3 Marks Question

Question 13 Marks

Suppose the total cost C(x) (in millions) for manufacturing x air-planes per year is given by the function
$C(x)=6+\sqrt{4 x+4} \quad 0 \leq x \leq 30$
a) Find the marginal cost at a production level of x air-planes per year.
(b) Find the marginal cost at a production level of 15 and 24 air-planes per year, and interpret the results.

Answer

(a) The marginal cost at a production level of x air-planes is
$C(x)=\frac{d}{d x}(6+\sqrt{4 x+4})$
$=\frac{2}{\sqrt{4 x+4}}$

(b) The marginal cost at a production level of 15 air-planes is
$C(15)=\frac{2}{\sqrt{4(15)+4}}=0.25$
At a production level of 15 air-planes per year, the total cost is increasing at the rate of ₹ 250,000 per one air-plane.
The marginal cost at a production level of 24 air-planes is
$C(24)=\frac{2}{\sqrt{4(24)+4}}=0.2$
At a production level of 24 air-planes per year, the total cost is increasing at the rate of at the rate of ₹ 200,000 per one air-plane.

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Question 23 Marks

A tank with rectangular base and rectangular sides open at the top is to be constructed so that its depth is 3 m and volume is $75 m^3$. If building of tank costs ₹ 100 per square metre for the base and ₹ 50 per square metre for the sides, find the cost of least expensive tank.

Answer

Let 1, b, h be the length, breadth and depth of the tank, respectively.
$\therefore \quad 1 \times b \times 3=75$
$\Rightarrow 1 \times b=25$
Let C be the cost, then
$C=100(l \times b)+50 \times 2[h(b+l)]$
$=100\left(l \times \frac{25}{l}\right)+300\left(\frac{25}{l}+l\right)$
$=2500+300\left(\frac{25}{l}+l\right)$
Differentiating w.rt. l,
$\frac{d C}{d l}=0+300\left(\frac{-25}{l^2}+1\right)$
For maximum and minimum cost,
$\frac{d C}{d l}=0$
$\Rightarrow \quad 300\left(-\frac{25}{l^2}+1\right)=0$
$\Rightarrow \quad l^2=25$ or $l= \pm 5$
Getting $\left(\frac{d^2 C}{d l^2}\right)=300\left(\frac{50}{l^3}\right)$
$\Rightarrow\left(\frac{d^2 C}{d l^2}\right)_{a t l=5}=\frac{15000}{125}>0$
i.e., C is minimum, when l = 5
$\Rightarrow \quad b=5$
$\therefore \quad C=100(25)+300(10)$
= 2 ,500+3,000
= 5 ,500
Hence, the minimum cost is ₹5,500.

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Question 33 Marks

A jet of enemy is flying along the curve $y=x^2+2$and a soldier is placed at the point (3, 2) . Find the minimum distance between the soldier and the jet.

Answer

Let P(x,y) be the position of the jet and the soldier is placed at A(3, 2).
$\Rightarrow \quad A P=\left|\sqrt{(x-3)^2+(y-2)^2}\right|$ .....(i)
As given, $y=x^2+2$
$\Rightarrow \quad y-2=x^2$ .....(ii)
$\therefore \quad A P^2=(x-3)^2+x^4=z$ (say)
$\therefore \quad \frac{d z}{d x}=2(x-3)+4 x^3$
For maximum and minimum value,
$\frac{d z}{d x}=0$
$\therefore 2 x-6+4 x^3=0$
Put x = 1
2-6+4=0
$\therefore x-1$ is a factor
and $\frac{d^2 z}{d x^2}=12 x^2+2$
Now, $\frac{d^2 z}{d x^2}($ at $x=1)>0$
$\therefore z$ is minimum, when $x=1, y=1+2=3$ ....[from (ii)]
$\therefore$ Minimum distance $=\left|\sqrt{(1-3)^2+(3-2)^2}\right|$
$=\sqrt{5}$ units $\quad[$ from $(i)] 1$

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Question 43 Marks

An open tank with a square base and vertical sides is to be constructed from a metal sheet so as to hold a given quantity of water. Show that the cost of material will be least when depth of the tank is half of its width.

Answer

Let side of base = x and depth of tank = y
$\begin{array}{c}V=x^2 y \Rightarrow y=\frac{V}{x^2} \\ (V=\text { Quantity of water }=\text { constant })\end{array}$
Cost of material is least when area of sheet used is minimum.
$A ($ Surface area of tank $)=x^2+4 x y=x^2+\frac{4 V}{x}$
$\begin{array}{l}\frac{d A}{d x}=2 x-\frac{4 V}{x^2} \\ \frac{d A}{d x}=0\end{array}$
$\Rightarrow \quad x^3=2 V$
$\Rightarrow \quad x^3=2 x^2 y$
$y=\frac{x^3}{2 x^2}=\frac{x}{2}$
[as $\left.V=x^2 y\right]$
$\frac{d^2 A}{d x^2}=2+\frac{8 V}{x^3}>0$,
$\therefore$ Area is minimum, thus cost is minimum when
$y=\frac{x}{2}$

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Question 53 Marks

Find the maximum value of $\left(\frac{1}{x}\right)^x$.

