Mathematics Test

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Mathematics Test - 7

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Weekly Quiz Competition

Question 1
1 / -0

If the two pairs of lines x² – 2mxy – y² = 0 and x² – 2nxy – y² = 0 are such that one of them represents the bisector of the angles between the other, then

A
mn + 1 = 0

B
mn – 1 = 0

C

$\frac{1}{m}+\frac{1}{n}=0$

D

$\frac{1}{m}-\frac{1}{n}=0$

Solution

Equation of bisectors of the angles between the lines is given by; $\mathrm{x}^{2}-2 \mathrm{mxy}-\mathrm{y}^{2}=0$

$\frac{x^{2}-y^{2}}{1-(-1)}=\frac{x y}{-m}$

$\mathrm{x}^{2}+\left(\frac{2}{\mathrm{m}}\right) \mathrm{xy}-\mathrm{y}^{2}=0 \ldots \ldots \ldots \ldots \ldots \ldots .(1)$

Given that $x^{2}-2 n x y-y^{2}=0 \ldots \ldots \ldots . .(2)$

Comparing (1) and (2) we get $\frac{2}{\mathrm{m}}=-2 \mathrm{n}$

$-\mathrm{mn}=1$

$\mathrm{nm}+1=0$

Hence option A is the correct answer.
Question 2
1 / -0

For any positive integer $n, \int \frac{d x}{x^{n+1}+x}$ is equal to :

A

$\frac{1}{n} \log _{e}\left(x^{n}+1\right)+c$

B

$\frac{1}{n} \log _{e}\left(\frac{1}{x^{n}+1}\right)+c$

C

$\frac{1}{n} \log _{e}\left(\frac{x^{n}}{x^{n}+1}\right)+c$

D

$\frac{1}{n+1} \log _{e}\left(\frac{x^{n}}{x^{n}+1}\right)+c$

Solution

Let $I=\int \frac{d x}{x\left(x^{n}+1\right)}$

$\Rightarrow \quad I=\int \frac{x^{n-1}}{x^{n}\left(x^{n}+1\right)} d x$

Put $x^{n}=t$

$\Rightarrow$ $n x^{n-1} d x=d t$

$\Rightarrow x^{n-1} d x=\frac{1}{n} d t$

$\therefore \quad I=\frac{1}{n} \int\left(\frac{1}{t}-\frac{1}{t+1}\right) d t$

$=\frac{1}{n} \log \left(\frac{t}{t+1}\right)=\frac{1}{n} \log \left(\frac{x^{n}}{x^{n}+1}\right)+c$

Hence option C is the correct answer.
Question 3
1 / -0

If $\overrightarrow{ a }$ is perpendicular to $\vec{b}$ and $\vec{c},|\vec{a}|=2,|\vec{b}|=3,|\vec{c}|=4$ and the angle between $\vec{b}$ and $\vec{c}$ is $\frac{2 \pi}{3},$ then $[\overrightarrow{ a } \overrightarrow{ b } \overrightarrow{ c }]$ is equal to :

A
4√3

B
6√3

C
12√3

D
18√3

Solution

$\because[\vec{a} \vec{b} \vec{c}]=\vec{a} \cdot(\vec{b} \times \vec{c})$

$=\vec{a} \cdot\left(|\vec{b}| \vec{c} \mid \sin \frac{2 \pi}{3} \hat{n}\right)$

$=|\vec{a}| \cdot|\vec{b} \| \vec{c}|\left(\sin \frac{2 \pi}{3}\right)=2 \times 3 \times 4 \times \frac{\sqrt{3}}{2}$

$=12 \sqrt{3}$

Hence option C is the correct answer.
Question 4
1 / -0

If f(x) = (x – 2)(x – 4)(x – 6)….(x – 2n), then f'(2) is equal to :

A
(-1)ⁿ 2^{n - 1}(n - 1)!

B
(-2)^{n - 1}(n - 1)!

C
(-2)ⁿ n!

D
(-1)^n-12ⁿ (n - 1)!

