point (2,5), which does not lie on this curve — y(2)=4 — so the tangent is taken at x=2.)

2. (a) With u=\sin2x, \dfrac{d}{du}\coth^{-1}u=\dfrac{1}{1-u^{2}}, so

(Strictly, \coth^{-1}u is defined only for |u|>1, while \sin2x never exceeds 1; the printed question asks for the formal result, 2\sec2x.) (b) As x\to\infty the form is \dfrac{\infty}{\infty}. Apply L'Hospital's rule:

Divide by x^{4}:

3. (a) V=\dfrac43\pi r^{3} and S=4\pi r^{2}, so \dfrac{dV}{dr}=4\pi r^{2} and \dfrac{dS}{dr}=8\pi r:

(b) 4\pi r^{2}=10\pi at that instant. Since \dfrac{dV}{dt}=4\pi r^{2}\dfrac{dr}{dt}:

4. (a) Differentiate y^{2}=8x with respect to t: 2y\dfrac{dy}{dt}=8\dfrac{dx}{dt}. The abscissa and ordinate change at the same rate, \dfrac{dy}{dt}=\dfrac{dx}{dt}, so 2y=8, y=4. Then 16=8x gives x=2. The point is (2,4) and indeed y=2x. ∎ (b) V=x^{3} so dV=3x^{2}\,dx=3(100)(0.01)=3\ \text{cm}^{3}:

(Here dy is the differential f'(x)\,dx and \Delta y the actual change; for small dx they are nearly equal, which is why the percentage error is computed as \dfrac{dy}{y}\times100.)

5. (a) \dfrac{d}{dx}\text{cosech}^{-1}x=-\dfrac{1}{|x|\sqrt{1+x^{2}}}, x\neq0. (b) If f(a)=g(a)=0 (or both tend to \infty) and \lim\dfrac{f'(x)}{g'(x)} exists, then \displaystyle\lim_{x\to a}\frac{f(x)}{g(x)}=\lim_{x\to a}\frac{f'(x)}{g'(x)}. (c) \dfrac{dy}{dx}\to\infty, equivalently \dfrac{dx}{dy}=0. (d) \displaystyle\int\frac{dx}{x^{2}-a^{2}}=\frac{1}{2a}\ln\Big|\frac{x-a}{x+a}\Big|+C. (e) \dfrac{dy}{dx}+Py=Q, with P and Q functions of x alone.

1. (a) With u=\tan x and \dfrac{d}{du}\sinh^{-1}u=\dfrac{1}{\sqrt{1+u^{2}}}:

valid where \cos x>0, so that \sqrt{\sec^{2}x}=\sec x. (b) The form is \dfrac00. Apply L'Hospital's rule twice:

(c) The point (2,2) lies on the curve: 8-6=2. The slope is \dfrac{dy}{dx}=3x^{2}-3=9 at x=2.

1. (a) A rational fraction \dfrac{f(x)}{g(x)} is proper if the degree of the numerator is less than the degree of the denominator, for example \dfrac{x+1}{x^{2}+3x+2}. (b) The denominator factorises: x^{3}+3x^{2}+2x=x(x+1)(x+2). The fraction is proper, so

Multiply through by x(x+1)(x+2) and use the cover-up rule at the roots:

2. (a) x^{3}-x^{2}-2x=x(x-2)(x+1) and

By the cover-up rule: A=\dfrac{1}{(-2)(1)}=-\dfrac12, B=\dfrac{1}{(2)(3)}=\dfrac16, C=\dfrac{1}{(-1)(-3)}=\dfrac13.

(b) 9x^{2}+1=9\big(x^{2}+\tfrac19\big), so with a=\dfrac13:

3. x^{3}+4x=x(x^{2}+4), and the quadratic factor is irreducible:

Put x=0: 4=4A, A=1. Compare x^{2}: 2=A+B, B=1. Compare x: -1=C.

using \int\dfrac{x}{x^{2}+4}dx=\dfrac12\ln(x^{2}+4) and \int\dfrac{dx}{x^{2}+4}=\dfrac12\tan^{-1}\dfrac x2.

