5 Marks Questions

Question 15 Marks

Prove the following identity: $\cos ^3 2 x+3 \cos 2 x=4\left(\cos ^6 x-\sin ^6 x\right)$.

Answer

We have to prove that $\cos ^3 2 x+3 \cos 2 x=4\left(\cos ^6 x-\sin ^6 x\right)$
Let us consider $\text{RHS} =4\left(\cos ^6 x-\sin ^6 x\right)$
$=4\left(\left(\cos ^2 x\right)^3-\left(\sin ^2 x\right)^3\right)$
$=4\left(\cos ^2 x-\sin ^2 x\right)\left(\cos ^4 x+\sin ^4 x+\cos ^2 x \sin ^2 x\right) \ldots\left\{\because a^3-b^3=(a-b)\left(a^2+b^2+a b\right)\right\}$
$=4 \cos 2 x\left(\cos ^4 x+\sin ^4 x+\cos ^2 x \sin ^2 x+\cos ^2 x \sin ^2 x-\cos ^2 x \sin ^2 x\right) \ldots\left\{\because \cos 2 x=\cos ^2 x-\sin ^2 x\right\}$
$=4 \cos 2 x\left(\cos ^4 x+\sin ^4 x+2 \cos ^2 x \sin ^2 x-\cos ^2 x \sin ^2 x\right)$
$\left.=4 \cos 2 x\left\{\left(\cos ^2 x\right)^2+\left(\sin ^2 x\right)^2+2 \cos ^2 x \sin ^2 x-\cos ^2 x \sin ^2 x\right)\right\} \ldots\left\{\because a^2+b^2+2 a b=(a+b)^2\right\}$
$\left.=4 \cos 2 x\left\{\left(\cos ^2 x+\sin ^2 x\right)^2-\cos ^2 x \sin ^2 x\right)\right\} \ldots\left\{\because \cos ^2 x+\sin ^2 x=1\right\}$
$=4 \cos 2 x\left\{(1)^2-\frac{1}{4}\left(4 \cos ^2 x \sin ^2 x\right)\right\}$
$=4 \cos 2 x\left\{(1)^2-\frac{1}{4}(2 \cos x \sin x)^2\right\} \ldots\{\because \sin 2 x=2 \sin x \cos x\}$
$=4 \cos 2 x\left\{(1)^2-\frac{1}{4}(2 \sin 2 x)^2\right\}$
$=4 \cos 2 x\left(1-\frac{1}{4} \sin ^2 2 x\right) \ldots\left\{\because \sin ^2 x=1-\cos ^2 x\right\}$
$=4 \cos 2 x\left(1-\frac{1}{4}\left(1-\cos ^2 2 x\right)\right)$
$\left.=4 \cos 2 x\left(1-\frac{1}{4}+\frac{1}{4} \cos ^2 2 x\right)\right)$
$\left.=4 \cos 2 x\left(\frac{3}{4}+\frac{1}{4} \cos ^2 2 x\right)\right)$
$\left.=4\left(\frac{3}{4} \cos 2 x+\frac{1}{4} \cos ^3 2 x\right)\right)$
$=3 \cos 2 x+\cos ^3 2 x$
$\text{RHS = LHS}$
Hence Proved.

