Question 12 Marks
Convert the decimal number 13 to the equivalent binary number.
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Question 22 Marks
If $\mathrm{y}=\mathrm{ae}^{2 \mathrm{x}}+\mathrm{be}^{-\mathrm{x}}$, prove that $\frac{d^{2} y}{d x^{2}}-\frac{d y}{d x}-2 \mathrm{y}=0$.
Answer
View full question & answer→Given: $\mathrm{y}=\mathrm{ae}^{2 \mathrm{x}}+\mathrm{be} \mathrm{e}^{-\mathrm{x}} . .$.. (1)
Differentiate equation (1), w.r.t. 'x', we get
$\begin{array}{l}\frac{d y}{d x}=\frac{d}{d x}\left(a e^{2 x}\right)+\frac{d}{d x}\left(b e^{-x}\right) \\ \frac{d y}{d x}=2 a e^{2 x}-b e ^{- x } \ldots . \text { (2) }\end{array}$
Again differentiate equation (2), w.r.t. 'x'
$\Rightarrow \frac{d^{2} y}{d x^{2}}=4 a e^{2 x}+b e^{-x}$ ....(3)
(3) - (2) $\Rightarrow$
$\Rightarrow \quad \frac{d^{2} y}{d x^{2}}-\frac{d y}{d x}=4 \mathrm{ae}^{2 \mathrm{x}}-2 \mathrm{ae}^{2 \mathrm{x}}+\mathrm{be}^{-\mathrm{x}}+\mathrm{be}^{-\mathrm{x}}$
$\Rightarrow \quad \frac{d^{2} y}{d x^{2}}-\frac{d y}{d x}=2\left(\mathrm{ae}^{2 \mathrm{x}}+\mathrm{be}^{-\mathrm{x}}\right)$
using (1)
$\Rightarrow \quad \frac{d^{2} y}{d x^{2}}-\frac{d y}{d x}-2 y=0$
Differentiate equation (1), w.r.t. 'x', we get
$\begin{array}{l}\frac{d y}{d x}=\frac{d}{d x}\left(a e^{2 x}\right)+\frac{d}{d x}\left(b e^{-x}\right) \\ \frac{d y}{d x}=2 a e^{2 x}-b e ^{- x } \ldots . \text { (2) }\end{array}$
Again differentiate equation (2), w.r.t. 'x'
$\Rightarrow \frac{d^{2} y}{d x^{2}}=4 a e^{2 x}+b e^{-x}$ ....(3)
(3) - (2) $\Rightarrow$
$\Rightarrow \quad \frac{d^{2} y}{d x^{2}}-\frac{d y}{d x}=4 \mathrm{ae}^{2 \mathrm{x}}-2 \mathrm{ae}^{2 \mathrm{x}}+\mathrm{be}^{-\mathrm{x}}+\mathrm{be}^{-\mathrm{x}}$
$\Rightarrow \quad \frac{d^{2} y}{d x^{2}}-\frac{d y}{d x}=2\left(\mathrm{ae}^{2 \mathrm{x}}+\mathrm{be}^{-\mathrm{x}}\right)$
using (1)
$\Rightarrow \quad \frac{d^{2} y}{d x^{2}}-\frac{d y}{d x}-2 y=0$
Question 32 Marks
If $\mathrm{y}=\log \{\sqrt{x-1}-\sqrt{x+1}\}$, show that $\frac{d y}{d x}=\frac{-1}{2 \sqrt{x^{2}-1}}$.