4. (a) Put e^{x}=t, so e^{x}dx=dt and dx=\dfrac{dt}{t}:

∎ (b) \dfrac{1}{x^{2}-1}=\dfrac12\Big[\dfrac{1}{x-1}-\dfrac{1}{x+1}\Big], so

Note the order: \dfrac{x-1}{x+1}, not \dfrac{x+1}{x-1} — the distractor in the 2080 key.

5. Put t=\tan\dfrac x2, so \cos x=\dfrac{1-t^{2}}{1+t^{2}} and dx=\dfrac{2\,dt}{1+t^{2}}:

1. (a) Put t=\tan\dfrac x2, so a+b\cos x=\dfrac{(a+b)+(a-b)t^{2}}{1+t^{2}} and dx=\dfrac{2\,dt}{1+t^{2}}:

Case a>b: a-b>0, so the denominator is a sum of squares:

Case a<b: a-b<0, the denominator is a difference of squares and the result is logarithmic:

Case a=b: a(1+\cos x)=2a\cos^{2}\dfrac x2, so the integral is \dfrac{1}{2a}\displaystyle\int\sec^{2}\dfrac x2\,dx=\dfrac1a\tan\dfrac x2+C. So the integral has an inverse-tangent form, a logarithmic form, or a tangent form according as a>b, a<b or a=b.

(b) Here a=3>b=2: \sqrt{a^{2}-b^{2}}=\sqrt{5} and \sqrt{\dfrac{a-b}{a+b}}=\dfrac{1}{\sqrt{5}}:

(c) Write the numerator in terms of the derivative of the denominator, \dfrac{d}{dx}(x^{2}+4x+13)=2x+4:

and, completing the square x^{2}+4x+13=(x+2)^{2}+9,

1. (a) Divide by \sin x: \dfrac{dy}{dx}+\cot x\cdot y=x. So P=\cot x and

(The left side is exactly \dfrac{d}{dx}(y\sin x) — that is why the I.F. is \sin x.) (b) Rewrite as a linear equation:

because \dfrac{d}{dx}\big(xe^{-x}\big)=(1-x)e^{-x}. Therefore

Check: y'=\dfrac{(1+Ce^{x})(1+x)-(x+Ce^{x})}{(1+x)^{2}}; substituting into (1+x)y'-\dfrac{x(1+x)}{1+x}\,y reproduces 1-x. ✓

2. (a) Separate the variables:

or equivalently \dfrac{y-x}{1+xy}=c. (b) \dfrac{dy}{dx}=e^{x}e^{-y}, so e^{y}\,dy=e^{x}\,dx:

3. (a) Example: (2xy+3)\,dx+(x^{2}+4y)\,dy=0. Here M=2xy+3, N=x^{2}+4y, and

This equality is the characteristic of an exact equation: M\,dx+N\,dy is then the total differential of some F(x,y). Solution: \int M\,dx=x^{2}y+3x (treating y as constant); the terms of N free of x are 4y, giving 2y^{2}:

(b) \dfrac{dy}{dx}=\dfrac{x^{2}+y^{2}}{2xy} is homogeneous of degree 0. Put y=vx, \dfrac{dy}{dx}=v+x\dfrac{dv}{dx}:

4. (a) The highest derivative is \dfrac{d^{3}s}{dt^{3}}, so the order is 3. Square to remove the radical:

The highest derivative appears to the power 2, so the degree is 2. (b) This is linear: \dfrac{dv}{dt}+kv=g\cos\alpha, P=k, I.F. =e^{kt}:

This formula fails when k=0, because it divides by k. The equation itself still has a solution then: \dfrac{dv}{dt}=g\cos\alpha gives v=gt\cos\alpha+C. So it is the closed form that does not exist for k=0, not the solution of the equation (an answer the examiner's scheme for this item may phrase simply as "k\neq0").

