Mathematics Test

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Mathematics Test - 10

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Weekly Quiz Competition

Question 1
1 / -0

If $A = [\begin{matrix} 1 & 2 \\ 2 & 3 \end{matrix}]$ and A² - kA - I₂ = o, where I₂: is the 2 × 2 identity matrix, then what is the value of k?

A
4

B
-4

C
8

D
-8

Solution

Given:

$A^{2}-k A+I_{2}=0$ and $A=\left[\begin{array}{ll}1 & 2 \\ 2 & 3\end{array}\right]$

$\Rightarrow\left[\begin{array}{ll}1 & 2 \\ 2 & 3\end{array}\right]\left[\begin{array}{ll}1 & 2 \\ 2 & 3\end{array}\right]-k\left[\begin{array}{ll}1 & 2 \\ 2 & 3\end{array}\right]+\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]=0$

$\left[\begin{array}{cc}5 & 8 \\ 8 & 13\end{array}\right]-\left[\begin{array}{cc}k & 2 k \\ 2 k & 3 k\end{array}\right]+\left[\begin{array}{cc}1 & 0 \\ 0 & 1\end{array}\right]=0$

$\left[\begin{array}{cc}5-\mathrm{k}+1 & 8-2 \mathrm{k}+0 \\ 8-2 \mathrm{k}+0 & 13-3 \mathrm{k}+1\end{array}\right]=\left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right]$

$\Rightarrow 8-2 k=0$

$\Rightarrow 8=2 k$

$k=4$
Question 2
1 / -0

If $x$ is positive, the first negative term in the expansion of $(1+x)^{27 / 5}$ is:

A
$5^{ th}$ term

B
$6^{ th}$ term

C
$7^{ th}$ term

D
$8^{ th}$ term

Solution

We know, $T_{r+1}=\frac{n(n-1) \ldots(n-r+1)}{r !} x^{r}$
Therefore, for firest negative value,
$\Rightarrow$$n-r+1<0$, $n=\frac{27}{5}$
$\Rightarrow \frac{27}{ 5}+1<\mathrm{r}$
or,
$\Rightarrow$$\mathrm{r}>6$
Here $\mathrm{r}=7$
So, $T_{r+1}<0$
Therefore, $\mathrm{T}_{\mathrm{r}+1}=\mathrm{T}_{7+1}=\mathrm{T}_{8}=$ 8th term is negative.
Question 3
1 / -0

Let [x] denote the greatest integer function. What is the number of solutions of the equation x² - 4x + [x] = 0 in the interval [0, 2] ?

A
Zero (N solution)

B
One

C
Two

D
Three

Solution

We have to find number of solution of the equation x²-4x+[x] = 0 in the interval [0, 2]

We are dividing [0, 2] in two intervals

i.e. [0, 1] & [1, 2]

we know that
Question 4
1 / -0

What is the solution to differential equation $\frac{d x}{d y}+\frac{x}{y}=0$?

A
$\frac{x}{y}=c$

B
xy=c

C
y=cx

D
None of these

Solution

Given:

$\frac{d x}{d y}+\frac{x}{y}=0$

On solving, we get-

$\Rightarrow \frac{d x}{d y}=-\frac{x}{y}$

$\Rightarrow \frac{d x}{x}=-\frac{d y}{y}$

$\Rightarrow \ln x=-\ln y+k$ or

$\Rightarrow \ln x=\ln \frac{1}{y}+\ln c$

$\Rightarrow \ln x=\ln \frac{c}{y}$

$\Rightarrow x=\frac{c}{y}$

$\Rightarrow x y=c$
Question 5
1 / -0

Which of the following is equal to $\tan ^{-1}\left(\frac{8 x \sqrt{x}}{1-16 x^{3}}\right)?$

A
$2 \tan ^{-1}(4 x \sqrt{x})$

B
$\tan ^{-1}(8 x \sqrt{x})$

C
$\tan ^{-1}(4 x \sqrt{x})$

D
$\tan ^{-1}(4 x)$

Solution

Given,

$\tan ^{-1}\left(\frac{8 x \sqrt{x}}{1-16 x^{3}}\right)$...(i)

Rewrite the equation (i) as follows:

$\tan ^{-1}\left(\frac{8 x \cdot \sqrt{x}}{1-16 x^{3}}\right)=\tan ^{-1}\left(\frac{2(4 x \sqrt{x})}{1-(4 x \sqrt{x})^{2}}\right)$

On comparing equation (i) with $2 \tan ^{-1} \mathrm{x}=\tan ^{-1}(\frac{2 \mathrm{x}}{1-\mathrm{x}^{2}}))$ we get,

$=2 \tan ^{-1}(4 x \sqrt{x})$
Question 6
1 / -0

The solution to the differential equation dy = (1 + y²) dx is:

A
y = tan x + c

B
y = tan(x + c)

C
tan^-1 (y + c) = x

D
tan^-1 (y + x) = 2x

Solution

Given equation dy = (1 + y²) dx

On solving, we get-

$\frac{d y}{\left(1-y^{2}\right)}=d x$

Integrating both sides

$\int \frac{d y}{\left(1+y^{2}\right)}=\int d x$

$\tan ^{-1}(y)=x+c$

$y=\tan (x+c)$
Question 7
1 / -0

What is the number of ways that 4 boys and 3 girls can be seated so that boys and girls alternate?

