Mathematics Test - 25

Equality of complex numbers:

Two complex numbers \(z_{1}=x_{1}+i y_{1}\) and \(z_{2}=x_{2}+i y_{2}\) are equal if and only if \(x_{1}=x_{2}\) and \(y_{1}=y_{2}\).

or \(\operatorname{Re}\left(z_{1}\right)=\operatorname{Re}\left(z_{2}\right)\) and \(\operatorname{Im}\left(z_{1}\right)=\operatorname{lm}\left(z_{2}\right)\)

Given: \({x}+{iy}=\frac{3+4 {i}}{2-{i}}\)

\(\Rightarrow {x}+{iy}=\frac{3+4 {i}}{2-{i}} \times \frac{2+{i}}{2+{i}}\)

\(\Rightarrow {x}+{iy}=\frac{6+11 {i}+4 {i}^{2}}{4-{i}^{2}}\)

As we know \({i}^{2}=-1\)

\(\Rightarrow {x}+{iy}=\frac{6+11 {i}-4}{4+1}\)

\(\Rightarrow {x}+{iy}=\frac{2+11 {i}}{5}=\frac{2}{5}+{i} \frac{11}{5}\)

Comparing real and imaginary parts, we get,

\({x}=\frac{2}{5}\) and \({y}=\frac{11}{5}\)

Here \(f(x)\) is continuous at \(x=c\) if, \(L H L=R H L=\) value of \(f(c)\) i.e., \(\lim _{x \rightarrow c} f(x)=\lim _{x \rightarrow c^{+}} f(x)=f(c)\)

Now,

\( f\left(\frac{\pi}{2}\right)=\frac{\lambda \pi}{2} \ldots(i) \)

\(R H L=\lim _{x \rightarrow \frac{\pi}{2}+} f(x)=\lim _{x \rightarrow \frac{\pi}{2}+} \cos 2 \mathrm{x} \)

\( =\cos 2\left(\frac{\pi}{2}\right)=-1 \ldots(\text { ii })\)

Equating \((i)\) and \((i i)\),

\(\frac{\lambda \pi}{2}=-1 \)

\(\therefore \lambda=-\frac{2}{\pi}\)

We have,

\(e^{x}=(1-x)\left(B_{0}+B_{1} x+B_{2} x^{2}+\ldots+B_{n-1} x^{n-1}+B_{n} x^{n}+\ldots\right)\)

By the expansion of e^x, we get,

\(1+\frac{x}{1 !}+\frac{x^{2}}{2 !}+\ldots+\frac{x^{n}}{n !}+\ldots\)

\(=(1-x)\left(B_{0}+B_{1} x+B_{2} x^{2}+\ldots+B_{n-1} x^{n-1}+B_{n} x^{n}+\ldots\right)\)

Equating the coefficient of xⁿ on both sides, we get,

\(B_{n}-B_{n-1}=\frac{1}{n !}\)

\(A(\vec{a}), B(\vec{b}), C(\vec{c})\) are the vertices of the triangle \(A B C\).

\(D\) is the midpoint of \(BC\)

\(\Rightarrow D =\left(\frac{\overrightarrow{ b }+\overrightarrow{ c }}{2}\right)\)

\(E\) is the midpoint of \(CA\)

\(\Rightarrow E=\left(\frac{\vec{a}+\vec{c}}{2}\right)\)

\(F\) is the midpoint of \(A B\)

\(\Rightarrow F =\left(\frac{\overrightarrow{ a }+\overrightarrow{ b }}{2}\right)\)

Now, \(AD , BE\) and \(CF\) are the medians of triangle \(ABC\)

All three medians intersects at point \(P\).

\(\because P\) divides \(AD\) in the ratio \(2: 1\) and other medians also intersect at \(P\).

\(\Rightarrow P\) divides all three medians in the ratio \(2: 1\)

\(\Rightarrow P\) is the centroid of the triangle \(ABC\).

