Mathematics Test - 8

From the given figure we have

According to Lami,s theorem,

\(\frac{A}{\sin \alpha}=\frac{B}{\sin \beta}=\frac{C}{\sin \gamma}\)

\(\frac{T_{1}}{\sin (120^{\circ})}=\frac{T_{2}}{\sin (150^{\circ})}=\frac{T_{3}}{\sin (90^{\circ})} \)

Then we can write,

\(\frac{T_{1}}{\sin (120^{\circ})}=\frac{T_{2}}{\sin (150^{\circ})}\)

\(\frac{T_{1}}{\frac{\sqrt3}{2}}=\frac{T_{2}}{\frac{1}{2}}\)

\(\frac{T_{1}}{T_{2}}=\sqrt{3}\)

Again, \(\frac{T_{1}}{\sin (120^{\circ})}=\frac{T_{3}}{\sin (90^{\circ})} \)

\(\frac{T_{1}}{\frac{\sqrt3}{2}}=\frac{T_{3}}{1} \)

\(\frac{T_{1}}{T_{3}}=\frac{\sqrt{3}}{2}\)

By solving the above equation, we get

\(\frac{T_{1}}{T_{2}}=\sqrt{3}\) and \(\frac{T_{1}}{T_{3}}=\frac{\sqrt{3}}{2}\)

\(x^8+12 x^5-4 x^4+10 x^3-6 x+5\)

If \(f\) and \(g\) are both differentiable, then

\(\frac{d}{d x}[f(x)-g(x)]=\frac{d}{d x} f(x)-\frac{d}{d x} g(x)\)

As we know,

\(\frac{d}{d x}\left(x^n\right)=n x^{n-1}\)

Differentiating with respect to \(x\), we get

\(\frac{d}{d x}\left(x^8+12 x^5-4 x^4+10 x^3-6 x+5\right)=\frac{d}{d x}\left(x^8\right)+12 \frac{d}{d x}\left(x^5\right)-4 \frac{d}{d x}\left(x^4\right)+10 \frac{d}{d x}\left(x^3\right)-6 \frac{d}{d x}(x)+\frac{d}{d x}(5)\)

\(=8 x^7+12\left(5 x^4\right)-4\left(4 x^3\right)+10\left(3 x^2\right)-6(1)+0 \)

\(=8 x^7+60 x^4-16 x^3+30 x^2-6\)

Given,

\((2 x)^{2 y}=4 e^{2 x-2 y}\)

As we know,

\(\log _{b}\left(M^{p}\right)=p \log _{b}(M)\)

\(\log _{b}(M N)=\log _{b}(M)+\log _{b}(N)\)

Taking \(log\) both sides, we get

\(\ln (2 x)^{2 y}= \ln 4 e^{2 x-2 y}\)

\( \Rightarrow\ln (2 x)^{2 y}=\ln 4+\ln (2 x-2 y)\)

\(\Rightarrow2 y \ln (2 x)=\ln 2^2+(2 x-2 y)\)

\(\Rightarrow 2 y \ln (2 x)=2 \ln 2+(2 x-2 y)...(1)\)

\(\Rightarrow y \ln (2 x)=\ln 2+(x-y)\)

\(\Rightarrow y(1+\ln 2 x)=x+\ln 2\)

\(\therefore y=\frac{x+\ln 2}{1+\ln (2 x)}....(2)\)

As we know,

\(\frac{d}{d x}\left(\ln x\right)=\frac{1}{x }\)

If \(f\) and \(g\) are both differentiable, then

\(\frac{d}{d x}[f(x) g(x)]=f(x) \frac{d}{d x}[g(x)]+g(x) \frac{d}{d x}[f(x)]\)

Differentiating equation \((1)\) with respect to \(x\), we get

\(2 y \times \frac{1}{2 x} \times 2+2 \ln (2 x) \frac{d y}{d x}=0+2-2 \frac{d y}{d x}\)

