Mathematics Test - 39

Given,

\(\frac{d y}{d x}=e^{-3 y}\)

\(\Rightarrow \frac{d y}{e^{-3 y}}=d x \)

\(\Rightarrow e^{3 y} d y=d x\)

On integrating both side

We get,

\(\int e^{3 y} d y=\int d x \)

\(\Rightarrow \frac{e^{3 y}}{3}=x+C \ldots \ldots(\mathrm{i}) \)

\(\Rightarrow \frac{e^{2(0)}}{3}=5+C \quad\left(\because e^{0}=1\right) \)

\(\Rightarrow C+5=\frac{1}{3} \)

\(\Rightarrow C=\frac{1}{3}-5 \)

\(\Rightarrow C=\frac{-14}{3}\)

Substituting \(y=5\) and \(C=\frac{-14}{5}\),

We get,

\(\frac{e^{15}}{3}=x-\frac{14}{3}\)

\(\Rightarrow x=\frac{e^{15}+14}{3}\)

Given,

\(f(x)=\frac{x^{2}-5 x-6}{x^{2}+5 x-6}\)

Here, we have to find the value of \(x\) for which \(f(x)\) is not continuous.

So, if any function is not continuous at \(x=a\) then \(\lim _{x \rightarrow a} f(x)=l \neq f(a)\)

So, for the function \(f(x)\) if denominator is \(0\) at \(x=\) a then we can say that \(f(a)\) is infinite and limit cannot exist.

Let's find the value of \(x\) for which the denominator of \(f(x)\) is \(0\).

\( x^{2}+5 x-6=0\)

\(\Rightarrow(x+6)(x-1)=0\)

\(\Rightarrow x=-6,1\)

Area \(=\left|\int_{-1}^1 x\right| x|d x|=\left|\int_{-1}^0 x\right| x|d x|+\left|\int_0^1 x\right| x|d x| \)

\( =\left|\int_{-1}^0 x(-x) d x\right|+\left|\int_0^1 x \cdot x d x\right|=\left|\int_{-1}^0-x^2 d x\right|+\left|\int_0^1 x^2 d x\right| \)

\( =\left[\frac{-x^3}{3}\right]_{-1}^0+\left[\frac{x^3}{3}\right]_0^1=\frac{2}{3}\)

Given,

\(x^{2}<4\)

\(\Rightarrow x^{2}-4<0\)

\(\Rightarrow(x-2) \times(x+2)<0\)

\(\Rightarrow-2

\(\Rightarrow x \in(-2,2)\)

Given,

The distance of the point \((k+2,2 k+3)\) from the line \(x+3 y=7\) is \(\frac{4}{\sqrt{10}}\).

Let \(P=(k+2,2 k+3)\)

\(\Rightarrow x_{1}=k+2\) and \(y_{1}=2 k+3\)

Here, \(a=1\) and \(b=3\)

Now substitute \(x_{1}=k+2\) and \(y_{1}=2 k+3\) in the equation \(x+3 y-7=0\) we get,

\(\Rightarrow\left|x_{1}+3 \cdot y_{1}-7\right|=|7 k+4|\)

\(\Rightarrow \sqrt{a^{2}+b^{2}}\)

\(=\sqrt{1^{2}+3^{2}}\)

\(=\sqrt{10}\)

As we know that, the perpendicular distance \(d\) from \(P\left(x_{1}, y_{1}\right)\) to the line \(a x+b y+c\) \(=0\) is given by:

\(d=\left|\frac{a x_{1}+b y_{1}+c}{\sqrt{a^{2}+b^{2}}}\right|\)

\( d=\left|\frac{7 k+4}{\sqrt{1^{2}+3^{2}}}\right|\)

\(=\frac{4}{\sqrt{10}} \)

\(\Rightarrow|7 k+4|\)

\(=4\)

\(\Rightarrow k=0 \text { or }\frac{-8}{ 7}\)

Given,

\(\operatorname{cosec} \theta=\sqrt{2 \sqrt{2 \sqrt{2}}} \ldots \ldots \ldots\)

Let \(x=\sqrt{2 \sqrt{2 \sqrt{2}}} \ldots \ldots \ldots\)

On squaring both sides we get,

\(x^{2}=2 \sqrt{2 \sqrt{2}} \ldots \ldots \ldots\)

\(\Rightarrow x^{2}=2 x \quad(\because x=\sqrt{2 \sqrt{2 \sqrt{2}}} \ldots \ldots \ldots .) \)

\(\Rightarrow x^{2}-2 x=0\)

\(\Rightarrow x(x-2)=0 \)

\(\Rightarrow x=2 \text { OR } x=0 \)

\(\text { If } x=2 \text { : } \)

\(\operatorname{cosec} \theta=2\)

\(\Rightarrow \theta=30^{\circ}\)

\(\therefore \tan \theta=\tan 30^{\circ}=\frac{1}{\sqrt{3}}\)

