Mathematics Test-36

Given,

The length of the perpendicular from the origin to a line is 7 and the line makes an angle of 150 degrees with the positive direction of the \(x\)-axis.

We know that, If \(p\) is the length of the normal from origin to a line and \(\alpha\) is the angle made by the normal with the positive direction of \(x\)-axis then the equation of the line is given by:

\(x \cos \alpha+y \sin \alpha=p\)

Now, the equation of a line is

\(x \times \cos 150^{\circ} +y \times \sin 150^{\circ} =7\)

\(\Rightarrow x \times \cos (180^{\circ}-30^{\circ})+y \times \sin (180^{\circ}-30^{\circ}) =7\)\(\quad\quad(\because cos (180^{\circ}-\theta) = cos \theta,sin (180^{\circ}-\theta) = sin \theta)\)

\(\Rightarrow x \times \cos 30^{\circ}+y \times \sin 30^{\circ}=7\)

\(\Rightarrow \frac{\sqrt{3} x}{2}+\frac{y}{2}=7\) \(\quad(\because \cos 30^{\circ}=\frac{\sqrt{3}}{2} \text { and } \sin 30^{\circ}\)=\(\frac{1}{2})\)

\(\Rightarrow \sqrt{3} x+y=7 \times 2\)

\(\Rightarrow \sqrt{3} x+y=14\)

Riya may have any one of \(365\) days of the year as her birthday.

Similarly Kajal may have any one of \(365\) days as her birthday.

Total number of ways in which Riya and Kajal may have their birthday are:

\(365 \times 365\)

Then Riya and Kajal may have same birthday on any one of \(365\) days.

Probability \(= \frac{Number\ of\ favourable\ outcomes}{Total\ number\ of\ outcomes}\)

Therefore number of ways in which Riya and Kajal may have same birthday are:

\(=\frac{365}{365 \times 365} \)

\(=\frac{1}{365}\)

\( x d y-2 y d x=0 \)

\( \Rightarrow x d y=2 y d x \)

\( \Rightarrow \frac{d y}{y}=2 \frac{d x}{x}\)

Integrating both sides, we get

\( \Rightarrow \int \frac{ dy }{ y }=2 \int \frac{ dx }{ x } \)

\( \Rightarrow \ln y =2 \ln x +\ln c \)

\(\Rightarrow \ln y =\ln x ^2+\ln c \)

\( \Rightarrow \ln y -\ln x ^2=\ln c \)

\( \Rightarrow \ln \frac{ y }{ x ^2}=\ln c \)

\(\therefore y = x ^2 c\)

Since the line \(y=4 x+c\) touches the ellipse \(\frac{x^{2}}{4}+y^{2}=1\), we substitute \(y\) as \(4 x+c\) into the ellipse.

We get \(\frac{x^{2}}{4}+16 x^{2}+8 c x+c^{2}=1\)

\(\Rightarrow x^{2}+64 x^{2}+32 cx+4 c^{2}-4=0\)

The discriminant of this quadratic should be zero since we have only one value of \(x\).

\(\Rightarrow(32 c)^{2}=4 \times 65 \times\left(4 c^{2}-4\right)\)

\(\Rightarrow 1024 c^{2}=16 \times 65 \times\left(c^{2}-1\right)\)

\(\Rightarrow 64 c^{2}=65 \times\left(c^{2}-1\right)\)

\(\Rightarrow 64 c^{2}=65 c^{2}-65\)

\(\Rightarrow c^{2}=65\)

\(c=\pm\sqrt{65}\)

Thus, we have two values of \(c\).

Let A denotes, the set of Americans who like cheese and B denotes those who like apples.

Let the population of America be 100, then

\(n(A)=63\) and \(n(B)=76\)

Now. \(n(A \cup B)=n(A)+n(B)-n(A \cap B)\)

\(N(A \cup B)=63+76-n(A \cap B)\)

\(N(A \cap B)=139-n(A \cup B)\)

But \(n(A \cup B) \leq 100\)

Se. \(139-n(A \cup B) \geq 139-100\)

\(139-n(A \cup B) \geq 39\)

\(n(A \cap B) \geq 39\)

Now \(n (A \cap B) \leq n(A)\) and \(n(A \cap B) \leq n(B)\)

\(n(A \cap B) \leq 63\) and \(n(A \cap B) \leq 76\)

\(n(A \cap B) \leq 63\)

From Eqs. (i) and (ii)

\(39 \leq n(A \cap B) \leq 63\)

\(39 \leq x \leq 63\)

Thus,39% ≤ x ≤ 63% is the correct answer.