A
12

B
72

C
120

D
144

Solution

Given:

4 boys and 3 girls can be seated so that boys and girls alternate

The only pattern in which they can sit according to the given situation is

B G B G B G B

4 boys can be seated in 4! Ways

Girl can be seated in 3! Ways

Required number of ways,

= 4! × 3!

= 144
Question 8
1 / -0

If $5 \times^{ n } P _{3}=4 \times^{( n +1)} P _{3}$, find $n$?

A
10

B
11

C
12

D
14

Solution

${ }^{n} P_{3}=n \times(n-1) \times(n-2)$

${ }^{(n+1)} P_{3}=(n+1) \times n \times(n-1)$

Now, $5 \times n \times(n-1) \times(n-2)=4 \times(n+1) \times n \times(n-1)$

or, $5(n-2)=4(n+1)$

or, $5 n-10=4 n+4$

or, $5 n-4 n=4+10$

So, $n=14$
Question 9
1 / -0

Given function $f(x)=\left(\frac{e^{2 z}-1}{e^{2 z}+1}\right)$ is:

A
Increasing

B
Decreasing

C
Even

D
None of these

Solution

$\Rightarrow$$f(x)=\frac{e^{2 z}-1}{e^{2 z}+1}$
$\Rightarrow$$f(-x)=\frac{e^{-2 z}-1}{e^{-2 z}+1}=\frac{1-e^{2 z}}{1+e^{2 z}}$
$\Rightarrow$$f(x)=-\frac{e^{2 x}-1}{e^{2 x}+1}=-f(x)$
${f}(\mathrm{x})$ is an odd function.
Again
$\Rightarrow$$f(x)=\frac{e^{2 x}-1}{e^{2 x}+1} \Rightarrow f^{\prime}(x)=\frac{4 e^{2 x}}{\left(1+e^{2 x}\right)^{2}}>$$0 ~\forall~n \in R$
$\mathrm{f}(\mathrm{x})$ is an increasing function.
Question 10
1 / -0

If $\sin ^{-1} \frac{3}{x}+\sin ^{-1} \frac{9}{x}=\frac{\pi}{2}$ then what is the value of $x$ ?

A
1

B
$3 \sqrt{10}$

C
0

D
$\sqrt{10}$

Solution

Here, $\sin ^{-1} \frac{3}{x}+\sin ^{-1} \frac{9}{x}=\frac{\pi}{2}$

Let,

$\sin ^{-1} \frac{9}{x}=y \cdots(1)$

$\Rightarrow \sin y=\frac{9}{x}$

$\therefore \cos ^2 y=1-\sin ^2 y$

$=1-\left(\frac{9}{x}\right)^2$

$=1-\frac{81}{x^2}$

$\cos y=\frac{\sqrt{x^2-81}}{x}$

$\Rightarrow y=\cos ^{-1} \frac{\sqrt{x^2-81}}{x}$

So, $\sin ^{-1} \frac{9}{x}=\cos ^{-1} \frac{\sqrt{x^2-81}}{x}$

$\sin ^{-1} \frac{3}{x}+\cos ^{-1} \frac{\sqrt{x^2-81}}{x}=\frac{\pi}{2}$

Now we know $\sin ^{-1} \mathrm{y}+\cos ^{-1} \mathrm{y}=\frac{\pi}{2}$

So,$\frac{3}{x}=\frac{\sqrt{x^2-81}}{x}$

$9=x^2-81$

$\Rightarrow x^2=90$

$x=3 \sqrt{10}$

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Re-Attempt Weekly Quiz Competition

Mathematics Test - 10

Result

Mathematics Test - 10

Score

-

Rank

-

Question 1

4

-4

8

-8

Solution

Question 2

\(5^{ th}\) term

\(6^{ th}\) term

\(7^{ th}\) term

\(8^{ th}\) term

Solution

Question 3

Zero (N solution)

One

Two

Three

Solution

Question 4

\(\frac{x}{y}=c\)

xy=c

y=cx

None of these

Solution

Question 5

\(2 \tan ^{-1}(4 x \sqrt{x})\)

\(\tan ^{-1}(8 x \sqrt{x})\)

\(\tan ^{-1}(4 x \sqrt{x})\)

\(\tan ^{-1}(4 x)\)

Solution

Question 6

y = tan x + c

y = tan(x + c)

tan-1 (y + c) = x

tan-1 (y + x) = 2x

Solution

Question 7

12

72

120

144

Solution

Question 8

10

11

12

14

Solution

Question 9

Increasing

Decreasing

Even

None of these

Solution

Question 10

1

\(3 \sqrt{10}\)

0

\(\sqrt{10}\)

Solution

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tan^-1 (y + c) = x

tan^-1 (y + x) = 2x