5 Marks Questions

Question 15 Marks

i. If $f(x)=\left\{\begin{array}{cl}|x|+1, & x<0 \\ 0, & x=0 \\ |x|-1, & x>0\end{array}\right.$, for what values (s) of a does $\lim _{x \rightarrow a} f(x)$ exist?
ii. Find the derivative of the function $\cos \left(x-\frac{\pi}{8}\right)$ from the first principle.

Answer

i. $f(x)=\left\{\begin{array}{cc}|x|+1, & x<0 \\ 0, & x=0 \\ |x|-1, & x>0\end{array}\right.$
At x = 0,
$\begin{array}{l}\text { RHL }=\lim _{x \rightarrow 0^{+}} f(x)=\lim _{h \rightarrow 0} f(0+h) \\ =\lim _{h \rightarrow 0}|0+h|-1 \\ =-1\end{array}$
$LHL =\lim _{h \rightarrow 0^{-}} f(x)=\lim _{h \rightarrow 0} f|0-h|$
$\begin{array}{l}=\lim _{h \rightarrow 0}|0-h|+1 \\ =\lim _{h \rightarrow 0}-(0-h)+1 \\ =\lim _{h \rightarrow 0} h+1 \\ =0+1=1 \\ \Rightarrow \text { RHL } \neq \text { LHL } \\ \Rightarrow \text { At } x=0, \text { limi does not exist. }\end{array}$
Hence, $\lim _{x \rightarrow a} f(x)$ exists for all $a \neq 0$.
ii. Let $f(x)=\cos \left(x-\frac{\pi}{8}\right)$
By using first principle of derivative
We have,
$\begin{array}{l}f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} \\ \Rightarrow f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{\cos \left(x+h-\frac{\pi}{8}\right)-\cos \left(x-\frac{\pi}{8}\right)}{b}\left[\because f(x)=\cos \left(x-\frac{\pi}{8}\right)\right] \\ =\lim _{h \rightarrow 0} \frac{-2 \sin \left(\frac{\left.x+h-\frac{\pi}{8}+x-\frac{\pi}{8}\right)}{2}\right) \sin \left(\frac{x+h-\frac{\pi}{8}-x+\frac{\pi}{8}}{2}\right)}{h} \\ {\left[\because \cos C-\cos D=-2 \sin \left(\frac{C+D}{2}\right) \sin \left(\frac{C-D}{2}\right)\right]}\end{array}$
$\begin{array}{l}=\lim _{h \rightarrow 0} \frac{-2 \sin \frac{2 x-2\left(\frac{\pi}{8}\right)+h}{2}}{2 \times \frac{h}{2}} \\ =-\sin \frac{2 x-2\left(\frac{\pi}{8}\right)+0}{2} \times 1\left[\therefore \lim _{x \rightarrow 0} \frac{\sin \frac{h}{2}}{\frac{h}{2}}=1\right] \\ =-\sin \frac{2\left(x-\frac{\pi}{8}\right)}{2} \\ \Rightarrow f^{\prime}(x)=-\sin \left(x-\frac{\pi}{8}\right)\end{array}$

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Question 25 Marks

Two dice are thrown. The events A, B, C, D, E and F are described as follows: A = Getting an even number on the first die.
B = Getting an odd number on the first die.
C = Getting at most 5 as a sum of the numbers on the two dice.
D = Getting the sum of the numbers on the dice greater than 5 but less than 10.
E = Getting at least 10 as the sum of the numbers on the dice.
F = Getting an odd number on one of the dice.
Describe the following events: A and B, B or C, B and C, A and E, A or F, A and F.

