Answers to Calculus Practical 2

Question 1 §

$$f(x)=\frac{2x}{3x^2+1}⟹f(-x)=\frac{-2x}{3x^2+1}:\text{odd}$$

$$f(x)=\frac{3x^2}{3x^4+4}⟹f(-x)=\frac{3x^2}{3x^4+4}:\text{even}$$

$$f(x)=1+x^3-x^5⟹f(-x)=1-x^3+x^5:\text{neither}$$

$$f(x)=\ln (2x^6-x^4+2)⟹f(-x)=\ln (2x^6-x^4+2):\text{even}$$

$$g(t)=\frac{2t^3+t-1}{3t^2+5}⟹f(-x)=\frac{-2t^3-t-1}{3t^2+5}:\text{neither}$$

$$h(x)=\frac{a{x}^3+bx}{c{x}^2+d}⟹f(-x)=\frac{-a{x}^3-bx}{c{x}^2+d}:\text{odd}$$

Question 2 §

Prove ${\cosh }^{2}x-{\sinh }^{2}x=1$:

$${\cosh }^{2}x-{\sinh }^{2}x=1:LHS=(\frac{e^x+e^{-x}}{2}{)}^2-(\frac{e^x-e^{-x}}{2}{)}^2,RHS=1$$

$$⟹LHS=\frac{e^{2x}+2+e^{-2x}}{4}-\frac{e^{2x}-2+e^{-2x}}{4}$$

$$⟹LHS=\frac{4}{4},RHS=1\therefore LHS=RHS■$$

Hence, prove that ${\text{sech}}^{2}x=1-{\tanh }^{2}x$ and ${\text{cosech}}^{2}x={\coth }^{2}x-1$:

$$\frac{{\cosh }^{2}x}{{\cosh }^{2}x}-\frac{{\sinh }^{2}x}{{\cosh }^{2}x}=\frac{1}{{\cosh }^{2}x}$$

$$⟹1-{\tanh }^{2}x={\text{sech}}^{2}x■$$

$$\frac{{\cosh }^{2}x}{{\sinh }^{2}x}-\frac{{\sinh }^{2}x}{{\sinh }^{2}x}=\frac{1}{{\sinh }^{2}x}$$

$$⟹{\coth }^{2}x-1={\text{cosech}}^{2}x■$$

Question 3 §

$i^2=-1$ $i^3=-i$ $i^4=1$ $i^5=i$ $i^{100}=1$ $i^{505}=i$ $i^{-1}=\frac{1}{i}=-i$ $i^{-7}=-\frac{1}{i}=i$

Question 4 §

$z_1=-2+i,z_2=1-3i$, such that:

$\Re (z_1)=-2$ $\Im (z_1)=1$

$\Re (z_2)=1$ $\Im (z_2)=-3$

$|z_1|=\sqrt{(-2{)}^2+(1{)}^2}=\sqrt{5}$ $|z_2|=\sqrt{(1{)}^2+(-3{)}^2}=\sqrt{10}$

$\angle (z_1)=-{\tan }^{-1}(\frac{1}{2})$ $\angle (z_2)=-{\tan }^{-1}(3)=-\frac{\pi }{4}$

Question 5 §

$z_1=3-2i,z_2=1+4i$, such that

$z_1+z_2=4+2i$ $z_1-z_2=2-6i$

$z_1{z}_2=11+10i$ $\frac{z_1}{z_2}=-\frac{5+14i}{17}=-\frac{5}{17}-\frac{14}{17}i$

$\frac{z_2}{z_1}=\frac{-5+14i}{13}=-\frac{5}{13}+\frac{14}{13}i$ $\overline{z_1}\overline{z_2}=(3+2i)(1-4i)=11-10i$

Question 6 §

$1+i:(r,\theta )=(\sqrt{2},\frac{\pi }{4})$ $-1+i:(r,\theta )=(\sqrt{2},\frac{3\pi }{4})$ $-1-i:(r,\theta )=(\sqrt{2},\frac{5\pi }{4})$ $\sqrt{3}+i:(r,\theta )=(2,\frac{\pi }{6})$ $\sqrt{3}-i:(r,\theta )=(2,\frac{11\pi }{6})$ $-\sqrt{3}-i:(r,\theta )=(2,\frac{4\pi }{3})$

Question 7 §

$\frac{1+i}{i(2-3i)}+\frac{2}{i}=\frac{i(1+i)+2i(2-3i)}{3i-2}=\frac{5i+5}{3i-2}$

$$⟹\frac{5i+5}{3i-2}=\frac{25i-5}{13}=-\frac{5}{13}+\frac{25}{13}i$$

$\frac{1}{3+2i}+\frac{1}{2-i}=\frac{41}{65}+\frac{3}{65}i$

Question 8 §

Show that the product of two odd functions, $f(x)$ and $g(x)$, is always an even function.

Let $f(x),g(x)$ be such that $f(-x)=-f(x),g(-x)=-g(x)$. Thus, if we take the product of the two functions: $h(x)=f(x)g(x)$, prove that $h(-x)=h(x)$.

$$h(-x)=f(-x)\cdot g(-x)=-f(x)\cdot -g(x)=f(x)\cdot g(x)=h(x)$$

$$\therefore h(-x)=h(x)■$$