Mathematics Test - 13

Let \(x_{1}, x_{2}, \ldots \ldots \ldots ., x_{n}\) be the n terms.
\(\therefore \bar{x}=\frac{x_{1}+x_{2}+\ldots+x_{n}}{n}\)
lew mean \(=\frac{\left(x_{1}+1\right)+\left(x_{2}+2\right)+\ldots+\left(x_{n}+n\right)}{n}\)
\(=\frac{\left(\mathrm{x}_{1}+\mathrm{x}_{2}+\ldots+\mathrm{x}_{\mathrm{n}}\right)+(1+2+\ldots+\mathrm{n})}{\mathrm{n}}\)
\(=\frac{\left(\mathrm{x}_{1}+\mathrm{x}_{2}+\ldots+\mathrm{x}_{\mathrm{n}}\right)}{\mathrm{n}}+\frac{\mathrm{n}(\mathrm{n}+1) / 2}{\mathrm{n}}=\overline{\mathrm{x}}+\frac{\mathrm{n}+1}{2}\)

Let \(\mathrm{E}_{\mathrm{i}}(\mathrm{i}=1,2,3,4)\) denote the event that the \(\mathrm{i}^{\text {th }}\) person answers the question
correctly.
\(\mathrm{P}(\) at least one of them answers correctly \()\) \(=\mathrm{P}\left(\mathrm{E}_{1} \cup \mathrm{E}_{2} \cup \mathrm{E}_{3} \cup \mathrm{E}_{4}\right)=1-\mathrm{P}\left(\mathrm{E}_{1} \cap \mathrm{E}_{2} \cap \mathrm{E}_{3} \cap \mathrm{E}_{4}\right)\)
\(=1-\mathrm{P}\left(\mathrm{E}_{1}\right) \mathrm{P}\left(\mathrm{E}_{2}^{\prime} \mathrm{P}\left(\mathrm{E}_{3}^{\prime}\right) \mathrm{P}\left(\mathrm{E}_{4}^{\prime}\right)=1-\left(1-\frac{1}{6}\right)^{4}=1-\left(\frac{5}{6}\right)^{4}\right.\)

\(\int \frac{2 x-3}{\left(x^{2}+x-1\right)^{2}} d x\)
\(=\int\left(\frac{2 x+1}{\left(x^{2}+x+1\right)^{2}}-\frac{4}{\left(x^{2}+x+1\right)^{2}}\right) d x\)
\(=\int \frac{2 x+1}{\left(x^{2}+x+1\right)^{2}} d x-\frac{4}{\left(x^{2}+x+1\right)^{2}} d x\)
\(=I_{1}-I_{2}\)
\(u=x^{2}+x+1 \rightarrow d x=\frac{1}{2 x+1} d x\)
\(I_{1}=\int \frac{1}{u^{2}} d u=-\frac{1}{u}=-\frac{1}{x^{2}+x+1}\)
\(I_{2}=\int \frac{1}{x^{2}+x+1} d x=\int \frac{1}{\left(\left(x+\frac{1}{2}\right)^{2}+\frac{3}{4}\right)^{2}} d x=16 \int \frac{1}{\left((2 x+1)^{2}+3\right)^{2}} d x\)

Now, \(y=\sin x, \frac{d y}{d x}=\cos x\left(\frac{d y}{d x}\right)_{(h, k)}=\cosh\)
The equation of the tangent at \((h, k)\) is \(y-k=(\cos h)(x-h)\)
This passes through (0,0) . \(\therefore-k=-h \cosh \frac{k}{h}=\cosh \ldots(i 1)\)
From (i) and (ii):
\(k^{2}+\frac{k^{2}}{h^{2}}=1\)
\(h^{2}-k^{2}=k^{2} h^{2}\)
Hence, the locus of \((\mathrm{h}, \mathrm{k})\) is \(\mathrm{x}^{2}-\mathrm{y}^{2}=\mathrm{x}^{2} \mathrm{y}^{2}\)

\(3 f(x)+b f(1 / x)=x\)
Substituting \(1 / x\) in place of \(x\) \(3 f(1 / x)+b f(x)=1 / x\)
Above is a system of equation in \(f(x)\) and \(f(1 / x)\). Operating \((i) \times 3-(\) ii \() \times b\) :
\(\left(9-b^{2}\right) f(x)=3 x-\frac{b}{x} f(x)=\frac{3 x^{2}-b}{x\left(9-b^{2}\right)}\)
But \(f(2)=0\) \(0=\frac{(12-b)}{2\left(9-b^{2}\right)}\)
\(b=12\)