1. (a) "Rate of increase of depth proportional to the square root of the depth":

(b) Separate the variables and integrate:

Initially t=0 and h=n, so C=2\sqrt{n}:

∎ (c) Put h=16, t=6: \sqrt{16}=\dfrac{6k}{2}+\sqrt{n}, so

∎

1. The largest coefficient in column 1 is 3 (row 3), so interchange R_1\leftrightarrow R_3 (partial pivoting):

Eliminate x with R_2\to3R_2-R_1 and R_3\to3R_3-2R_1:

Eliminate y with R_3\to7R_3+R_2: 12z=36, so z=3. Back-substitute:

Check: 2+2+3=7 ✓; 1+6+6=13 ✓; 3+4+3=10 ✓.

2. Add slack variables: 3x+4y+s_1=12, -x+y+s_2=2, and Z-7x-y=0.

Most negative in the Z-row: -7, so x enters. Ratios: \dfrac{12}{3}=4 (the entry -1 is ignored), so s_1 leaves; pivot =3. Divide R_1 by 3, then R_2\to R_2+R_1, Z\to Z+7R_1:

No negative entry remains in the Z-row, so the solution is optimal:

Check at the corner points of the feasible region: (0,0)\to0; (4,0)\to28; \big(\tfrac47,\tfrac{18}{7}\big)\to6.57; (0,2)\to2. The maximum is 28 at (4,0). ✓

3. Slack form: 2x+3y+s_1=15, 3x+y+s_2=12, Z-25x-35y=0.

y enters (-35). Ratios: \dfrac{15}{3}=5, \dfrac{12}{1}=12; s_1 leaves, pivot 3.

x enters (-\frac53). Ratios: \dfrac{5}{2/3}=7.5, \dfrac{7}{7/3}=3; s_2 leaves, pivot \frac73.

All Z-row entries are non-negative:

Check: 2(3)+3(3)=15 ✓, 3(3)+3=12 ✓, 25(3)+35(3)=180 ✓.

4. Clear fractions: \dfrac x3+\dfrac y4\leq1\Rightarrow4x+3y\leq12 and \dfrac x6+\dfrac y3\leq1\Rightarrow x+2y\leq6.

x enters (-7); ratios \dfrac{12}{4}=3, \dfrac61=6; s_1 leaves, pivot 4.

No negatives remain:

Check: corner points (0,0)\to0, (3,0)\to21, (1.2,2.4)\to20.4, (0,3)\to15. The maximum is 21. ✓

1. (a) Convergence: in each row the diagonal entry exceeds the sum of the others (10>1+1, 10>2+1, 10>2+2), so the matrix is diagonally dominant and the iteration converges. Solve each equation for its diagonal unknown:

Working for iteration 1: x=\dfrac{12-0-0}{10}=1.2; y=\dfrac{13-2(1.2)-0}{10}=1.06 (uses the new x); z=\dfrac{14-2(1.2)-2(1.06)}{10}=0.948. The iterates approach (1,1,1), which satisfies all three equations exactly.

(b) The printed 2082 question lists three constraints, 6x+7y\leq42, x+3y\leq42 and x+3y\leq9; the middle one is redundant (implied by x+3y\leq9), so two constraints are used here. Slack form: 6x+7y+s_1=42, x+3y+s_2=9, P-10x-3y=0.

x enters (-10). Ratios: \dfrac{42}{6}=7, \dfrac91=9; s_1 leaves, pivot 6.

No negatives remain:

Check: corner points (0,0)\to0, (7,0)\to70, \big(\tfrac{63}{11},\tfrac{12}{11}\big)\to60.5, (0,3)\to9. ✓

1. The two strings and the line joining the supports form a triangle with sides 4, 3, 5. Since 3^{2}+4^{2}=5^{2}, the angle at the body is 90^{\circ}. Let T_{1} be the tension in the 4 m string and T_{2} in the 3 m string. The 4 m string makes an angle A with the horizontal and the 3 m string an angle B, where

Horizontal equilibrium: T_{1}\cos A=T_{2}\cos B, so \dfrac45T_{1}=\dfrac35T_{2}, T_{1}=\dfrac34T_{2}. Vertical equilibrium: T_{1}\sin A+T_{2}\sin B=78:

Check: the strings are perpendicular, so T_{1}^{2}+T_{2}^{2} must equal 78^{2}: 46.8^{2}+62.4^{2}=2190.24+3893.76=6084=78^{2}. ✓