\(\Rightarrow 2[1+\ln (2 x)] \frac{d y}{d x}=2-\frac{2 y}{x}\)

\(\Rightarrow[1+\ln (2 x)] \frac{d y}{d x}=1-\frac{y}{x}....(3)\)

Put equation \((2)\) in equation \((3)\), we get

\(\Rightarrow[1+\ln (2 x)] \frac{d y}{d x}=1-\frac{\frac{x+\ln 2}{1+\ln (2 x)}}{x}\)

\(\Rightarrow[1+\ln (2 x)] \frac{d y}{d x}=1-\frac{1}{x}\left(\frac{x+\ln 2}{1+\ln 2 x}\right)\)

\(\Rightarrow[1+\ln (2 x)]^{2} \frac{d y}{d x}=\frac{x+x \ln (2 x)-x-\ln 2}{x}\)

\(\Rightarrow[1+\ln (2 x)]^{2} \frac{d y}{d x}=\frac{x \ln (2 x)-\ln 2}{x}\)

\(\Rightarrow[1+\log_e (2 x)]^{2} \frac{d y}{d x}=\frac{x \log_e (2 x)-\log_e 2}{x}\)

Given,

The Objective function is,

\(\mathrm{Z}=3 {x}-4 {y}\)

At \( (0,0) \),

\(\mathrm{Z}=3 \times 0-4 \times 0=0\)

At \( (5,0) \),

\(\mathrm{Z}=3 \times 5-4 \times 0=15\)

At \( (6,5) \),

\(\mathrm{Z}=3 \times 6-4 \times 5=-2\)

At \( (6,8) \),

\(\mathrm{Z}=3 \times 6-4 \times 8=-14\)

At \( (4,10) \),

\(\mathrm{Z}=3 \times 4-4 \times 10=-28\)

At \( (0,8) \),

\(\mathrm{Z}=3 \times 0-4 \times 8=-32\)

Maximum of \(\mathrm{Z}\) occurs at\( (5,0) \).

The horizontal force is, \(\Sigma F_{x}=0\)

And vertical force is, \(\Sigma F_{y}=0\)

Given,

\(\Sigma F_{x}=-T_{A} \cos 45^{\circ}+T_{B} \cos 45^{\circ}=0\)

\(\therefore T_{\mathrm{A}}=\mathrm{T}_{\mathrm{B}} \ldots \ldots .(i)\)

\(\Sigma F_{y}=T_{A} \sin 45^{\circ}+T_{B} \sin 45^{\circ}-W=0\)

\(\therefore\left(T_{A}+T_{B}\right) \sin 45^{\circ}=W\)

\(2 T_{A} \times \sin 45^{\circ}=W\)

\(\therefore T_{A}=\frac{W}{\sqrt{2}}\)

Given,

\(P_{3}=1000 {N}\)

\(P_{1}, P_{2}\) and \(P_{3}\) are the three forces lying on the same plane and passing through a common point.

\(\alpha=\) Angle between \(P_{2}\) and \(P_{3} = 120^{\circ}\)

\(\beta=\) Angle between \(P_{1}\) and \(P_{3} = 150^{\circ}\)

\(\gamma=\) Angle between \(P_{1}\) and \(P_{2} = 90^{\circ}\)

Applying Lami’s theorem:

\(\frac{P_{1}}{\sin \alpha}=\frac{P_{2}}{\sin \beta}=\frac{P_{3}}{\sin \gamma}\)

\(\frac{P_{1}}{\sin \alpha}=\frac{P_{3}}{\sin \gamma}\)

\(\Rightarrow \frac{P_{1}}{\sin 120}=\frac{1000}{\sin 90^{\circ}}\)

\(\Rightarrow P_{1}=1000 \times \sin 120^{\circ} \quad(\because \sin 90^{\circ}=1)\)

\(\therefore P_{1}=866 \mathrm{~N}\)

