Mathematics Test - 40

Given:

\(\frac{d^{n}}{d^{n} x}\left(x e^{2 x}\right)\)

For \(n=1\)

\(\frac{d}{d x}\left(x e^{2 x}\right)=2 x e^{2 x}+e^{2 x}\)

\(=(2 x+1) e^{2 x} \)

\(=2^{n-1}(2 x+n) e^{2 x}\)

Let \(P(m)\) be true so that:

\(\frac{d^{m}}{d^{m} x}\left(x e^{2 x}\right)=2^{m-1}(2 x+m) e^{2 x}\)

For \(P(m+1)=\) We should get \(2^{m}(2 x+m+1) e^{2 x}\)

\(\frac{d^{m+1}}{d^{m+1} x}\left(x e^{2 x}\right)=\frac{d}{d x}\left(\frac{d^{m}}{d^{m} x}\left(x e^{2 x}\right)\right) \)

\(=\frac{d}{d x} 2^{m-1}\left[(2 x+m) e^{2 x}\right] \)

\(=2^{m}(2 x+m) e^{2 x}+e^{2 x} \)

\(=2^{m}(2 x+m+1) e^{2 x}\)

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

\(\int \sec ^2 x d x=\tan x+C\)

\(\int \sec x \tan x d x=\sec x+c\)

Calculation:

Let \(I=\int \frac{\sin x}{(\cos x)^3} d x\)

\(=\int \tan x \sec ^2 x d x\)

Let \(\tan x=t\)

\(\Rightarrow \sec ^2 x d x=d t\)

Therefore, the integral becomes.

\(=\int \mathrm{t} d \mathrm{t}\)

\(=\frac{\mathrm{t}^2}{2}+\mathrm{C}\)

Re-substitute \(t=\tan x\).

Thus,

\(=\frac{\tan ^2 x}{2}+C\)

Given, \(\frac{3-|x|}{4-|x|} \geq 0\)

\(\Rightarrow 3-|x| \leq 0\) and \(4-|x|<0\)

or \(3-|x| \geq 0\) and \(4-|x|>0\)

\(\Rightarrow |x| \geq 3\) and \(|x|>4\)

\(\Rightarrow x \in(-\infty,-4) \cup[-3,3] \cup(4, \infty)\)

\( \text { Given, slope }=\frac{x^2-4 x+y+8}{x-2} \\\)

\( \Rightarrow \frac{d y}{d x}=\frac{x^2-4 x+y+8}{x-2}=\frac{(x-2)^2+(y+4)}{(x-2)} \\\)

\( =(x-2)+\frac{y+4}{x-2}\)

Let \((\mathrm{x}-2)=\mathrm{t} \Rightarrow \mathrm{dx}=\mathrm{dt}\)

and \((y+4)=u \Rightarrow d y=d u\)

\(\therefore \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{du}}{\mathrm{dt}}\)

Now, \(\frac{d y}{d x}=(x-2)+\frac{(y+4)}{(x-2)}\)

\(\Rightarrow \frac{\mathrm{du}}{\mathrm{dt}}=\mathrm{t}+\frac{\mathrm{u}}{\mathrm{t}} \Rightarrow \frac{\mathrm{du}}{\mathrm{dt}}-\frac{\mathrm{u}}{\mathrm{t}}=\mathrm{t}\)

Here, integrating factor (IF) \(=1 / \mathrm{t}\)

\(\Rightarrow \mathrm{u} \cdot\left(\frac{1}{\mathrm{t}}\right)=\int \mathrm{t}\left(\frac{1}{\mathrm{t}}\right) \mathrm{d} \mathrm{t} \Rightarrow \mathrm{u} / \mathrm{t}=\mathrm{t}+\mathrm{c}\)

\(\Rightarrow \frac{(\mathrm{y}+4)}{(\mathrm{x}-2)}=(\mathrm{x}-2)+\mathrm{c}\)

\(\because\) It passes through origin [i.e. \((0,0)]\), then

\(\therefore \frac{4}{-2}=-2+\mathrm{c}\)

\(\Rightarrow-2+2=c \Rightarrow c=0\)

Hence, \(\frac{(y+4)}{(x-2)}=(x-2)+0\)

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

Clearly, this curve passes through \((5,5)\) as it satisfies the equation.

