Mathematics Test - 8

We know that:

\(\underset{{{x \rightarrow a}}}{\lim}\left[\frac{f(x)}{g(x)}\right]=\frac{\underset{{{x \rightarrow a}}}{\lim}f(x)}{\underset{{{x \rightarrow a}}}{\lim}g(x)}\), provided \(\underset{{{x \rightarrow a}}}{\lim} g(x) \neq 0\) 

Given that:

\(\underset{{{x \rightarrow 0}}}{\lim}\frac{\tan 2 x}{e^{2 x}-1}\)

\(=\underset{{{x \rightarrow 0}}}{\lim} \frac{\frac{\tan 2 x}{2 x} \times 2 x}{\frac{e^{2 x}-1}{2 x} \times 2 x}\)

\(=\frac{\underset{{{x \rightarrow 0}}}{\lim} \frac{\tan 2 x}{2 x}}{\underset{{{x \rightarrow 0}}}{\lim} \frac{e^{2 x}-1}{2 x}}\)

As we know that:

\(\underset{{{{x} \rightarrow 0}}}{\lim} \frac{\tan {x}}{{x}}=1\) 

and, \(\underset{{{{x} \rightarrow 0}}}{\lim} \frac{{e}^{x}-1}{{x}}=1\)

Therefore, 

\(\underset{{{{x} \rightarrow 0}}}{\lim} \frac{\tan 2 {x}}{2 {x}}=1\) 

and, \(\underset{{{{x} \rightarrow 0}}}{\lim} \frac{{e}^{2 x}-1}{2 {x}}=1\)

Therefore, 

\(\underset{{{{x} \rightarrow 0}}}{\lim}\frac{\tan 2 {x}}{{e}^{2 x}-1}=\frac{1}{1}=1\)

\(L_1=\frac{x}{1}=\frac{y-1}{2}=\frac{z-2}{3}=\lambda\)

\(\mathrm{M}(\lambda, 1+2 \lambda, 2+3 \lambda)\)

\(\overrightarrow{\mathrm{PM}}=(\lambda-1) \hat{\mathrm{i}}+(1+2 \lambda) \hat{\mathrm{j}}+(3 \lambda-5) \hat{\mathrm{k}}\)

\(\overrightarrow{\mathrm{PM}} \text { is perpendicular to line } \mathrm{L}_1\)

\(\overrightarrow{\mathrm{PM}} \cdot \overrightarrow{\mathrm{b}}=0 \quad(\overrightarrow{\mathrm{b}}=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})\)

\(\Rightarrow \lambda-1+4 \lambda+2+9 \lambda-15=0\)

\(14 \lambda=14 \Rightarrow \lambda=1\)

\(\therefore \mathrm{M}=(1,3,5)\)

\(\overrightarrow{\mathrm{Q}}=2 \overrightarrow{\mathrm{M}}-\overrightarrow{\mathrm{P}}[\mathrm{M} \text { is midpoint of } \overrightarrow{\mathrm{P}} \& \overrightarrow{\mathrm{Q}}]\)

\(\overrightarrow{\mathrm{Q}}=2 \hat{\mathrm{i}}+6 \hat{\mathrm{j}}+10 \hat{\mathrm{k}}-\hat{\mathrm{i}}-7 \hat{\mathrm{k}}\)

\(\vec{Q}=\hat{i}+6 \hat{j}+3 \hat{k}\)

\(\vec{Q}=\hat{i}+6 \hat{j}+3 \hat{k}\)

\(\therefore(\alpha, \beta, \gamma)=(1,6,3)\)

Required line having direction cosine (l,m,n)

l2+m2+n2=1

\(\Rightarrow l^2+\left(-\frac{1}{2}\right)^2+\left(-\frac{1}{\sqrt{2}}\right)^2=1\)

\(l^2=\frac{1}{4}\)

∴l= \(\frac{1}{2}\)[Line make acute angle with x-axis]

Equation of line passing through (1,6,3) will be

\(\overrightarrow{\mathrm{r}}=(\hat{\mathrm{i}}+6 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})+\mu\left(\frac{1}{2} \hat{\mathrm{i}}-\frac{1}{2} \hat{\mathrm{j}}-\frac{1}{\sqrt{2}} \hat{\mathrm{k}}\right)\)

Option (3) satisfying for µ=4

A vector in the direction of the required line can be obtained by cross product of

\(\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 5 \\ -1 & 3 & 2\end{array}\right|\)

\( =-9 \hat{i}-9 \hat{j}+9 \hat{k}\)

Required line,

\(\overrightarrow{\mathrm{r}}=(5 \hat{i}-4 \hat{j}+3 \hat{k})+\lambda(-9 \hat{i}-9 \hat{j}+9 \hat{k}) \\\)

