Mathematics Test - 35

Maximize

\(\mathrm{Z}=(\mathrm{x}+\mathrm{y})\)

\(\Rightarrow \mathrm{x}-\mathrm{y} \leq-1, \mathrm{y}-\mathrm{x} \leq 0, \mathrm{x}, \mathrm{y} \leq 0\)

put \(\mathrm{x}=0\) in \(\mathrm{x}-\mathrm{y}=-1 ..............\) (i)

\(\mathrm{y}=1\)

put \(\mathrm{x}=1\) in equation (i)

\(\mathrm{y}=2\)

put \(\mathrm{x}=-1\)

\(-1-\mathrm{y}=-1\)

\(\mathrm{y}=0\)

\(\mathrm{y}-\mathrm{x}=0\).......(ii)

put \(\mathrm{x}=0\) in equation (ii)

\(\mathrm{y}=0\)

put \(\mathrm{x}=1\)

\(\mathrm{y}=1\)

put \(\mathrm{x}=2\)

\(\mathrm{y}=2\)

put \(\mathrm{x}=-1\)

\(\mathrm{y}=-1\)

No feasible region, hence, no maximum value.

\(\frac{x}{a} \cos \theta+\frac{y}{b} \sin \theta=1\)

Or, \(b x\cos \theta+a y \sin \theta-a b=0\).......(1)

Length of the perpendicular from point \(\left(\sqrt{a^{2}-b^{2}} ; 0\right)\) to line (1) is

\(P_{1}=\frac{|b \cos \theta\left(\sqrt{a^{2}-b^{2}}\right)+a \sin \theta(0)-a b \mid}{\sqrt{b^{2} \cos ^{2} \theta+a^{2} \sin ^{2} \theta}}=\frac{\mid b \cos \theta \sqrt{a^{2}-b^{2}}-a b|}{\sqrt{b^{2} \cos ^{2} \theta+a^{2} \sin ^{2} \theta}}\)....(2)

Length of the perpendicular from point \(\left(-\sqrt{a^{2}-b^{2}}, 0\right)\) to line \((2)\) is

\(P_{2}=\frac{|b \cos \theta\left(-\sqrt{a^{2}-b^{2}}\right)+a \sin \theta(0)-a b|}{\sqrt{b^{2} \cos ^{2} \theta+a^{2} \sin ^{2} \theta}}=\frac{|b \cos \theta (-\sqrt{a^{2}-b^{2}})-a b \mid}{\sqrt{b^{2} \cos ^{2} \theta+a^{2} \sin ^{2} \theta}}\)

\(P_{2}=\frac{|b \cos \theta\left(-\sqrt{a^{2}-b^{2}}\right)+a \sin \theta(0)-a b|}{\sqrt{b^{2} \cos ^{2} \theta+a^{2} \sin ^{2} \theta}}=\frac{|b \cos \theta \sqrt{a^{2}-b^{2}}+a b \mid}{\sqrt{b^{2} \cos ^{2} \theta+a^{2} \sin ^{2} \theta}}\)....(3)

On multiplying equations (2) and (3) we get,

\(P_{1} P_{2}=\frac{|(b \cos \theta \sqrt{a^{2}-b^{2}}-a b)(b \cos \theta \sqrt{a^{2}-b^{2}}+a b) \mid}{\left(\sqrt{b^{2} \cos ^{2} \theta+a^{2} \sin ^{2} \theta}\right)^{2}}\)

\(=\frac{\left|\left(b \cos \theta \sqrt{a^{2}-b^{2}}-a b\right)\left(b \cos \theta \sqrt{a^{2}-b^{2}}+a b\right)\right|}{\left(b^{2} \cos ^{2} \theta+a^{2} \sin ^{2} \theta\right)}\)

\(=\frac{\left|\left(b \cos \theta \sqrt{a^{2}-b^{2}}\right)^{2}-(a b)^{2}\right|}{\left(b^{2} \cos ^{2} \theta+a^{2} \sin ^{2} \theta\right)}\)\(\quad(\because (a+b)(a-b) = a^{2} -b^{2})\)

\(=\frac{\left|b^{2} \cos ^{2} \theta\left(a^{2}-b^{2}\right)-a^{2} b^{2}\right|}{\left(b^{2} \cos ^{2} \theta+a^{2} \sin ^{2} \theta\right)}\)

\(=\frac{\left|a^{2} b^{2} \cos ^{2} \theta-b^{4} \cos ^{2} \theta-a^{2} b^{2}\right|}{b^{2} \cos ^{2} \theta+a^{2} \sin ^{2} \theta}\)

\(=\frac{b^{2}\left|a^{2} \cos ^{2} \theta-b^{2} \cos ^{2} \theta-a^{2}\right|}{b^{2} \cos ^{2} \theta+a^{2} \sin ^{2} \theta}\)

\(=\frac{b^{2}\left|a^{2} \cos ^{2} \theta-b^{2} \cos ^{2} \theta-a^{2} \sin ^{2} \theta-a^{2} \cos ^{2} \theta\right|}{b^{2} \cos ^{2} \theta+a^{2} \sin ^{2} \theta}\) \(\quad\quad\because[\sin ^{2} \theta+\cos ^{2} \theta=1]\)

\(=\frac{b^{2}\left|-\left(b^{2} \cos ^{2} \theta+a^{2} \sin ^{2} \theta\right)\right|}{b^{2} \cos ^{2} \theta+a^{2} \sin ^{2} \theta}\)

\(=\frac{b^{2}\left(b^{2} \cos ^{2} \theta+a^{2} \sin ^{2} \theta\right)}{\left(b^{2} \cos ^{2} \theta+a^{2} \sin ^{2} \theta\right)}\)

