Mathematics Test - 25

Given set \(=\{1,2,3, \ldots \ldots . .50\}\)

\(\mathrm{P}(\mathrm{A})=\) Probability that number is multiple of 4

\(P(B)=\) Probability that number is multiple of 6

\(\mathrm{P}(\mathrm{C})=\) Probability that number is multiple of 7

Now,

\(\mathrm{P}(\mathrm{A})=\frac{12}{50}, \mathrm{P}(\mathrm{B})=\frac{8}{50}, \mathrm{P}(\mathrm{C})=\frac{7}{50}\)

again

\(\begin{aligned} & P(A \cap B)=\frac{4}{50}, P(B \cap C)=\frac{1}{50}, P(A \cap C)=\frac{1}{50} \\ & P(A \cap B \cap C)=0\end{aligned}\)

Thus

\( P(A \cup B \cup C)=\frac{12}{50}+\frac{8}{50}+\frac{7}{50}-\frac{4}{50}-\frac{1}{50}-\frac{1}{50}+0 \\\)

\( =\frac{21}{50}\)

\(\left|P^{-1} A P-2 I\right|=\left|P^{-1} A P-2 P^{-1} P\right| \\\)

\( =\left|P^{-1}(A-2 I) P\right| \\\)

\( =\left|P^{-1}\right||A-2 I||P| \\\)

\( =|A-2 I| \\\)

\(=\left|\begin{array}{ccc}0 & 1 & 2 \\ 6 & 0 & 11 \\ 3 & 3 & 0\end{array}\right|=69\)

So, Prime factor of 69 is \(3 \& 23\)

So, sum \(=26\)

Given function is,

\(\begin{aligned} & f(x)=\left\{\begin{array}{cc}|x|+[x], & -1 \leq x<1 \\ x+|x|, & 1 \leq x<2 \\ x+[x], & 2 \leq x \leq 3\end{array}\right. \\ & =\left\{\begin{array}{cc}-x-1, & -1 \leq x<0 \\ x, & 0 \leq x<1 \\ 2 x, & 1 \leq x<2 \\ x+2, & 2 \leq x<3 \\ 6, & x=3\end{array}\right. \\ & \end{aligned}\)

\(\Rightarrow \mathrm{f}(-1)=0, \mathrm{f}\left(-1^{+}\right)=0 \\\)

\( \mathrm{f}\left(0^{-}\right)=-1, \mathrm{f}(0)=0, \mathrm{f}\left(0^{+}\right)=0 \\\)

\( \mathrm{f}\left(1^{-}\right)=1, \mathrm{f}(1)=2, \mathrm{f}\left(1^{+}\right)=2 \\\)

\( \mathrm{f}\left(2^{-}\right)=4, \mathrm{f}(2)=4, \mathrm{f}\left(2^{+}\right)=4 \\\)

\( \mathrm{f}\left(3^{-}\right)=5, \mathrm{f}(3)=6\)

\(\mathrm{f}(\mathrm{x})\) is discontinuous at \(\mathrm{x}=\{0,1,3\}\)

Hence, \(f(x)\) is discontinuous at only three points.

\(\begin{aligned} & \vec{a}=\lambda \hat{i}+2 \hat{j}-3 \hat{k} \\ & \vec{b}=\hat{i}-\lambda \hat{j}+2 \hat{k} \\ & \Rightarrow(\vec{b}-\vec{a}) \times\vec{a}+\vec{b}) \times(\vec{a} \times \vec{b}=8 \hat{i}-40 \hat{j}-24 \hat{k} \\ & \Rightarrow\vec{a}-\vec{b}) \cdot(\vec{a}+\vec{b}(\vec{a} \times \vec{b})=8 \hat{i}-40 j-24 \hat{k} \\ & \Rightarrow 8(\vec{a} \times \vec{b})=8 \hat{i}-40 \hat{j}-24 \hat{k}\end{aligned}\)

Now, \(\vec{a} \times \vec{b}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ \lambda & 2 & -3 \\ 1 & -\lambda & 2\end{array}\right|\)

\(\begin{aligned} & =(4-3 \lambda) \hat{i}-(2 \lambda+3) \hat{j}+\left(-\lambda^2-2\right) \hat{k} \\ & \Rightarrow \lambda=1 \\ & \therefore \vec{a}=\hat{i}+2 \hat{j}-3 \hat{k} \\ & \vec{b}=\hat{i}-\hat{j}+2 \hat{k} \\ & \Rightarrow \vec{a}+\vec{b}=2 \hat{i}+\hat{j}-\hat{k}, \vec{a}-\vec{b}=3 \hat{j}-5 \hat{k}\end{aligned}\)

