Mathematics Test - 6

Given,

\(y=\left|\begin{array}{cc}3 x & 4 x^2 \\ -1 & x^2\end{array}\right|\)

\(y=3 x\left(x^2\right)-\left(-4 x^2\right)\)

\( y=3 x^3+4 x^2\)

As we know,

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

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

\(\frac{d y}{d x}=3\left(3 x^2\right)+4(2 x)\)

\(\frac{d y}{d x}=9 x^2+8 x\)

\(\frac{d y}{d x}=x(9 x+8)\)

Given,

The objective function is ,

\(F=4 x+6 y\)

At\((0,2)\),

\(F=4 \times 0+6 \times 2=12\)

At \((3,0)\),

\(F=4 \times 3+6 \times 0=12\)

At \((6,0)\),

\(F=4 \times 6+6 \times 0=24\)

At \((0,5)\),

\(F=4 \times 0+6 \times 5=30\)

Thus, the maximum value of \(\mathrm{F}=30\)

The minimum value of \(\mathrm{F}=12\)

And the value of the maximum value of \(\mathrm{F}\) - minimum value of \(\mathrm{F}\) \(=30-12\)

\(=18\)

When a point \({P}\left({x}_{1}, {y}_{1}\right)\) and slope 'm' is give the equation of line is:

Equation of line is:

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

Equation of pair of tangents to given curve is:

\({T}^{2}={SS}_{1}\)

\(S:\) Given curve

\({S}_{1}\): Given curve at the \({P}\left({x}_{1}, {y}_{1}\right)\)

\(T:\) Tangent equation

Given Pair of tangents are \(y+6=m_{1}(x+a)\) and \(y+b=m_{2}(x+a)\) the point is \((-a,-b)\)

Parabola equation is \({y}^{2}=4 {ax}\)

\(T=y(-b)-2 a(x-a) \quad\left[y\left(y_{1}\right)-2 a\left(x-x_{1}\right)\right]\)

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

\({S}_{1}=(-{b})^{2}-4 {a}(-{a})={b}^{2}+4 {a}^{2}\)

\(\left[-b y-2 a x+2 a^{2}\right]^{2}=\left(y^{2}-4 a x\right)\left(b^{2}+4 a^{2}\right)\)

\(a x^{2}-a y^{2}+b x y+b^{2} x+2 a^{2} x-a b y+a^{3}=0 \quad\quad\ldots \ldots(1)\)

From the given equations:

\(\left[y+b-m_{1}(x+a)\right]\left[(y+b)-m_{2}(x+a)\right]\)

\((y+b)^{2}+m_{1} m_{2}(x+a)^{2}-m_{2}(y+b)(x+a)-m_{1}(x+a)(y+b)=0\)

\({y}^{2}+{b}^{2}+2 {b} {y}+{m}_{1} {m}_{2} {x}^{2}+{m}_{1} {m}_{2} {a}^{2}+2 {m}_{1} \mathrm{~m}_{2} \times {a}-({y}+{b})({x}+{a})\left({m}_{1}+{m}_{2}\right)=0\quad\quad \ldots \ldots(2)\)

Compare (1) and (2),

\(-a y^{2}=y^{2}\)

\(-a=1\)

\(a=-1\)

\(a x^{2}=m_{1} m_{2} x^{2}\)

\(a=m_{1} m_{2}\)

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

A non-empty subset \({H}\) of a group \(({G}, *)\) is a group of \({G}\) iff,

\(\Rightarrow {a}, {b} \in {H} \)

\(\Rightarrow {a} * {b} \in {H}\)

\(\Rightarrow {a} \in {H}\)

\(\Rightarrow \frac{1}{ {a}} \in {H}\)

\( \Rightarrow {a}^{-1}=\frac{1 }{{a}}\)

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:

If G is a group, then the subgroup consisting of G itself is the improper subgroup of G. All other subgroups are proper subgroups

In any group, the number of improper subgroups is 2.

