Mathematics Test

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Mathematics Test - 23

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Weekly Quiz Competition

Question 1
1 / -0

Find the general solution of given differential equation \(\frac{x d y}{d x}+3 y=4 x^3 ?\)

A
\(x^3 \cdot y=\frac{2}{3} \cdot x^5+c\)

B
\(x^3 \cdot y=3 x^6+c\)

C
\(x^3 \cdot y=\frac{2}{3} \cdot x^6+c\)

D
None of the above

Solution

Given that,

\(\Rightarrow \frac{ xdy }{ dx }+3 y =4 x ^3\)

Now,

\(\Rightarrow \frac{ dy }{ dx }+\frac{3 y }{ x }=4 x ^2\)

By comparing with \(\frac{ dy }{ dx }+ Py = Q\)

\( \Rightarrow P=\frac{3 } x \text { and } Q=4 x^2 \)

\( \Rightarrow \text { I.F. }=e^{\int P d x}=e^{\int \frac{3}{x} d x} \)

\(\Rightarrow \text { I.F. }=e^{3 \ln x} \)

\( \Rightarrow \text { I.F. }=e^{\ln x^3}\)

\( \Rightarrow \text { I.F. }=x^3\left(\because e^{\ln x}=x\right)\)

Now general solution will be,

\( \Rightarrow y \cdot( I \cdot F \cdot)=\int( Q \cdot( I \cdot F \cdot)) dx + c \)

\( \Rightarrow y \cdot\left( x ^3\right)=\int\left(4 x ^2 \cdot\left( x ^3\right)\right) dx + c\)

\( \Rightarrow x ^3 \cdot y =\int 4 x ^5 dx + c \)

\( \Rightarrow x ^3 \cdot y =4 \frac{ x ^6}{6}+ c \)

\( \Rightarrow x^3 \cdot y=\frac{2}{3} \cdot x^6+c\)
Question 2
1 / -0

Find the distance between the planes 2x - 3y + 6z - 5 = 0 and 6x - 9y + 18z + 20 = 0.

A
\(\frac{2}{3}\)

B
\(\frac{5}{3}\)

C
\(\frac{1}{3}\)

D
None of these

Solution

Given,

Two planes \(2 x-3 y+6 z-5=0\) and \(6 x-9 y+18 z+20=0\)

The plane \(6 x-9 y+18 z+20=0\) can be re-written as \(2 x-3 y+6 z+\frac{20 }{ 3}=0\).

So, the planes \(2 x-3 y+6 z-5=0\) and \(2 x-3 y+6 z+\frac{20 }{ 3}=0\) are parallel.

As we know that, distance between two parallel planes \(ax +b y+c z+d_{1}=0\) and \(a x+b y+c z+d_{2}=0\) is given by,

\(\left|\frac{d_{1}-d_{2}}{\sqrt{a^{2}+b^{2}+c^{2}}}\right|\)

\(\therefore d_{1}=-5, d_{2}=\frac{20 }{ 3}, a=2, b=-3\) and \(c=6\)

So, the distance between the given parallel planes \(=\left|\frac{-5-\frac{20}{3}}{\sqrt{2^{2}+(-3)^{2}+6^{2}}}\right|\)

\(=\left|\frac{\frac{-15-20}{3}}{\sqrt{4+9+36}}\right|\)

\(=\left|\frac{\frac{-35}{3}}{\sqrt{49}}\right|\)

\(=\left|\frac{-35}{7\times 3}\right|\)

\(=\frac{5}{3}\) units
Question 3
1 / -0

The region represented by \(\{\mathrm{z}=\mathrm{x}+\mathrm{iy} \in \mathrm{C}:|\mathrm{z}|-\operatorname{Re}(\mathrm{z}) \leq 1\}\) is also given by the inequality:

A
\(y^2 \geq 2(x+1)\)

B
\(y^2 \leq 2\left(x+\frac{1}{2}\right)\)

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

D
\(y^2 \geq x+1\)

Solution

\(\because|z|-\operatorname{Re}(z) \leq 1(\because z=x+i y) \\\)

\(\Rightarrow \sqrt{x^2+y^2}-x \leq 1 \Rightarrow \sqrt{x^2+y^2} \leq 1+x \\\)

\(\Rightarrow x^2+y^2 \leq 1+x^2+2 x \\\)

\(\Rightarrow y^2 \leq 1+2 x \Rightarrow y^2 \leq 2\left(x+\frac{1}{2}\right)\)
Question 4
1 / -0

Let \(y=\log _e\left(\frac{1-x^2}{1+x^2}\right),-1

A
732

B
746

C
742

D
736

Solution

\(y=\log _e\left(\frac{1-x^2}{1+x^2}\right) \\\)

\(\frac{d y}{d x}=y^{\prime}=\frac{-4 x}{1-x^4}\)