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Question 25 Marks

Prove that: $\tan 20^{\circ} \tan 30^{\circ} \tan 40^{\circ} \tan 80^{\circ}=1$

Answer

$LHS =\tan 20^{\circ} \tan 30^{\circ} \tan 40^{\circ} \tan 80^{\circ}$
$=\frac{1}{\sqrt{3}}\left(\tan 20^{\circ} \tan 40^{\circ} \tan 80^{\circ}\right) \quad\left[\because \tan 30^{\circ}=\frac{1}{\sqrt{3}}\right]$
$=\frac{\left(\sin 20^{\circ} \sin 40^{\circ} \sin 80^{\circ}\right)}{\left(\cos 20^{\circ} \cos 40^{\circ} \cos 80^{\circ}\right) \sqrt{3}}$
$=\frac{\left(2 \sin 20^{\circ} \sin 40^{\circ}\right) \sin 80^{\circ}}{\sqrt{3}\left(2 \cos 20^{\circ} \cos 40^{\circ}\right) \cos 80^{\circ}}$
Applying
$\Rightarrow 2 \sin A \sin B =\cos ( A - B )-\cos ( A + B ) \text { and } 2 \cos A \cos B =\cos ( A + B )+\cos ( A - B ), \text { we get }$
$=\frac{\left[\cos \left(40^{\circ}-20^{\circ}\right)-\cos \left(20^{\circ}+40^{\circ}\right)\right] \sin 80^{\circ}}{\left[\cos \left(20^{\circ}+40^{\circ}\right)+\cos \left(40^{\circ}-20^{\circ}\right)\right] \cos 80^{\circ} \sqrt{3}}$
$=\frac{\left(\cos 20^{\circ}-\cos 60^{\circ}\right) \sin 80^{\circ}}{\sqrt{3}\left(\cos 60^{\circ}+\cos 20^{\circ}\right) \cos 80^{\circ}}$
$=\frac{\left(\cos 20^{\circ}-\frac{1}{2}\right) \sin 80^{\circ}}{\sqrt{3}\left(\frac{1}{2}+\cos 20^{\circ}\right) \cos 80^{\circ}}$
$=\frac{2 \cos 20^{\circ} \sin 80^{\circ}-\sin 80^{\circ}}{\sqrt{3}\left(\cos 80^{\circ}+2 \cos 20^{\circ} \cos 80^{\circ}\right)}$
Now,
$\Rightarrow 2 \sin A \cos B =\sin ( A + B )+\sin ( A - B )$
$=\frac{\sin \left(80^{\circ}+20^{\circ}\right)+\sin \left(80^{\circ}-20^{\circ}\right)-\sin 80^{\circ}}{\sqrt{3}\left[\cos 80^{\circ}+\cos \left(20^{\circ}+80^{\circ}\right)+\cos \left(80^{\circ}-20^{\circ}\right)\right]}$
$=\frac{\sin 100^{\circ}+\sin 60^{\circ}-\sin 80^{\circ}}{\sqrt{3}\left(\cos 80^{\circ}+\cos 100^{\circ}+\cos 60^{\circ}\right)}$
$=\frac{\sin 100^{\circ}+\sin 60^{\circ}-\sin \left(180^{\circ}-100^{\circ}\right)}{\sqrt{3}\left(\cos 80^{\circ}+\cos \left(180^{\circ}-80^{\circ}\right)+\cos 60^{\circ}\right)}$
$=\frac{\sin 100^{\circ}+\frac{\sqrt{3}}{2}-\sin 100^{\circ}}{\sqrt{3}\left(\cos 80^{\circ}-\cos 80^{\circ}+\cos 60^{\circ}\right)}$
$=\frac{\frac{\sqrt{3}}{2}}{\sqrt{3}\left(\frac{1}{2}\right)}=1=\text { RHS }$

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Question 35 Marks

Find the sum of the following series up to n terms:
$i. 5 + 55 + 555 + ……$
$ii. 6 + .66 + .666 + …..$