2. Let the plane make angle \alpha with the horizontal and let the weight be R. Force A parallel to the plane. Resolving along the plane, A=R\sin\alpha. Force B horizontal (parallel to the base). Resolving along the plane, B\cos\alpha=R\sin\alpha, so B=R\tan\alpha. Then

∎

3. Before the interchange: X at A, Y at B. The resultant X+Y acts at P between A and B with X\cdot AP=Y\cdot PB. Since AP+PB=AB,

After the interchange: Y at A, X at B. The new point P' satisfies Y\cdot AP'=X\cdot P'B, so

Both points lie on AB measured from A, so the displacement is

∎ (Because X>Y, the resultant moves towards B.)

4. u=120, \theta=30^{\circ}, \sin\theta=\dfrac12, g=10. (a) Time to the highest point: \dfrac{u\sin\theta}{g}=\dfrac{120\times0.5}{10}=6\ \text{s}. (b) Time of flight: 2\times6=12\ \text{s}. (c) Range: \dfrac{u^{2}\sin2\theta}{g}=\dfrac{14400\times\sin60^{\circ}}{10}=1440\times0.8660=1247\ \text{m} (=720\sqrt{3}\ \text{m}). (d) Greatest height: \dfrac{u^{2}\sin^{2}\theta}{2g}=\dfrac{14400\times0.25}{20}=180\ \text{m}. (e) Maximum range at \theta=45^{\circ}: \dfrac{u^{2}}{g}=\dfrac{14400}{10}=1440\ \text{m}.

5. (a) m=0.025 kg, u=250\ \text{m s}^{-1}, final v=0, t=0.02 s. Retardation:

(Equivalently s=\dfrac{u+v}{2}\,t=125\times0.02=2.5 m. The resisting force is ma=0.025\times12\,500=312.5 N.) (b) Let the forces be P at distance 0.5 m and Q at distance 2.5-0.5=2.0 m from the line of action of the resultant. Moments about the resultant balance: P(0.5)=Q(2.0), so P=4Q. And P+Q=30:

1. (a) Lami's theorem: if three forces P, Q, R acting at a point keep it in equilibrium, each force is proportional to the sine of the angle between the other two:

where \alpha is the angle between Q and R, \beta between R and P, \gamma between P and Q. Proof. By the triangle law the three forces, taken in order, form the sides of a closed triangle. The sides of this triangle are parallel to P, Q, R, and the interior angles are 180^{\circ}-\alpha, 180^{\circ}-\beta, 180^{\circ}-\gamma (the angle between two forces and the interior angle of the force triangle are supplementary). By the sine rule on this triangle,

and \sin(180^{\circ}-\theta)=\sin\theta, which gives Lami's theorem. ∎ (b) Let T_{1} act at 30^{\circ} to the horizontal and T_{2} at 60^{\circ}. The angle between the two strings is 180^{\circ}-30^{\circ}-60^{\circ}=90^{\circ}. The angle between T_{1} and the weight (vertical) is 90^{\circ}+30^{\circ}=120^{\circ}, and between T_{2} and the weight is 90^{\circ}+60^{\circ}=150^{\circ}. By Lami:

Check (resolving): horizontal 50\cos30^{\circ}=43.3=86.6\cos60^{\circ} ✓; vertical 50\sin30^{\circ}+86.6\sin60^{\circ}=25+75=100 ✓.

2. (a) Take the origin at the point of projection, x horizontal and y upward. Initial velocity components: u\cos\theta and u\sin\theta. Horizontal motion is uniform; vertical motion has acceleration -g:

Time of flight: the particle returns to the ground when y=0, t\neq0:

Greatest height: at the top the vertical velocity u\sin\theta-gt=0, at t=\dfrac{u\sin\theta}{g}. Substituting,

Range: horizontal distance in time T:

(b) u=60, \theta=30^{\circ}, g=10:

1. (a) For a,b,c\in\mathbb{Z} under +: Closure: a+b\in\mathbb{Z}. Associativity: (a+b)+c=a+(b+c). Identity: 0\in\mathbb{Z} and a+0=0+a=a. Inverse: for each a, -a\in\mathbb{Z} and a+(-a)=0. Commutativity: a+b=b+a. All five hold, so (\mathbb{Z},+) is an abelian group. ∎ (b) Under multiplication closure, associativity and the identity 1 hold, but inverses fail: the inverse of 2 would be \tfrac12\notin\mathbb{Z}. (Only 1 and -1 have inverses, and 0 has none.) So (\mathbb{Z},\times) is not a group.