Equation of line with Z-axis is,

\(\overrightarrow{\mathrm{b}}=\hat{\mathrm{k}}\)

Equation of plane is x -3y + 5z -8 = 0

or in vector form \(, \overrightarrow{\mathrm{n}}=\hat{\mathrm{i}}-3 \hat{\mathrm{j}}+5 \hat{\mathrm{k}}\)

\(\therefore \sin \alpha=\left|\frac{\overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{n}}}{|\overrightarrow{\mathrm{b}}||\overrightarrow{\mathrm{n}}|}\right|\)

\(\Rightarrow \sin \alpha=\frac{(\hat{\mathrm{i}}-3 \hat{\mathrm{j}}+5 \hat{\mathrm{k}}) \cdot(0 \hat{\mathrm{i}}+0 \hat{\mathrm{j}}+\hat{\mathrm{k}})}{\sqrt{1^{2}+(-3)^{2}+5^{2}} \cdot \sqrt{1^{2}}}=\frac{5}{\sqrt{35}}\)

\(\Rightarrow \sin \alpha=\frac{5}{\sqrt{35}}\)

\(\Rightarrow \alpha=\sin ^{-1}\left(\frac{5}{\sqrt{35}}\right)\)

\(\Rightarrow \alpha=\frac{5}{\sqrt{35}}\)

A box contains 3 white and 2 black balls.

Total balls = 3 + 2 = 5

So probability of first ball to be black \( = \frac{2}{5}\) and

Probability of second ball to be also black \(=\frac{1}{4}\) (∵second time we have 2 -1 = 1 black and 5 - 1 = 4 total balls)

∴ The probability that both the balls are black \(=\frac{ 2}{5} × \frac{1}{4}\)

\(= \frac{1}{10}\)

When, a coin is tossed

Outcome = S = {H, T}

When a die is rolled,

S = {1, 2, 3, 4, 5, 6}

Now, a coin is tossed and a die is rolled, and we need only head on the coin

So, S = {(H, 1), (H, 2), (H, 3), (H, 4), (H, 5), (H, 6)}

n(S) = 6

Given,

\(y=(\tan x)^{\tan x^{\tan x} }...(1)\)

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 on both sides, we get

\(\log y=\log (\tan x)^{\tan x^{\tan x}}\)

\(\log y=(\tan x)^{\tan x} \log (\tan x)\)

Again, taking log on both sides, we get

\(\log (\log y)=\log (\tan x)^{\tan x}+\log (\log (\tan x))\)

\(\log (\log y)=\tan x \log (\tan x)+\log (\log (\tan x))\)

As we know,

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

\(\frac{d}{d x} \tan x=\sec ^{2} x\)

\(\frac{d}{d x} \log x=\frac{1}{x}\)

Now, differentiating with respect to \(x\), we get

\(\frac{1}{\log y} \cdot \frac{1}{y} \cdot \frac{d y}{d x}=\tan x\left(\frac{\sec ^{2} x}{\tan x}\right)+\sec ^{2} x \log (\tan x)+\frac{1}{\log (\tan x)} \cdot \frac{1}{\tan x} \cdot \sec ^{2} x\)

\(\frac{d y}{d x}=y \log y\left[\tan x\left(\frac{\sec ^{2} x}{\tan x}\right)+\sec ^{2} x \log (\tan x)+\frac{1}{\log (\tan x)} \cdot \frac{1}{\tan x} \cdot \sec ^{2} x\right]\)

At \(x=\frac{\pi}{4}\),

From equation \((1)\), we get

\(y=(\tan \frac{\pi}{4})^{\tan \frac{\pi}{4}^{tan\frac{\pi}{4}} }\)

\(y=1^{1^{1}}=1\)

\(\log y=\log 1=0\)

So,\(\frac{d y}{d x} _{(x= \frac{x}{4})}=0\)