Mathematics Test - 21

We know that:

If \(f(x)=x^{n}\), then \(f'(x)=n x^{n-1}\)

\(f(x)=\log x\), then \(f'(x)=\frac{1}{x}\)

\(f(x)=\) Constant, then \(f'(x)=0\)

\(\log m^{n}=n \log m\)

Given here:

\(\mathrm{f}(\mathrm{x})=\log \mathrm{x}^{3}+2 \mathrm{x}^{2}-3 \mathrm{x}\)

\(\Rightarrow \mathrm{f}(\mathrm{x})=3 \log \mathrm{x}+2 \mathrm{x}^{2}-3 \mathrm{x}\)

\(\Rightarrow \mathrm{f}^{\prime}(\mathrm{x})=3 \frac{\mathrm{d}}{\mathrm{dx}} \log \mathrm{x}+\frac{\mathrm{d}}{\mathrm{dx}} 2 \mathrm{x}^{2}-\frac{\mathrm{d}}{\mathrm{dx}} 3 \mathrm{x}\)

\(\Rightarrow \mathrm{f}^{\prime}(\mathrm{x})=\frac{3}{x}+4 \mathrm{x}-3\)

Putting x = 3 in above, we get:

\(\Rightarrow \mathrm{f}^{\prime}(3)=\frac{3}{3}+4(3)-3\)

\(\Rightarrow \mathrm{f}^{\prime}(3)=10\)

Curve 1: \(y=\sin x=f(x)\) (say)

Curve 2: Lines \({x}=-\frac{\pi}{3}\) and \({x}=\frac{\pi}{3}\)

It can be drawn as follows:

According to the figure the sum of area curve \(\mathrm{OAB}\) and curve OCD.

Here, \(\mathrm{OAB}\) and \(\mathrm{OCD}\) are equal and limit 0 to \(\frac{\pi}{3}\).

So,  Area \(=2 \times\) area of \(\mathrm{OAB}\).

The area between the curves \(y_{1}=f(x)\) and \(y_{2}=g(x)\) is given by: 

Area enclosed \(=\left|\int_{\mathrm{x}_{1}}^{\mathrm{x}_{2}}\left(\mathrm{y}_{1}-\mathrm{y}_{2}\right) \mathrm{dx}\right|\)

Where, \(\mathrm{x}_{1}\) and \(\mathrm{x}_{2}\) are the intersections of curves \(\mathrm{y}_{1}\) and \(\mathrm{y}_{2}\)

Now, the required area (A) is,

Area of \(O A B=\left|\int_{\mathrm{x}_{1}}^{\mathrm{x}_{2}} \mathrm{f}(\mathrm{x}) \mathrm{d} \mathrm{x}\right|\)

\(=\left|\int_{0}^{\frac{\pi}{3}} \sin x ~d x\right|\)

\(=\left|[-\cos x]_{0}^{\frac{\pi}{3}}\right|\)

\(=\left|-\cos \frac{\pi}{3}+\cos 0^{\circ}\right|\)

\(=\left|1-\frac{1}{2}\right|\) \(=\frac{1}{2}\)

\(\therefore\) Shaded Area \(=2 \times \frac{1}{2}=1\)

We know that:

Let \(A\) and \(B\) be any two sets.

\(A \times B=\{(x, y)\) where \(x\) in \(A\) and \(y\) in \(B\}\)

Let \({A}\) and \({B}\) are two non-empty sets.

Consider, \(A=\{1,2,3\}\) and \(B=\{2,3\}\)

Here, in \({A}\) and \({B}\), there are 2 common elements.

Now,

\({A} \times {B}=\{(1,2),(1,3),(2,2),(2,3),(3,2),(3,3)\}\)

And, \({B} \times {A}=\{(2,1),(2,2),(2,3),(3,1),(3,2),(3,3)\}\)

In \(A \times B\) and \(B \times A\), there are \(4(=2^{2})\) i.e., common elements \(=n^{2}\)

Therefore, if \({A}\) and \({B}\) are two non-empty sets having \({n}\) elements in common, then \({n}^{2}\) common elements in the sets \(A \times B\) and \(B \times A\).