Answer
View full question & answer→Here, $\mathrm{y}=\log \{\sqrt{x-1}-\sqrt{x+1}\}$
Differentiate it with respect to x we get,
$\frac{d y}{d x}=\frac{d}{d x} \log \{\sqrt{x-1}-\sqrt{x+1}\}$
$=\frac{1}{(\sqrt{x-1}-\sqrt{x+1})} \frac{d}{d x}(\sqrt{x-1}-\sqrt{x+1})$ [Using chain rule]
$=\frac{1}{(\sqrt{x-1}-\sqrt{x+1})}\left[\frac{d}{d x} \sqrt{x-1}-\frac{d}{d x} \sqrt{x+1}\right]$
$=\frac{1}{(\sqrt{x-1}-\sqrt{x+1})}\left[\frac{1}{2}(x-1)^{\frac{-1}{2}}-\frac{1}{2}(x+1)^{\frac{-1}{2}}\right]$
$=\frac{1}{2} \frac{1}{(\sqrt{x-1}-\sqrt{x+1})}\left(\frac{1}{\sqrt{x-1}}-\frac{1}{\sqrt{x+1}}\right)$
$=\frac{1}{2} \frac{1}{(\sqrt{x-1}-\sqrt{x+1})}\left\{\frac{-(\sqrt{x-1}-\sqrt{x+1})}{(\sqrt{x-1})(\sqrt{x+1})}\right\}$
$=\frac{-1}{2}\left(\frac{1}{(\sqrt{x-1})(\sqrt{x+1})}\right)$
$=\frac{-1}{2 \sqrt{x^{2}-1}}$
So, $\frac{d y}{d x}=\frac{-1}{2 \sqrt{x^{2}-1}}$
Differentiate it with respect to x we get,
$\frac{d y}{d x}=\frac{d}{d x} \log \{\sqrt{x-1}-\sqrt{x+1}\}$
$=\frac{1}{(\sqrt{x-1}-\sqrt{x+1})} \frac{d}{d x}(\sqrt{x-1}-\sqrt{x+1})$ [Using chain rule]
$=\frac{1}{(\sqrt{x-1}-\sqrt{x+1})}\left[\frac{d}{d x} \sqrt{x-1}-\frac{d}{d x} \sqrt{x+1}\right]$
$=\frac{1}{(\sqrt{x-1}-\sqrt{x+1})}\left[\frac{1}{2}(x-1)^{\frac{-1}{2}}-\frac{1}{2}(x+1)^{\frac{-1}{2}}\right]$
$=\frac{1}{2} \frac{1}{(\sqrt{x-1}-\sqrt{x+1})}\left(\frac{1}{\sqrt{x-1}}-\frac{1}{\sqrt{x+1}}\right)$
$=\frac{1}{2} \frac{1}{(\sqrt{x-1}-\sqrt{x+1})}\left\{\frac{-(\sqrt{x-1}-\sqrt{x+1})}{(\sqrt{x-1})(\sqrt{x+1})}\right\}$
$=\frac{-1}{2}\left(\frac{1}{(\sqrt{x-1})(\sqrt{x+1})}\right)$
$=\frac{-1}{2 \sqrt{x^{2}-1}}$
So, $\frac{d y}{d x}=\frac{-1}{2 \sqrt{x^{2}-1}}$
Question 42 Marks
Workout for the day of week on the given date 13th August 1975.
Answer
View full question & answer→We know that:
In 1600 years, we have 0 odd days.
In 300 years, we have 1 odd day.
So, in 1900 years there $0+1=1$ odd day.
Now from 1901 to 1974, there are 18 leap years and 56 non-leap years.
So, the number of odd days is $18 \times 2+56 \times 1=92$ odd days or $13 \times 7+1$
$=1$ odd day (up to 31st Dec. 1974).
Since 1975 is a non-leap year, so from 1st Jan. 1975 to 13th Aug. 1975, we have
31 (Jan.) + 28 (Feb.) +31 (Mar.) +30 (Apr.) +31 (May) +30 (Jun.) +31 (Jul.) + 13 days
$=225$ days $=32 \times 7+1=1$ odd day
So, the total number of odd days $=1+1+1=3$ odd days.
$\therefore$ Day on 13th Aug. 1975 was Wednesday
In 1600 years, we have 0 odd days.
In 300 years, we have 1 odd day.
So, in 1900 years there $0+1=1$ odd day.
Now from 1901 to 1974, there are 18 leap years and 56 non-leap years.
So, the number of odd days is $18 \times 2+56 \times 1=92$ odd days or $13 \times 7+1$
$=1$ odd day (up to 31st Dec. 1974).
