Mathematics Test

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Mathematics Test - 11

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Weekly Quiz Competition

Question 1
4 / -1

$\int \frac{1}{{x}}(\log {x}) {dx}$ is equal to:

A
$\log (\log x)+C$

B
$\frac{1}{2}(\log x)^{2}+C$

C
$-\cot \frac{x}{2}+C$

D
$\frac{x}{2}+C$

Solution

${I}=\int \frac{1}{{x}}(\log {x}) {dx}$
Let $f({x})=\log {x}$ and ${g}({x})=1 / {x}$
$\therefore {f}^{\prime}({x})=\frac{{d}}{{dx}} \log {x}=\frac{1}{{x}}$ and $\int {g}({x}) {dx}=\int \frac{1}{{x}} {dx}=\log {x} .$
${I}=\int {f}({x}) {g}({x}) {d} {x}={f}({x}) \int {g}({x}) {d} {x}-\int {f}^{\prime}({x})\left[\int {g}({x}) {d} {x}\right] {d} {x}$
$\Rightarrow {I}=\log {x} \times \log {x}-\int \frac{1}{{x}}(\log {x}) {dx}$
$\Rightarrow {I}=(\log {x})^{2}-1+{C}$
$\Rightarrow 2 {I}=(\log {x})^{2}+{C}$
$\Rightarrow {I}=\frac{1}{2}(\log {x})^{2}+{C}$
Question 2
4 / -1

In linear programming, lack of points for a solution set is said to:

A
have no feasible solution

B
have a feasible solution

C
have single point method

D
have infinte point method

Solution

If there is no point in the feasible set, there is no feasible solution of the linear programming model.
In linear programming, lack of points for a solution set is said to have no feasible solution
Question 3
4 / -1

The area of the region bounded by the curve y = x² and the line y = 16

A
$\frac{64}{3}$ sq. units

B
$\frac{32}{3}$ sq. units

C
$\frac{256}{3}$ sq. units

D
$\frac{128}{3}$ sq. units

Solution

Given,
y= x² and y = 16,
Area of region bounded by the curve, A = ∫xdy
From figure we can see that, dy varies from 0 to 16
⇒ A = ∫xdy
$\Rightarrow A = 2 \int_{0}^{16} (\sqrt{y}) d y$
$\Rightarrow A = 2 x \frac{2}{3} {|y^{\frac{3}{2}}|}_{0}^{16}$
$\Rightarrow A = \frac{256}{3}$
Question 4
4 / -1

Let $g(x)=1+x-[x]$
$f(x)=\left\{\begin{array}{l}-1, x<0 \\ 0, x=0 \\ 1, x>0\end{array}\right.$ then the value of $\mathrm{f}(\mathrm{g}(\mathrm{x}))$ is:

A
$\left\{\begin{array}{l}2+x, 0 \leq x \leq 1 \\ 2-x, 1

B
\(\left\{\begin{array}{l}2-x, 0

C
\(\left\{\begin{array}{l}2-x, 0 \leq x \leq 1 \\ 2+x, 1 \leq x \leq 2 \\ 4+x, 2

D
None of these

Solution

Let \(g(x)=1+x-[x] $.....(i)
$f(x)=\left\{\begin{array}{l}-1, x<0 \\ 0, x=0 \\ 1, x>0\end{array}\right.$.......(ii)
We have to find the value of $f(g(x))$.
Now,$g(x)=1+x-[x]$or $g(x)=1+\{x\}$
Since $0<\{x\}<1 \Rightarrow g(x)$, is greater than 1 for all $x \in R$.
So, $f(g(x))=1$ [Using (ii)]
Question 5
4 / -1

One kind of cake requires $200 \mathrm{~g}$ of flour and $25 \mathrm{~g}$ of fat, and another kind of cake requires $100 \mathrm{~g}$ of flour and $50 \mathrm{~g}$ of fat. Find the maximum number of cakes which can be made from $5 \mathrm{~kg}$ of flour and $1 \mathrm{~kg}$ of fat assuming that there is no shortage of the other ingredients used in making the cakes.

A
Maximum number of cakes $=30,20$ of kind one and $10$ cakes of another kind

B
Maximum number of cakes $=32,20$ of kind one and $12$ cakes of another kind

C
Maximum number of cakes $=34,27$ of kind one and $7$ cakes of another kind

D
Maximum number of cakes $=33,22$ of kind one and $11$ cakes of another kind

Solution

Let number of first kind of cake is $\mathrm{X}$ and another kind of cake is $\mathrm{Y}$.