There are a total of \(7\) red \(+4\) blue \(=11\) balls.

Probability of drawing \(1\) red ball \(=\frac{{ }^{7} C_{1}}{{ }^{11} C_{1}}=\frac{7}{11}\)

Probability of drawing \(1\) blue ball \(=\frac{{ }^{4} C_{1}}{{ }^{11} C_{1}}=\frac{4}{11}\)

Probability of drawing (\(1\) red) AND (\(1\) blue) ball \(=\frac{7}{11} \times \frac{4}{11}=\frac{28}{121}\)

Similarly, Probability of drawing (\(1\) blue) AND (\(1\) red) ball \(=\frac{4}{11} \times \frac{7}{11}=\frac{28}{121}\)

Probability of getting the balls of different colors \(=\frac{28}{121}+\frac{28}{121}\)

\(=\frac{56}{121}\)

\(S_n=\left(\frac{1-q^{n+1}}{1-q}\right), T_n=\frac{1-\left(\frac{q+1}{2}\right)^{n+1}}{1-\left(\frac{q+1}{2}\right)} \\\)

\(\Rightarrow T_{100}=\frac{1-\left(\frac{q+1}{2}\right)^{101}}{1-\left(\frac{q+1}{2}\right)} \\\)

\( S n=\frac{1}{1-q}-\frac{q^{n+1}}{1-q}, T_{100}=\frac{2^{101}-(q+1)^{101}}{2^{100}(1-q)}\)

Now, \({ }^{101} \mathrm{C}_1+{ }^{101} \mathrm{C}_2 \mathrm{~S}_1+{ }^{101} \mathrm{C}_3 \mathrm{~S}_2+\ldots+{ }^{101} \mathrm{C}_{101} \mathrm{~S}_{100}\)

\( =\left(\frac{1}{1-q}\right)\left({ }^{101} \mathrm{C}_2+\ldots+{ }^{101} \mathrm{C}_{101}\right) \\\)

\( -\frac{1}{1-q}\left({ }^{101} \mathrm{C}_2 \mathrm{q}^2+{ }^{101} \mathrm{C}_3 \mathrm{q}^3+\ldots+{ }^{101} \mathrm{C}_{101} \mathrm{q}^{101}\right)+101 \\\)

\(=\frac{1}{1-\mathrm{q}}\left(2^{101}-1-101\right)-\left(\frac{1}{1-\mathrm{q}}\right)\left((1+q)^{101}-1\right. \\\)

\( \left.-{ }^{101} \mathrm{C}_1 \mathrm{q}\right)+101 \\\)

\( =\frac{1}{1-\mathrm{q}}\left[2^{101}-102-(1+\mathrm{q})^{101}+1+101 \mathrm{q}\right]+101 \\\)

\( =\frac{1}{1-\mathrm{q}}\left[2^{101}-101+101 \mathrm{q}-(1+\mathrm{q})^{101}\right]+101 \\\)

\( =\left(\frac{1}{1-\mathrm{q}}\right)\left[2^{101}-(1+\mathrm{q})^{101}\right]=2^{100} \mathrm{~T}_{100}\)

Hence, by comparison \(\alpha=2^{100}\)

A triangular region in the 3rd quadrant

Given inequalities x ≥ 0 , y ≥ 0 , 2x + y + 6 ≥ 0

Now take x = 0, y = 0 and 2x + y + 6 = 0

when x = 0, y = -6

when y = 0, x = -3

So, the points are A(0, 0), B(0, -6) and C(-3, 0)

So, the graph of the inequations x ≤ 0 , y ≤ 0 , and 2x + y + 6 ≥ 0 is a triangular region in the 3rd quadrant.

Given:

\(\lim _{\mathrm{x} \rightarrow 3} \frac{\mathrm{x}^{4}-81}{\mathrm{x}^{3}-27}\)

On checking the limit by putting \(x=3\) in above equation, we get \((\frac{0}{0})\) form.

