Mathematics Test-13

The equation of line is,

\(\Rightarrow \frac{(x-1)}{\cos \theta}=\frac{(y+2)}{\sin \theta}=\frac{(z-3)}{0}\)

The plane is perpendicular to line.

\(\therefore\) The direction ratio at normal of the plane is \((\cos \theta, \sin \theta, 0)\).

\(\Rightarrow\) It is given that, the plane passes through \(x\)-axis then \((0,0,0)\) lies on the plane.

\(\therefore\) The equation of the plane is

\(\Rightarrow \cos \theta(x-0)+\sin \theta(y-0)+0(z-0)=0 \)

\(\Rightarrow \cos \theta \cdot x+\sin \theta \cdot y=0\)

\(\therefore\) The equation of plane is \(x \cdot \cos \theta+y \cdot \sin \theta=0\).

Given,

First term of AP = a = 10

Last term of AP = l = 50

Sum of n terms of an AP = 300

As we know,

sum of \(n\) terms of an \(A P=S_{n}=\frac{n}{2} \times[a+l]\)

\(\Rightarrow 300=\frac{\mathrm{n}}{2}(10+50)\)

\(\Rightarrow 300=\frac{\mathrm{n}}{2} \times 60\)

\(\Rightarrow 5=\frac{\mathrm{n}}{2}\)

\(\therefore \mathrm{n}=10\)

Given,

\(n(X)=17, n(Y)=23, n(X \cup Y)=38\)

We know that,

\(\Rightarrow\)\(n(X \cup Y)=n(X)+n(Y)-n(X \cap Y)\)

\(\therefore 38=17+23-n(X \cap Y)\)

\(\Rightarrow\)\(n(X \cap Y)=40-38=2\)

\(\Rightarrow\)\(n(X \cap Y)=2\)

Given:

\(\left(\frac{i}{5}+\frac{5}{i}\right)\)

We know,

\(z=x+i y\)

\(\arg (z)=\arg (x+i y)=\tan ^{-1}(\frac{y }{x})\)

Therefore, the argument \(\theta\) is represented as:

\(\theta=\tan ^{-1}(\frac{{y} }{ {x}})\)

\(\tan \left(\frac{\pi}{2}\right)=\) infinity

\(i^{2}=-1\)

Let \(z=\left(\frac{i}{5}+\frac{5}{i}\right)\)

\(z=\frac{i}{5}+\frac{5 i}{i^{2}}\)

\(z=\frac{i}{5}-5 i\)

\(=\frac{-24}{5} {i}\)

So, \(\arg (z)=\tan ^{-1}\left(\frac{\frac{-24} { 5}}{0}\right)\)

\(=-\tan ^{-1} \infty\)

\(=-\frac{\pi}{2}\)

Given:

\(6 x^2+4 x-a=0\)

The roots of the given quadratic equation are real and equal.

So, the discriminant will be \(0\).

\(b^2-4 a c=0\)

\((4)^2-4 \times 6 \times(-a)=0\)

\(16+24 a=0\)

\(2+3 a=0\)

\(3 a+2=0\)

Given:

\(p=\sec \theta-\tan \theta\) and \(q=\operatorname{cosec} \theta+\cot \theta\)

\({p}=\sec \theta-\tan \theta\)

\(=\frac{1}{\cos \theta}-\frac{\sin \theta}{\cos \theta}\)

\(=\frac{1-\sin \theta}{\cos \theta} \)

\({q}=\operatorname{cosec} \theta+\cot \theta\)

\(=\frac{1}{\sin \theta}+\frac{\cos \theta}{\sin \theta}\)

\(=\frac{1+\cos \theta}{\sin \theta} \)

\({p}+{q}({p}-1)=\frac{1-\sin \theta}{\cos \theta}+\frac{1+\cos \theta}{\sin \theta}\left(\frac{1-\sin \theta}{\cos \theta}-1\right) \)

\(=\frac{1-\sin \theta}{\cos \theta}+\frac{1+\cos \theta}{\sin \theta}\left(\frac{1-\sin \theta-\cos \theta}{\cos \theta}\right) \)

\(=\frac{1-\sin \theta}{\cos \theta}+\frac{\left(1-\sin \theta-\cos \theta+\cos \theta-\sin \theta \cos \theta-\cos ^{2} \theta\right)}{\sin \theta \cos \theta} \)

\(=\frac{\left(\sin \theta-\sin ^{2} \theta\right)+\left(1-\sin \theta-\cos \theta+\cos \theta-\sin \theta \cos \theta-\cos ^{2} \theta\right)}{\sin \theta \cos \theta} \)

\(=\frac{-\sin \theta \cos \theta}{\sin \theta \cos \theta} \)

\(=-1\)

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,

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

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

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

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

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

Let the smaller of the two consecutive odd positive integers be \(x\). Then, the second integer will be \(x+2\). According to the question,

\(x^2+(x+2)^2 =290\)

\(\Rightarrow x^2+x^2+4 x+4 =290\)

\(\Rightarrow 2 x^2+4 x-286 =0\)

\(\Rightarrow x^2+2 x-143 =0\)

Which is a quadratic equation in \(x\).

Using the quadratic formula, we get

\(x=\frac{-2 \pm \sqrt{4+572}}{2}=\frac{-2 \pm \sqrt{576}}{2}=\frac{-2 \pm 24}{2}\)

\(\Rightarrow x=11\) or \(x=-13\)

But \(x\) is given to be an odd positive integer.

Therefore, \(x \neq-13, x=11\).

Thus, the two consecutive odd integers are \(11\) and \(13\).

Given: \(\mathrm{A}=\{1,4\}, \mathrm{B}=\{2,3\}, \mathrm{C}=\{3,5\}\)

Here, we have to find \((A \times B) \cap(A \times C)\)

As we know that, \(A \times B=\{(a, b) \mid a \in {A}\) and \(b \in B\}\)

\(\Rightarrow A \times B=\{(1,2),(1,3),(4,2),(4,3)\}\quad\quad\)......(1)

\(\Rightarrow A \times C=\{(1,3),(1,5),(4,3),(4,5)\}\quad\quad\)......(2)

From (1) and (2) we can say that,

\(\Rightarrow(A \times B) \cap(A \times C)=\{(1,3),(4,3)\}\)