5 Marks Questions

Question 15 Marks

Show that the semi$-$vertical angle of a cone of maximum volume and given slant height is $\tan ^{-1} \sqrt{2}$ or $\cos ^{-1} \frac{1}{\sqrt{3}}$.

Answer

Let $\alpha$ be the semi$-$vertical angle of a cone $\text{VAB}$ of given slant height $1$ .
In $\triangle \text{AOV},$

$\cos \alpha=\frac{V O}{V A}$ and $\sin \alpha=\frac{O A}{V A}$
$\Rightarrow \cos \alpha=\frac{V O}{l}$ and $\sin \alpha=\frac{O A}{l}$
$\Rightarrow V O=1 \cos a, OA=1 \sin \alpha$
Let $V$ be the volume of the cone, Then,
$V=\frac{1}{3} \pi(OA)^2(VO)$
$\Rightarrow V=\frac{1}{3} \pi(1 \sin \alpha)^2(1 \cos \alpha)$
$\Rightarrow V=\frac{1}{3} \pi 1^3 \sin ^2 \alpha \cos \alpha$
$\Rightarrow \frac{d V}{d \alpha}=\frac{\pi}{3} 1^3\left(-\sin ^3 \alpha+2 \sin \alpha \cos ^2 \alpha\right)$
$\Rightarrow \frac{d V}{d \alpha}=\frac{\pi l^3}{3} \sin \alpha\left(-\sin ^2 \alpha+2 \cos ^2 \alpha\right) \ldots$
The critical points of $V$ are given by $\frac{d V}{d \alpha}=0$.
$\because \frac{d V}{d \alpha}=0$
$\Rightarrow \frac{\pi l^3}{3} \sin \alpha\left(-\sin ^2 \alpha+2 \cos ^2 \alpha\right)=0$
$\Rightarrow 2 \cos ^2 \alpha=\sin ^2 \alpha$
$\Rightarrow \tan ^2 \alpha=2 $
$\Rightarrow \tan \alpha=\sqrt{2}\ [\because \text { a is acute } \because \sin \alpha \neq 0]$
$\because \cos \alpha=\frac{1}{\sqrt{1+\tan ^2 \alpha}}=\frac{1}{\sqrt{3}}[\because \tan \alpha=\sqrt{21}]$
Differentiating $(i)$ with respect to a, we get
$\frac{d^2 V}{d \alpha^2}=\frac{\pi l^3}{3}\left(-3 \sin ^2 \alpha \cos \alpha+2 \cos ^3 \alpha-4 \sin ^2 \alpha \cos \alpha\right)$
$=\frac{\pi l^3}{3} \cos ^3 \alpha\left(2-7 \tan ^2 \alpha\right)$
$\because\left(\frac{d^2 V}{d \alpha^2}\right)_{\tan \alpha=\sqrt{2}}$$=\frac{1}{3} \pi l^3\left(\frac{1}{\sqrt{3}}\right)^3(2-7 \times 2)=\frac{-4 \pi l^3}{3 \sqrt{3}}<0$
Thus$, V$ is maximum, when $\tan \alpha=\sqrt{2}$ or $\alpha=\tan ^{-1} \sqrt{2}$ i.e. when the semi$-$vertical angle of the cone is $\tan ^{-1} \sqrt{2}$.

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Question 25 Marks

Show that the right circular cylinder of given surface and maximum volume is such that its height is equal to the diameter of the base.

Answer

$s=2 \pi r h+2 \pi r^2$
$\Rightarrow \frac{s-2 \pi r^2}{2 \pi r}=h$
Now volume of cylinder is, $v=\pi r^2 h$
$\Rightarrow v=\pi \cdot r^2\left(\frac{s-2 \pi r^2}{2 \pi r}\right)$
$\Rightarrow v=\frac{1}{2}\left[s r-2 \pi r^3\right]$
Now$, \frac{d v}{d r}=\frac{1}{2}\left[s-6 \pi r^2\right]$
$\Rightarrow \frac{d^2 v}{d r^2}=\frac{1}{2}[0-12 \pi r]$
For maximum/minimum
$\frac{d v}{d r}=0$
$\Rightarrow s=6 \pi r^2$
From equ $(1)$
$\Rightarrow 2 \pi r h+2 \pi r^2=6 \pi r^2$
$\Rightarrow r=\frac{h}{2}$
${\left[\frac{d^2 v}{d r^2}\right]_{r=\frac{4}{2}}=\frac{1}{2}\left[0-12 \pi \times \frac{h}{2}\right]}$
$=-3 \pi h<0$
$\Rightarrow s$ is maximum at $r=\frac{h}{2}$
Hence $h=2 r$

