Mathematics Test - 33

Given,

\(\left|\begin{array}{ll}x & 2 \\ 18 & x\end{array}\right|=\left|\begin{array}{cc}6 & 2 \\ 3 x & 6\end{array}\right|\)

\(\Rightarrow x^{2}-36=36-6 x\)

\(\Rightarrow x^{2}+6 x-72=0\)

\(x=\frac{-6 \pm \sqrt{6^{2}+4 \times 72}}{2}\)

\(x=\frac{-6 \pm \sqrt{36(1+8)}}{2}\)

\(x=-3 \pm 9\)

\(x=6\) or \(-12\)

\(\Rightarrow x d y=y d x+y^2 d y \)

\(\Rightarrow x d y-y d x=y^2 d y \)

\(\Rightarrow \frac{x d y-y d x}{y^2}=d y \)

\(\because d\left(\frac{x}{y}\right)=\frac{y d x-x d y}{y^2}=\frac{x d y-y d x}{y^2}=-d\left(\frac{x}{y}\right) \)

\(\Rightarrow-d\left(\frac{x}{y}\right)=d(y)\)

Integrate both side

\(\Rightarrow \quad-\frac{x}{y}=y+c \Rightarrow\) General solution

Given that \(y(1)=1\)

So by putting \(x =1\) and \(y =1\)

\(\Rightarrow-1 / 1=1+c \)

\(\Rightarrow c=-2 \)

\(\Rightarrow-\frac{x}{y}=y-2 \)

\(\Rightarrow y^2-2 y+x=0 \Rightarrow\) particular solution

Now the value of \(y(-3)\) put \(x=-3\)

\(\Rightarrow y ^2-2 y -3=0\)

Solve the quadratic equation

\(\Rightarrow y^2-3 y+y-3=0 \)

\(\Rightarrow y(y-3)+1(y-3)=0\)

\(\Rightarrow( y -3)( y +1)=0\)

So two values of \(y\) we will get \(y=3, y=-1\)

But in our question given that \(y>0\) so the answer is \(y=3\)

We know that, Standard formula of \(y-\)axis parabola is:

\(x^{2}=4ay\)

Let, Vertex of parabola be \((0,k)\)

Then,  \((x-0)^{2}=4 a(y-k)\)

\(\Rightarrow x^{2}=4 a y-4 a k\)

Taking derivative on both side,

We get, \(\frac{d}{d x}\left(x^{2}\right)=\frac{d}{d x}(4 a y-4 a k)\)

\(\Rightarrow 2 x=4 a \frac{d y}{d x}\)

\(\Rightarrow \frac{1}{x} \frac{d y}{d x}=\frac{1}{2 a}\)

Again taking derivative on both side,

We get, \(\frac{d}{dx}\left(\frac{1}{x} \frac{dy}{dx}\right)=\frac{{d}}{dx}\left(\frac{1}{2 z}\right)\)

\(\Rightarrow \frac{1}{x} \frac{{d}^{2}y}{dx^{2}}+\frac{dy}{dx}\left(\frac{-1}{x^{2}}\right)=0\)

\(\Rightarrow x \times \frac{d^{2} y}{dx^{2}}-\frac{dy}{dx}=0\)

Let us first draw the graph of \(y=\cos x\) in the interval \(0

Let the required area of the positive enclosed region be A.

Using the formula of the area under the curve as, 

\(\mathrm{A}=\left|\int_{\mathrm{a}}^{\mathrm{b}} \mathrm{f}(\mathrm{x})-\mathrm{g}(\mathrm{x}) \mathrm{d} \mathrm{x}\right|\),

\(\Rightarrow \mathrm{A}=\left|\int_{0}^{\frac{\pi}{2}} \operatorname{cosxd} \mathrm{x}\right|\)

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

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

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

\(=\left|1-0\right|\) \(=1\)

Given,

\(\frac{3(x-2)}5 \geq \frac{5(2-x)}3\)

⇒ 3(x – 2) × 3 ≥ 5(2 – x) × 5

⇒ 9(x – 2) ≥ 25(2 – x)

⇒ 9x – 18 ≥ 50 – 25x

⇒ 9x – 18 + 25x ≥ 50

⇒ 34x – 18 ≥ 50

⇒ 34x ≥ 50 + 18

⇒ 34x ≥ 68

\(\Rightarrow x \geq \frac{68}{34}\)

⇒ x ≥ 2

⇒ x ∈ [2, ∞)

Let \(x\) is the amount of goods A and \(y\) is the amount of goods B.