Solution

$\because f(x)=(x-2)(x-4)(x-6) \ldots .(x-2 n)$
Taking log on both sides, we get $\log f(x)=\log (x-2)+\log (x-4)$
$+\ldots+\log (x-2 n)$
On differentiating w.r.t. $x$, we get $\frac{1}{f(x)} f^{\prime}(x)=\frac{1}{(x-2)}+\frac{1}{(x-4)}$
$+\ldots+\frac{1}{(x-2 n)}$
$f(x)=(x-4)(x-6) \ldots(x-2 n)$
$+(x-2)(x-6) \ldots(x-2 n)$
$+\ldots+(x-2)(x-6) \ldots(x-2(n-1))$
$\therefore \quad f(2)=(-2)(-4) \ldots(2-2 n)$
$=(-2)^{(2)}(1 \cdot 2 \cdot \ldots(n-1))=(-2)^{n-1}(n-1) !$

Hence option B is the correct answer.
Question 5
1 / -0

The solution of the differential equation $x \frac{d y}{d x}+2 y=x^{2}$ is :

A

$y=\frac{x^{2}+c}{4 x^{2}}$

B

$y=\frac{x^{2}}{4}+c$

C

$y=\frac{x^{4}+c}{x^{2}}$

D

$y=\frac{x^{4}+c}{4 x^{2}}$

Solution

$x \frac{d y}{d x}+2 y=x^{2}$
$\therefore \frac{d y}{d x}+\frac{2}{x} y=x$
Integrating factor $=e^{\int \frac{2}{x} d x}=x^{2}$
Required solution is $y \cdot x^{2}=\int x^{3} d x=\frac{x^{4}}{4}+c^{1}=\frac{x^{4}+c}{4}$
$\therefore y=\frac{x^{4}+c}{4 x^{2}}$

Hence option D is the correct answer.
Question 6
1 / -0

The differential equation of all non - horizontal lines in a plane is :

A
d²y/dx² = 0

B
dx/dy = 0

C
dy/dx = 0

D
d²x/dy²= 0

Solution

Let $y=m x+c$ represents all non horizontal lines in a plane.

$\therefore \quad \frac{d y}{d x}=m$ and $\frac{d^{2} y}{d x^{2}}=0$

Hence option A is the correct answer.
Question 7
1 / -0

The equation of the sphere concentric with the sphere 2x² + 2y² + 2z² - 6x + 2y - 4z = 1 and double its radius is :

A
x² + y² + z² - x + y - z = 1

B
x² + y² + z² - 6x + 2y - 4z = 1

C
2x² + y² + z² - 6x + 2y - 4z - 15 = 0

D
2x² + 2y² + 2z² - 6x + 2y - 4z - 25 = 0

Solution

Equation of sphere is $2 x^{2}+2 y^{2}+2 z^{2}-6 x+2 y-4 z-1=0$
Radius of sphere is $\sqrt{\frac{9}{4}+\frac{1}{4}+\frac{4}{4}+\frac{1}{2}}=2$
Equation of family of concentric sphere is
$x^{2}+y^{2}+z^{2}-3 x+y-2 z+\lambda=0$....(i)
$\therefore$ According to question,
$\sqrt{\frac{9}{4}+\frac{1}{4}+1-\lambda}=4$
$\Rightarrow \quad \frac{14}{4}-\lambda=16$
$\Rightarrow \quad \lambda=\frac{14}{4}-16=-\frac{25}{2}$
$\therefore$ From Eq. (i)
$x^{2}+y^{2}+z^{2}-3 x+y-2 z-\frac{25}{2}=0$
$\Rightarrow 2 x^{2}+2 y^{2}+2 z^{2}-6 x+2 y-4 z-25=0$

Hence option D is the correct answer.
Question 8
1 / -0

Let f : R be a differentiable function and f(1) = 4. Find the value of
\begin{equation}\lim _{x \rightarrow 1} \int_{4}^{f(x)} \frac{2 t}{x-1} d t, \text { if } f^{\prime}(1)=2\end{equation}

A
16

B
8

C
4

D
2

Solution

$\lim _{x \rightarrow 1} \frac{\int_{4}^{f(x)} 2 t}{x-1}$

$=\lim _{x \rightarrow 1} \frac{2 f(x) \cdot f^{\prime}(x)}{1}$

$=2 f(1) \cdot f^{\prime}(1)=2 \cdot 4 \cdot 2=16$

Hence option A is the correct answer.
Question 9
1 / -0

If three natural numbers between 1 and 100 are selected randomly, then the probability that all are divisible by both 2 and 3 is :