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Question 35 Marks

Three shopkeepers $\text{A, B}$ and $C$ go to a store to buy stationery. A purchases $12$ dozen notebooks, $5$ dozen pens and $6$ dozen pencils. $B$ purchases $10$ dozen notebooks, $6$ dozen pens and $7$ dozen pencils. $C$ purchases $11$ dozen notebooks, $13$ dozen pens and $8$ dozen pencils. A notebook costs $40$ paise, a pen costs ₹ $1.25$ and a pencil costs $35$ paise. Use matrix multiplication to calculate each individual's bill.

Answer

Given the purchase details of three shopkeeper $A, B$, and $C.$
$A: 12$ dozen notebooks, $5$ dozen pens, and $6$ dozen pencils
$B: 10$ dozen notebooks, $6$ dozen pens, and $7$ dozen pencils
$C: 11$ dozen notebooks, $13$ dozen pens, and $8$ dozen pencils
Hence, the items purchased by $A, B,$ and $C$ can be represented in matrix $X$ of order $3 \times 3$ where the rows denoting the e shopkeeper and columns denoting the number of dozens of items as -
$X=\left[\begin{array}{ccc}12 & 5 & 6 \\10 & 6 & 7 \\11 & 13 & 8\end{array}\right]$
The price of each of the items is also given.
Cost of one notebook $=40$ paise
$\Rightarrow$ Cost of one dozen notebooks $=12 \times 40$ paise
$\Rightarrow$ Cost of one dozen notebooks $=480$ paise
$\therefore$ Cost of one dozen notebooks $=₹ 4.80$
Cost of one pen $=₹ 1.25$
$\Rightarrow$ Cost of one dozen pens $=12 \times ₹ 1.25$
$\therefore$ Cost of one dozen pens =₹ $15$
Cost of one pencil $= 35$ paise
$\Rightarrow$ Cost of one dozen notebooks $=12 \times 35$ paise
$\Rightarrow$ Cost of one dozen notebooks $=420$ paise
$\therefore$ Cost of one dozen notebooks $= ₹ 4.20$
Hence, the cost of purchasing one dozen of the items can be represented in matrix form with each row corresponding to an item as
$Y=\left[\begin{array}{c}4.80 \\15 \\4.20\end{array}\right]$
Now, the individual bill for each shopkeeper can be found by taking the product of the matrices $X$ and $Y$ .
The product is feasible because the order of $X$ is $3 \times 3$ and that of $Y$ is $3 \times 1$, which makes the number of columns of $x$ is equal to the number of the columns of $Y$.The product is given as follows.
$\begin{array}{l}X Y=\left[\begin{array}{lll}12 & 5 & 6 \\10 & 6 & 7 \\11 & 13 & 8\end{array}\right]\left[\begin{array}{c}4.80 \\15 \\4.20\end{array}\right]\end{array}$
$\Rightarrow X Y=\left[\begin{array}{c}12 \times 4.80+5 \times 15+6 \times 4.20 \\10 \times 4.80+6 \times15+7 \times 4.20 \\11 \times 4.80+13 \times 15+8 \times 4.20\end{array}\right]$
$\Rightarrow\left[\begin{array}{c}57.60+75+25.20 \\48+90+29.40 \\52.80+195+33.60 \end{array}\right]$
$\therefore X Y=\left[\begin{array}{l}157.80 \\167.40 \\281.40 \end{array}\right]$
Thus, the bills of shopkeepers $A, B$ and $C$ are ₹ $157.80$, ₹ $167.40$ and ₹ $281.40$ respectively.

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Question 45 Marks

Find the area of the region $\left\{(x, y): 0 \leqslant y \leqslant\left(x^2+1\right), 0 \leqslant y \leqslant(x+1), 0 \leqslant x \leqslant 2\right\}$

Answer

There are three curves respectively,
$C_1=\left\{(x, y): 0 \leq y \leq x^2+1\right\}$
$C_2=\{(x, y): 0 \leq y \leq x+1\}$
$C_3=\{(x, y): 0 \leq x \leq 2\}$
The points of intersection of, $y=x^2$ and $y=x+1 y$ are $A(0,1)$ and $B(1,2)$.
The required area of shaded region $=\int_0^1 y_1 d x+\int_1^2 y_2 d x$
where $y _1= x ^2+1$ and $y _2= x +1$
$\therefore A=\int_0^1\left(x^2+1\right) d x+\int_1^2(x+1) d x$
$=\left[\frac{x^3}{3}+x\right]_0^1+\left[\frac{x^2}{2}+x\right]_1^2$
$=\left(\frac{4}{3}-0\right)+\left(4-\frac{3}{2}\right)=\frac{8+24-9}{6}=\frac{23}{6}$ sq. units