2. Composition table (row \times column):

Closure: every entry is in the set. Associativity: inherited from multiplication of complex numbers. Identity: 1 (its row and column repeat the headings). Inverses: 1^{-1}=1, (-1)^{-1}=-1, i^{-1}=-i, (-i)^{-1}=i (each row contains 1). Commutativity: the table is symmetric about the leading diagonal. So it is an abelian group of order 4. ∎

3. Continuity at x=\frac12: from the left, \dfrac12-\dfrac12=0; from the right, \big(\dfrac12-\dfrac12\big)^{2}=0; and f\big(\tfrac12\big)=0. So f is continuous on [0,1]. Differentiability at x=\frac12: left derivative =\dfrac{d}{dx}\big(\tfrac12-x\big)=-1; right derivative =\dfrac{d}{dx}\big(\tfrac12-x\big)^{2}=-2\big(\tfrac12-x\big)=0 at x=\tfrac12. They differ, so f is not differentiable at x=\tfrac12, a point inside (0,1). The hypothesis of the theorem fails. Does the conclusion hold anyway? \dfrac{f(1)-f(0)}{1-0}=\dfrac{\frac14-\frac12}{1}=-\dfrac14. For x<\tfrac12, f'=-1\neq-\tfrac14. For x>\tfrac12, f'=2x-1=-\tfrac14 gives x=\tfrac38, which is not greater than \tfrac12. So no point c exists: f does not obey the Mean Value Theorem on [0,1].

4. (a) f is a polynomial, so continuous and differentiable everywhere. f(0)=0 and f(2)=0, so f(0)=f(2). Rolle's conditions hold. f(x)=x^{3}-3x^{2}+2x, so f'(x)=3x^{2}-6x+2=0:

Both 1-\frac{1}{\sqrt{3}}\approx0.423 and 1+\frac{1}{\sqrt{3}}\approx1.577 lie in (0,2). Rolle's theorem is verified. (b) f(x)=x^{2} is continuous and differentiable. \dfrac{f(4)-f(1)}{4-1}=\dfrac{16-1}{3}=5 and f'(c)=2c=5 gives c=2.5\in(1,4). Verified.

5. h=\dfrac{1-0}{4}=0.25 and y=\dfrac{1}{1+x^{2}}:

Trapezoidal rule:

Simpson's rule:

The exact value is \dfrac{\pi}{4}=0.785398. Simpson's error is about 6\times10^{-6}, the trapezoidal error about 2.6\times10^{-3}: Simpson's rule is far more accurate for the same strips.

1. (a) Rolle's theorem: if f is continuous on [a,b], differentiable on (a,b) and f(a)=f(b), there is at least one c\in(a,b) with f'(c)=0. Proof. Because f is continuous on a closed interval it attains a greatest value M and a least value m. If M=m, f is constant and f'(c)=0 for every c. Otherwise at least one of M,m differs from f(a)=f(b), so is attained at an interior point c\in(a,b); say f(c)=M. Then f(c+h)-f(c)\leq0 for small h, so the quotient \dfrac{f(c+h)-f(c)}{h} is \leq0 for h>0 and \geq0 for h<0. Since f'(c) exists, its limit from both sides must be equal, which forces f'(c)=0. The case f(c)=m is identical. ∎ (b) f is a polynomial, so continuous and differentiable. f(1)=1-4+3=0 and f(3)=9-12+3=0. Then f'(x)=2x-4=0 gives c=2\in(1,3). Verified. (c) Take any x_1<x_2 in [a,b]. f is continuous and differentiable on [x_1,x_2], so by the Mean Value Theorem there is c\in(x_1,x_2) with

since f'(c)=0. So f(x_2)=f(x_1) for all x_1,x_2: f is constant. ∎