5 Marks Questions

Question 15 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

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.
$\begin{array}{l}\therefore n(E)={ }^6 C_1 \times{ }^4 C_2 \\ \therefore P(E)=\frac{{ }^6 C_1 \times{ }^4 C_2}{{ }^{18} C_3}=\frac{6 \times 4 \times 3}{2} \times \frac{3 \times 2}{18 \times 17 \times 16} \\ P(E)=\frac{3}{68}\end{array}$
ii. Let E be the event that two blue balls and one red ball was drawn.
$\begin{array}{l}\therefore n(E)={ }^8 C_2 \times{ }^6 C_1 \\ \therefore P(E)=\frac{{ }^8 C_2 \times{ }^6 C_1}{{ }^{18} C_3}=\frac{8 \times 7}{2} \times 6 \times \frac{3 \times 2 \times 1}{18 \times 17 \times 16}=\frac{7}{34} \\ P(E)=\frac{7}{34}\end{array}$
iii. Let E be the event that one of the ball must be red.
$\begin{array}{l}\therefore E=\{(R, W, B) \text { or }(R, W, W) \text { or }(R, B, B)\} \\
\therefore n(E)={ }^6 C_1 \times{ }^4 C_1 \times{ }^8 C_1+{ }^6 C_1 \times{ }^4 C_2+{ }^6 C_1 \times{ }^8 C_2 \\
\therefore P(E)=\frac{{ }^6 C_1 \times{ }^4 C_1 \times{ }^8 C_1+{ }^6 C_1 \times{ }^4 C_2+{ }^6 C_1 \times{ }^8 C_2}{{ }^{18} C_3}=\frac{6 \times 4 \times 8+\frac{6 \times 4 \times 3}{2 \times 1}+\frac{6 \times 8 \times 7}{2 \times 1}}{\frac{18 \times 17 \times 16}{3 \times 2 \times 1}} \\
=\frac{396}{816}=\frac{33}{68}\end{array}$

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Question 25 Marks

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

Answer

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

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Question 35 Marks

If $\cos x=-\frac{3}{5}$ and $x$ lies in the IIIrd quadrant, find the values of $\cos \frac{x}{2}, \sin \frac{x}{2}$ and $\sin 2 x$.

Answer

We have to find the values of $\cos \frac{x}{2}, \sin \frac{x}{2}$ and $\sin 2 x$.
It is given that $\cos x=-\frac{3}{5}$ and $x$ lies in the IIIrd quadrant
We know,
$
\begin{array}{l}
\cos 2 x=2 \cos ^2 x-1 \\
\cos x=2 \cos ^2 \frac{x}{2}-1 \\
-\frac{3}{5}=2 \cos ^2 \frac{x}{2}-1 \ldots\left[\because \cos x=-\frac{3}{5}\right] \\
2 \cos ^2 \frac{x}{2}=-\frac{3}{5}+1 \\
2 \cos ^2 \frac{x}{2}=\frac{2}{5} \\
\cos ^2 \frac{x}{2}=\frac{1}{5} \\
\cos \frac{x}{2}= \pm \frac{1}{\sqrt{5}}
\end{array}
$
Since,
$
x \in\left(\pi, \frac{3 \pi}{2}\right) \Rightarrow \frac{x}{2} \in\left(\frac{\pi}{2}, \frac{3 \pi}{4}\right)
$$\cos \frac{x}{2}$ will be negative in 3rd quadrant
So,
$
\cos x=-\frac{1}{\sqrt{5}}
$
We know,
$
\begin{array}{l}
\cos 2 x=1-2 \sin ^2 x \\
\cos x=1-2 \sin ^2 \frac{x}{2} \ldots\left[\because \cos x=-\frac{3}{5}\right] \\
-\frac{3}{5}=1-2 \sin ^2 \frac{x}{2} \\
2 \sin ^2 \frac{x}{2}=\frac{3}{5}+1 \\
2 \sin ^2 \frac{x}{2}=\frac{8}{5}
\end{array}
$
$
\begin{array}{l}
\sin ^2 \frac{x}{2}=\frac{4}{5} \\
\sin \frac{x}{2}= \pm \frac{2}{\sqrt{5}}
\end{array}
$
Since,
$
x \in\left(\pi, \frac{3 \pi}{2}\right) \Rightarrow \frac{\pi}{2} \in\left(\frac{\pi}{2}, \frac{3 \pi}{4}\right)
$
$\sin \frac{x}{2}$ will be positive in 2nd quadrant
So,
$
\sin \frac{x}{2}=\frac{2}{\sqrt{5}}
$
We know,
$
\begin{array}{l}
\sin ^2 x+\cos ^2 x=1 \\
\sin ^2 x=1-\cos ^2 x \\
\sin ^2 x=1-\left(-\frac{3}{5}\right)^2 \ldots\left[\because \cos x=-\frac{3}{5}\right] \\
\sin ^2 x=1-\frac{9}{25} \\
\sin ^2 x=\frac{25-9}{25} \\
\sin ^2 x=\frac{16}{25} \\
\sin x= \pm \frac{4}{5}
\end{array}
$
Since,
$
x \in\left(\pi, \frac{3 \pi}{2}\right)
$
$\sin x$ will be negative in 3rd quadrant
So,
$
\sin x=-\frac{4}{5}
$
Now,
$
\begin{array}{l}
\sin 2 x=2(\sin x)(\cos x) \ldots\left[\because \cos x=-\frac{3}{5} \text { and } \sin x=-\frac{4}{5}\right] \\
\sin 2 x=2 \times-\frac{4}{5} \times-\frac{3}{5} \\
\sin 2 x=\frac{24}{25}
\end{array}
$
Hence, values of $\cos \frac{x}{2}, \sin \frac{x}{2}, \sin 2 x$ are $-\frac{1}{\sqrt{5}}, \frac{2}{\sqrt{5}}$ and $\frac{24}{25}$

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Question 45 Marks

Find the three numbers in GP, whose sum is 52 and sum of whose product in pairs is 624.

