2 Marks Questions

Question 12 Marks

Convert the decimal number to the binary number: 250

Answer

Given decimal number is 250

Put all the reminders together in reverse order.
The required binary number is 11111010

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Question 22 Marks

Differentiate the function with respect to $\mathrm{x}: \log \left\{\mathrm{x}+2+\sqrt{x^{2}+4 x+1}\right\}$

Answer

Let $\mathrm{y}=\log \left\{\mathrm{x}+2+\sqrt{x^{2}+4 x+1}\right\}$
Differentiate both side with respect to x we get,
$\frac{d y}{d x}=\frac{d}{d x} \log \left[\mathrm{x}+2+\sqrt{x^{2}+4 x+1}\right]$
$=\frac{1}{\left[x+2+\sqrt{x^{4}+4 x+1}\right]} \frac{d}{d x}\left[\mathrm{x}+2+\left(x^{2}+4 x+1\right)^{\frac{1}{2}}\right]$ [using chain rule]
$=\frac{1}{\left[x+2+\sqrt{x^{4}+4 x+1}\right]} \times\left[1+0+\frac{1}{2}\left(x^{2}+4 x+1\right)^{-\frac{1}{2}} \frac{d}{d x}\left(\mathrm{x}^{2}+4 \mathrm{x}+1\right)\right]$
$=\frac{1+\frac{2 x+4}{2\left(\sqrt{x^{2}+4 x+1}\right)}}{\left[x+2+\sqrt{x^{4}+4 x+1}\right]}$
$=\frac{\sqrt{x^{2}+4 x+1}+x+2}{\left[x+2+\sqrt{x^{2}+4 x+1}\right] \times \sqrt{x^{2}+4 x+1}}$
$=\frac{1}{\sqrt{x^{2}+4 x+1}}$
So, $\frac{d}{d x} \log \left[\mathrm{x}+2+\sqrt{x^{2}+4 x+1}\right]=\frac{1}{\sqrt{x^{2}+4 x+1}}$

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Question 32 Marks

Find the derivative of the given function: $\frac{a x+b}{c x+d}$

Answer

Let us assume, $\mathrm{f}(\mathrm{x})=\frac{a x+b}{c x+d}$
Let $\mathrm{u}=\mathrm{ax}+\mathrm{b}$ and $\mathrm{v}=\mathrm{cx}+\mathrm{d}$
$\therefore \mathrm{f}(\mathrm{x})=\frac{u}{v}$
So, $\mathrm{f}^{\prime}(\mathrm{x})=\left(\frac{u}{v}\right)^{\prime}$
$\mathrm{f}^{\prime}(\mathrm{x})=\frac{u^{\prime} v-v^{\prime} u}{v^{2}}$
Finding $u$ and $v$
$\mathrm{u}=\mathrm{ax}+\mathrm{b}$
$u^{\prime}=a+0$
= a
$\mathrm{v}=\mathrm{cx}+\mathrm{d}$
$\mathrm{v}^{\prime}=\mathrm{c}+0$
= c
$\mathrm{f}^{\prime}(\mathrm{x})=\left(\frac{u}{v}\right)^{\prime}$
$=\frac{u^{\prime} v-v^{\prime} u}{v^{2}}$
$=\frac{a(c x+d)-c(a x+b)}{(c x+d)^{2}}$
$=\frac{a c x+a d-a c x-c b}{(c x+d)^{2}}$
$=\frac{a c x+a d-a c x-c b}{(c x+d)^{2}}$
$=\frac{a d-c b}{(c x+d)^{2}}$
Hence, $\mathrm{f}^{\prime}(\mathrm{x})=\frac{a d-c b}{(c x+d)^{2}}$

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Question 42 Marks

A can do a piece of work in 80 days. He works at it for 10 days and then $B$ alone finishes the remaining work in 42 days. In how much time will $A$ and $B$, working together, finish the work?

Answer

Work done by $A$ in 10 days $=\frac{1}{80} \times 10=\frac{1}{8}$
Remaining work $=\left(1-\frac{1}{8}\right)=\frac{7}{8}$
Now, $\frac{7}{8}$ work is done by B in 42 days
Whole work will be done by B in $\left(42 \times \frac{8}{7}\right)=48$ days
$\therefore$ A's 1 day's work $=\frac{1}{80}$ and B's 1 days' work $=\frac{1}{48}$
$\therefore(\mathrm{A}+\mathrm{B})$ 's 1 days' work $=\frac{1}{80}+\frac{1}{48}=\frac{8}{240}=\frac{1}{30}$
Hence, both will finish the work in 30 days.

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Question 52 Marks

If $A=\{x: x=2 n, n \in Z\}$ and $B=\{x: x=3 n, n \in Z\}$, then find $A \cap B$.

Answer

$A \cap B=\{x: x=2 n, n \in Z\} \cap\{x: x=3 n, n \in Z\}$
$=\{\ldots-6,-4,-2,0,2,4,6 \ldots\} \cap\{\ldots,-9-6,-3,0,3,6,9, \ldots\}$
$=\{\ldots,-6,0,6,12, \ldots\}$
$=\{x: x=6 n, n \in Z\}$

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Question 62 Marks

In a survey of 425 students in a school, it was found that 115 drink apple juice, 160 drink orange juice and 80 drink both apple as well as orange juice. How many drink neither apple juice nor orange juice?

Answer

Let $\mathrm{U}=$ set of all students surveyed;
$A=$ set of all students who drink apple juice and $B=$ set of all students who drink orange juice.
Then, we know that $n(U)=425, n(A)=115, n(B)=160$ and $n(A \cap B)=80$.
$\therefore$ We have, $\mathrm{n}(\mathrm{A} \cup \mathrm{B})=\mathrm{n}(\mathrm{A})+\mathrm{n}(\mathrm{B})-\mathrm{n}(\mathrm{A} \cap \mathrm{B})=(115+160-80)=195$.
Set of students who drink neither apple juice nor orange juice $=\left(A^{\prime} \cap B^{\prime}\right)=(A \cup B)^{\prime}$
$\Rightarrow \mathrm{n}\left\{(\mathrm{A} \cup \mathrm{B})^{\prime}\right\}=\mathrm{n}(\mathrm{U})-\mathrm{n}(\mathrm{A} \cup \mathrm{B})=(425-195)=230$.
Therefore, 230 students drink neither apple juice nor orange juice.

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Question 72 Marks

In an examination the average of 150 students is 25 . The average of passed students is 40 and average of failed students is 15 . Find the number of students who passed the examination.

Answer

Average of 150 students $=25$
$\therefore$ Total of 150 students $=150 \times 25=3750 \quad\ldots(ⅰ)$
Let number of passed students be x
$\therefore$ Average of x students $=40$
$\therefore$ Total of x students $=40 \mathrm{x} \quad\ldots(ⅰⅰ)$
Average of $(150-x)$ students $=15$
$\therefore$ Total of $(150-x)$ students $=(150-x) \times 15 \quad\ldots(ⅰⅰⅰ)$
From ( 0 , (ii) and (iii), we get
$40 x+(150-x) 15=3750$
$\Rightarrow 40 \mathrm{x}+2250-15 \mathrm{x}=3750 \Rightarrow 25 \mathrm{x}=1500 \Rightarrow \mathrm{x}=60$
$\therefore$ Passed students $=60$

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Applied Maths 2 Marks Questions

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