Answer

A = Getting an even number on the first die.
A = {(2, 1),(2,2),(2,3),(2,4),(2,5),(2,6),(4,1),(4,2),(4,3),(4,4),(4,5),(4,6),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)}
B = Getting an odd number on the first die.
B = {(1, 1),(1,2),(1,3),(1,4),(1,5),(1,6),(3,1),(3,2),(3,3),(3,4),(3,5),(3,6),(5,1),(5,2),(5,3),(5,4),(5,5),(5,6)}
C = Getting at most 5 as sum of the numbers on the two dice.
C = {(1, 1),(1,2),(1,3),(1,4),(2,1),(2,2),(2,3),(3,1),(3,2),(4,1)}
D = Getting the sum of the numbers on the dice >5 but <10.
D = {(1, 5),(1,6),(2,4),(2,5),(2,6),(3,3),(3,4),(3,5),(3,6),(4,2),(4,3),(4,4),(4,5),(5,1),(5,2),(5,4),(6,1),(6,2),(6,2),(6,3)}
E = Getting at least 10 as the sum of the numbers on the dice.
E = {(4, 6),(5,5),(5,6),(6,4),(6,5),(6,6)}
F = Getting an odd number on one of the dice.
F = {(1, 1),(1,2),(1,3),(1,4),(1,5),(1,6),(3,1),(3,2),(3,3),(3,4)(3,5),(3,6),(5,1),(5,2),(5,3),(5,4),(5,5),(5,6),(2,1),(2,3),(2,5),(4,1),(4,3),(4,5),(6,1),(6,3),(6,5)}

$A$ and $B$ are mutually exclusive events, thus $A \cap B=\varnothing$
$B \cup C$ = {(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(3,1),(3,2),(3,3),(3,4),(3,5),(3,6)(5,1),(5,2),(5,3),(5,4),(5,5),(5,6),(2,1),(2,2),(2,3),(4,1)}
$B \cap C$ = {(1,1),(1,2),(1,3),(1,4),(3,1),(3,2)}
$A \cap E$ = {(4,6),(6,4),(6,5),(6,6)}
$A \cup F$ = {(2,1),(2,2),(2,3),(2,4),(2,5),(2,6), (4,1),(4,2),(4,3),(4,4),(4,5)
(4,6), (6,1),(6,2),(6,3),(6,4),(6,5),(6,6), (1,1),(1,2),(1,3),(1,4),(1,5),(1,6), (3,1)}
{(3,2),(3,3),(3,4),(3,5),(3,6), (5,1),(5,2), (5,3),(5,4),(5,5),(5,6)}
$A \cap F$ = {(2,1),(2,3),(2,5),(4,1),(4,3),(4,5),(6,1),(6,3),(6,5)}

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Question 35 Marks

Prove that: $4 \sin A \sin \left(60^{\circ}- A \right) \sin \left(60^{\circ}+ A \right)=\sin 3 A$.
Hence deduce that: $\sin 20^{\circ} \times \sin 40^{\circ} \times \sin 60^{\circ} \times \sin 80^{\circ}=\frac{3}{16}$

Answer

$\begin{array}{l}L H S=4 \sin A \times \sin \left(60^{\circ}-A\right) \times \sin \left(60^{\circ}+A\right) \\ =2 \sin A\left[2 \sin \left(60^{\circ}-A\right) \sin \left(60^{\circ}+A\right)\right] \\ =2 \sin A\left[\cos \left\{\left(60^{\circ}- A \right)-\left(60^{\circ}+ A \right)\right\}-\cos \left\{\left(60^{\circ}- A \right)+\left(60^{\circ}+ A \right)\right\}\right] \\ {[\because 2 \sin A \times \sin B =\cos ( A - B )-\cos ( A + B )]} \\ =2 \sin A\left[\cos (-2 A)-\cos 120^{\circ}\right] \\ =2 \sin A\left[\cos 2 A-\cos 120^{\circ}\right][\because \cos (-\theta)=\cos \theta] \\ =2 \sin A \times \cos 2 A-2 \sin A \times \cos 120^{\circ} \\ =[\sin ( A +2 A)+\sin ( A -2 A)]-2 \sin A\left(-\frac{1}{2}\right)\end{array}$
$\begin{array}{l}{\left[\because 2 \sin A \times \cos B=\sin (A+B)+\sin (A-B) \text { and } \cos 120^{\circ}=-\frac{1}{2}\right]} \\ =\sin 3 A+\sin (-A)+\sin A \\ =\sin 3 A-\sin A+\sin A=\sin 3 A=\text { RHS }[\because \sin (-\theta)=-\sin \theta] \\ \therefore \text { LHS }=\text { RHS }\end{array}$
Hence proved.
Now, $4 \sin A \sin \left(60^{\circ}-A\right) \times \sin \left(60^{\circ}+A\right)=\sin 3 A$
On putting $A =20^{\circ}$, we get
$4 \sin 20^{\circ} \times \sin \left(60^{\circ}-20^{\circ}\right) \sin \left(60^{\circ}+20^{\circ}\right)=\sin 3 \times\left(20^{\circ}\right)$
$\begin{array}{l}\Rightarrow 4 \sin 20^{\circ} \times \sin 40^{\circ} \times \sin 80^{\circ}=\sin 60^{\circ}=\frac{\sqrt{3}}{2} \\ \Rightarrow \sin 20^{\circ} \times \sin 40^{\circ} \times \sin 80^{\circ}=\frac{\sqrt{3}}{8} \\ \Rightarrow \sin 20^{\circ} \times \sin 40^{\circ} \times \frac{\sqrt{3}}{2} \times \sin 80^{\circ}=\frac{\sqrt{3}}{8} \times \frac{\sqrt{3}}{2}\end{array}$
[multiplying both sides by $\frac{\sqrt{3}}{2}$ ]
$\therefore \sin 20^{\circ} \times \sin 40^{\circ} \times \sin 60^{\circ} \times \sin 80^{\circ}=\frac{3}{16}\left[\because \frac{\sqrt{3}}{2}=\sin 60^{\circ}\right]$