Answer

$i. S _{ n }=5+55+555+\ldots \ldots \ldots .$. up to $n$ terms
$=5[1+11+111+\ldots \ldots . . \text { up to } n \text { terms }]$
$=\frac{5}{9}[9+99+999+\ldots \ldots . \text { up ton terms }]$
$=\frac{5}{9}\left[(10-1)+\left(10^2-1\right)+\left(10^3-1\right)+\ldots \ldots . \text { up to } n \text { terms }\right]$
$=\frac{5}{9}\left[\frac{10\left(10^n-1\right)}{10-1}-n\right]$
$=\frac{5}{9}\left[\frac{10}{9}\left(10^n-1\right)-n\right]$
$=\frac{50}{81}\left(10^n-1\right)-\frac{5}{9} n$
$ii. S _{ n }=.6+.66+.666+........$ up to n terms
$=6[.1+.11+.111+\ldots \ldots \ldots . \text { up to } n \text { terms }]$
$=\frac{6}{9}[.9+.99+.999+\ldots \ldots \text { up to } n \text { terms }]$
$=\frac{6}{9}\left[\frac{9}{10}+\frac{99}{100}+\frac{999}{1000}+\ldots \ldots . . \text { up to } n \text { terms }\right]$
$=\frac{6}{9}\left[\left(1-\frac{1}{10}\right)+\left(1-\frac{1}{10^2}\right)+\left(1-\frac{1}{10^3}\right) \ldots \ldots . \text { up to } n \text { terms }\right]$
$=\frac{6}{9}\left[n-\left(\frac{1}{10}+\frac{1}{10^2}+\frac{1}{10^3}+\ldots \ldots \ldots \text { up to } n \text { terms }\right)\right]$
$=\frac{2}{3}\left[n-\frac{\frac{1}{10}\left(1-\frac{1}{10^2}\right)}{1-\frac{1}{10}}\right]$
$=\frac{2}{3}\left[n-\frac{1}{9}\left(1-\frac{1}{10^2}\right)\right]$
$=\frac{2 n}{3}-\frac{2}{27}\left(1-\frac{1}{10^2}\right)$

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Question 45 Marks

Differentiate $\frac{\cos x}{x}$ from first principle.

Answer

We have to find the derivative of $f(x)=\frac{\omega s x}{x}$
Derivative of a function $f ( x )$ is given by $f ^{\prime}( x )=\underset{{h \rightarrow 0}}{\lim}=\frac{f(x+h)-f(x)}{h}$ (where $h$ is a very small positive number)
$\therefore$ Derivative of $f ( x )=\frac{\cos x}{x}$ is given as $f ^{\prime}( x )=\underset{{h \rightarrow 0}}{\lim}=\frac{f(x+h)-f(x)}{h}$
$\Rightarrow f ( x )=\underset{{h \rightarrow 0}}{\lim} \frac{\frac{\frac{\cos (x+h)}{z+h}-\frac{\cos z}{z}}{h}}{h}$
$\Rightarrow f ( x )=\lim _{ h \rightarrow 0} \frac{\frac{z \cos (x+h)-(x+h) \cos x}{z(x+h)}}{h}=\underset{{h \rightarrow 0}}{\lim} \frac{x \cos (x+h)-(x+h) \cos x}{h(x)(x+h)}$
Using the algebra of limits we have