Again,

\(\frac{d^2 y}{d x^2}=y^{\prime \prime}=\frac{-4\left(1+3 x^4\right)}{\left(1-x^4\right)^2}\)

Again

\(y^{\prime}-y^{\prime \prime}=\frac{-4 x}{1-x^4}+\frac{4\left(1+3 x^4\right)}{\left(1-x^4\right)^2}\)

at \(\mathrm{x}=\frac{1}{2}\),

\(\mathrm{y}^{\prime}-\mathrm{y}^{\prime \prime}=\frac{736}{225}\)

Thus \(225\left(y^{\prime}-y^{\prime \prime}\right)=225 \times \frac{736}{225}=736\)
Question 5
1 / -0

If \(n \in N,\) then \(121^{n}-25^{n}+1900^{n}-(-4)^{n}\) is divisible by which one of the following?

A
\(1904\)

B
\(2000\)

C
\(2002\)

D
\(2006\)

Solution

\((121)^{n}-25^{n}+1900^{n}-(-4)^{n}\)
For \(n=1\)

\(=121-25+1900-(-4)\)
\(=121-25+1900+4\)
\(=2025-25\)
\(=2000\)
Question 6
1 / -0

The function \(f(x)=\sqrt{\cos (\sin x)}+\sin ^{-1}\left(\frac{1+x^{2}}{2 x}\right)\) is defined for:

A
\(x \in\{-1,1\}\)

B
\(x \in[-1,1]\)

C
\(x \in R\)

D
\(x \in(-1,1)\)

Solution

Given function is \(\sqrt{\cos (\sin x)}+\sin ^{-1}\left(\frac{1+x^{2}}{2 x}\right)\)
Range of \(\sin x\) is \([-1,1] .\) so, \(\cos (-{ve})\) also positive value.
\(\Rightarrow\) Defined for all real number .....(1)
\(\sin ^{-1}(x)\) is defined for \(-1 \leq x \leq 1\)
\(-1 \leq \frac{1+x^{2}}{2 x} \leq 1\)
Case (1) \(x^{2}+1 \geq-2 x\)
\((x+1)^{2} \geq 0\)
Always positive and this is equal to zero when \(x=-1\)
case (2) \(1+x^{2} \leq 2 x\)
\((x-1)^{2} \leq 0\)
This is equal to zero when \({x}=1\)
The function domain will be inter section of \((1)\) and \((2)\) function
For function (2) \(x \in\{-1,1\}\)
The final answer for this is \({x} \in\{-1,1\}\)
Question 7
1 / -0

Given the sets \(A =\{2,3,4,5,6,7\}, B =\{6,7,8\}\) and \(C =\{1,5,8,9\}\) then find \(A \cap(B \cup C)\) and number of elements in \(A \cap(B \cup C)\)

A
\(\{6,7,8\}, 3\)

B
\(\{5,6,7\}, 3\)

C
\(\{4,5,6,7\}, 4\)

D
\(\{4,5,6\}, 3\)

Solution

Given:

\(A=\{2,3,4,5,6,7\}, B=\{6,7,8\}\) and \(C=\{1,5,8,9\}\)

Let, \(P=(B \cup C)\)

\(=\{6,7,8\} \cup\{1,5,8,9\}\)

\(=\{1,5,6,7,8,9\}\)

\(\Rightarrow A \cap P=\{2,3,4,5,6,7\} \cap\{1,5,6,7,8,9\}\)

\(\Rightarrow A \cap(B \cup C)=\{5,6,7\}\)

Therefore, number of elements \(=3\)
Question 8
1 / -0

Let \(\overrightarrow{O A}=\vec{a}, \overrightarrow{O B}=12 \vec{a}+4 \vec{b}\) and \(\overrightarrow{O C}=\vec{b}\), where \(O\) is the origin. If \(S\) is the parallelogram with adjacent sides \(\mathrm{OA}\) and \(\mathrm{OC}\), then \(\frac{\text { area of the quadrilateral } O A B C}{f S}\) is equal to:

A
6

B
10

C
7

D
8

Solution

\( \text { Area of parallelogram, } \mathrm{S}=|\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}| \\\)

\( \text { Area of quadrilateral }=\operatorname{Area}(\triangle \mathrm{OAB})+\operatorname{Area}(\triangle \mathrm{OBC}) \\\)

\( =\frac{1}{2}\{|\overrightarrow{\mathrm{a}} \times(12 \overrightarrow{\mathrm{a}}+4 \overrightarrow{\mathrm{b}})|+|\overrightarrow{\mathrm{b}} \times(12 \overrightarrow{\mathrm{a}}+4 \overrightarrow{\mathrm{b}})|\} \\\)

\(=8|(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}})| \\\)

\( \text { Ratio }=\frac{8|(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}})|}{|(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}})|}=8\)
Question 9
1 / -0

If \(\sec ^{2} \theta+\tan ^{2} \theta=7\) then, find the value of \(\sec \theta\):