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Question 45 Marks

Prove that $\cos \frac{2 \pi}{15} \cdot \cos \frac{4 \pi}{15} \cdot \cos \frac{8 \pi}{15} \cdot \cos \frac{16 \pi}{15}=\frac{1}{16}$

Answer

$\begin{array}{l}\text { LHS }=\cos \frac{2 \pi}{15} \cdot \cos \frac{4 \pi}{15} \cdot \cos \frac{8 \pi}{15} \cdot \cos \frac{16 \pi}{15} \\ =\cos \frac{2 \pi}{15} \cos 2\left(\frac{2 \pi}{15}\right) \cos 4\left(\frac{2 \pi}{15}\right) \cos 8\left(\frac{2 \pi}{15}\right) \\ \text { Put } \frac{2 \pi}{15}=\alpha \\ \Rightarrow \text { LHS }=\cos \alpha \cdot \cos 2 \alpha \cdot \cos 4 \alpha \cdot \cos 8 \alpha\end{array}$
$=\frac{2 \sin \alpha[\cos \alpha \cdot \cos 2 \alpha \cdot \cos 4 \alpha \cdot \cos 8 \alpha]}{2 \sin \alpha}$ [multiplying numerator and denominator by $2 \sin \alpha$ ]
$\begin{array}{l}=\frac{(2 \sin \alpha \cdot \cos \alpha) \cdot \cos 2 \alpha \cdot \cos 4 \alpha \cdot \cos 8 \alpha}{2 \sin \alpha} \\ =\frac{2(\sin 2 \alpha \cdot \cos 2 \alpha \cdot \cos 4 \alpha \cdot \cos 8 \alpha)}{2(2 \sin \alpha)}[\because 2 \sin \alpha \cos \alpha=\sin 2 \alpha \text { and multiplying numerator and denominator by } 2] \\ =\frac{(2 \sin 2 \alpha \cdot \cos 2 \alpha) \cdot \cos 4 \alpha \cdot \cos 8 \alpha}{4 \sin \alpha} \\ =\frac{2(\sin 4 \alpha \cdot \cos 4 \alpha) \cos 8 \alpha}{2(4 \sin \alpha)}[\because \cdot 2 \sin \alpha \cos \alpha=\sin 2 \alpha \text { and multiplying numerator and denominator by } 2] \\ =\frac{2(\sin 8 \alpha \cdot \cos 8 \alpha)}{2(8 \sin \alpha)} \\ =\frac{\sin 16 \alpha}{16 \sin \alpha}=\frac{\sin (15 \alpha+\alpha)}{16 \sin \alpha} \\ \text { Now, } 15 \alpha=2 \pi \\ =\frac{\sin (2 \pi+\alpha)}{16 \sin \alpha}=\frac{\sin \alpha}{16 \sin \alpha}=\frac{1}{16}=\text { RHS } \\ \therefore \text { LHS = RHS }\end{array}$
Hence proved.