Given, \(\mathrm{f}(0)=1 \ldots\) (i)

\( f(2)=e^2 \ldots \text { (ii) } \\\)

\( f^{\prime}(x)=f^{\prime}(2-x)\)

Integrating w.r.t. \(\mathrm{x}\),

\(f(x)=-f(2-x)+C\)

Put \(\mathrm{x}=0\)

\(f(0)=-f(2)+C\)

\(\Rightarrow 1=-e^2+C\) [from Eqs. (i) and (ii)]

\( \Rightarrow \mathrm{C}=1+\mathrm{e}^2 \\\)

\( \therefore \mathrm{f}(\mathrm{x})=-\mathrm{f}(2-\mathrm{x})+1+\mathrm{e}^2 \\\)

\( \text { or } \mathrm{f}(\mathrm{x})+\mathrm{f}(2-\mathrm{x})=1+\mathrm{e}^2 \ldots\)

Let \(I=\int_0^2 f(x) d x \ldots\) (iv)

Also, \(I=\int_0^2 f(2-x) d x \ldots(v)\)

Now, adding Eqs. (iv) and (v),

\( 2 I=\int_0^2[f(x)+f(2-x)] d x \\\)

\( 2 I=\int_0^2\left(1+\mathrm{e}^2\right) d x \text { [from Eq. (iii)] } \\\)

\( 2 I=2\left(1+\mathrm{e}^2\right) \\\)

\( \therefore I=\left(1+\mathrm{e}^2\right)\)

We know that:

\({ }^{{n}} {P}_{{r}}=\frac{{n} !}{({n}-{r}) !}={n} \times({n}-1) \times \ldots. \times({n}-{r}+1)\)

Given that:

\({ }^{10} {P}_{{r}}=5040\)

\(\Rightarrow{ }^{10} {P}_{{r}}=10 \times 504\)

\(\Rightarrow{ }^{10} {P}_{{r}}=10 \times 9 \times 56\)

\(\Rightarrow{ }^{10} {P}_{{r}}=10 \times 9 \times 8 \times 7\)

\(\Rightarrow{ }^{10} {P}_{{r}}=10 \times(10-1) \times(10-2) \times(10-4+1)\)

On comparing the above expression with \({ }^{{n}} {P}_{{r}}={n} \times({n}-1) \times \ldots \ldots \times({n}-{r}+1)\), we get:

\(r=4\)

Given: \(x=\cos \theta+i \sin \theta\)

Then \(x^{n}=(\cos \theta+i \sin \theta)^{n}\)

\(\Rightarrow x^{n}=(\cos n \theta+i \sin n \theta) \quad \ldots\)(i)

\(\Rightarrow \frac{1}{x^{n}}=\frac{1}{(\operatorname{cosn} \theta+i \sin n \theta)^{1}}=(\cos n \theta+i \sin n \theta)^{- 1}=\cos n \theta-i \operatorname{sinn} \theta \quad \ldots\)(ii)

On adding equation (i) and (ii) we get,

\(\Rightarrow x^{n}+\frac{1}{x^{n}}=\cos n \theta+i \sin n \theta+\cos n \theta-i \sin n \theta=2 \operatorname{cos} n \theta\)

Given:

A basket contains \(2\) white, \(3\) red and \(4\) black balls, two balls are drawn at random.

Total number of balls \(=2+3+4=9\)

We know that:

Probability \(=\frac{\text { Favorable Outcome }}{\text { Total Outcome }}\)

\(P(E)=\frac{n(E)}{n(S)}\)

Let \(S\) be the sample space.

Let \(E=\) Event of drawing \(2\) balls, none of them is black

Number of all combinations of \(n\) things, taken \(r\) at a time is given by \({ }^{n} C_{r}=\frac{n !}{(r) !(n-r) !}\)

Total number of ways of drawing \(2\) balls out of \(9\) balls

\(n(S)={ }^{9} C_{2}\)

There are four black balls in the total nine balls. Total number of non-black balls \(=9-4=5\)

Number of ways of drawing \(2\) balls out of \(5\) balls, if none of them is black

\(n(E)={ }^{5} C_{2}\)

\(\Rightarrow P(E)=\frac{n(E)}{n(S)}=\frac{{ }^{5} C_{2}}{{ }^{9} C_{2}}\)

\(=\frac{\frac{5 !}{2 ! 3 !}}{\frac{9}{2 ! 7 !}}\)

\(=\frac{\frac{5 \times 4 \times 3 \times 2 \times 1}{2 \times 1 \times 3 \times 2 \times 1}}{\frac{9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{2 \times 1 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}}\)

\(=\frac{\left\{\frac{(5 \times 4)}{(2 \times 1)}\right\}}{\left\{\frac{(9 \times 8)}{(2 \times 1)}\right\}}\)

\(=\frac{10}{36}\)

\(=\frac{5}{18}\)

\(\therefore\) The probability of not any ball being drawn is black is \(\frac{5}{18}\).