Mathematics Test - 30

Let,

\({L}=\lim _{{x} \rightarrow \infty} \frac{{x}^{3}+3 {x}^{2}+6 {x}+5}{{x}^{3}+2 {x}+6}\) ....(1)

Putting \(x=\infty\) in equation (1), we get:

\(L=\frac{\pm \infty}{\pm \infty}\)

Since, we know that:

\(\frac{d}{dx}(x^n)= nx^{n-1}\) and differentiation of constant term is \(0.\)

Differentiating numerator and denominator with respect to \(x\) in equation (1), we get:

\({L}=\lim _{{x} \rightarrow \infty} \frac{3 {x}^{2}+6 {x}+6}{3 {x}^{2}+2}\) ....(2)

Again putting \(x=\infty\) in equation (2), we get: 

\(L=\frac{\pm \infty}{\pm \infty}\)

Differentiating numerator and denominator with respect to \(x\) in equation (2), we get:

\({L}=\lim _{{x} \rightarrow \infty} \frac{6 {x}+6}{6 {x}}\) ....(3)

Again putting \(x=\infty\) in equation (3), we get 

\(L=\frac{\pm \infty}{\pm \infty}\)

Again differentiating numerator and denominator with respect to \(x\) in equation (3), we get:

\({L}=\lim _{{x} \rightarrow \infty} \frac{6}{6}\)

\(=\lim _{{x} \rightarrow \infty} 1\) ....(4)

Applying limit \({x} \rightarrow \infty\) in equation (4), we get:

\( {L}=1\)

Given,

\(\frac{d y}{d x}=x\left[\frac{y^2}{x^2}+\frac{\phi\left(\frac{y^2}{x^2}\right)}{\phi^{\prime}\left(\frac{y^2}{x^2}\right)}\right] \ldots \ldots(i)\)

Let \(\mathrm{t}=\frac{\mathrm{y}}{\mathrm{x}}\)

\(\Rightarrow \mathrm{y}=\mathrm{xt} \\\)

\(\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{t}+\frac{\mathrm{dt}}{\mathrm{dx}}\)

\(\therefore\) Eq. (i) becomes

\( t\left(t+x \frac{d t}{d x}\right)=\left(t^2+\frac{\phi\left(t^2\right)}{\phi^{\prime}\left(t^2\right)}\right) \\\)

\( \Rightarrow x t \frac{d t}{d x}=\frac{\phi\left(t^2\right)}{\phi^{\prime}\left(t^2\right)} \\\)

\( \Rightarrow \frac{t \phi^{\prime}\left(t^2\right)}{\phi\left(t^2\right)} d t=\frac{d x}{x}\)

Integrating both sides

\(\int \frac{t \phi^{\prime}\left(t^2\right)}{\left.\phi t^2\right)} d t=\int \frac{d x}{x}\)

Let \(\phi\left(t^2\right)=u\)

\(\Rightarrow t \phi^{\prime}\left(t^2\right) d t=\frac{d u}{2} \\\)

\(\therefore \frac{1}{2} \int \frac{d u}{u}=\int \frac{d x}{x} \\\)

\(\Rightarrow \frac{1}{2} \ln u=\ln x+C \\\)

\(\Rightarrow \frac{1}{2} \ln \phi\left(t^2\right)=\ln x+C \\\)

\(\Rightarrow \frac{1}{2} \ln \left(\phi\left(\frac{y^2}{x^2}\right)\right)=\ln x+C\)

If \(x=1, y=-1\), then \(C=\frac{1}{2} \ln (\phi(1))\)

\(\therefore \frac{1}{2} \ln \left(\phi\left(\frac{\mathrm{y}^2}{\mathrm{x}^2}\right)\right)=\ln \mathrm{x}+2 \ln (\phi(1))\)

If \(x=2\), then \(\ln \left(\phi\left(\frac{y^2}{4}\right)\right)=\ln 4+\ln [\phi(1)]\)

\(\operatorname{Or} \phi\left(\frac{y^2}{4}\right)=4 \phi(1)\)

\(\begin{aligned} & \mathrm{AA}^{\mathrm{T}}=\left|\begin{array}{cc}\frac{1}{5} & \frac{2}{\sqrt{5}} \\ \frac{-2}{\sqrt{5}} & \frac{1}{\sqrt{5}}\end{array}\right|\left|\begin{array}{cc}\frac{1}{\sqrt{5}} & -\frac{2}{\sqrt{5}} \\ \frac{2}{\sqrt{5}} & \frac{1}{\sqrt{5}}\end{array}\right| \\ & \mathrm{AA}^{\mathrm{T}}\left|\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right|=1\end{aligned}\)

\(\begin{aligned} & \text { Now, } Q^2=A^T B A A^T B A \\ & \Rightarrow Q^2=A^T B^2 A \\ & \text { Again, } Q^3=\left(A^T B A\right)\left(A^T B^2 A\right)=A^T B^3 A \\ & \text { Similarly, } \\ & Q^{2021}=A^T B^{2021} A \\ & A Q^{2021} A^T=A\left(A^T B^{2021} A\right) A^T \\ & =\left(A A^T\right) B^{2021}\left(A A^T\right)=B^{2021}\end{aligned}\)

\(\begin{aligned} & B=\left|\begin{array}{ll}1 & 0 \\ i & 1\end{array}\right| \\ & B^2=\left|\begin{array}{cc}1 & 0 \\ 2 i & 1\end{array}\right| \text {, similarly } B^{2021}=\left|\begin{array}{cc}1 & 0 \\ 2021 i & 1\end{array}\right| \\ & \left(B^{2021}\right)^{-1}=\frac{\operatorname{adj}\left(B^{2021}\right)}{\left|B^{2021}\right|}=\left(\begin{array}{cc}1 & 0 \\ -2021 i & 1\end{array}\right)\end{aligned}\)

Given,

25, -125, 625, -3125, …….