A
4/105

B
4/33

C
4/35

D
4/1155

Solution

The numbers are divisible by both 2 and 3 means no.'s divisible by $6 .$

No. of no.'s divisible by 6 in 1 to 100 are 16

No. of ways of selecting 3 of them are $16_{\mathrm{C}_{3}}$

Probability $=\frac{16_{\mathrm{C}_{3}}}{100_{\mathrm{C}_{3}}}=\frac{16 \times 15 \times 14}{100 \times 99 \times 98}$

$=\frac{4}{1155}$

Hence option D is the correct answer.
Question 10
1 / -0

The sum to $n$ terms of the series $\frac{4}{3}+\frac{10}{9}+\frac{28}{27}+\ldots \ldots .$ is :

A

$\frac{3^{n}(2 n+1)+1}{2\left(3^{n}\right)}$

B

$\frac{3^{n}(2 n+1)-1}{2\left(3^{n}\right)}$

C

$\frac{3^{n} n-1}{2\left(3^{n}\right)}$

D

$\frac{3^{n}-1}{2\left(3^{n}\right)}$

Solution

upto terms
$= n + \frac{1}{3} (1 + \frac{1}{3} + \frac{1}{3^{2}} + . . . . . n t e r m)$

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Re-Attempt Weekly Quiz Competition

Mathematics Test - 7

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Mathematics Test - 7

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SHARING IS CARING

Question 1

mn + 1 = 0

mn – 1 = 0

\(\frac{1}{m}+\frac{1}{n}=0\)

\(\frac{1}{m}-\frac{1}{n}=0\)

Solution

Question 2

\(\frac{1}{n} \log _{e}\left(x^{n}+1\right)+c\)

\(\frac{1}{n} \log _{e}\left(\frac{1}{x^{n}+1}\right)+c\)

\(\frac{1}{n} \log _{e}\left(\frac{x^{n}}{x^{n}+1}\right)+c\)

\(\frac{1}{n+1} \log _{e}\left(\frac{x^{n}}{x^{n}+1}\right)+c\)

Solution

Question 3

4√3

6√3

12√3

18√3

Solution

Question 4

(-1)n 2n - 1(n - 1)!

(-2)n - 1(n - 1)!

(-2)n n!

(-1)n-12n (n - 1)!

Solution

Question 5

\(y=\frac{x^{2}+c}{4 x^{2}}\)

\(y=\frac{x^{2}}{4}+c\)

\(y=\frac{x^{4}+c}{x^{2}}\)

\(y=\frac{x^{4}+c}{4 x^{2}}\)

Solution

Question 6

d2y/dx2 = 0

dx/dy = 0

dy/dx = 0

d2x/dy2= 0

Solution

Question 7

x2 + y2 + z2 - x + y - z = 1

x2 + y2 + z2 - 6x + 2y - 4z = 1

2x2 + y2 + z2 - 6x + 2y - 4z - 15 = 0

2x2 + 2y2 + 2z2 - 6x + 2y - 4z - 25 = 0

Solution

Question 8

16

8

4

2

Solution

Question 9

4/105

4/33

4/35

4/1155

Solution

Question 10

\(\frac{3^{n}(2 n+1)+1}{2\left(3^{n}\right)}\)

\(\frac{3^{n}(2 n+1)-1}{2\left(3^{n}\right)}\)

\(\frac{3^{n} n-1}{2\left(3^{n}\right)}\)

\(\frac{3^{n}-1}{2\left(3^{n}\right)}\)

Solution

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(-1)ⁿ 2^{n - 1}(n - 1)!

(-2)^{n - 1}(n - 1)!

(-2)ⁿ n!

(-1)^n-12ⁿ (n - 1)!

d²y/dx² = 0

d²x/dy²= 0

x² + y² + z² - x + y - z = 1

x² + y² + z² - 6x + 2y - 4z = 1

2x² + y² + z² - 6x + 2y - 4z - 15 = 0

2x² + 2y² + 2z² - 6x + 2y - 4z - 25 = 0