The position vector of the centroid of a triangle \(ABC\) with position vectors of the vertices being \(\vec{a}, \vec{b}\) and \(\vec{c}\) respectively is given by:

\(P=\frac{\vec{a}+\vec{b}+\vec{c}}{3}\)

Given,

\(\sin ^{-1} x-\cos ^{-1} x=\sin ^{-1}\left ( \frac{1}{2}\right)\)

\(\Rightarrow \sin ^{-1} x-\cos ^{-1} x=\frac{\pi}{6}\)

\(\Rightarrow \sin ^{-1} x-\left(\frac{\pi}{2}-\sin ^{-1} x\right)=\frac{\pi}{6}\)

\(\Rightarrow 2 \sin ^{-1} x=\frac{\pi}{6}+\frac{\pi}{2}=\frac{2 \pi}{3}\)

\(\Rightarrow \sin ^{-1} x=\frac{\pi}{3}\)

\( \Rightarrow x=\frac{\sqrt{3}}{2}\)

Thus, the given equation has only one solution.

Direction cosine: It is the cosines of each angles that the directed line makes with the \(x\)-axis, \(y\)-axis, and \(z\)-axis i.e., \(a, \beta\) and \(\gamma\) respectively.

Represented as: \(\{l, m, n\}\)

Where,

\(l=\cos \alpha, m=\cos \beta\) and \(n=\cos \gamma\)

Also, \(l^{2}+m^{2}+n^{2}=1\)

Given that:

Direction cosines of a line \(\langle 0,1,0\rangle\)

Therefore,

\(l=\cos \alpha=0\)

\(\Rightarrow \alpha=90^{\circ}\)

\(m=\cos \beta=1\)

\(\Rightarrow \beta=0^{\circ}\)

\(n=\cos \gamma=0\)

\(\Rightarrow \gamma=90^{\circ}\)

So, cosine angle along \(y\)- axis is zero. Therefore, the line is parallel to \(y\)-axis.

We know , the coefficient of middle term in the expansion of \((1+\beta x)^{4}={ }^{4} C_{2} \beta^{2}\)

Similarly the coefficient of middle term in the expansion of \((1-\beta x)^{6}={ }^{6} C_{3}(-\beta)^{3}\)

As it is given \({ }^{4} C_{2} \beta^{2}={ }^{6} C_{3}(-\beta)^{3}\)

\(\Rightarrow { }^{4} C_{2} \beta^{2}={ }^{6} C_{3}(-\beta)^{3} \)

\(\Rightarrow { }^{4} C_{2} \beta^{2}=-{ }^{6} C_{3} \beta^{3} \)

\(\Rightarrow { }^{4} C_{2}+{ }^{6} C_{3} \beta=0\)

\(\Rightarrow \beta=-\frac{{ }^{4} C_{2}}{{ }^{6} C_{3}} \)

\(\Rightarrow \beta=-\frac{\left(\frac{4 !}{2 ! 2 !}\right)}{\left(\frac{6 !}{3 ! 3 !}\right)}\)

\(\Rightarrow \beta=-\frac{\left(\frac{4 \times 3 \times 2 !}{2 \times 2 !}\right)}{\left(\frac{6 \times 5 \times 4 \times 3 !}{3 \times 2 \times 1 \times 3 !}\right)} \)

\(\Rightarrow \beta=-\frac{6}{20} \)

\(\Rightarrow \beta=-\frac{3}{10}\)

Let,

\(y=\sin [(n+1) A] \sin [(n+2) A]+\cos [(n+1) A] \cos [(n+2) A]\)

\(\Rightarrow y=\cos [(n+1) A] \cos [(n+2) A]+\sin [(n+1) A] \sin [(n+2) A]\)

\(\because \cos (A-B)=\cos A \cdot \cos B-\sin A \cdot \sin B\)

\(\Rightarrow y=\cos [(n+1) A-(n+2) A]\)

\(\Rightarrow y=\cos (-A)\)

\(\because \cos (-\theta)=\cos \theta\)

\(\Rightarrow y=\cos A\)