\(=b^{2}\)

\(\sin ^{-1} x+\sin ^{-1} y=\frac{5 \pi}{6}\)

We know that:

\(\sin ^{-1} x+\cos ^{-1} x=\frac{\pi}{2}\)

\(\tan ^{-1} x+\cot ^{-1} x=\frac{\pi}{2}\)

Let, \(\cos ^{-1} x+\cos ^{-1} y=a\)

Adding the two equation we get,

\(\left(\sin ^{-1} x+\cos ^{-1} x\right)+\left(\sin ^{-1} y+\cos ^{-1} y\right)=\frac{5 \pi}{6}+a\)

\(\Rightarrow \frac{\pi}{2}+\frac{\pi}{2}=\frac{5 \pi}{6}+a\)

\(\Rightarrow \pi-\frac{5 \pi}{6}=a \Rightarrow a=\frac{\pi}{6}\)

So, \(\cos ^{-1} x+\cos ^{-1} y=\frac{\pi}{6}\)

Given,

The distances of \(\mathrm{P}(x, y)\) from \(\mathrm{A}(4,1)\) and \(\mathrm{B}(-1,4)\) are equal.

The distance ' \(d\) ' between two points \(\left(x_{1}, y_{1}\right)\) and \(\left(x_{2}, y_{2}\right)\) is obtained by using the Pythagoras' Theorem: \(d^{2}=\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}\).

Using the formula for the distance between two points:

\(\mathrm{AP}^{2}=(x-4)^{2}+(y-1)^{2}\)

\(\mathrm{BP}^{2}=(x+1)^{2}+(y-4)^{2}\)

Since the distances are equal, we have:

\(\mathrm{AP}^{2}=\mathrm{BP}^{2} \)

\(\Rightarrow(x-4)^{2}+(y-1)^{2}=(x+1)^{2}+(y-4)^{2}\)

\(\Rightarrow x^{2}-8 x+16+y^{2}-2 y+1=x^{2}+2 x+1+y^{2}-8 y+16 \)

\(\Rightarrow 10 x=6 y\)

\(\Rightarrow 5 x=3 y\)

Section formula is given by \(\frac{m A+n B}{m+n}\).

Given: Position vector of \(\mathrm{A}\) and \(\mathrm{B}\) are \(\vec{a}\) and \(\vec{b}\) respectively.

Let \(M\) be the mid-point of \(A B\), such that \(A M=M B\), it means that \(M\) divides \(A B\) in the ratio \(1: 1\), so by section formula, we get,

\(\frac{(1) \vec{a}+(1) \vec{b}}{1+1}=\frac{1}{2}(\vec{a}+\vec{b})\)

Thus, we conclude that If the position vectors of \({A}\) and \({B}\) are \(\vec{a}\) and \(\vec{b}\) respectively, then the position vector of mid-point of \({AB}\) is \(\frac{1}{2}(\vec{a}+\vec{b})\).

We know that the area of a circle of radius units is given by \(A =\pi r ^{2}\) sq.units.

\(\therefore \frac{ dA }{ dt }=\frac{ d }{ dt }\left(\pi r ^{2}\right)=2 \pi r \left(\frac{ dr }{ dt }\right)\)

Now, \(\left[\frac{ dA }{ dt }\right]_{ r =4}=8 \pi(0.01)\) m\(^{2}/\) sec \(=0.08 \pi\) m\(^{2}/\)sec

Given:

\(\mathrm{z}=2 \mathrm{x}+5 \mathrm{y}\)

\(2 \mathrm{x}+4 \mathrm{y} \leq 8 \)

\(\Rightarrow \mathrm{x}+2 \mathrm{y} \leq 4\)

\(3 x+y \leq 6, x+y \leq 4, x \geq 0, y \geq 0\)

Draw the lines \(\mathrm{x}+2 \mathrm{y}=4\) (passes through (4,0\(),(0,2))\)

\(3 \mathrm{x}+\mathrm{y}=6\) (passes through \((2,0),(0,6)\) and

\(\mathrm{x}+\mathrm{y}=4\) (passes through \((4,0),(0,4)\)

Shade the region satisfied by the given inequations;

The shaded region in the figure gives the feasible region determined by the given inequations.

Solving \(3 \mathrm{x}+\mathrm{y}=6\) and \(\mathrm{x}+2 \mathrm{y}=4\) simultaneously, we get

\(\mathrm{x}=\frac{8}{5}\) and \(\mathrm{y}=\frac{6}{5}\)

We observe that the feasible region \(\mathrm{OABC}\) is a convex polygon and bounded and has corner points.

\(\mathrm{O}(0,0), \mathrm{A}(2,0), \mathrm{B}\left(\frac{8}{5}, \frac{6}{5}\right), \mathrm{C}(0,2)\)

The optimal solution occurs at one of the corner points.

At \(\mathrm{O}(0,0), \mathrm{z}=2×0+5×0=0\)

At \(\mathrm{A}(2,0),~~ \mathrm{z}=2×2+5×0=4\)

At B \(\left(\frac{8}{5}, ~~\frac{6}{5}\right), z=2 × \frac{8}{5}+5 × \frac{6}{5}=\frac{46}{5}\)

At \(\mathrm{C}(0,2),~~ \mathrm{z}=2×0+5×2=10\)

Therefore, z maximum value at \(\mathrm{C}\) and maximum value \(=10\)

Let \(f(x)=y\)

\(\therefore f^{-1}(y)=x\)

Now, \(y=2 x+6\)

\(2 x=y-6\)

\(x=\frac{y-6}{2}\)

\(\therefore f^{-1}(y)=\frac{y}{2}-3\)

\(f^{-1}(x)=\frac{x}{2}-3\)