Here,

First term, a = 25

Common ratio, r =\(\frac{-125}{25}\) = -5

As we know that, if a₁, a₂, …., a_n is a GP then the general term is given by:

a_n = a × r^{n - 1}

Where a is the first term and r is the common ratio.

The general term is:

\(a_{n}=25 \times(-5)^{n-1}\)

\(=(-1)^{n-1} 5^{n+1}\)

Given,

\(\sin ^{-1}\left(2 x \sqrt{1-x^{2}}\right)\)

The domain of inverse sin function , \(\sin x\) is \(\in [-1,1]\)

Domain of the function is calculated as follows:

\(\sin ^{-1}\left(2 x \sqrt{1-x^{2}}\right)\)

\(-1 \leq 2 x \sqrt{1-x^{2}} \leq 1\)

\(-\frac{1}{2} \leq x \sqrt{1-x^{2}} \leq \frac{1}{2}\)

\(x^{2}\left(1-x^{2}\right) \leq \frac{1}{4}\)

Let, \(x^{2} =t\)

\(t\left(1-t\right) \leq \frac{1}{4}\)

\(t-t^{2}-\frac{1}{4} \leq 0\)

\(\left(t-\frac{1}{2}\right)^{2} \leq 0\)

\(t \leq \frac{1}{2}\)

\(x^{2} \leq \frac{1}{{2}}\)

\(x\leq \frac{1}{\sqrt{2}}\)

\(x \in\left[-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right]\)

\(\begin{aligned} & p=\frac{k \cot 2 \alpha}{\sqrt{\operatorname{cosec}^2 \alpha+\sec ^2 \alpha}} \\ & \Rightarrow q=\frac{k \sin 2 \alpha}{\sqrt{\sin ^2 \alpha+\cos ^2 \alpha}} \\ & \Rightarrow p=\frac{k\left(\frac{\cos 2 \alpha}{\sin 2 \alpha}\right)}{\sqrt{\frac{\sin ^2 \alpha+\cos ^2 \alpha}{\sin ^2 \alpha \cos ^2 \alpha}}}=\frac{k \cos 2 \alpha}{\sin 2 \alpha} \\ & \Rightarrow p=\left(\frac{k}{2}\right) \cos 2 \alpha \\ & \Rightarrow q=k \sin 2 \alpha \\ & \Rightarrow \cos 2 \alpha=(2 p / k) \\ & \Rightarrow \sin 2 \alpha=(q / k) \\ & \Rightarrow \sin 2 \alpha+\cos ^2 2 \alpha=1 \\ & \Rightarrow \frac{4 p^2}{k^2}+\frac{q^2}{k^2}=1 \\ & \Rightarrow 4 p^2+q^2=k^2\end{aligned}\)

Given: 

\(y=\sin x+\cos x\) in the interval \(0

Let the required area be \(\mathrm{A}\).

Using the formula of the area under the curve as:

\(\mathrm{A}=\left|\int_{\mathrm{a}}^{\mathrm{b}} \mathrm{f}(\mathrm{x})+\mathrm{g}(\mathrm{x}) \mathrm{dx}\right|\)

\(=\left|\int_{\mathrm{a}}^{\mathrm{b}} \mathrm{f}(\mathrm{x}) \mathrm{dx}\right|+\left|\int_{\mathrm{a}}^{\mathrm{b}} \mathrm{g}(\mathrm{x}) \mathrm{d} \mathrm{x}\right|\)

\(\Rightarrow \mathrm{A}=\left|\int_{0}^{\frac{\pi}{2}}(\sin \mathrm{x}+\cos \mathrm{x}) \mathrm{d} \mathrm{x}\right|\)

\(=\left|\int_{0}^{\frac{\pi}{2}}(\sin \mathrm{x}) \mathrm{dx}\right|+\left|\int_{0}^{\frac{\pi}{2}}(\cos \mathrm{x}) \mathrm{dx}\right|\)

\(=\left|[-\cos \mathrm{x}]_{0}^{\frac{\pi}{2}}\right|+\left|[\sin \mathrm{x}]_{0}^{\frac{\pi}{2}}\right|\)

Using the value of \(\cos 0=1, \sin 0=0, \cos \frac{\pi}{2}=0\) and \(\sin \frac{\pi}{2}=1\)

\(\Rightarrow \mathrm{A}=\left|-\cos \frac{\pi}{2}+\cos 0\right|+\left|\sin \frac{\pi}{2}+\sin 0\right|\)

\(=|0+1|+|1-0|\)

\(=1+1\)

\(=2\)

Given diameter of spherical balloon \(D=\frac{3}{2}(4 x+3)\)

Radius \(R =\frac{ D }{2}=\frac{3}{4}(4 x +3)\)

Rate of change of radius \(R\) wrt \(x =\frac{ dR }{ dx }=\frac{ d }{ dx }\left[\frac{3}{4}(4 x +3)\right]\)

\( \frac{ dR }{ dx }=\frac{3}{4} \times \frac{ d }{ dx }(4 x +3) \)

\( \frac{ dR }{ dx }=\frac{3}{4} \times 4=3\)

Volume of the spherical balloon \(V=\frac{4 \pi}{3} \times R^3\)

Rate of change of radius \(V\) wrt \(x=\frac{d V}{d x}=\frac{d V}{d R} \times \frac{d R}{d x}\)

\(\frac{ dV }{ dx }=\frac{ d }{ dR }\left[\frac{4 \pi}{3} \times R ^3\right] \times \frac{ dR }{ dx } \)

\( \frac{ dV }{ dx }=\frac{4 \pi}{3} \times 3 R ^2 \times 3=12 \pi R ^2\)