So, total flour required $=200 \mathrm{X}+100 \mathrm{Y} \mathrm{g}$

And total fat required $=25 \mathrm{X}+50 \mathrm{Y} \mathrm{g}$

Since, maximum flour available is $5 \mathrm{~kg}=5000 \mathrm{~g}$

$\therefore 200 \mathrm{X}+100 \mathrm{Y} \leq 5000$

$\Rightarrow 2 \mathrm{X}+\mathrm{Y} \leq 50$.....(1)

Also, maximum fat available is $1 \mathrm{~kg}=1000 \mathrm{~g}$

$\therefore 25 \mathrm{X}+50 \mathrm{Y} \leq 1000$

$\Rightarrow \mathrm{X}+2 \mathrm{Y} \leq 40$..........(2)

Since, quantity of cakes can't be negative.

$\therefore \mathrm{X} \geq 0, \mathrm{Y} \geq 0$.....(3)

We have to maximize number of cakes that can be made.

So, Objective function is $\mathrm{Z}=\mathrm{X}+\mathrm{Y}$

After plotting all the constraints given by equation (1), (2) and (3), we got the feasible region as shown in the image.

Corner point $\mathrm{A}(\mathrm{0}, 20)$

Value of $Z=0+20=20$

Corner point $\mathrm{B}(20,10)$,

Value of $Z=20+10=30$

Corner point $\mathrm{C}(25,0)$,

Value of $Z=25+0=25$

So, maximum cake that can be made is $30$, where first kind of cake will be $20$ and second kind of cake will be $10$.
Question 6
4 / -1

Let $f(x)=a x^{2}+b x+c$ where $a, b, c$ are rational, and $f: Z \rightarrow Z$ where $Z$ is the set of integer. Then $a+b$ is:

A
A negative integer

B
An integer

C
Non-integral rational number

D
None of these

Solution

$f: Z \rightarrow Z$ such that$f(x)=a x^{2}+b x+c,$ where $a, b, c \in Q$
We have to find the value of $a+b$.
Now,For $x=0, f(0) ~= c=$ Integer $[\because f(0) \in Z$ because $f: Z \rightarrow Z]$
$\Rightarrow {c}$ is an integer.
For $x=1, f(1)=a+b+c=k($ say $)[$ where $f(1)=k \in Z]$
$\Rightarrow a+b=k-c$
since $k \in Z, c \in Z \Rightarrow k-c \in Z$
$\Rightarrow a+b \in Z$$\Rightarrow a+b$ is an integer.
Question 7
4 / -1

What is the mean deviation from the mean of the numbers $10,9,21,16,24 ?$

A
$5.2$

B
$5.0$

C
$4.5$

D
$4.0$

Solution

Given numbers are $10,9,21,16,24 .$
Total numbers $=5$
Mean $(\bar{x})=\frac{\text { Sum of all the observations }}{\text { Total number of observations }}$
$=\frac{10+9+21+16+24}{5}$
$=\frac{80}{5}=16$
We know that,
Mean deviation from mean $=\frac{\sum_{i=1}^{i=n}\left|x_{i}-\bar{x}\right|}{n}$
$=\frac{|10-16|+|9-16|+|21-16|+|16-16|+|24-16|}{5}$
$=\frac{6+7+5+0+8}{5}$
$=\frac{26}{5}=5.2$
Question 8
4 / -1

A box contains 2 blue caps, 4 red caps, 5 greens caps, and 1 yellow cap. If four caps are picked at random, the probability that none of them are green is:

A
$\frac{7}{99}$

B
$\frac{7}{12}$

C
$\frac{5}{99}$

D
$\frac{5}{12}$

Solution

The total number of caps $=2+4+5+1=12$
The number of ways for selecting four caps $={ }^{12} \mathrm{C}_{4}$
The total number of caps other than green $=12-5=7$
The number of ways for selecting four caps other than green $={ }^{7} \mathrm{C}_{4}$
The probability of selecting four caps and none of them are green P(A')$=\frac{\text { The number of ways for selecting four caps other than green }}{\text { The number of ways for selecting four caps }}$
$ \mathrm{P}\left(\mathrm{A}^{\prime}\right)=\frac{{ }^{7} \mathrm{C}_{4}}{{ }^{12} \mathrm{C}_{4}}$
$\mathrm{P}\left(\mathrm{A}^{\prime}\right)=\frac{7 !(12-4) ! 4 !}{4 !(7-4) ! 12 !}$
$\mathrm{P}\left(\mathrm{A}^{\prime}\right)=\frac{7}{99}$
Question 9
4 / -1

Find the cartesian equation of the line which passes through the point $(-1,2,1)$ and parallel to the line $\frac{x-1}{2}=\frac{y-2}{4}=\frac{z-7}{3}$.

