Mathematics Test - 10

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

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

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

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

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

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

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

Given:

y = x³ – 3x + 2

Therefore, y’ = 3x²-3.

For a point of absolute maximum or minimum, y’ = 0

Hence, x = ±1

Let y = f(x)

Therefore, f(0) = 0³-3(0) +2 = 2

f(1) = 1³– 3(1)+2 = 0

f(2) = 2³ – 3(2) +2 = 4

Hence, f(x) achieves a maximum value of 4 when x = 2.

$z_1=1+\sqrt{3}i=2e\frac{\pi i}{3}$

Since $\angle z_{1}o{z}_3=\frac{2\pi }{3}$  and $\angle z_{2}o{z}_3=\frac{2\pi }{3}$ 

$z_2=z_1{e}^{2\pi i/3}$ 

and $z_3=z_{2}e\frac{2\mathrm{\pi i}}{3}$ 

$\Rightarrow z_2=2e^{\pi i}=2(\cos \pi +i\sin \pi )=-2$

and $z_3=2e\frac{5\pi i}{3}$

$=2\left[\cos \left(2\pi -\frac{\pi }{3}\right)+i\sin \left(2\pi -\frac{\pi }{3}\right)\right]$

$=2\left[\cos \left(2\pi -\frac{\pi }{3}\right)+i\sin \left(2\pi -\frac{\pi }{3}\right)\right]$

$=1-\sqrt{3}i$

$\frac{d^{2}y}{d{x}^2}+\cos \left(\frac{dy}{dx}\right)=0$

1. As the given differential equation is not a polynomial equation in derivatives, the degree of this equation is not defined

So, this statement is true

2. Here Highest derivative:$\frac{d^{2}y}{d{x}^2}$

∴ Order = 2

Let \(1+\log x=t\)\(\quad \dots (1)\)

Differentiating w.r.t. \(x\), we get

\(\frac{1}{x} d x=d t\)

Now,

New limit to \(eq^{n}(1)\)

\(x=1 \Rightarrow t=1 \)

\(x=2 \Rightarrow t=1+\log 2\)

\(\therefore \int_{1}^{2} \frac{1}{x(1+\log x)^{2}} d x =\int_{1}^{1+\log 2} \frac {xd t}{x\times t^{2}} \)

\(=\int_{1}^{1+\log 2} \frac{d t}{t^{2}} \)

\(=\left[\frac{-1}{t}\right]_{1}^{1+\log 2} \)

After put the limit, We get

\(=\left[\frac{-1}{1+\log 2}-\frac{(-1)}{1}\right] \)

\(=\left[\frac{-1}{1+\log 2}+1\right] \)

\( =\left[\frac{-1+1+\log 2}{1+\log 2}\right] \quad[\because \log e=1] \)

\( =\left[\frac{\log 2}{1+\log 2}\right] \quad~\quad[\log a+\log b=\log a b] \)

\( =\frac{\log 2}{\log e+\log 2} \)

\( =\frac{\log 2}{\log 2e}\)

Let the first element of A.P. is 'a' and common difference is 'd'.

According to the question,

T25 = T15 + 70

As we know,

\(a_{n}=a+(n-1) d\)

⇒ a + (25 – 1)d = a + ( 15 – 1)d + 70

⇒ 24d – 14d = 70

⇒ 10d = 70

⇒ d = 7

∴ The common difference is 7.

As we know,

The scalar triple product of three vectors is zero if any two of them are equal

If \(\vec{a}\) is perpendicular to the vectors \(\vec{b}\) then \(\vec{a} \cdot \vec{b}=0\)

Given: \(\vec{c}=\vec{a}+\vec{b}\), where \(|\vec{a}|=|\vec{b}| \neq 0\)

Statement 1: \(\vec{c}\) is perpendicular to \((\vec{a}-\vec{b})\).

First let's find out \(\vec{c} \cdot(\vec{a}-\vec{b})=(\vec{a}+\vec{b}) \cdot(\vec{a}-\vec{b})\)

\(\Rightarrow(\vec{a}+\vec{b}) \cdot(\vec{a}-\vec{b})=|\vec{a}|^{2}-\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{a}-|\vec{b}|^{2}\)

As we know that, \(\vec{a} \cdot \vec{b}=\vec{b} \cdot \vec{a}\)

\(\Rightarrow(\vec{a}+\vec{b}) \cdot(\vec{a}-\vec{b})=|\vec{a}|^{2}-|\vec{b}|^{2}\)

\(\because\) It is given that \(|\vec{a}|=|\vec{b}| \neq 0\)

\(\Rightarrow(\vec{a}+\vec{b}) \cdot(\vec{a}-\vec{b})=|\vec{a}|^{2}-|\vec{b}|^{2}=0 \)

\(\Rightarrow \vec{c} \cdot(\vec{a}-\vec{b})=0\)

So, statement 1 is true.

Statement 2: \(\vec{c}\) is perpendicular to \(\vec{a} \times \vec{b}\).

First let's find out \(\vec{c} \cdot(\vec{a} \times \vec{b})=(\vec{a}+\vec{b}) \cdot(\vec{a} \times \vec{b})\)

\(\Rightarrow(\vec{a}+\vec{b}) \cdot(\vec{a} \times \vec{b})=\vec{a} \cdot(\vec{a} \times \vec{b})+\vec{b} \cdot(\vec{a} \times \vec{b})\)

As we know that, the scalar triple product of three vectors is zero if any two of them are equal \(\Rightarrow(\vec{a}+\vec{b}) \cdot(\vec{a} \times \vec{b})=0\)

So, statement 2 is true.

Given,

\(\cos \left\{\cos ^{-1} \frac{4}{5}+\cos ^{-1} \frac{12}{13}\right\}\)

\(=\cos ^{-1}\left[\frac{4}{5} \times \frac{12}{13}-\sqrt{1-\left(\frac{4}{5}\right)^{2}} \sqrt{1-\left(\frac{12}{13}\right)^{2}}\right] \) \(\quad\quad (\because cos^{-1}x + cos^{-1}y = cos^{-1}(xy - \sqrt{1−x^{2}} × \sqrt{1−y^{2}}))\)

\(=\cos ^{-1}\left[\frac{48}{65}-\frac{3}{5} \times \frac{5}{13}\right] \)

\(=\cos ^{-1}\left[\frac{48}{65}-\frac{15}{65}\right] \)

\(=\cos ^{-1} \frac{33}{65}\)

Now,

\(\cos \left\{\cos ^{-1} \frac{4}{5}+\cos ^{-1} \frac{12}{13}\right\} \)

\(=\cos \left(\cos ^{-1} \frac{33}{65}\right) \)

\(=\frac{33}{65} \quad\left(\because \cos \left(\cos ^{-1} x\right)=x\right)\)

Concept:

Two events are independent if the incidence of one event does not affect the probability of the other event.

If A and B are two independent events in a sample space S, then P (A ∩ B) = P (A) P (B)

Calculation:

Event A is choosing a yellow pencil first, and Event B is choosing a yellow pencil second.

Initially, there are 12 pencils, 7 of which are yellow.

Probability the first pencil is yellow = P(A) =\(\frac{7}{12}\)

If a yellow pencil is chosen, there will be 11 pencils left, 6 of which are yellow.

Probability the second pencil is yellow = P(B) =\(\frac{6}{11}\)

Given: Two pencils are chosen at random from the box without replacement.

So events are independent of each other.

Probability they are both yellow = P(A ∩ B) = P(A) × P(B)

\(=\frac{7}{12} \times \frac{6}{11}=\frac{7}{22}\)