Let \(\triangle=\left|\begin{array}{ccc}\sin ^{2} x & \cos ^{2} x & 1 \\ \cos ^{2} x & \sin ^{2} x & 1 \\ -10 & 12 & 2\end{array}\right|\)

Applying \(\mathrm{C_{1} \rightarrow C_{1}+C_{2}}\), we get:

\(\triangle=\left|\begin{array}{ccc}\sin^2x +\cos^2x & \cos ^{2} x & 1 \\ \cos^2x+\sin^2x & \sin ^{2} x & 1 \\ -10+12 & 12 & 2\end{array}\right| \)

\(\triangle=\left|\begin{array}{ccc}1 & \cos ^{2} x & 1 \\ 1 & \sin ^{2} x & 1 \\ 2 & 12 & 2\end{array}\right| \quad\left(\because \sin ^{2} x+\cos ^{2} x=1\right)\)

As we know, if two rows or two columns of a determinant are identical the value of the determinant is zero.

Here \(\mathrm{C}_{1}\) and \(\mathrm{C}_{3}\) are identical.

So, \(\triangle=0\)

The graph of the profit function is called an iso profit line. It is called this because iso means same or equal and the profit anywhere on the line is the same.

So, an iso-profit lines represents an infinite number of solutions all of which yield the same profit.

Given,

\(\sin \theta+\sin 3 \theta+\sin 5 \theta+\operatorname{sin} 7 \theta\)

\(=(\sin 3 \theta+\sin \theta)+(\sin 7 \theta+\sin 5 \theta)\)

\(=2 \sin \left(\frac{3 \theta+\theta}{2}\right) \cos \left(\frac{3 \theta-\theta}{2}\right)+2 \sin \left(\frac{7 \theta+5 \theta}{2}\right) \cos \left(\frac{7 \theta-5 \theta}{2}\right)\)\(\quad (\because\sin x+\sin y=2 \sin \left(\frac{x+y}{2}\right) \cos \left(\frac{x-y}{2}\right))\)

\(=(2 \sin 2 \theta \cos \theta)+(2 \sin 6 \theta \cos \theta)\)

\(=2 \cos \theta(\sin 2 \theta+\sin 6 \theta)\)

\(=2 \cos \theta(\sin 6 \theta+\sin 2 \theta)\)

\(=2 \cos \theta \times 2 \sin \left(\frac{6 \theta+2 \theta}{2}\right) \cos \left(\frac{6 \theta-2 \theta}{2}\right)\)\(\quad (\because\sin x+\sin y=2 \sin \left(\frac{x+y}{2}\right) \cos \left(\frac{x-y}{2}\right))\)

\(=2 \cos \theta \times 2 \sin 4 \theta \cos 2 \theta\)

\(=4 \cos \theta \cos 2 \theta \sin 4 \theta\)

\(\therefore\) \(\sin \theta+\sin 3 \theta+\sin 5 \theta+\operatorname{sin} 7 \theta\) \(=4 \cos \theta \cos 2 \theta \sin 4 \theta\)

Given, a man speaks truth 2 out of 3 times. He picks one of the natural numbers in the set \(S=\{1,2,3,4,5,6,7\}\) and reports that it is even.

\({E}_{1}=\) the event of an even number is picked.

\(E_{2}=\) the event of an odd number is picked.

\(\mathrm{E}=\) the event a man reports that it is even

A man speaks truth \(2\) out of \(3\) times.