Answer

Let, $y=\left(\frac{1}{x}\right)^x$
$\Rightarrow \log y=x \cdot \log \frac{1}{x}$
$\therefore \frac{1}{y} \cdot \frac{d y}{d x}=x \cdot \frac{1}{\frac{1}{x}} \cdot\left(-\frac{1}{x^2}\right)+\log \frac{1}{x} \cdot 1$
$=-1+\log \frac{1}{x}$
$\therefore \quad \frac{d y}{d x}=\left(\log \frac{1}{x}-1\right)\left(\frac{1}{x}\right)^x$
Now, $\quad \frac{d y}{d x}=0$
$\Rightarrow \quad \log \frac{1}{x}=1=\log e$
$\Rightarrow \quad \frac{1}{x}=e$
$\Rightarrow \quad x=\frac{1}{e}$
Hence, the maximum value of $f\left(\frac{1}{e}\right)=(e)^{v / e}$.

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Question 63 Marks

If x is real find the minimum value of $x^2-8 x+17$.

Answer

Let, $f(x)=x^2-8 x+17$
On differentiating with respect to x, we get
$f^{\prime}(x)=2 x-8$
So, $f^{\prime}(x)=0$
$\Rightarrow \quad 2 x-8=0$
$\Rightarrow \quad 2 x=8$
$\therefore \quad x=4$
Now, Again on differentiating with respect to x, we get
$f^{\prime \prime}(x)=2>0, \forall x$
So, x = 4 is the point of local minima.
Minimum value of f(x) at x = 4
$f(4)=4 \times 4-8 \times 4+17=1$
Alter :
$f(x)=x^2-8 x+17$
$=\left(x-4^2\right)+1$
$\therefore$ Minimum value is 1

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Question 73 Marks

Find the intervals in which the function $f(x)=-2 x^3$$-9 x^2-12 x+1$is (i) Strictly increasing (ii) strictly decreasing.

Answer

$f(x)=-2 x^3-9 x^2-12 x+1$
Now, $f^{\prime}(x)=-6 x^2-18 x-12$
$=-6\left[x^2+3 x+2\right]$
$=-6\left[x^2+2 x+x+2\right]$
$f^{\prime}(x)=-6(x+1)(x+2)$
$\Rightarrow$ Intervals are $(-\infty,-2),(-2,-1)$ and $(-1, \infty)$
Getting $f^{\prime}(x)>0$ in $(-2,-1)$ and $f^{\prime}(x)<0$ in $(-\infty,-2)$$\cup(-1, \infty)$
$\Rightarrow f(x)$ is strictly increasing in $(-2,-1)$
and strictly decreasing in $(-\infty,-2) \cup(-1, \infty)$

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Question 83 Marks

Determine for what values of x, the function $f(x)=x^3+\frac{1}{x^3} \quad(x \neq 0)$ is strictly increasing or strictly decreasing.

Answer

Here, $f^{\prime}(x)=3 x^2-3 x^{-4}$
$=\frac{3\left(x^6-1\right)}{x^4}$
$=\frac{3\left(x^4+x^2+1\right)}{x^4}(x+1)(x-1)$
Critical points are - 1 and 1.
or $f^{\prime}(x)>0$ if $x>1$ or $x<-1$, and $f^{\prime}(x)<0$ if -1< x < 1
$\left(\because \frac{3\left(x^4+x^2+1\right)}{x^4}\right.$ always + ve $)$

Hence, f(x) is strictly increasing for x > 1
or x < - 1 ; and strictly decreasing in $(-1,0) \cup(0,1) $

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Question 93 Marks

Find the intervals in which the function $\frac{x^4}{4}-x^3-5 x^2+24 x+12$ is (a) strictly increasing (b) strictly decreasing.

Answer

$f(x)=\frac{x^4}{4}-x^3-5 x^2+24 x+12$
$f^{\prime}(x)=x^3-3 x^2-10 x+2$
= (x - 2)(x - 4)(x + 3)
$f^{\prime}(x)=0 \Rightarrow x=-3,2,4$
sign of f'(x) :

$\therefore f(x)$ is strictly increasing on $(-3,2) \cup(4, \infty)$ and $f(x)$ is strictly decreasing on $(-\infty,-3) \cup(2,4)$

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Question 103 Marks

Find the intervals in which the function $f(x)=3 x^4-4 x^3-12 x^2+5$ is (a) strictly increasing (b) strictly decreasing.

Answer

$f^{\prime}(x)=12 x^3-12 x^2-24 x$
= 12x(x + 1)(x - 2)
$f^{\prime}(x)>0, \forall x \in(-1,0) \cup(2, \infty)$
$f^{\prime}(x)<0, \forall x \in(-\infty,-1) \cup(0,2)$

$\therefore f(x)$ is strictly increasing in $(-1,0) \cup(2, \infty)$
and strictly decreasing in $(-\infty,-1) \cup(0,2)$

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Question 113 Marks

Find the intervals in which
$f(x)=\frac{3}{10} x^4-\frac{4}{5} x^3-3 x^2+\frac{36}{5} x+11$ is
(a) strictly increasing
(b) strictly decreasing.

Answer

$f^{\prime}(x)=\frac{12}{10} x^3-\frac{12}{5} x^2-6 x+\frac{36}{5}$
$=\frac{6}{5}(x-1)(x+2)(x-3)$

f'(x) = 0 at x = 1, - 2, 3
$\therefore$ Intervals are $(-\infty,-2),(-2,1),(1,3)$ and $(3, \infty)$
$\because f^{\prime}(x)>0$ for $(-2,1) \cup(3, \infty)$
f(x) is strictly increasing in (-2, 1) and $(3, \infty)$
$\because f^{\prime}(x)<0$ for $(-\infty,-2)$ and $(1,3)$
f(x) is strictly decreasing in $(-\infty,-2) \cup(1,3)$

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