\(\frac{\cot \theta}{\cot \theta-\cot 3 \theta}+\frac{\tan \theta}{\tan \theta-\tan 3 \theta}\)

\(\frac{\cot \theta \tan \theta-\cot \theta \tan 3 \theta+\cot \theta \tan \theta-\tan \theta \cot 3 \theta}{(\cot \theta-\cot 3 \theta)(\tan \theta-\tan 3 \theta)} \quad\quad\quad \quad [\tan \theta\cot \theta =1]\)

\(\Rightarrow \frac{1-\cot \theta \tan3 \theta+1-\tan \theta \cot 3 \theta}{\cot \theta \tan \theta-\cot \theta \tan 3 \theta-\tan \theta \cot 3 \theta+\cot 3 \theta \tan 3 \theta}\)

\(\quad=\frac{2-\cot \theta \tan 3 \theta-\tan \theta \cot 3 \theta}{1-\cot \theta \tan 3 \theta-\tan \theta \cot 3 \theta+1}\)

\(\quad=\frac{2-\cot \theta \tan 3 \theta-\tan \theta \cot 3 \theta}{2-\cot \theta \tan 3 \theta-\tan \theta \cot 3 \theta}=1\)

As we know that,

In \(\triangle \mathrm{ALO}\) and \(\triangle \mathrm{O C M}\),

\(\mathrm{O A=O C}\) [radius of the same circle]

\(\angle \mathrm{O A L=\angle O C M}\) [Given]

\(\mathrm{\angle O L A=\angle O M C=90^{\circ}}\) [length of bisector of chord is \(\perp r\) to the chord]

\(\Delta\mathrm{ A L O \cong \triangle O C M}\) [Angle-Angle-Side Postulate]

\(\therefore \mathrm{A L=C M}\) [by Corresponding parts of Congruent triangles]

\(\mathrm{B L=B M}\) [\(\mathrm{L}\) and \(\mathrm{M}\) are the midpoints of \(\mathrm{A B}\) and \(\mathrm{B C}\) respectively]

Given,

Number of history books \(=3\)

Number of Science books \(=4\)

Number of Maths books \(=2\)

According to the question,

A number of ways history books can be arranged \(=3 !=3 \times 2 \times 1\) \(=6\)

A number of ways science books can be arranged \(=4 !=4 \times 3 \times 2 \times 1\) \(=24\)

A number of ways maths books can be arranged \(=2 !\) \(=2\)

A number of ways all three books can be arranged \(=3 !\) \(=3 \times 2 \times 1\) \(=6\)

\(\therefore\) Total number of ways \(=6 \times 6 \times 24 \times 2\) \(=1728\)

\(\sin 2 \mathrm{~A}=2 \sin \mathrm{A}\)

If \(\mathrm{A}=0\), then \(\sin (2 \times 0)=2 \sin 0^{\circ}\)

\(\Rightarrow \sin 0^{\circ}=2 \sin 0^{\circ}=0=2 \times 0=0\)

Which is correct

\(\therefore \mathrm{A}=0^{\circ}\)

If \(\mathrm{A}=30^{\circ}\), then

\(\begin{aligned} & \sin \left(2 \times 30^{\circ}\right)=2 \sin 30^{\circ} \\ \Rightarrow & \sin 60^{\circ}=2 \sin 30^{\circ} \\ \Rightarrow & \frac{\sqrt{3}}{2}=2 \times \frac{1}{2}=1 \end{aligned}\)

Which is not true

Similarly we can prove that

\(\mathrm{A}=45^{\circ}\) or \(\mathrm{A}=60^{\circ}\) are not true

\(\therefore\) Only \(A=0\) is true