Mathematics IIA: Interactive Paper
Review the Maths IIA exam paper below. Navigate sections using the tabs. Click "Show Solution" to reveal the step-by-step answer. For statistics questions, use the "Visualize Data" button to see the dataset plotted.
1. Write the conjugate of $(3+4i)(2-3i)$
Step 1: Simplify the complex number by multiplication.
$$ z = (3+4i)(2-3i) = 6 - 9i + 8i - 12i^2 $$Since $i^2 = -1$, we substitute:
$$ z = 6 - i - 12(-1) = 6 - i + 12 = 18 - i $$Step 2: Find the conjugate.
The conjugate of $a+bi$ is $a-bi$. Therefore, $\bar{z} = 18 + i$.
2. If $(a+ib)^2 = x+iy$ find $x^2+y^2$
Take the modulus of both sides:
$$ |(a+ib)^2| = |x+iy| $$Using property $|z^n| = |z|^n$:
$$ |a+ib|^2 = \sqrt{x^2+y^2} $$ $$ (\sqrt{a^2+b^2})^2 = \sqrt{x^2+y^2} $$ $$ a^2+b^2 = \sqrt{x^2+y^2} $$Squaring both sides gives the result:
$$ x^2+y^2 = (a^2+b^2)^2 $$3. Find the value of $(1-i)^8$
Rewrite as powers of squares:
$$ (1-i)^8 = ((1-i)^2)^4 $$Calculate inner square:
$$ (1-i)^2 = 1 - 2i + i^2 = 1 - 2i - 1 = -2i $$Raise to 4th power:
$$ (-2i)^4 = (-2)^4 \cdot i^4 = 16 \cdot 1 = 16 $$4. Find the quadratic equation whose roots are $7 \pm 2\sqrt{5}$
Let roots be $\alpha, \beta = 7 \pm 2\sqrt{5}$.
Sum ($S$): $(7+2\sqrt{5}) + (7-2\sqrt{5}) = 14$
Product ($P$): $(7+2\sqrt{5})(7-2\sqrt{5}) = 49 - 20 = 29$
Equation is $x^2 - Sx + P = 0$:
$$ x^2 - 14x + 29 = 0 $$5. If -1, 2, $\alpha$ are the roots of $2x^3+x^2-7x-6=0$ then find $\alpha$
Using relation between roots and coefficients for $ax^3+bx^2+cx+d=0$:
$$ \text{Sum of roots} = -\frac{b}{a} = -\frac{1}{2} $$Given roots are $-1, 2, \alpha$:
$$ -1 + 2 + \alpha = -\frac{1}{2} $$ $$ 1 + \alpha = -\frac{1}{2} $$ $$ \alpha = -\frac{3}{2} $$6. Find the number of ways of arranging letters of the word 'INTERMEDIATE'
Total letters ($n$) = 12.
Repeated letters: I(2), T(2), E(3).
Formula: $\frac{n!}{p!q!r!...}$
$$ \text{Ways} = \frac{12!}{2! \cdot 2! \cdot 3!} = \frac{12!}{24} $$7. Find the number of ways of selecting 4 boys and 3 girls from a group of 8 boys and 5 girls
Select 4 boys from 8: $\binom{8}{4} = 70$
Select 3 girls from 5: $\binom{5}{3} = 10$
Total ways:
$$ 70 \times 10 = 700 $$8. Prove that $C_0+2.C_1+4.C_2+ \dots +2^n.C_n=3^n$
Binomial expansion:
$$ (1+x)^n = C_0 + C_1 x + C_2 x^2 + \dots + C_n x^n $$Substitute $x=2$:
$$ (1+2)^n = C_0 + 2C_1 + 2^2C_2 + \dots + 2^nC_n $$ $$ 3^n = C_0 + 2C_1 + 4C_2 + \dots + 2^nC_n $$Hence proved.
