Question

Answer the following questions:
1. If S={0, 1, 2, 3, 4, 5, 6, 7, 8, 9), A = {1, 3, 5), B = {1, 2, 3, 4, 5) and C = {4, 6, 8) the find (An B) nC'.
2. If cosec x =-13/12
and x does not lie in third quadrant then find all other trigonometric ratios.

3. Find the modulus and amplitude of the complex number
1-3i/1 +2i.
4. If A and B are the roots of the equation ax2 + bx +c= 0, then form an equation whose roots are
A+1/B , B+1/A.​

Answer 1

(1)

Given:

• S = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}
• A = {1, 3, 5}
• B = {1, 2, 3, 4, 5}
• C = {4, 6, 8}

To find: (A ∩ B) ∩ C'

Solution:

• A ∩ B = {1, 3, 5}
• C' = S - C = {0, 1, 2, 3, 5, 7, 9}
• Then (A ∩ B) ∩ C' = {1, 3, 5}

Answer: (A ∩ B) ∩ C' = {1, 3, 5}

(2)

Given: $cosecx=-\frac{13}{12}$

To find: all trigonometric ratios

Solution:

• Here $cosecx=-\frac{13}{12}$

• In the third quadrant, sin, cos, cosec, sec are negative and tan, cot ratios are positive.

• $sinx=\frac{1}{cosecx}=\frac{1}{-\frac{13}{12}}=-\frac{12}{13}$

• $cosx=-\frac{\sqrt{13^{2}-12^{2}}}{13}=-\frac{\sqrt{169-144}}{13}\\=-\frac{\sqrt{25}}{13}=-\frac{5}{13}$

• $secx=\frac{1}{cosx}=\frac{1}{-\frac{5}{13}}=-\frac{13}{5}$

• $tanx=\frac{sinx}{cosx}=\frac{-\frac{12}{13}}{-\frac{5}{13}}=\frac{12}{5}$
• $cotx=\frac{1}{tanx}=\frac{1}{\frac{12}{5}}=\frac{5}{12}$

Answer:

1. $sinx=-\frac{12}{13}$
2. $cosx=-\frac{5}{13}$
3. $tanx=\frac{12}{5}$
4. $cosecx=-\frac{13}{12}$
5. $secx=-\frac{13}{5}$
6. $cotx=\frac{5}{12}$

(3)

Given: $\frac{1-3i}{1+2i}$

To find:

• modulus & amplitude

Solution:

• Now $z=\frac{1-3i}{1+2i}$

• $=\frac{(1-3i)(1-2i)}{(1+2i)(1-2i)}$

• $=\frac{1-5i+6i^{2}}{1-4i^{2}}$

• $=\frac{1-5i-6}{1+4}$

• $=\frac{-5-5i}{5}$
• $=-1-i$

• $\Rightarrow z=-1-i$

• Thus $|z|=\sqrt{(-1)^{2}+(-1)^{2}}\\=\sqrt{1+1}=\sqrt{2}$

• Now amplitude $\theta=tan^{-1}\frac{-1}{-1}-\pi$

• $\Rightarrow \theta=tan^{-1}(1)-\pi$

• $\Rightarrow \theta=\frac{\pi}{4}-\pi$

• $\Rightarrow \theta=-\frac{3\pi}{4}$

• Note: $z=-1-i=\sqrt{2}(cos\frac{5\pi}{4}+isin\frac{5\pi}{4})$

• Here $\frac{5\pi}{4}$ does not give the principal argument of $-1-i$ since $\theta=\frac{5\pi}{4}$ does not satisfy the relatioon $-\pi<\theta\leqslant\pi$.

Answer:

1. modulus is $\sqrt{2}$
2. amplitude is $(-\frac{3\pi}{4})$

(4)

Given:

• $A$ and $B$ are roots of
• $\quad ax^{2}+bx+c=0$

To find: an equation whose roots are $A+\frac{1}{B}$ and $B+\frac{1}{A}$

Solution:

Since $A$ and $B$ are the roots of the given equation $ax^{2}+bx+c=0$, we get

• $A+B=-\frac{b}{a}$
• $AB=\frac{c}{a}$

So the equation whose roots are $A+\frac{1}{B}$ and $B+\frac{1}{A}$ is given by

$\quad \{x-(A+\frac{1}{B})\}\{x-(B+\frac{1}{A})\}=0$

$\Rightarrow x^{2}-(A+B+\frac{1}{A}+\frac{1}{B})x-(A+\frac{1}{B})(B+\frac{1}{A})=0$

$\Rightarrow x^{2}-(A+B+\frac{A+B}{AB})x-(AB+\frac{1}{AB}+2)=0$

$\Rightarrow x^{2}-(-\frac{b}{a}+\frac{-\frac{b}{a}}{\frac{c}{a}})x-(\frac{c}{a}+\frac{1}{\frac{c}{a}}+2)=0$

$\Rightarrow x^{2}-(-\frac{b}{a}-\frac{b}{c})x-(\frac{c}{a}+\frac{a}{c}+2)=0$

$\Rightarrow x^{2}+\frac{b(a+c)}{ac}x-\frac{c^{2}+a^{2}+2ac}{ac}=0$

$\Rightarrow acx^{2}+b(a+c)x-(a+c)^{2}=0$

Answer: the required equation is

• $acx^{2}+b(a+c)x-(a+c)^{2}=0$