$\Rightarrow f ( x )=\lim _{ h \rightarrow 0} \frac{x \cos (x+h)-(x+h) \cos x}{h} \times \underset{{h \rightarrow 0}}{\lim} \frac{1}{x(x+h)}$
$\Rightarrow f ( x )=\lim _{ h \rightarrow 0} \frac{x \cos (x+h)-(x+h) \cos x}{h} \times \frac{1}{x(x+0)}$
$\Rightarrow f ( x )=\frac{1}{ x ^2} \lim _{ h \rightarrow 0} \frac{x \cos (x+h)-(x+h) \cos x}{h}$
$\Rightarrow f ( x )=\frac{1}{ x ^2} \lim _{ h \rightarrow 0} \frac{x \cos (x+h)-x \cos x-h \cos x}{h}$
Using the algebra of limits, we have:
$\Rightarrow f ( x )=\frac{1}{ x ^2}\left\{\underset{{h \rightarrow 0}}{\lim} \frac{-h \cos x}{h}+\underset{{h \rightarrow 0}}{\lim} \frac{x \cos (x+h)-x \cos x}{h}\right\}$
$\Rightarrow f ( x )=\frac{1}{ x ^2}\left\{-\underset{{h \rightarrow 0}}{\lim} \cos x+\underset{{h \rightarrow 0}}{\lim} \frac{x(\cos (x+h)-\cos x)}{h}\right\}$
Using the algebra of limits we have:
$\therefore f ^{\prime}( x )=-\frac{\cos x}{x^2}+\frac{1}{x} \underset{{h \rightarrow 0}}{\lim} \frac{\cos (x+h)-\cos x}{h}$
We can't evaluate the limits at this stage only as on putting value it will take $\frac{0}{0}$ form.
So, we need to do little modifications.
Use: $\cos A -\cos B =-2 \sin \left(\frac{(A+B)}{2}\right) \sin \left(\frac{(A-B)}{2}\right)$
$\therefore f^{\prime}(x)=-\frac{\cos x}{x^2}+\frac{1}{x} \underset{{h \rightarrow 0}}{\lim} \frac{-2 \sin \left(\frac{2 x+h}{2}\right) \sin \left(\frac{h}{2}\right)}{h}$
$\Rightarrow f^{\prime}(x)=-\frac{\cos x}{x^2}+\frac{1}{x} \underset{{h \rightarrow 0}}{\lim} \frac{\sin \left(x+\frac{h}{2}\right) \sin \left(\frac{h}{2}\right)}{\frac{h}{2}}$
Using algebra of limits:
$\Rightarrow f^{\prime}(x)=-\frac{\cos x}{x^2}+\frac{1}{x} \underset{{h \rightarrow 0}}{\lim} \frac{\sin \left(\frac{h}{2}\right)}{\frac{h}{2}} \times \underset{{h \rightarrow 0}}{\lim} \sin \left(x+\frac{h}{2}\right)$
By using the formula we get : $\lim _{x \rightarrow 0} \frac{\sin x}{x}=1$
$\Rightarrow f^{\prime}(x)=-\frac{\cos x}{x^2}+\frac{1}{x} \underset{{h \rightarrow 0}}{\lim} \sin \left(x+\frac{h}{2}\right)$
Put the value of $h$ to evaluate the limit:
$\therefore f^{\prime}(x)=-\frac{\cos x}{x^2}+\frac{1}{x} \times \sin (x+0)=-\frac{\cos x}{x^2}-\frac{\sin x}{x}$
Hence,
Derivative of $f(x)=(\cos x) / x$ is $-\frac{\cos x}{x^2}-\frac{\sin x}{x}$