Answer

Let the three numbers in GP be $\frac{a}{r}$, a, ar.
Sum of three numbers = 52 [given]
$\begin{array}{l}\Rightarrow \frac{a}{r}+ a + ar =52 \\ \Rightarrow a \left(\frac{1}{r}+1+r\right)=52 \ldots \text { (i) }\end{array}$
And sum of product in pair = 624
$\begin{array}{l}\Rightarrow \frac{a}{r} \times a + a \times ar +\frac{a}{r} \times ar =624 \\ \Rightarrow a ^2\left(\frac{1}{r}+r+1\right)=624 \ldots \text { (ii) }\end{array}$
On dividing Eqs. (ii) by (i), we get
$a =\frac{624}{52} \Rightarrow a =12$
On putting a = 12 in Eq. (i), we get
$12\left(\frac{1}{r}+r+1\right)=52$
$\begin{array}{l}\Rightarrow \frac{r^2+r+1}{r}=\frac{52}{12} \Rightarrow \frac{r^2+r+1}{r}=\frac{13}{3} \\ \Rightarrow 3 r^2+3 r+3=13 r \\ \Rightarrow 3 r^2-10 r+3=0 \\ \Rightarrow(3 r-1)( r -3)=0 \\ \Rightarrow r =\frac{1}{3} \text { or } r =3\end{array}$
When $r=\frac{1}{3}$, then numers are $\frac{12}{\frac{1}{3}}, 12,12 \times \frac{1}{3}$ i.e., $36,12,4$.
When $r=3$, then numbers are $\frac{12}{3}, 12,12 \times \frac{1}{3}$.i.e, $4,12,36$.

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Question 55 Marks

Find the derivative of x sinx from first principle.

Answer

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

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Question 65 Marks

Evaluate : $\lim _{x \rightarrow \sqrt{10}} \frac{\sqrt{7-2 x}-(\sqrt{5}-\sqrt{2})}{x^2-10}$

Answer

We have,
$\begin{array}{l}\lim _{x \rightarrow \sqrt{10}} \frac{\sqrt{7-2 x}-(\sqrt{5}-\sqrt{2})}{x^2-10} \\ =\lim _{x \rightarrow \sqrt{10}} \frac{\sqrt{7-2 x}-\sqrt{(\sqrt{5}-\sqrt{2})^2}}{x^2-10}\left(\text { form } \frac{0}{0}\right) \\ =\lim _{x \rightarrow \sqrt{10}} \frac{\sqrt{7-2 x}-\sqrt{7-2 \sqrt{10}}}{x^2-10}\left(\text { form } \frac{0}{0}\right) \\ =\lim _{x \rightarrow \sqrt{10}} \frac{\sqrt{7-2 x}-\sqrt{7-2 \sqrt{10}}}{x^2-10} \times \frac{\sqrt{7-2 x}+\sqrt{7-2 \sqrt{10}}}{\sqrt{7-2 x}+\sqrt{7-2 \sqrt{10}}} \\ =\lim _{x \rightarrow \sqrt{10}} \frac{(7-2 x)-(7-2 \sqrt{10})}{(x-\sqrt{10})(x+\sqrt{10})\{\sqrt{7-2 x}+\sqrt{7-2} \sqrt{10}\}} \\ =\lim _{x \rightarrow \sqrt{10}} \frac{-2 x+2 \sqrt{10}}{(x-\sqrt{10})(x+\sqrt{10})\{\sqrt{7-2 x}+\sqrt{7-2} \sqrt{10}\}}\end{array}$
$\begin{array}{l}=\lim _{x \rightarrow \sqrt{10}} \frac{-2(x-\sqrt{10})}{(x-\sqrt{10})(x+\sqrt{10})\{\sqrt{7-2 x}+\sqrt{7-2 \sqrt{10}}\}} \\ =\lim _{x \rightarrow \sqrt{10}} \frac{-2}{(x+\sqrt{10})\{\sqrt{7-2 x}+\sqrt{7-2 \sqrt{10}}\}} \\ =\lim _{x \rightarrow \sqrt{10}} \frac{-2}{2 \sqrt{10}\{\sqrt{7-2 \sqrt{10}}+\sqrt{7-2 \sqrt{10}}\}} \\ =\frac{-1}{\sqrt{10} \times 2 \times \sqrt{7-2 \sqrt{10}}}=\frac{-1}{2 \sqrt{10}(\sqrt{5}-\sqrt{2})}\left[\because(\sqrt{5}-\sqrt{2})^2=7-2 \sqrt{10}\right] \\ =\frac{-1}{2 \sqrt{10}} \times \frac{(\sqrt{5}+\sqrt{2})}{3}=-\frac{(\sqrt{5}+\sqrt{2})}{6 \sqrt{10}}\end{array}$

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