\(\frac{P_{2}}{\sin \beta}=\frac{P_{3}}{\sin \gamma}\)

\(\Rightarrow P_{2}=1000 \times \sin 150(\because \sin 90^{\circ}=1)\)

\(\therefore \mathrm{P}_{2}=500 \mathrm{~N}\)

Therefore, Chain \(1(P _2)=500 \mathrm{~N}\), Chain \(2(\mathrm{P}_{1})=866 \mathrm{~N}\)

Let, \(z=\frac{1-\mathrm{i}}{1+\mathrm{i}}\)

or \(\mathrm{z}=\frac{1-\mathrm{i}}{1+\mathrm{i}} \times \frac{1-\mathrm{i}}{1-\mathrm{i}}=\frac{1+\mathrm{i}^{2}-2 \mathrm{i}}{1-\mathrm{i}^{2}}\)

⇒ z = -\(\frac{2i}{2}\) = -i

x + iy = -i

so, x = 0 and y = -1

We know that, \(\operatorname{Arg}(z)=\tan ^{-1}\left(\frac{y}{x}\right)\)

\(\operatorname{Arg}(z)=\tan ^{-1}\left(\frac{-1}{0}\right)=-\tan ^{-1}\left(\frac{1}{0}\right)\)

\(\operatorname{Arg}(z)=-\tan ^{-1}(\infty)=-\frac{\pi}{2}\)

Equation of parabola is \({y}^{2}=4 {ax}\).

The given equation is in \(x y-\)plane.

It is a Parabola with the horizontal axis of symmetry and vertex in the origin.

The value of \(x\) and \(y\) are the coordinates in the \(xy\) plane.

The parabola gives a direct relation between \(x\) and \(y\).

Each and every point of the parabola must satisfy this relation.

Also, parametric equations represent each point of the parabola as a function of parameter 't'.

A parametric representation is:

\({x}={at}^{2}\)

\({y}=2.at\)

For every value of \(t \in R\) parametric equation yield a point \(p(t)=\left(a t^{2}, 2 a t\right)\) on the parabola.

Given,

\((a+\sqrt{2} b \cos x)(a-\sqrt{2} b \cos y)=a^{2}-b^{2}\)

As we know,

\(\frac{d}{dx} \sin x = \cos x\)

\(\frac{d}{dx} \cos x = -\sin x\)

Differentiation of a constant is always \(0\).

Differentiating both sides with respect to \(x\), we get

\(\Rightarrow(-\sqrt{2} b \sin x)(a-\sqrt{2} b \cos y)+(a+\sqrt{2} b \cos x)(\sqrt{2} b \sin y) \frac{d y}{d x}=0\)

\(\Rightarrow \frac{d y}{d x}=\frac{(\sqrt{2} b \sin x)(a-\sqrt{2} b \cos y)}{(a+\sqrt{2} b \cos x)(\sqrt{2} b \sin y)}\)

\(\therefore\left[\frac{d y}{d x}\right]_{\left(\frac{\pi}{4}, \frac{\pi}{4}\right)}=\frac{\left(\sqrt{2} b \sin \frac{\pi}{4}\right)\left(a-\sqrt{2} b \cos \frac{\pi}{4}\right)}{\left(a+\sqrt{2} b \cos \frac{\pi}{4}\right)\left(\sqrt{2} b \sin \frac{\pi}{4}\right)}\)

As we know,

\(\cos \frac{\pi}{4} =\sin \frac{\pi}{4} = \frac{1}{\sqrt 2}\)

\(\Rightarrow\left[\frac{d y}{d x}\right]_{\left(\frac{\pi}{4}, \frac{\pi}{4}\right)}=\frac{a-b}{a+b}\)

\(\Rightarrow\left[\frac{d x}{d y}\right]_{\left(\frac{\pi}{4}, \frac{\pi}{4}\right)}=\frac{a+b}{a-b}\)