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Question 55 Marks

The Sum of two no. is 6 times their geometric mean, show that no. are in the ratio $(3+3 \sqrt{2}):(3-2 \sqrt{2})$

Answer

$\begin{array}{l}a+b=6 \sqrt{a b} \\ \frac{a+b}{2 \sqrt{a b}}=\frac{3}{1}\end{array}$
by C and D
$\begin{array}{l}\frac{a+b+2 \sqrt{a b}}{a+b-2 \sqrt{a b}}=\frac{3+1}{3-1} \\
\frac{(\sqrt{a}+\sqrt{b})^2}{(\sqrt{a}-\sqrt{b})^2}=\frac{2}{1} \\
\frac{\sqrt{a}+\sqrt{b}}{\sqrt{a}-\sqrt{b}}=\frac{\sqrt{2}}{1}\end{array}$
again by C and D
$\begin{array}{l}\frac{\sqrt{a}+\sqrt{b}+\sqrt{a}-\sqrt{b}}{\sqrt{a}+\sqrt{b}-\sqrt{a}-\sqrt{b}}=\frac{\sqrt{2}+1}{\sqrt{2}-1} \\ \frac{2 \sqrt{a}}{2 \sqrt{b}}=\frac{\sqrt{2}+1}{\sqrt{2}-1} \\ \frac{a}{b}=\frac{(\sqrt{2}+1)^2}{(\sqrt{2}-1)^2} \text { (on squaring both sides) } \\ \frac{a}{b}=\frac{2+1+2 \sqrt{2}}{2+1-2 \sqrt{2}} \\ \frac{a}{b}=\frac{3+2 \sqrt{2}}{3-2 \sqrt{2}}\end{array}$
$a: b=(3+2 \sqrt{2}):(3-2 \sqrt{2})$

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Question 65 Marks

Find the derivative of (sinx + cosx) from first principle.

Answer

We have, $f(x)=\sin x+\cos x$
By using first principle of derivative
$\begin{array}{l}f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} \\ \therefore f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{\sin (x+h)+\cos (x+h)-\sin x-\cos x}{h} \\ =\lim _{h \rightarrow 0} \frac{[\sin x \cdot \cos h+\cos x \cdot \sin h+\cos x \cdot \cos h-\sin x \cdot \sin h-\sin x-\cos x]}{h}\end{array}$
$(\because \sin (x+y)=\sin x \cos y+\cos x \sin y$ and $\cos (x+y)=\cos x \cos y-\sin x \sin y)$
$\begin{array}{l}=\lim _{h \rightarrow 0} \frac{\sin h(\cos x-\sin x)+\sin x(\cos h-1)+\cos x(\cos h-1)}{h} \\ =\lim _{h \rightarrow 0} \frac{\sin h}{h}(\cos x-\sin x)+\lim _{h \rightarrow 0} \frac{\sin x(\cos h-1)}{h}+\lim _{h \rightarrow 0} \frac{\cos x(\cos h-1)}{h} \\ =1 \cdot(\cos x-\sin x)+\lim _{h \rightarrow 0} \sin x\left[\frac{-(1-\cos h)}{h}\right]+\lim _{h \rightarrow 0} \cos x\left[\frac{-(1-\cos h)}{h}\right]\left[\because \lim _{x \rightarrow 0} \frac{\sin x}{x}=1\right] \\ =(\cos x-\sin x)-\sin x \cdot \lim _{h \rightarrow 0}\left(\frac{1-\cos h}{h}\right)-\cos x \cdot \lim _{h \rightarrow 0}\left(\frac{1-\cos h}{h}\right) \\ =(\cos x-\sin x)-\sin x \cdot \lim _{h \rightarrow 0} \frac{2 \sin ^2 \frac{h}{2}}{h \times \frac{h}{4}} \times \frac{h}{4}-\cos x \cdot \lim _{h \rightarrow 0} \frac{2 \sin ^2 \frac{h}{2}}{h \times \frac{h}{4}} \times \frac{h}{4} \\ =(\cos x-\sin x)-\sin x \cdot 2 \cdot \frac{1}{4} \lim _{\frac{h}{2} \rightarrow 0}\left(\frac{\sin \frac{h}{2}}{\frac{h}{2}}\right)^2 \times h-\cos x \cdot 2 \cdot \frac{1}{4} \lim _{\frac{h}{2} \rightarrow 0}\left(\frac{\sin \frac{h}{2}}{\frac{h}{2}}\right)^2 h \\ =(\cos x-\sin x)-\frac{1}{2} \cdot \sin x \cdot(1) \times 0-\cos x \cdot \frac{1}{2} \cdot(1) \times 0\left[\because \lim _{x \rightarrow 0} \frac{\sin x}{x}=1\right]\end{array}$
= (cos x - sin x) - 0 - 0
= cos x - sin x

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