\(\frac{1}{(1-x)(3-x)}=\frac{1}{2}\left[\frac{1}{1-x}-\frac{1}{3-x}\right]\)
\(=\frac{1}{2}\left[(1-x)^{-1}-\frac{1}{3}\left(1-\frac{x}{3}\right)^{-1}\right]\)
\(=\frac{1}{2}\left[1+x+x^{2}+x^{3}+\ldots+x^{n}+\ldots\right.\)
\(-\frac{1}{3}\left\{1+\frac{x}{3}+\left(\frac{x}{3}\right)^{2}+\left(\frac{x}{3}\right)^{3}+\ldots\right\}\)
Sum of coefficient of \(x^{n}\) is \(\frac{1}{2}\left[1-\frac{1}{3} \cdot \frac{1}{3^{n}}\right]=\frac{3^{n+1}-1}{2\left(3^{n+1}\right)}\)

\(1+\frac{x}{a_{1}}+\frac{x\left(x+a_{1}\right)}{a_{1} a_{2}}+\frac{\left.x\left(x+a_{1}\right) x+a_{2}\right)}{a_{1} a_{2} a_{3}}+\ldots=\frac{x+a_{1}}{a_{1}}+\frac{x\left(x+a_{1}\right)}{a_{1} a_{2}}+\frac{x\left(x+a_{1}\right)\left(x+a_{2}\right)}{a_{1} a_{2} a_{3}}+\ldots\)
\(=\left(\frac{x+a_{1}}{a_{1}}\right)\left[1+\frac{x}{a_{2}}\right]+\frac{x\left(x+a_{1}\right)\left(x+a_{2}\right)}{a_{1} a_{2} a_{3}}+\ldots\)
\(=\frac{\left.\left(x+a_{1}\right) x+a_{2}\right)}{a_{1} a_{2}}+\frac{x\left(x+a_{1}\right) x+a_{2}}{a_{1} a_{2} a_{3}}+\ldots\)
\(=\frac{\left.\left(x+a_{1}\right) x+a_{2}\right)}{a_{1} a_{2}}\left[1+\frac{x}{a_{3}}\right]+\ldots\)
\(=\frac{\left(\mathrm{x}+\mathrm{a}_{1}\right)\left(\mathrm{x}+\mathrm{a}_{2}\right)\left(\mathrm{x}+\mathrm{a}_{3}\right)}{\mathrm{a}_{1} \mathrm{a}_{2} \mathrm{a}_{3}}+\ldots\)
\(=\frac{\left(\mathrm{x}+\mathrm{a}_{1}\right)\left(\mathrm{x}+\mathrm{a}_{2}\right) \cdot\left(\mathrm{x}+\mathrm{a}_{3}\right) \ldots\left(\mathrm{x}+\mathrm{a}_{\mathrm{n}}\right)}{\mathrm{a}_{1} \mathrm{a}_{2} \mathrm{a}_{3} \ldots \ldots \mathrm{a}_{\mathrm{n}}}\)
\(=\frac{\left.\left(\mathrm{x}+\mathrm{a}_{1}\right) \mathrm{x}+\mathrm{a}_{2}\right) \cdot\left(\mathrm{x}+\mathrm{a}_{3}\right) \ldots\left(\mathrm{x}+\mathrm{a}_{\mathrm{n}}\right)}{\mathrm{a}_{1} \mathrm{a}_{2} \mathrm{a}_{3} \ldots \ldots \mathrm{a}_{\mathrm{n}}}\)

Given, \(\sin y=x \sin (a+y)\)
\(\Rightarrow x=\frac{\sin y}{\sin (a+y)}\)
On differentiating with respect to \(y\), we get \(\frac{d x}{d y}=\frac{\sin (a+y) \cos y-\sin y \cos (a+y)}{\sin ^{2}(a+y)}\)
\(=\frac{\sin (a+y-y)}{\sin ^{2}(a+y)}\)
\(\Rightarrow \frac{d y}{d x}=\frac{\sin ^{2}(a+y)}{\sin a}\)

Let \(I=\int_{0}^{\pi} \frac{1}{1+\sin x} d x\)
\(=\int_{0}^{\pi} \frac{1}{1+\frac{1}{1+\tan ^{2} \frac{x}{2}}} d x\)
Put \(1+\tan \frac{x}{2}=t \Rightarrow \frac{1}{2} \sec ^{2} \frac{x}{2}=d t\)
\(=\int_{0}^{\pi} \frac{\sec ^{2} \frac{x}{2}}{\left(1+\tan \frac{x}{2}\right)^{2}} d x\)
\(\left.\therefore I=\int_{1}^{\infty} \frac{2 d t}{1+t^{2}}=-\frac{2}{t}\right]_{1}^{\infty}=2\)