\(\Rightarrow(\vec{a}+\vec{b}) \times(\vec{a}-\vec{b})=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & -1 \\ 0 & 3 & -5\end{array}\right|=2 \hat{i}+10 \hat{j}+6 \hat{k}\)

∴ required answer =4+100+36=140

Given:

\(P(n): 1 \times 3+2 \times 3^{2}+3 \times 3^{3}+\ldots+n \times 3^{n} \) \(=\frac{(2 n-1) 3^{n+1}+3}{4} \)

For \( n=1,\)

\(\text {L H.S }=1 \times 3=3\) and R.H.S. \(=\frac{(2 \times 1-1) 3^{1+1}+3}{4}=\frac{3^{2}+3}{4}=\frac{12}{4}=3\)

Thus \(\mathrm{P}(1)\) is true.

\(P(n)\) be true for some \(n=k\).

\(1 \times 3+2 \times 3^{2}+3 \times 3^{3}+\ldots+k \times 3^{k}\) \(=\frac{(2 k-1) 3^{k+1}+3}{4}\)

Now, \(P(n)\) is true for \(n=k+1\).

\(1 \times 3+2 \times 3^{2}+3 \times 3^{3}+\ldots+(k+1) \times 3^{k+1}\) \(=\frac{(2 k+1) 3^{k+2}+3}{4}\)

Adding \((k+1) \times 3^{k+1}\) on both sides, we get

\(1 \times 3+2 \times 3^{2}+3 \times 3^{3}+\ldots+k \times 3^{k}+(k+1) \times 3^{k+1} \) \(=\frac{(2 k-1) 3^{k+1}+3}{4}+(k+1) \times 3^{k+1} \)

\(=\frac{(2 k-1) 3^{k+1}+3 \mid 4(k+1) 3^{k+1}}{4} \)

\(=\frac{3^{k+1}[2 k-1+4(k+1)]+3}{4} \)

\(=\frac{3^{k+1}(6 k+3)+3}{4} \)

\(=\frac{3^{(k+1)+1}(2 k+1)+3}{4} \)

\(=\frac{(2 k+1) 3^{k+2}+3}{4}\)

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

By the principle of mathematical induction, statement P(n) is true for all natural numbers.

Given,

\(4 x+3<6 x+7\)

Subtracting 3 from both sides.

\(4 x+3-3<6 x+7-3\)

\(\Rightarrow 4 x<6 x+4\)

Subtracting \(6 x\) from both sides,

\(4 x-6 x<6 x+4-6 x \)

\(\Rightarrow-2 x<4 \text { or }\)

\(\Rightarrow x>-2\) i.e., all the real numbers greater than \(-2\), are the solutions of the given inequality.

So, the solution set is \((-2, \infty)\), i.e. \(x \in(-2, \infty)\)

Given, \(\sqrt{\frac{1+\cos \theta}{1-\cos \theta}}\)

\(\Rightarrow \sqrt{\frac{1+\cos \theta}{1-\cos \theta}}=\sqrt{\left.\frac{(1+\cos \theta) \times(1+\cos \theta))}{(1-\cos \theta) \times(1+ \cos \theta) }\right)} \quad \text { (By rationalization) }\)

\(\Rightarrow \sqrt{\frac{1+\cos \theta}{1-\cos \theta}}=\sqrt{\frac{(1+\cos \theta)^{2}}{1-\cos ^{2} \theta}}\)

\(\Rightarrow \sqrt{\frac{1+\cos \theta}{1-\cos \theta}}=\sqrt{\frac{(1+\cos \theta)^{2}}{\sin ^{2} \theta}}\)\(\quad \left(\because 1-\cos ^{2} x=\sin ^{2} x\right)\)

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

\(\Rightarrow \sqrt{\frac{1+\cos \theta}{1-\cos \theta}}=\frac{1+\cos \theta}{\sin  \theta}\)

\(\Rightarrow \sqrt{\frac{1+\cos \theta}{1-\cos \theta}}\) = \(\frac{1}{\sin \theta}+\frac{\cos \theta}{\sin \theta}\)

\(\Rightarrow \sqrt{\frac{1+\cos \theta}{1-\cos \theta}}\) = \(\operatorname{cosec} \theta+\cot \theta\) \(\quad\quad (\because\operatorname{cosec} x=\frac{1}{\sin x},\cot x=\frac{\cos x}{\sin x})\)

B={x:x∈(−∞,−1]∪[5,∞)}
A∩B={x:x∈(−2,−1]}
A∪B={x:x∈(−∞,2)∪[5,∞)}
A−B={x:x∈(−1,2)}
B−A={x:x∈(−∞,−2]∪[5,∞)}