To find: The probability that it is actually even i.e. \(P\left(E_{1} \mid E\right)\)

\(\mathrm{P}\left(\mathrm{E}_{1} \mid \mathrm{E}\right)=\frac{\mathrm{P}\left(\mathrm{E} \mid \mathrm{E}_{1}\right) \cdot \mathrm{P}\left(\mathrm{E}_{1}\right)}{\mathrm{P}\left(\mathrm{E} \mid \mathrm{E}_{1}\right) \cdot \mathrm{P}\left(\mathrm{E}_{1}\right)+\mathrm{P}\left(\mathrm{E} \mid \mathrm{E}_{2}\right) \cdot \mathrm{P}\left(\mathrm{E}_{2}\right)}\)

\(\Rightarrow \mathrm{P}\left(\mathrm{E}_{1} \mid \mathrm{E}\right)=\frac{\left(\frac{2}{3}\right) \cdot\left(\frac{3}{7}\right)}{\left(\frac{2}{3}\right) \cdot\left(\frac{3}{7}\right)+\left(\frac{1}{3}\right) \cdot\left(\frac{4}{7}\right)}\)

\(\Rightarrow \mathrm{P}\left(\mathrm{E}_{1} \mid \mathrm{E}\right)=\frac{6}{10}\)

\(\Rightarrow \Rightarrow\left(\mathrm{E}_{1} \mid \mathrm{E}\right)=\frac{3}{5}\)

Thus, a man speaks truth 2 out of 3 times. He picks one of the natural numbers in the set \(S=\{1,2,3,4,5,6,7\}\) and reports that it is even. The probability that it is actually even is \(\frac{3}{5}\).

If \(\tan \left(\frac{\pi}{9}\right), \mathrm{x}, \tan \left(\frac{7 \pi}{18}\right)\) are in AP.

So, \(x=\frac{1}{2}\left[\tan \frac{\pi}{9}+\tan \left(\frac{7 \pi}{18}\right)\right]\)

( \(\because\) if \(a, b, c\) are in \(A P\), so, \(b=\frac{a+c}{2}\) )

And \(\tan \left(\frac{\pi}{9}\right), \mathrm{y}_1 \tan \left(\frac{5 \pi}{18}\right)\) are in AP.

Now, \(\mathrm{x}-2 \mathrm{y}=\frac{1}{2}\left[\tan \frac{\pi}{9}+\tan \frac{7 \pi}{18}\right]-\left(\tan \frac{\pi}{9}+\tan \frac{5 \pi}{18}\right)\)

\(\Rightarrow|x-2 y|=\left|\frac{\cot \frac{\pi}{9}-\tan \frac{\pi}{9}}{2}-\tan \frac{5 \pi}{18}\right| \quad\left\{\begin{array}{c}\because \tan \frac{5 \pi}{18}=\cot \frac{2 \pi}{9} \\ \text { and } \tan \frac{7 \pi}{18}=\cot \frac{\pi}{9}\end{array}\right\}\)

\(=\left|\cot \frac{2 \pi}{9}-\cot \frac{2 \pi}{9}\right|=0\)

\(\left[\because \cot 2 \mathrm{~A}=\frac{2 \cot ^2 \mathrm{~A}^{-1}}{2 \cot \mathrm{A}}\right]\)

\(\lim _{x \rightarrow \frac{\pi}{2}} \frac{\tan \left(\frac{\pi}{4}-\frac{x}{2}\right) \cdot(1-\sin x)}{(\pi-2 x)^3}\)

Let \(x=\frac{\pi}{2}+y ; y \rightarrow 0\)

\(=\lim _{y \rightarrow 0} \frac{\tan \left(-\frac{y}{2}\right) \cdot(1-\cos y)}{(-2 y)^3} \\\)

\( =\lim _{y \rightarrow 0} \frac{-\tan \frac{y}{2} 2 \sin ^2 \frac{y}{2}}{(-8) \cdot \frac{y^3}{8} \cdot 8}\left[\because 1-\cos \theta=2 \sin ^2 \frac{\theta}{2}\right] \\\)

\( =\lim _{y \rightarrow 0} \frac{1}{32} \frac{\tan \frac{y}{2}}{\left(\frac{y}{2}\right)} \cdot\left[\frac{\sin y / 2}{y / 2}\right]^2=\frac{1}{32}\left[\because \lim _{\theta \rightarrow 0} \frac{\sin \theta}{\theta}=\lim _{\theta \rightarrow 0} \frac{\tan \theta}{\theta}=1\right]\)