9. Find the mean deviation about median for the data 4, 6, 9, 3, 10, 13, 2
Visualize Data1. Sort Data: 2, 3, 4, 6, 9, 10, 13 ($n=7$)
2. Median ($M$): 4th term = 6.
3. Deviations $|x_i - M|$:
|2-6|=4, |3-6|=3, |4-6|=2, |6-6|=0, |9-6|=3, |10-6|=4, |13-6|=7
4. Mean Deviation:
$$ \text{MD} = \frac{\sum |x_i - M|}{n} = \frac{4+3+2+0+3+4+7}{7} = \frac{23}{7} \approx 3.28 $$10. Find the variance for the discrete data 6, 7, 10, 12, 13, 4, 8, 12
Visualize Data1. Mean ($\bar{x}$): Sum = 72, $n=8$. $\bar{x} = 9$.
2. Squared Deviations $(x_i - 9)^2$:
9, 4, 1, 9, 16, 25, 1, 9. Sum = 74.
3. Variance ($\sigma^2$):
$$ \sigma^2 = \frac{74}{8} = 9.25 $$11. If $z=3-5i$ then show that $z^3-10z^2+58z-136=0$
Rearrange $z-3 = -5i$ and square:
$$ (z-3)^2 = (-5i)^2 \implies z^2-6z+9 = -25 $$ $$ z^2-6z+34 = 0 $$Divide the polynomial by $z^2-6z+34$:
$$ z^3-10z^2+58z-136 = z(z^2-6z+34) - 4(z^2-6z+34) $$Since $z^2-6z+34=0$, the entire expression becomes 0.
12. Determine the range of $\frac{x^2+x+1}{x^2-x+1}$ for $x \in R$
Let $y = \text{expression}$. Rearrange to quadratic in $x$:
$$ x^2(y-1) - x(y+1) + (y-1) = 0 $$Since $x \in R$, Discriminant $\Delta \ge 0$:
$$ (y+1)^2 - 4(y-1)^2 \ge 0 $$ $$ -3y^2 + 10y - 3 \ge 0 \implies 3y^2 - 10y + 3 \le 0 $$Roots are $1/3$ and $3$. Thus range is $[1/3, 3]$.
13. Find the rank of the word 'EAMCET'
Alphabetical: A, C, E, E, M, T.
- Start with A: $5!/2! = 60$
- Start with C: $5!/2! = 60$
- Start with E:
- EAC... ($3! = 6$)
- EAE... ($3! = 6$)
- EAMCET (1)
14. Selecting cricket team (11 players) from 7 batsmen, 6 bowlers (min 5 bowlers).
Case 1 (5 Bowlers, 6 Batsmen):
$$ \binom{6}{5} \times \binom{7}{6} = 6 \times 7 = 42 $$Case 2 (6 Bowlers, 5 Batsmen):
$$ \binom{6}{6} \times \binom{7}{5} = 1 \times 21 = 21 $$Total Ways: $42 + 21 = 63$.
15. Resolve $\frac{2x^2+3x+4}{(x-1)(x^2+2)}$ into partial fractions.
Form: $\frac{A}{x-1} + \frac{Bx+C}{x^2+2}$
Find $A$ (cover-up at $x=1$): $9/3 = 3$. So $A=3$.
Compare coeff $x^2$: $2 = A+B \implies B = -1$.
Compare constant: $4 = 2A - C \implies C = 2$.
Answer:
$$ \frac{3}{x-1} + \frac{-x+2}{x^2+2} $$16. Resolve $\frac{3x+7}{x^2-3x+2}$ into partial fractions.
Denominator: $(x-1)(x-2)$.
$$ \frac{A}{x-1} + \frac{B}{x-2} $$Find $A$ ($x=1$): $10/(-1) = -10$.
Find $B$ ($x=2$): $13/1 = 13$.
Answer:
$$ \frac{-10}{x-1} + \frac{13}{x-2} $$17. If x is real, prove $\frac{x}{x^2-5x+9}$ lies between 1 and $-1/11$.
Set to $y$ and form quadratic in $x$:
$$ yx^2 - (5y+1)x + 9y = 0 $$Discriminant $\ge 0$:
$$ (5y+1)^2 - 36y^2 \ge 0 $$ $$ -11y^2 + 10y + 1 \ge 0 \implies 11y^2 - 10y - 1 \le 0 $$Roots are $1, -1/11$. Since $\le 0$, $y$ is between roots.