We know that:

L-Hospital Rule as:

\(\lim _{x \rightarrow c} \frac{f(x)}{g(x)}=\lim _{x \rightarrow c} \frac{f^{\prime}(x)}{g^{\prime}(x)}\)

\(\therefore\) By using the L-Hospital rule, we get,

\(\lim _{x \rightarrow 3} \frac{x^{4}-81}{x^{3}-27}=\lim _{x \rightarrow 3} \frac{\frac{d}{dx}(x^{4}-81)}{\frac{d}{dx}(x^{3}-27)}\)

\(=\lim _{x \rightarrow 3} \frac{4 x^{3}}{3 x^{2}}\quad\) \((\because \frac{d}{dx}(x^n)= nx^{n-1}\) and differentiation of constant term is \(0.)\)

\(=\lim _{x \rightarrow 3} \frac{4 x}{3}\)

Putting the value of \(x \rightarrow3\) in above equation, we get

\(=\frac{4(3)}{3}\)

\(=4\)

Given,

Two parallel line\(p(x+y)+q=0\) and \(p(x+y)-r=0\)

The distance between the parallel lines \(ax+by+c_{1}=0\) and \(a x+b y+c_{2}=0\) is given by:

\(d=\left|\frac{c_{1}-c_{2}}{\sqrt{a^{2}+b^{2}}}\right|\)

Here, we have to find the distance between the parallel lines \(p(x+y)+q=0\) and \(p(x\) \(+y)-r=0\)

The given equations of line can be re-written as: \(p x+p y+q=0\) and \(p x+p y-r=0\)

By comparing the equations of the given line with \(ax+by+\mathrm{c}_{1}=0\) and \(ax+by+\) \(c_{2}=0\) we get,

\(a=p, b=p, c_{1}=q\) and \(c_{2}=-r\)

As we know that, the distance between the parallel lines is given by:

\(d=\left|\frac{c_{1}-c_{2}}{\sqrt{a^{2}+b^{2}}}\right|\)

\(d=\left|\frac{q+r}{\sqrt{p^{2}+p^{2}}}\right|\)

\(=\frac{|q+r|}{\sqrt{2} p}\)

Distance between the parallel lines \(p(x+y)+q=0\) and \(p(x+y)-r=0\) is\(\frac{|q+r|}{\sqrt{2} p}\).

let \(a_1\) be any natural number

\( a_1, a_1+1, a_1+2, \ldots, a_1+99 \text { are values of } \\\)

\( \bar{x}=\frac{a_1+\left(a_1+1\right)+\left(a_1+2\right)+\ldots+a_1+99}{100} \\\)

\(=\frac{100 a_1+(1+2+\ldots .+99)}{100}=a_1+\frac{99 \times 100}{2 \times 100} \\\)

\(=a_1+\frac{99}{2} \\\)

\( \text { Mean deviation about mean }=\frac{\sum_{1=1}^{100}\left|x_i-\bar{x}\right|}{100} \\\)

\(=\frac{2\left(\frac{99}{2}+\frac{97}{2}+\frac{95}{2}+\ldots+\frac{1}{2}\right)}{100} \\\)

\(=\frac{1+3+\ldots .99}{100} \\\)

\(=\frac{50}{2}[1+99] \\\)

\(=25\)

So, it is true for every natural number ' \(a_1\) '

Given:

\(\mu_{1}=60.125\)

\(\mu_{2}=61.345\)

\(\mu_{3}=62.688\)

Skewness coefficient is denoted by \(\beta_{1}\)

\(\beta_{1}=\frac{(\mu_{3})^{2}}{(\mu_{2})^{3}}\)

\(=\frac{(62.688)^{2}}{(61.345)^{3}}\)

\(\Rightarrow \beta_{1}=0.0170\)

Absolute skewness measurement \(=\gamma_{1}\)

\(\Rightarrow \gamma_{1}=\sqrt{( \beta_{1})}\)

\(=\sqrt{ 0.0170}\)

\(=0.130\)

\(\therefore\) Absolute skewness measurement is\(0.130\).