\(\vec{r}=(5 \hat{i}-4 \hat{j}+3 \hat{k})+\lambda(\hat{i}+\hat{j}-\hat{k})\)

Now distance of (0,2,−2)

\(\mathrm{P} \cdot \mathrm{V} \text { of } \mathrm{P} \equiv(5+\lambda) \hat{i}+(\lambda-4) \hat{\mathrm{j}}+(3-\lambda) \hat{\mathrm{k}} \\\)

\( \overrightarrow{\mathrm{AP}}=(5+\lambda) \hat{i}+(\lambda-6) \hat{\mathrm{j}}+(5-\lambda) \hat{\mathrm{k}} \\\)

\( \overrightarrow{\mathrm{AP}} \cdot(\hat{\mathrm{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}})=0 \\\)

\( 5+\lambda+\lambda-6-5+\lambda=0 \\\)

\( \lambda=2 \\\)

\( |\overrightarrow{\mathrm{AP}}|=\sqrt{49+16+9} \\\)

\( |\overrightarrow{\mathrm{AP}}|=\sqrt{74}\)

We know that:

\({ }^{n} C_{r}=\frac{n !}{r !(n-r) !}\)

A particular player should not be included,

We have to select 5 players from (8 - 1) = 7 players.

Therefore, required number of ways,

\(={ }^{7} {C}_{5}\)

\(=\frac{7 !}{5 !(7-5) !}=21\)

Total number of results in tossing a coin \(6\) times \(n(S)=2^{6}=64\)

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

\(\therefore\) The number of times that the coin is tossed \(n(E)=6\) times the probability of the coin being tossed \(3\) times

\(\Rightarrow n(E)={ }^{6} C_{3}\)

\(=\frac{6 !}{3 ! \times 3 !}\)

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

\(=20\)

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

\(=\frac{20}{64}\)

\(=\frac{5}{16}\)

Given,

The objective function is ,

\(F=4 x+6 y\)

At\((0,3)\),

\(F=4 \times 0+6 \times 3=18\)

At \((3,2)\),

\(F=4 \times 3+6 \times 2=24\)

At \((6,0)\),

\(F=4 \times 6+6 \times 0=24\)

At \((5,5)\),

\(F=4 \times 5+6 \times 5=50\)

Thus, the maximum value of \(\mathrm{F}=50\)

Given: Number of observation \((N) =20\)

\(\sum x_{i}=1000\) and \(\sum x_{i}^{2}=84000\)

Variance \(=\frac{\sum x_{ i }^{2}}{ N }-\frac{\left(\sum x _{ i }\right)^{2}}{ N ^{2}}\)

\(=\frac{84000}{20}-\frac{1000^{2}}{20^{2}}\)

\(=4200-2500\)

\(=1700\)

Since \(\mathrm{f}(\mathrm{x})\) is given to be continuous at \(\mathrm{x}=0\),

\(\lim _{\mathrm{x} \rightarrow 0} \mathrm{f}(\mathrm{x})=\mathrm{f}(0)\)

Also, \(\lim _{x \rightarrow a^{+}} f(x)=\lim _{x \rightarrow a^{-}} f(x)\) because \(f(x)\) is same for \(x>0\) and \(x<0\).

\(\therefore \lim _{x \rightarrow 0} f(x)=f(0)\)

We know that:

\(\lim _{x \rightarrow 0} \frac{\sin x}{x}=1\)

\(\lim _{x \rightarrow 0} \frac{e^{x}-1}{x}=1\)

Therefore,

\(\Rightarrow \lim _{x \rightarrow 0} \frac{\sin 3 x}{e^{2 x}-1}=k-2\)

Multiplying and dividing by 3x in numerator and by 2x in denominator, we get:

\(\Rightarrow \lim _{x \rightarrow 0} \frac{\frac{\sin 3 x}{3 x} \times 3 x}{\frac{e^{2 x}-1}{2 x} \times 2 x}=k-2\)

\(\Rightarrow \frac{3}{2}=k-2\)

\(\Rightarrow k=\frac{7}{2}\)

The given equation is,

\(\frac{d y}{d x}=y\left(1-3 x^{2}\right)\)

\(\Rightarrow\frac{d y}{y}=\left(1-3 x^{2}\right) d x\)

Integrating both side, we get,

\(\int \frac{d y}{y}=\int\left(1-3 x^{2}\right) d x\)

We know that,

\(\int\frac{dx}{x}=\log x\) and \(\int x^n dx=\frac{x^{n+1}}{n+1}\)

\(\therefore\log y=x-x^{3}+c\)

We also know that,

If \(\log x = n\) then \(x=e^n\)

\(\therefore y=e^{x-x^{3}+c}\)

\(\Rightarrow y=e^{x-x^{3}}. e^{c}\)

\(\Rightarrow y=C e^{x\left(1-x^{2}\right)}\quad\)\([\)Let \(e^c=C]\)