Since 1975 is a non-leap year, so from 1st Jan. 1975 to 13th Aug. 1975, we have
31 (Jan.) + 28 (Feb.) +31 (Mar.) +30 (Apr.) +31 (May) +30 (Jun.) +31 (Jul.) + 13 days
$=225$ days $=32 \times 7+1=1$ odd day
So, the total number of odd days $=1+1+1=3$ odd days.
$\therefore$ Day on 13th Aug. 1975 was Wednesday
Question 52 Marks
A and B are brothers. C and D are sisters. A's son is D's brother. How B is related to C?
Question 62 Marks
Answer
View full question & answer→i. It is evident from the Venn diagram that the policemen $x$ is a thief also. So, the given statement is not true. Hence, its truth value is 'F'.
ii. From the Venn-diagram, we find that $\mathrm{P} \cap \mathrm{T} \neq \Phi$. So, there are some thieves who are also policemen. Hence, the above statement is not true. So, its truth value is 'F'.
iii. It is evident from the Venn-diagram that there are some human beings who are neither policemen nor thieves. So, the above statement is not true and its truth value is 'F'.
iv. Clearly, policemen $x$ is a thief also. Therefore, the given statement is true and its truth value is 'T'.
ii. From the Venn-diagram, we find that $\mathrm{P} \cap \mathrm{T} \neq \Phi$. So, there are some thieves who are also policemen. Hence, the above statement is not true. So, its truth value is 'F'.
iii. It is evident from the Venn-diagram that there are some human beings who are neither policemen nor thieves. So, the above statement is not true and its truth value is 'F'.
iv. Clearly, policemen $x$ is a thief also. Therefore, the given statement is true and its truth value is 'T'.
Question 72 Marks
The average monthly savings of a company was ₹ 12 lakh for the first 3 months, ₹ 12.5 lakh for the next 4 months and ₹ 31.2 lakh during the next 5 months of a year. Total expenditure during the year is 78 lakh. Find the average monthly earnings of the company.
Answer
View full question & answer→Given, average monthly savings for the first 3 months $=₹ 12$ lakh
$\Rightarrow$ total savings for the first 3 months $=₹(3 \times 12)$ lakh $=₹ 36$ lakh.
Given, average monthly savings for the next 4 months $=₹ 12.5$ lakh
$\Rightarrow$ total savings for the next 4 months $=₹(4 \times 12.5)$ lakh $=₹ 50$ lakh.
Also, given average monthly savings for the next 5 months $=₹ 31.2$ lakh
$\Rightarrow$ total savings for the next 5 months $=₹(5 \times 31.2)$ lakh $=₹ 56$ lakh
$\therefore$ Total annual savings $=₹ 36$ lakh $+ ₹ 50$ lakh $+₹ 156$ lakh $=₹ 242$ lakh
Given total yearly expenditure $=₹ 78$ lakh
$\therefore$ Total yearly earnings $=₹ 242$ lakh $+ ₹ 78$ lakh $=₹ 320$ lakh
$\therefore$ Average monthly earnings $=₹ \frac{320}{12}$ lakh $=₹ 26.67$ lakh.
$\Rightarrow$ total savings for the first 3 months $=₹(3 \times 12)$ lakh $=₹ 36$ lakh.
Given, average monthly savings for the next 4 months $=₹ 12.5$ lakh
$\Rightarrow$ total savings for the next 4 months $=₹(4 \times 12.5)$ lakh $=₹ 50$ lakh.
Also, given average monthly savings for the next 5 months $=₹ 31.2$ lakh
$\Rightarrow$ total savings for the next 5 months $=₹(5 \times 31.2)$ lakh $=₹ 56$ lakh
$\therefore$ Total annual savings $=₹ 36$ lakh $+ ₹ 50$ lakh $+₹ 156$ lakh $=₹ 242$ lakh
Given total yearly expenditure $=₹ 78$ lakh
$\therefore$ Total yearly earnings $=₹ 242$ lakh $+ ₹ 78$ lakh $=₹ 320$ lakh
$\therefore$ Average monthly earnings $=₹ \frac{320}{12}$ lakh $=₹ 26.67$ lakh.