Mathematics Test - 15

Given,

\(\sin 2 \theta=\cos 3 \theta, 0<\theta<\pi / 2\)

As we know that, \(\sin 2 \theta=2 \sin \theta \cos \theta\) and \(\cos 3 \theta=4 \cos ^{3} \theta-3 \cos \theta\)

\(\Rightarrow 2 \sin \theta \cos \theta=4 \cos ^{3} \theta-3 \cos \theta \)

\(\Rightarrow 2 \sin \theta=4 \cos ^{2} \theta-3\)

\(\Rightarrow 2 \sin \theta=4\left(1-\sin ^{2} \theta\right)-3=4-4 \sin ^{2} \theta-3\)

\(\Rightarrow 4 \sin ^{2} \theta+2 \sin \theta-1=0\)

Comparing the above equation with quadratic equation \(a x^{2}+b x+c=0, a=4, b=2\) and \(c=-1\)

Now substituting the values in the quadratic formula \(\frac{x=\left(-b \pm \sqrt{b}^{2}-4 a c\right) }{2 a}\) we get,

\(\sin \theta=\frac{-2 \pm \sqrt{-2^{2}-4(4)(-1)}}{2(4)}\)

\(=\frac{-2 \pm \sqrt{4+16}}{8}\)

\(=\frac{-2 \pm \sqrt{20}}{8}\)

\(=\frac{-1 \pm \sqrt{5}}{4}\)

Thus, \(\sin \theta=\frac{-1 \pm \sqrt{5}}{4}\)

Since, \(0<\theta<\frac{\pi}{ 2} \Rightarrow \theta\) lies between \(0^{\circ}\) to \(90^{\circ} \Rightarrow\) all ratios are positive.

\(\Rightarrow \sin \theta=\frac{-1+\sqrt{5}}{4}\)

As we know that, \(\cos 2 \theta=\cos ^{2} \theta-\sin ^{2} \theta\)

\(=1-2 \sin ^{2} \theta\)

\(\Rightarrow \cos 2 \theta=1-2\left(\frac{-1+\sqrt{5}}{4}\right)^{2}\)

\(=\frac{1+\sqrt{5}}{4}\)

\(\therefore\) The the value of \(\cos 2 \theta \) is \(\frac{1+\sqrt{5}}{4}\)

The sum of the first 20 terms of the given series can be written as:

\(\sqrt{5}+\sqrt{20}+\sqrt{45}+\sqrt{80}+\ldots\)

\(=\sqrt{5}+2 \sqrt{5}+3 \sqrt{5}+4 \sqrt{5}+\ldots+20 \sqrt{5}\)

\(=\sqrt{5}(1+2+\ldots+20)\)

Sum of consecutive numbers from 1 to n:

\(1+2+3+\ldots+\mathrm{n}=\frac{\mathrm{n}(\mathrm{n}+1)}{2}\)

\(=\sqrt{5} \times \frac{20 \times 21}{2}\)

\(=210 \sqrt{5}\)

We have to find the total number of words formed when the vowels always come together.

Consider the three vowels A, U, E to be one letter V then total letters are D, G, H, T, R and V. So the number of letters becomes 6

The total number of words formed will be=number of ways the 6 letters can be arranged ×number of ways the 3 vowels can be arranged

On putting the given values we get,

⇒ The total number of words formed=6!×3!

We know \(n !=n \times(n-1) ! \times \ldots 3,2,1\)

\(\Rightarrow\) The total number of words formed \(=6 \times 4 \times 5 \times 3 \times 2 \times 1 \times 3 \times 2 \times 1\)

On multiplying all the numbers we get,

\(\Rightarrow\) The total number of words formed \(=24 \times 5 \times 6 \times 6\)

\(\Rightarrow\) The total number of words formed=120 \(\times 36\)

\(\Rightarrow\) The total number of words formed \(=4320\)

The number of words formed from 'DAUGHTER' such that all vowels are together is 4320 .

Let \(P(h, k)\) be any point on the locus which is equidistant from the point A(1, 3) and B(-2, 1).

According to question,

\(P A=P B\)

\(\sqrt{(h-1)^{2}+(k-3)^{2}}=\sqrt{(h+2)^{2}+(k-1)^{2}}\)..(i)

Squaring both side in equation (i) we get,

\(\Rightarrow(h-1)^{2}+(k-3)^{2}=(h+2)^{2}+(k-1)^{2} \)\(\quad\quad(\because (a\pm b)^{2} = a^{2}+b^{2} \pm 2ab)\)

\(\Rightarrow h^{2}-2 h+1+k^{2}-6 k+9=h^{2}+4 h+4+k^{2}-2 k+1\)

\(\Rightarrow-2 h-6 k+10=4 h-2 k+5 \)

\(\Rightarrow 6 h+4 k=5\)

\(\therefore\)The locus of \((h,k)\) is \(6 x+4 y=5\).