A
4

B
2

C
1

D
0

Solution

Given,

\(\sec ^{2} \theta+\tan ^{2} \theta=7\) ......(i)

Adding 1 both sides in eqaution (i) we get,

\(\sec ^{2} \theta+\tan ^{2} \theta+1=7+1\)

\(\Rightarrow \sec ^{2} \theta+\sec ^{2} \theta=8\)\(\quad\quad(\because \ 1+\tan ^{2} \theta=\sec ^{2} \theta)\)

\(\Rightarrow 2 \sec ^{2} \theta=8\)

\(\Rightarrow \sec ^{2} \theta=4\)

\(\Rightarrow\sec \theta=2\)

\(\therefore \sec \theta=2\)
Question 10
1 / -0

If the position vectors of the vertices \(A, B\) and \(C\) of a \(\triangle A B C\) are respectively \(4 \hat{i}+7 \hat{j}+8 \hat{k}, 2 \hat{i}+3 \hat{j}+4 \hat{k}\) and \(2 \hat{i}+5 \hat{j}+7 \hat{k}\), then the position vector of the point, where the bisector of \(\angle \mathrm{A}\) meets \(\mathrm{BC}\) is:

A
\(\frac{1}{2}(4 \hat{i}+8 \hat{j}+11 \hat{k})\)

B
\(\frac{1}{3}(6 \hat{i}+13 \hat{j}+18 \hat{k})\)

C
\(\frac{1}{4}(8 \hat{i}+14 \hat{j}+9 \hat{k})\)

D
\(\frac{1}{3}(6 \hat{i}+11 \hat{j}+15 \hat{k})\)

Solution

Suppose angular bisector of \(A\) meets \(B C\) at \(D\left(x_z, y_z z\right)\)

Using angular bisector theorem,

\(\frac{\mathrm{AB}}{\mathrm{AC}}=\frac{\mathrm{BD}}{\mathrm{DC}} \\\)

\(\frac{\mathrm{BD}}{\mathrm{DC}}=\frac{\sqrt{(4-2)^2+(7-3)^2+(8-4)^2}}{\sqrt{(4-2)^2+(7-5)^2+(8-7)^2}} \\\)

\( =\frac{\sqrt{2^2+4^2+4^2}}{\sqrt{2^2+2^2+1^2}}=\frac{6}{3}=2\)

\( \text { So, } \mathrm{D}(\mathrm{x}, \mathrm{y}, \mathrm{z}) \equiv\left(\frac{(2)(2)+(1)(2)}{2+1}, \frac{(2)(5)+(1)(3)}{2+1}\right. \\\)

\( \left.\frac{(2)(7)+(1)(4)}{2+1}\right) \\\)

\( \mathrm{D}(\mathrm{x}, \mathrm{y}, \mathrm{z}) \equiv\left(\frac{6}{3}, \frac{13}{3}, \frac{18}{3}\right)\)

Therefore, position vector of point \(P=\frac{1}{3}(6 \hat{i}+13 \hat{j}+18 \hat{k})\)

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Re-Attempt Weekly Quiz Competition

Mathematics Test - 23

Result

Mathematics Test - 23

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SHARING IS CARING

Question 1

\(x^3 \cdot y=\frac{2}{3} \cdot x^5+c\)

\(x^3 \cdot y=3 x^6+c\)

\(x^3 \cdot y=\frac{2}{3} \cdot x^6+c\)

None of the above

Solution

Question 2

\(\frac{2}{3}\)

\(\frac{5}{3}\)

\(\frac{1}{3}\)

None of these

Solution

Question 3

\(y^2 \geq 2(x+1)\)

\(y^2 \leq 2\left(x+\frac{1}{2}\right)\)

\(y^2 \leq x+\frac{1}{2}\)

\(y^2 \geq x+1\)

Solution

Question 4

732

746

742

736

Solution

Question 5

\(1904\)

\(2000\)

\(2002\)

\(2006\)

Solution

Question 6

\(x \in\{-1,1\}\)

\(x \in[-1,1]\)

\(x \in R\)

\(x \in(-1,1)\)

Solution

Question 7

\(\{6,7,8\}, 3\)

\(\{5,6,7\}, 3\)

\(\{4,5,6,7\}, 4\)

\(\{4,5,6\}, 3\)

Solution

Question 8

6

10

7

8

Solution

Question 9

4

2

1

0

Solution

Question 10

\(\frac{1}{2}(4 \hat{i}+8 \hat{j}+11 \hat{k})\)

\(\frac{1}{3}(6 \hat{i}+13 \hat{j}+18 \hat{k})\)

\(\frac{1}{4}(8 \hat{i}+14 \hat{j}+9 \hat{k})\)

\(\frac{1}{3}(6 \hat{i}+11 \hat{j}+15 \hat{k})\)

Solution

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