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Question 55 Marks

Differentiate If $y=\sqrt{\frac{\sec x-\tan x}{\sec x+\tan x}}$ show that $\frac{d y}{d x}=\sec x(\tan x+\sec x)$

Answer

We have to show that $\frac{d y}{d x}=(\sec x \tan x+\sec x)$
where, it is given that
$ y =\sqrt{\frac{\sec x-\tan x}{\sec x+\tan x}}$
$y =\sqrt{\frac{\frac{1}{\cos x} \frac{\sin z}{1}}{\cos x}+\frac{\sin x}{\cos x}}=\sqrt{\frac{1-\sin x}{1+\sin x}}$
$u =1-\sin x , v =1+\sin x , x=\frac{1-\sin x}{1+\sin x}$
if $z=\frac{u}{v}$
$\frac{d z}{d x}=\frac{v \times \frac{d a}{d x}-u \times \frac{d v}{d x}}{v^2}$
$=\frac{(1+\sin x) \times(-\cos x)-(1-\sin x) \times(\cos x)}{(1+\sin x)^2}$
$=\frac{-\cos x-\sin x \cos x-\cos x+\sin x \cos x}{(1+\sin x)^2}$
$=\frac{-2 \cos x}{(1+\sin x)^2}$
According to the chain rule of differentiation
$\frac{d y}{d x}=\frac{d y}{d x} \times \frac{d x}{d x}$
$=\left[-\frac{\cos x}{1} \times\left(\frac{1-\sin x}{1}\right)^{-\frac{1}{2}}\right] \times\left[\frac{1}{(1+\sin x)^{2-\frac{1}{2}}}\right]$
$=\left[\cos x \times(1+\sin x)^{-\frac{1}{2}}\right] \times(1-\sin x)^{-\frac{3}{2}} \times\left(\frac{1+\sin x}{1+\sin x}\right)^{\frac{3}{2}}$
Multiplying and dividing by $(1+\sin x)^{\frac{3}{2}}$
$=\left[\cos x \times(1+\sin x)^{\frac{2}{2}-\frac{1}{2}}\right] \times(1-\sin x)^{-\frac{2}{2}} \times\left(\frac{1}{1+\sin x}\right)^{\frac{3}{2}}$
$=\left[\cos x \times(1+\sin x)^{\frac{2}{2}-\frac{1}{2}}\right] \times(1-\sin x)^{-\frac{2}{2}} \times(1+\sin x)^{-\frac{2}{2}}$
$=\left[\cos x \times(1+\sin x)^1\right] \times\left(1-\sin ^2 x\right)^{-\frac{3}{2}}$
$=\left[\cos x \times(1+\sin x)^1\right] \times\left(\cos ^2 x\right)^{-\frac{3}{2}}$
$=\left[\cos x \times(1+\sin x)^1\right] \times\left(\cos ^2 x\right)^{-3}$
$=\left[(1+\sin x)^1\right] \times(\cos x)^{-3+1}$
$=\frac{1+\sin x}{\cos ^2 x}$
$=\frac{1}{\cos ^2 x} \times \frac{1+\sin x}{\cos ^2 x}$
$=\sec x\left(\frac{1}{\cos x}+\frac{\sin x}{\cos x}\right)$
$=\sec x(\sec x+\tan x)$
Hence proved

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Question 65 Marks

A bag contains $6$ red, $4$ white and $8$ blue balls. If three balls are drawn at random, find the probability that:
$i.$ one is red and two are white
$ii.$ two are blue and one is red
$iii.$ one is red.

Answer

Bag contains:
$6-$Red balls
$4-$White balls
$8-$Blue balls
Since three ball are drawn,
$\therefore n(S)={ }^{18} C_3$
$i.$ Let $E$ be the event that one red and two white balls are drawn.
$\therefore n(E)={ }^6 C_1 \times{ }^4 C_2$
$\therefore P(E)=\frac{{ }^6 C_1 \times{ }^4 C_2}{{ }^{18} C_3}=\frac{6 \times 4 \times 3}{2} \times \frac{3 \times 2}{18 \times 17 \times 16}$
$P(E)=\frac{3}{68}$
$ii.$ Let $E$ be the event that two blue balls and one red ball was drawn.
$\therefore n(E)={ }^8 C_2 \times{ }^6 C_1$
$\therefore P(E)=\frac{{ }^8 C_2 \times{ }^6 C_1}{{ }^{18} C_3}=\frac{8 \times 7}{2} \times 6 \times \frac{3 \times 2 \times 1}{18 \times 17 \times 16}=\frac{7}{34}$
$P(E)=\frac{7}{34}$
$iii.$ Let $E$ be the event that one of the ball must be red.
$\therefore E =\{( R , W , B )$ or $( R , W , W )$ or $( R , B , B )\}$
$\therefore n(E)={ }^6 C_1 \times{ }^4 C_1 \times{ }^8 C_1+{ }^6 C_1 \times{ }^4 C_2+{ }^6 C_1 \times{ }^8 C_2$
$\therefore P(E)=\frac{6 \times{ }^6 C_1 \times{ }^4 C_1 \times{ }^8 C_1+{ }^5 C_1 \times{ }^4 C_2+{ }^6 C_1 \times{ }^8 C_2}{{ }^{18} C_3}$
$=\frac{6 \times 4 \times 8+\frac{6 \times 4 \times 3}{2 \times 1}+\frac{6 \times 8 \times 7}{2 \times 1}}{\frac{18 \times 17 \times 16}{3 \times 2 \times 1}}$
$=\frac{396}{816}$
$=\frac{33}{68}$

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