Mathematics Test - 37

\(A=\{1,-1, i,-i\}\)

\(B=\{1,-1\}\)

\(C=\{i,-i\}\)

Now, \(B \cup C=\{1,-1, i,-i\}=A\)

Given: \(y = x ^2\) and \(x=1\)

As we know that equation of tangent is \(\frac{y-f(a)}{x-a}=f^{\prime}(a)\) for the function \(y = f ( x )\) at \(x\) \(= a\)

First find \(f ( a )\) and \(f ^{\prime}( a )\)

\( f(1)=1 \)

\(\Rightarrow f^{\prime}(x)=2 x \text { and } f^{\prime}(1)=2\)

As we know that equation of tangent is \(\frac{y-f(a)}{x-a}=f^{\prime}(a)\) for the function \(y = f ( x )\) at \(x\)\(= a \)

\( \frac{y-1}{x-1}=2\)

\( \Rightarrow y =2 x -1\)

Hence, the equation of tangent is \(y=2 x-1\).

Let \(\hat{x_i}+y \hat{j}+z \hat{k}\) be the required unit vector.

Since \(\hat{a}\) is perpendicular to \((2 \hat{i}-\hat{j}+2 \hat{k})\).

\(\therefore 2 \mathrm{x}-\mathrm{y}+2 \mathrm{z}=0\)

Since vector \(x \hat{i}+y \hat{j}+z \hat{k}\) is coplanar with the vector \(\hat{i}+\hat{j}-\hat{k}\) and

\(2 \hat{i}+2 \hat{j}-\hat{k} \\\)

\( \therefore x_i+y \hat{j}+z \hat{k}=p(\hat{i}+\hat{j}-\hat{k})+q(2 \hat{i}+2 \hat{j}-\hat{k})\)

where \(\mathrm{p}\) and \(\mathrm{q}\) are some scalars.

\(\Rightarrow x\hat{i}+y \hat{j}+z \hat{k}=(p+2 q) \hat{i}+(p+2 q) \hat{j}-(p+q) \hat{k} \)

\(\Rightarrow x=p+2 q, y=p+2 q, z=-p-q\)

Now from equation (i),

\( 2 p+4 q-p-2 q-2 p-2 q=0 \\\)

\(\Rightarrow-p=0 \Rightarrow p=0 \\\)

\( \therefore x=2 q, y=2 q, z=-q\)

Since vector \(\hat{x_i}+y \hat{j}+z \hat{k}\) is a unit vector, therefore

\( \left|\hat{x_i}+y \hat{j}+z \hat{k}\right|=1 \\\)

\( \Rightarrow \sqrt{x^2+y^2+z^2}=1 \\\)

\( \Rightarrow x^2+y^2+z^2=1 \\\)

\( \Rightarrow 4 q^2+4 q^2+q^2=1 \\\)

\( \Rightarrow 9 q^2=1 \Rightarrow q= \pm \frac{1}{3}\)

When \(\mathrm{q}=\frac{1}{3}\), then \(\mathrm{x}=\frac{2}{3}, \mathrm{y}=\frac{2}{3}, \mathrm{z}=-\frac{1}{3}\)

When \(\mathrm{q}=-\frac{1}{3}\), then \(\mathrm{x}=-\frac{2}{3}, \mathrm{y}=-\frac{2}{3}, \mathrm{z}=\frac{1}{3}\)

Here required unit vector is \(\frac{2 \hat{i}+2 \hat{j}-1 \hat{k}}{3}\)

or \(-\frac{2}{3} \hat{i}-\frac{2}{3} \hat{j}+\frac{1}{3} \hat{k}\)

Let \(C=\left[\begin{array}{cc}1 & 2 \\ -1 & -1\end{array}\right], D=\left[\begin{array}{cc}-1 & -2 \\ 1 & 1\end{array}\right]\)

\(\mathrm{DC}=\left[\begin{array}{cc}1 & 2 \\ -1 & -1\end{array}\right]\left[\begin{array}{cc}-1 & -2 \\ 1 & 1\end{array}\right]=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]=\mathrm{I}\)

B=CAD
Bn=⏟(CAD)(CAD)(CAD)......(CAD)n-times 
⇒Bn=CAnD...(1)

\(A^3=\left[\begin{array}{cc}1 & \frac{3}{51} \\ 0 & 1\end{array}\right]\)

Similarly \(A^n=A^3=\left[\begin{array}{cc}1 & \frac{n}{51} \\ 0 & 1\end{array}\right]\)

\(\mathrm{Bn}\left[\begin{array}{cc}1 & 2 \\ -1 & -1\end{array}\right]\left[\begin{array}{cc}1 & \frac{\mathrm{n}}{51} \\ 0 & 1\end{array}\right]\left[\begin{array}{cc}-1 & -2 \\ 1 & 1\end{array}\right]\)