A

$\frac{x-1}{2}=\frac{y-2}{4}=\frac{z-1}{3}$

B

$\frac{x+1}{2}=\frac{y-2}{4}=\frac{z-1}{3}$

C

$\frac{x+1}{1}=\frac{y-2}{2}=\frac{z-1}{7}$

D

$\frac{x-1}{1}=\frac{y-2}{2}=\frac{z-1}{7}$

Solution

Given,
The required line passes through the point $(-1,2,1)$ and is parallel to the line $\frac{x-1}{2}=\frac{y-2}{4}=\frac{z-7}{3}$
$\because$ The required line is parallel to the line $\frac{x-1}{2}=\frac{y-2}{4}=\frac{z-7}{3}$
So, both the lines will have same direction ratios.
${a}=2, {~b}=4, {c}=3$
$\because$ The required line passes through the point $(-1,2,1)$
$x_{1}=-1, y_{1}=2, z_{1}=1$
$\therefore$ The cartesian equation of the required line is:
$\frac{x-(-1)}{2}=\frac{y-2}{4}=\frac{z-1}{3}$
$\Rightarrow \frac{x+1}{2}=\frac{y-2}{4}=\frac{z-1}{3}$
Question 10
4 / -1

Find $\frac{d y}{d x}$ if $y=e^{\cos x}$:

A
$\sin x e^{\cos x}$

B
$-\sin x e^{\cos x}$

C
$-\cos x e^{\cos x}$

D
$\cos x e^{\cos x}$

Solution

$\frac{d}{d x}\left(e^{a x+b}\right)=e^{a x+b} \frac{d}{d x}(a x+b)=a e^{a x+b}$
$\frac{d(\cos x)}{d x}=-\sin x$
Given: $y=e^{\cos x}$
Here, we have to find $\frac{{dy}}{ {dx}}$
As we know that,
$\frac{d}{d x}\left(e^{a x+b}\right)=e^{a x+b} \frac{d}{d x}(a x+b)=a e^{a x+b}$
$\frac{d y}{d x}=e^{\cos x} \frac{d}{d x}(\cos x)=e^{\cos x}(-\sin x)=-\sin x e^{\cos x}$

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Mathematics Test - 11

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Mathematics Test - 11

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Question 1

\(\log (\log x)+C\)

\(\frac{1}{2}(\log x)^{2}+C\)

\(-\cot \frac{x}{2}+C\)

\(\frac{x}{2}+C\)

Solution

Question 2

have no feasible solution

have a feasible solution

have single point method

have infinte point method

Solution

Question 3

643 sq. units

323 sq. units

2563 sq. units

1283 sq. units

Solution

Question 4

\(\left\{\begin{array}{l}2+x, 0 \leq x \leq 1 \\ 2-x, 1

\(\left\{\begin{array}{l}2-x, 0

\(\left\{\begin{array}{l}2-x, 0 \leq x \leq 1 \\ 2+x, 1 \leq x \leq 2 \\ 4+x, 2

None of these

Solution

Question 5

Maximum number of cakes \(=30,20\) of kind one and \(10\) cakes of another kind

Maximum number of cakes \(=32,20\) of kind one and \(12\) cakes of another kind

Maximum number of cakes \(=34,27\) of kind one and \(7\) cakes of another kind

Maximum number of cakes \(=33,22\) of kind one and \(11\) cakes of another kind

Solution

Question 6

A negative integer

An integer

Non-integral rational number

None of these

Solution

Question 7

\(5.2\)

\(5.0\)

\(4.5\)

\(4.0\)

Solution

Question 8

\(\frac{7}{99}\)

\(\frac{7}{12}\)

\(\frac{5}{99}\)

\(\frac{5}{12}\)

Solution

Question 9

\(\frac{x-1}{2}=\frac{y-2}{4}=\frac{z-1}{3}\)

\(\frac{x+1}{2}=\frac{y-2}{4}=\frac{z-1}{3}\)

\(\frac{x+1}{1}=\frac{y-2}{2}=\frac{z-1}{7}\)

\(\frac{x-1}{1}=\frac{y-2}{2}=\frac{z-1}{7}\)

Solution

Question 10

\(\sin x e^{\cos x}\)

\(-\sin x e^{\cos x}\)

\(-\cos x e^{\cos x}\)

\(\cos x e^{\cos x}\)

Solution

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$\frac{64}{3}$ sq. units

$\frac{32}{3}$ sq. units

$\frac{256}{3}$ sq. units

$\frac{128}{3}$ sq. units