Mathematics Test - 9

Given:

\(\frac{\sin 7 x+6 \sin 5 x+17 \sin 3 x+12 \sin x}{\sin 6 x+5 \sin 4 x+12 \sin 2 x}\)

\(=\frac{\sin 7 x+\sin 5 x+5 \sin 5 x+5 \sin 3 x+12 \sin 3 x+12 \sin x}{\sin 6 x+5 \sin 4 x+12 \sin 2 x}\)

As we know,

\(\sin x+\sin y=2 \sin \left(\frac{x+y}{2}\right) \cos \left(\frac{x-y}{2}\right)\)

\(=\frac{2 \sin 6 x \cos x+10 \sin 4 x \cos x+24 \sin 2 x \cos x}{\sin 6 x+5 \sin 4 x+12 \sin 2 x}\)

\(=\frac{2 \cos x(\sin 6 x+5 \sin 4 x+12 \sin 2 x)}{\sin 6 x+5 \sin 4 x+12 \sin 2 x}\)

\(=2 \cos x\)

Substitute the line value in parabola equation, 

\(\Rightarrow 2 x+3=x^2 \)

\(\therefore x=-1,3\)

By Putting Value of x in \(y=x^2\) the required point of intersection is (-1,1) and (3,9)

Area bounded by curve \(=\int_{-1}^3 \)[ Top - Bottom ] dx

\(=\int_{-1}^3\left[(2 x+3)-x^2\right] d x \)

\( =\frac{32}{3} \) sq. unit

Thestandard error (SE)of a statistic is theapproximate standard deviation of a statistical sample population.

The standard error is a statistical term that measures the accuracy with which a sample distribution represents a population by using standard deviation.

In statistics,a sample mean deviates from the actual mean of a population; this deviation is the standard error of the mean.

Standard deviation (SD) and the estimated standard error of the mean (SEM) are used to present the characteristics of sample data and to explain statistical analysis results.

For a sample size 'n' and\(\bar{x}\),

The standard deviation (SD) is given by:

\(\sigma=\sqrt{\frac{\sum_{i=1}^{n}\left(x_{i}-\bar{x}\right)^{2}}{n-1}}\)

Variance \(=\sigma^{2}\)

Now the standard error is given by:

\(\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}\)

From the above formula, it is clear that the Standard error of the sample mean is the ratio of standard deviation and the square root of the number of observations.

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

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

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

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

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

Let \(\theta\) be the angle of inclination of the given line with the positive direction of \(x\) -axis in the anticlockwise sense.

Then its slope is given by \(m=\tan \theta\)

Given, slope is positive.

\(\Rightarrow \tan \theta<0\)

\(\Rightarrow \theta\) lies between \(0\) and \(180\) degree.

\(\Rightarrow \theta\) is an obtuse angle.

Given: \(\mathrm{f}(\mathrm{x})=\sqrt{1-\mathrm{e}^{-\mathrm{x}^{2}}}\)

\(\mathrm{f}^{\prime}(\mathrm{x})=\frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{f}(\mathrm{x}))\)

\(=\frac{\mathrm{d}}{\mathrm{dx}}\left(\sqrt{1-\mathrm{e}^{-\mathrm{x}^{2}}}\right)\)

\(=\frac{1}{2 \sqrt{1-\mathrm{e}^{-\mathrm{x}^{2}}}} \times \frac{\mathrm{d}}{\mathrm{dx}}\left(1-\mathrm{e}^{-\mathrm{x}^{2}}\right) \quad\left(\because \frac{\mathrm{d}}{\mathrm{dx}}(\sqrt{\mathrm{x}})=\frac{1}{2 \sqrt{\mathrm{x}}}\right)\)

\(\Rightarrow \mathrm{f}^{\prime}(\mathrm{x})=\frac{1}{2 \sqrt{1-\mathrm{e}^{-\mathrm{x}^{2}}}}\left(-\mathrm{e}^{-\mathrm{x}^{2}}\right) \frac{\mathrm{d}}{\mathrm{dx}}\left(-\mathrm{x}^{2}\right)\)

\(=\frac{\left(-\mathrm{e}^{-{x}^{2}}\right)}{2 \sqrt{1-{e}^{-{x}^{2}}}}(-2 {x})\)

\(=\frac{x \mathrm{e}^{-x^{2}}}{\sqrt{1-{e}^{-{x}^{2}}}}\)

Now, at \(x=0, f^{\prime}(x)=\frac{0 }{ 0}\)

So, \(f^{\prime}(x)\) is defined for all values of \(x\) except \(x=0\)

\(\Rightarrow f(x)\) is differentiable on \((-\infty, 0) \cup(0, \infty)\)

n observations are given by:

\(a_{1}, a_{2}, a_{3}, \ldots ..a_{m}+...a_{n}\)

Mean of "m" observation is "\(n\)"

\( \frac{a_{1}+a_{2}+a_{3} \ldots \ldots a_{m}}{m}=n \quad \quad \ldots\)(1)

Mean of remaining observation is \(m\) so number of such observation are \(n -3\)

\(\text { Mean }=\frac{a_{m+1}+\ldots+a_{n}}{n-m}=m \quad \quad \ldots\)(2)

By rearranging & adding equation (1) & (2)

\(a_{1}+a_{2}+\ldots \ldots a_{m+1} \ldots \ldots+a_{n}=n m+m(n-m)\)

\(=2 m n-m^{2}\)

So, total observation is "\(n\)"

So, mean is \(\frac{2 m n-m^{2}}{n}\)

\(=2 m-\frac{m^{2}}{n}\)