Objective function:

Maximum profit, \(Z=40x+50y\)

Subject to constraints:

Given,

Line \(x-3 y+5=0\)

The general equation of a line is \(y=m x+c\)

Where \(m\) is the slope and \(c\) is any constant

• The slope of parallel lines is equal.
• The slope of the perpendicular line have their product \(=-1\)

\(\Rightarrow y=\frac{1}{3} x+\frac{5}{3}\)

\(\Rightarrow \text { Slope }(m_{1})=\frac{1}{3} \text { and } c_{1}=\frac{5}{3}\)

Now for the slope of the perpendicular line \(m_2\)

\(m_{1} \times m_{2}=-1\)

\(\Rightarrow \frac{1}{3} \times m_{2}=-1\)

\(\Rightarrow m_{2}=-3\)

Perpendicular line has the slope \(-3\) and passes through \((2,-4)\)

\(\therefore\) Equation of the perpendicular line is

\(\left(y-y_{1}\right)=m\left(x-x_{1}\right) \)

\(\Rightarrow y-(-4)=-3(x-2)\)

\(\Rightarrow y+3 x-2=0\)

Given,

\(\cot ^{-1} \frac{1}{3}-2 \tan ^{-1} \frac{2}{3}\)

\(=\left[\frac{\pi}{2}-\tan ^{-1} \frac{1}{3}\right]-\tan ^{-1} \frac{(\frac{2}{3}+\frac{2}{3})}{(1-\frac{2}{3} \times \frac{2}{3})}\)\(\quad\quad(\because \cot ^{-1} x= \frac{\pi}{2}-\tan ^{-1} x, 2\tan ^{-1} x =\tan ^{-1} (\frac{2x}{1 -x^2}))\)

\(=\frac{\pi}{2}-\left[\tan ^{-1} \frac{12}{5}+\tan ^{-1} \frac{1}{3}\right]\)

\(=\frac{\pi}{2}-\left[\tan ^{-1} \frac{\frac{12}{3}+\frac{1}{1}}{1-\frac{12}{5} \times \frac{1}{3}}\right]\)\(\quad\quad(\because \tan ^{-1} x+\tan ^{-1} y =\tan ^{-1} \frac{x+y}{1 -xy} )\)

\(=\frac{\pi}{2}-\left[\tan ^{-1} \frac{41}{3}\right]\)

\(=\cot ^{-1} \frac{41}{3}\)

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

\(\mathrm{AA}^{\mathrm{T}}\left|\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right|=1\)

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

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

Given:

\(P(n): n(n+1)(n+5)=3 x, x \in N\)

For \(n=1,1 \times(1+1)(1+5)=12\), which is a multiple of \(3 .\)

\(\mathrm{So}, \mathrm{P}(1)\) is true.

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

\(k(k+1)(k+5)=3 x\), Where \(x \in N\).....(1)

Let \(P(k+1)\) be true for some \(n=k+1\).

\((k+1)(k+2)(k+6)=3 x, x \in N\).....(2)

From (1) and (2).

Now, \((k+1)(k+2)(k+6)-k(k+1)(k+5)\)

\(=(k+1)[(k+2)(k+6)-k(k+5)] \)

\(=(k+1)\left[k^{2}+8 k+12-k^{2}-5 k\right] \)

\(=(k+1)(3 k+12)\) \(=3 x\), where \(x \in N\)

Thus, \((k+1)(k+2)(k+6)\)

\(=k(k+1)(k+5)+3 x\)

\(=3 \mathrm{~m}+3 {x}\), which is multiple of \(3 .\)