Mathematics Test-9

Given, 

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

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

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

Now,

Integrating both sides, we get

\(\int {e}^{{y}} {dy}=\int {e}^{{x}} {dx}\)

\(\Rightarrow {e}^{{y}}={e}^{{x}}+{c}\)

Giving, \(|1-2 i|^{x}=5^{x}\)

We need to find the value of \(x\) for which should be an integer but not zero.

Let first find value of \(|1-2 i|\)

\(\Rightarrow\)\(|1-2 i|=\sqrt{(1)^{2}+(2)^{2}}=\sqrt{5} \quad\left[\because z=x+i y \Rightarrow|z|=\sqrt{x^{2}+y^{2}}\right]\)

So,

\(|1-2 i|^{x}=5^{x}\)

\(\Rightarrow(\sqrt{5})^{x}=5^{x}\)

\(\Rightarrow\)\((\sqrt{5})^{x}=(\sqrt{5})^{2 x}\)

Comparing powers,

\(\Rightarrow x=2 x\)

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

\(\Rightarrow-x=0\)

\(\Rightarrow x=0\)

\(A=\left[\begin{array}{ll}2 & 4 \\ 3 & 2\end{array}\right], B=\left[\begin{array}{cc}1 & 3 \\ -2 & 5\end{array}\right]\)

\(\therefore {A}+{B}=\left[\begin{array}{ll}2 & 4 \\ 3 & 2\end{array}\right]+\left[\begin{array}{cc}1 & 3 \\ -2 & 5\end{array}\right]\)

\(\Rightarrow {A}+{B}=\left[\begin{array}{ll}2+1 & 4+3 \\ 3-2 & 2+5\end{array}\right]\)

\(\Rightarrow {A}+{B}=\left[\begin{array}{ll}3 & 7 \\ 1 & 7\end{array}\right]\)

Given:

\(p + q =2\) and \(p ^{4}+ q ^{4}=272\)

Consider, \(\left(p^{2}+q^{2}\right)^{2}-2 p^{2} q^{2}=272\)

\(\left((p+a)^{2}-2 p q\right)^{2}-2 p^{2} q^{2}=272\)

\(\Rightarrow 16-16 p q+2 p^{2} q^{2}=272\)

\(\Rightarrow (p q)^{2}-8 p q-128=0\)

\(\Rightarrow (p q)^{2}-8 p q-128=0\)

\(\Rightarrow p q=\frac{8 \pm 24}{2}=16,-8\)

\(\therefore pq =16\)

Required equation: \(x ^{2}-(2) x +16=0\)

The odd numbers (outcomes) are \(1,2,3,4,5,6,7,8,9,10,11,12\) are \(1,3,5,7,9,11\).

Probability \(= \frac{Number\ of\ favourable\ outcomes}{Total\ number\ of\ outcomes}\)

Therefore probability that an odd number will come is:

\(\Rightarrow\) \(P(\)odd number\()=\frac{6}{12}\)

\(\Rightarrow P(\)odd number\()=\frac{1}{2}\)

Given,

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

Applying the column operations

\(C _1 \rightarrow C _1- C _2\) and \(C _2 \rightarrow C _2- C _3\)

\(A=\left|\begin{array}{ccc}a-b & b-c & c \\ b-a & c-b & a+b \\ a^2-b^2 & b^2-c^2 & c^2\end{array}\right|\)

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

Taking common (a-b) and (b-c) From first and second column respectively.

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

Applying column operation \(C_1 \rightarrow C_1-C_2\)

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

Now solving the determinant we get,

\(A=(a-b)(b-c)[(a-c) \times\{(a+b)-(-c)\}] \)

\(A=-(a-b)(b-c)(c-a)(a+b+c)\)

Suppose that the sides of the cube are unity and unit vector along \(O A, O B\) and \(O C\) are \(\hat{i}, \hat{j}, k\) respectively. \(OR, OS, OT\) are diagonals of a cube having corresponding vector \(a, 2 a\) and \(3 a\) (Magnitude) respectively.

\(\therefore\) Unit vector along \(O R=\frac{\hat{j}+\hat{k}}{\sqrt{2}}\)

\(\therefore\) Vector along \(\overrightarrow{O R}=a\left(\frac{\hat{j}+\hat{k}}{\sqrt{2}}\right)\) Similarly, vector along \(\overrightarrow{O S}=2 a\left(\frac{\hat{k}+\hat{i}}{\sqrt{2}}\right)\) and vector along \(\overrightarrow{O T}=3 a\left(\frac{\hat{i}+\hat{j}}{\sqrt{2}}\right)\)

\(\therefore\) Resultant \(R=\overrightarrow{O R}+\overrightarrow{O S}+\overrightarrow{O T}\)

\(=a\left(\frac{\hat{i}+\hat{k}}{\sqrt{2}}\right)+2 a\left(\frac{\hat{k}+\hat{i}}{\sqrt{2}}\right)+3 a\left(\frac{\hat{i}+\hat{j}}{\sqrt{2}}\right)\)

\(=\frac{5 a}{\sqrt{2}} \hat{i}+\frac{4 a}{\sqrt{2}} \hat{j}+\frac{3 a}{\sqrt{2}} \hat{k}\)

\(\therefore|R|=\sqrt{\frac{25 a^2}{2}+\frac{16 a^2}{2}+\frac{9 a^2}{2}}=5 a\)

Given:

\(\left(x^{3}-\frac{1}{x^{2}}\right)^{15}\)

Let \((r+1)^{t h}\) term be the constant term in the expansion of \(\left(x^{3}-\frac{1}{x^{2}}\right)^{15}\).

We know that in the binomial expansion of \((a+x)^{n}\), we have,

\(T_{r+1}={ }^{n} C_{r} x^{n-r} a^{r}\)

\(\therefore T_{r+1}={ }^{15} C_{r}\left(x^{3}\right)^{15-r}\left(-\frac{1}{x^{2}}\right)^{r}\)

\(T_{r+1}={ }^{15} C_{r} x^{45-5 r}(-1)^{r}\) is independent of \(x\). If:

\(45-5 r=0\)

\(\Rightarrow r=9\)

Thus, \(10^{\text {th }}\) term is independent of \(x\) and is given by

\(T_{10}={ }^{15} C_{9}(-1)^{9}\)

\(=-{ }^{15} C_{9}\)