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Question 55 Marks

Let $A =\{1,2,3\}$ and $R =\left\{( a , b ): a , b \in A\right.$ and $\left|a^2-b^2\right| \leq 5$. Write R as set of ordered pairs. Mention whether $R$ is i. reflexive
$ii.$ symmetric
$iii.$ transitive
Give reason in each case.

Answer

Given: $f : W \rightarrow W$ defined as $f(n)=\left\{\begin{array}{ll}n-1, & \text { if } n \text { is odd } \\ n+1, & \text { if } n \text { is even }\end{array}\right.$
Injectivity: Let $n, m$ be any two odd real numbers, then $f(n)=f(m)$
$\Rightarrow n-1=m-1$
$\Rightarrow n=m$
Again, let $n, m$ be any two even whole numbers, then $f(n)=f(m)$
$\Rightarrow n+1=m+1$
$\Rightarrow n=m$
If $n$ is even and $m$ is odd, then $n \neq m$.
Now $n$ is even implies $f(n)=n+1$ and $f(m)=m-1$.
Therefore, $f ( n ) \neq f ( m )$
Similarly $n$ is odd and $m$ is even gives $f(n) \neq f(m)$
Therefore in all cases $f$ is one$-$one.
Surjectivity: Let $n$ be an arbitrary whole number.
If $n$ is an odd number, then there exists an even whole number $(n+1)$ such that
$f(n+1)=n+1-1=n$
If $n$ is an even number, then there exists an odd whole number $(n-1)$ such that
$f(n-1)=n-1+1=n$
Therefore, every $n \in W$ has its pre-image in $W$.
So, $f: W \rightarrow W$ is a surjective. Thus $f$ is invertible and $f^{-1}$ exists.
For $f^{-1}: y=n-1$
$\Rightarrow n=y+1 \text { and } y=n+1 \Rightarrow n=y-1$
$\therefore f^{-1}(n)=\left\{\begin{array}{ll}n-1, & \text { if } n \text { is odd } \\ n+1, & \text { if } n \text { is even }\end{array}\right.$
Hence, $f^{-1}(y)=y$

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Question 65 Marks

Let $f: W \rightarrow W$ be defined as $f(n)=n-1$, if $n$ is odd and $f(n)=n+1$, if $n$ is even. Show that $f$ is invertible. Find the inverse of $f$. Here, $W$ is the set of all whole numbers.

Answer

Given: $f : W \rightarrow W$ defined as $f(n)=\left\{n-1, \text { if } n \text { is odd } n+1, \text { if } n \text { is even }\right.$
Injectivity: Let $n$, $m$ be any two odd real numbers, then $f(n)=f(m)$
$\Rightarrow n-1=m-1$
$\Rightarrow n=m$
Again, let $n$, $m$ be any two even whole numbers, then $f(n)=f(m)$
$\Rightarrow n+1=m+1$
$\Rightarrow n=m$
If $n$ is even and $m$ is odd, then $n \neq m$.
Now $n$ is even implies $f(n)=n+1$ and $f(m)=m-1$.
Therefore, $f ( n ) \neq f ( m )$
Similarly $n$ is odd and $m$ is even gives $f(n) \neq f(m)$
Therefore in all cases f is one-one.
Surjectivity: Let n be an arbitrary whole number.
If $n$ is an odd number, then there exists an even whole number $(n+1)$ such that
$f(n+1)=n+1-1=n$
If $n$ is an even number, then there exists an odd whole number $(n-1)$ such that
$f(n-1)=n-1+1=n$
Therefore, every $n \in W$ has its pre$-$image in $W$.
So, $f: W \rightarrow W$ is a surjective.
Thus f is invertible and $f ^{-1}$ exists.
For $f ^{-1}: y=n-1$
$\Rightarrow n=y+1 \text { and } y=n+1$
$\Rightarrow n=y-1 \therefore f^{-1}(n)=\left\{ n-1, \text { if } n \text { is odd } n+1, \text { if } n \text { is even } \right.$
Hence, $f^{-1}(y)=y$

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MATHS 5 Marks Questions

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