\(P\left(\frac{\bar{A}}{~B}\right)=\frac{P(\overline{A} \cap B)}{P(B)}\)

\(P(\bar{A} \cap \bar{B})=P(\overline{A \cup B})\)

\(=\frac{P(\overline{A \cup B})}{P(B)}=\frac{1-P(A \cup B)}{P(B)}\)

\(\sin x \frac{d y}{d x}+\frac{y}{\sin x}=x \sin x e^{\cot x}\)

\( \frac{d y}{d x}+\frac{y}{\sin ^2 x}=x \cdot e^{\cot x}\)

It is form of \(\frac{d y}{d x}+P y=Q\)

\( \text { I. F. }=e^{\int p d x} \)

\( \text { I. F. }=e^{\int \operatorname{cosec}^2 x d x}=e^{-\cot x}\)

The solution of the linear equation is given by:

\(y(I . F .)=\int Q(I . F .) d x+c \)

\( y e^{-\cot x}=\int x e^{\cot x} \cdot e^{-\cot x} d x+c \)

\(y e^{-\cot x}=\int x d x+c \)

\( y e^{-\cot x}=\frac{x^2}{2}+c\)

The given first-order ordinary differential equation is \(\left(1-x^2\right) \frac{d y}{d x}-x y=1\).

Dividing both sides by \(1- x ^2\), we can get it in the standard form.

\( \Rightarrow \frac{ dy }{ dx }+\left(\frac{- x }{1- x ^2}\right) y =\frac{1}{1- x ^2} \)

\(\therefore P =\frac{- x }{1- x ^2} .\)

Let's calculate \(\int Pdx\).

\(\int P d x=\int \frac{-x}{1-x^2} d x\)

Substituting \(1-x^2=t\), so that \(-2 x d x=d t\), we get:

\(\Rightarrow \int Pdx =\frac{1}{2} \int \frac{1}{ t } dt\)

Using \(\int \frac{1}{x} d x=\log x+C\) and ignoring the constant \(C\), we get:

\(\Rightarrow \int Pdx =\frac{1}{2} \log t\)

Back substituting \(1-x^2=t\), we get:

\( \Rightarrow \int Pdx =\frac{1}{2} \log \left(1- x ^2\right) \)

\( \Rightarrow \int Pdx =\log \sqrt{1- x ^2}\)

Now, the integrating factor will be:

\(F = e ^{\int P dx } \)

\( \Rightarrow F = e ^{\log \sqrt{1- x ^2}} \)

\( \Rightarrow F =\sqrt{1- x ^2} \) which is the required integrating factor.

Given: Mean of 100 observations is 50 and standard deviation is 10.

If 5 is added to each observation

∵ There are total 100 observations

500 is added to sum of the observations and now dividing it by the no. of observations, we get that the mean is increased by 5.

As we know that, the standard deviation of N observations is given by:

\(\sigma=\sqrt{\frac{1}{N} \times \sum_{i=1}^{N}\left(x_{i}-\mu\right)^{2}}\) where, \(\mu\) is the arithmetic mean

∵ Every observation increased by 5 and mean also increased by 5 and standard deviation is the square root of the difference of mean and each observation divided by the number of observations.

So, standard deviation will remain same.

Thus, new mean is 55 and new standard deviation is still 10.

a, ar, ar², ... is an infinite geometric progression, then the sum of infinite geometric series is given by:

\(S_{\infty}=\frac{a}{1-r},|r|<1\)

\(7^{\frac{1}{7}} \times 7^{\frac{1}{7^{2}}} \times 7^{\frac{1}{7^{3}}} \times \ldots\) upto \(\infty=7^{\frac{1}{7}+\frac{1}{7^2}+\frac{1}{7^{3}}+\ldots+\infty}\)

We know that,

The series \(\frac{1}{7}+\frac{1}{7^{2}}+\frac{1}{7^{3}}+\ldots+\infty\) is an infinite geometric series with the first term \(\mathrm{a}=\frac{1 } 7\) and common ratio \(\mathrm{r}=\frac1  7\).

\(\Rightarrow \frac{1}{7}+\frac{1}{7^{2}}+\frac{1}{7^{3}}+\ldots+\infty=\frac{\frac{1}{7}}{1-\frac{1}{7}}=\frac{1}{6}\)

\(\Rightarrow 7^{\frac{1}{7}} \times 7^{\frac{1}{7^{2}}} \times 7^{\frac{1}{7^{3}}} \times \ldots\) upto \(\infty=7^{\frac{1}{6}}\)