\(=\left[\begin{array}{cc}1 & \frac{\mathrm{n}}{51}+2 \\ -1 & -\frac{\mathrm{n}}{51}-1\end{array}\right]\left[\begin{array}{cc}-1 & -2 \\ 1 & 1\end{array}\right]\)

\(=\left[\begin{array}{cc}\frac{\mathrm{n}}{51}+1 & \frac{\mathrm{n}}{51} \\ -\frac{\mathrm{n}}{51} & 1-\frac{\mathrm{n}}{51}\end{array}\right]\)

\(\sum_{n=1}^{50} B^n=\left[\begin{array}{cc}25-50 & 25 \\ -25 & -25-50\end{array}\right]=\left[\begin{array}{cc}75 & 25 \\ -25 & 25\end{array}\right]\)

Sum of the elements =100

Given \(\left|\begin{array}{lll}a & a^2 & 1+a^3 \\ b & b^2 & 1+b^3 \\ c & c^2 & 1+c^3\end{array}\right|=0\)

\(\Rightarrow\left|\begin{array}{ccc}a & a^2 & 1 \\ b & b^2 & 1 \\ c & c^2 & 1\end{array}\right|+\left|\begin{array}{ccc}a & a^2 & a^3 \\ b & b^2 & b^3 \\ c & c^2 & c^3\end{array}\right|=0\)

\(\Rightarrow(1+a b c)\left|\begin{array}{lll}1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2\end{array}\right|=0\)

Given that\(\left(1, a, a^2\right),\left(1, b, b^2\right)\) and \(\left(1, c, c^2\right)\) are non-coplanar

\(\therefore\left|\begin{array}{lll}1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2\end{array}\right| \neq 0\) (given condition)

\(\therefore 1+a b c=0 \Rightarrow a b c=-1\)

Given:

\(\mathrm{P}(\mathrm{n}): 41^{\mathrm{n}}-14^{\mathrm{n}}\) is a multiple of 27.

\(\mathrm{P}(\mathrm{n})\) is true for \(\mathrm{n}=1\) since \(41^{1}-14^{1}=27\), which is a multiple of 27.

Let \(\mathrm{P}(\mathrm{k})\) be true for some positive integer \(\mathrm{k}\),

So, \(41^{\mathrm{k}}-14^{\mathrm{k}}\) is a multiple of 27.

\(41^{\mathrm{k}}-14^{\mathrm{k}}=27 \mathrm{~m}\), where \(\mathrm{m} \in \mathrm{N} \ldots \ldots \ldots\) (i)

We shall now prove that \(\mathrm{P}(\mathrm{k}+1)\) is true whenever \(\mathrm{P}(\mathrm{k})\) is true.

Consider\(41^{\mathrm{k}+1}-14^{\mathrm{k}+1} \)

\(=41^{\mathrm{k}} .41-14^{\mathrm{k}} \cdot 14 \)

\(=41\left(27 \mathrm{~m}+14^{\mathrm{k}}\right)-14^{\mathrm{k}} 14 .....[\mathrm{using}(\mathrm{i})] \)

\(=41.27 \mathrm{~m}+41.14^{\mathrm{k}}-14.14^{\mathrm{k}} \)

\(=41.27 \mathrm{~m}+27.14^{\mathrm{k}} \)

\(=27\left(41 \mathrm{~m}+14^{\mathrm{k}}\right)\)

\(=27 \times \mathrm{r}\), where \(\mathrm{r}=\left(41 \mathrm{~m}+14^{\mathrm{k}}\right)\) is a natural number

Therefore, \(41^{\mathrm{k}+1}-14^{\mathrm{k}+1}\) is a multiple if 27.

Thus, \(\mathrm{P}(\mathrm{k}+1)\) is true whenever \(\mathrm{P}(\mathrm{k})\) is true.

Given:

\(P(A)=0.3\) and \(P(A \cup B)=0.8\)

Formula used:

\(P(A \cup B)=P(A)+P(B)-P(A) \times P(B)\)

Two events are independent if the incidence of one event does not affect the probability of the other event.

We know that,

\(P(A \cup B)=P(A)+P(B)-P(A) \times P(B)\)

\(\Rightarrow 0.8=0.3+P(B)-0.3 \times P(B)\)

\(\Rightarrow 0.5=P(B)[1-0.3]=0.7 P(B)\)

\(\Rightarrow P(B)=\frac{0.5 }{0.7}\)

\(\therefore P(B)=\frac{5}{7}\)

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

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

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,

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

Multiplying and dividing by \(3 x\) in numerator and by \(2 x\) in denominator,

We get,

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

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

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

Given:

Number are \(15, 20, 18, 22\) and \(25\).

Mean \(=\frac{\sum x}{n}\)

\(=\frac{(15+20+18+22+25)}{5}\)

\(=\frac{100}{5}=20\)