18. Roots of $x^2-2x+4=0$. Show $\alpha^n+\beta^n=2^{n+1}\cos(n\pi/3)$.
Roots are $1 \pm i\sqrt{3}$. Polar form: $2\text{cis}(\pm \pi/3)$.
Using De Moivre's Theorem:
$$ \alpha^n + \beta^n = 2^n(\text{cis}(n\pi/3)) + 2^n(\text{cis}(-n\pi/3)) $$ $$ = 2^n [ (\cos \frac{n\pi}{3} + i\sin \dots) + (\cos \frac{n\pi}{3} - i\sin \dots) ] $$ $$ = 2^n (2\cos \frac{n\pi}{3}) = 2^{n+1}\cos(\frac{n\pi}{3}) $$19. Solve $2x^5+x^4-12x^3-12x^2+x+2=0$
Reciprocal eq (Class I, odd). One root is $-1$.
Synthetic division by -1 yields: $2x^4 - x^3 - 11x^2 - x + 2 = 0$.
Divide by $x^2$, set $y = x + 1/x$.
$$ 2(y^2-2) - y - 11 = 0 \implies 2y^2 - y - 15 = 0 $$Roots for $y$: $3, -5/2$. Solve quadratics:
- $y=3 \implies x^2-3x+1=0 \implies \frac{3\pm\sqrt{5}}{2}$
- $y=-5/2 \implies 2x^2+5x+2=0 \implies -2, -1/2$
20. Expansion coefficients proof $\frac{a_1}{a_1+a_2}+\dots$
Key Identity: $\frac{\binom{n}{r}}{\binom{n}{r}+\binom{n}{r+1}} = \frac{r+1}{n+1}$.
LHS becomes $\frac{r+1}{n+1} + \frac{r+3}{n+1} = \frac{2(r+2)}{n+1}$.
RHS becomes $2 \times \frac{r+2}{n+1}$.
LHS = RHS.
21. Sum of infinite series $1+\frac{1}{3}+\frac{1.3}{3.6}+\dots$
Match with $(1-x)^{-p/q}$.
Identify $p=1, q=2, x=2/3$.
$$ S = (1 - 2/3)^{-1/2} = (1/3)^{-1/2} = \sqrt{3} $$22. Find mean deviation about mean (Grouped Data)
| Marks | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 |
|---|---|---|---|---|---|
| f | 5 | 8 | 15 | 16 | 6 |
1. Midpoints ($x_i$): 5, 15, 25, 35, 45.
2. Mean ($\bar{x} = \frac{\sum fx}{\sum f}$): $\frac{25+120+375+560+270}{50} = \frac{1350}{50} = 27$.
3. Abs Deviations $|x_i - 27|$: 22, 12, 2, 8, 18.
4. $\sum f|x_i-\bar{x}| = 110 + 96 + 30 + 128 + 108 = 472$.
5. MD = $472 / 50 = 9.44$.
23. Find mean deviation about median (Discrete)
| $X_i$ | 6 | 9 | 3 | 12 | 15 | 13 | 21 | 22 |
|---|---|---|---|---|---|---|---|---|
| $f_i$ | 4 | 5 | 3 | 2 | 5 | 4 | 4 | 3 |
1. Arrange $x_i$ asc: 3, 6, 9, 12, 13, 15, 21, 22.
2. Cum Freq ($N=30$): Median is avg of 15th, 16th terms.
CF: 3, 7, 12, 14, 18 (includes 15-18th terms). So Median = 13.
3. Calc $\sum f|x_i - 13|$: $3(10)+4(7)+5(4)+2(1)+4(0)+5(2)+4(8)+3(9)$.
Sum = $30+28+20+2+0+10+32+27 = 149$.
MD = $149/30 \approx 4.97$.
24. If $t=\frac{4}{5}+\frac{4.6}{5.10}+\dots$ prove $9t=16$.
Series $1+t$ matches $(1-x)^{-n}$.
Using ratio of 1st/2nd terms: $\frac{n+1}{n} = 3/2 \implies n=2$.
Then $nx = 4/5 \implies 2x=4/5 \implies x=2/5$.
$1+t = (1-2/5)^{-2} = (3/5)^{-2} = 25/9$.
$t = 25/9 - 1 = 16/9 \implies 9t=16$.