Mathematics Test - 12

Maximizing \(z=x_1-x_2\)

Constraints \(x_1+x_2 \leq 10\)

\(x_1 \geq 0\)

\(x_2 \geq 0 \)

\( x_2 \leq 5\)

Corner points are: \((0,0)(5,0),(5,5),(0,10)\).

\(Z(0,0)=0-0=0 \)

\(Z(0,5)=0-5=-5\)

\( Z(5,5)=5-5=0\)

\( Z(10,0)=10-0=10\)

Maximum \({Z}={Z}(10,0)=10\)

Given:\(x^{2} d y+y^{2} d x=0\)

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

\(\Rightarrow \frac{\mathrm{dy}}{\mathrm{y}^{2}}=-\frac{\mathrm{dx}}{\mathrm{x}^{2}}\)

Integrating both sides, we get

\(\int \frac{\mathrm{dy}}{\mathrm{y}^{2}}=-\int \frac{\mathrm{dx}}{\mathrm{x}^{2}}\)

\(\frac{-1}{\mathrm{y}}=-\frac{-1}{\mathrm{x}}+\mathrm{c}\)

\(\frac{-1}{\mathrm{y}}=\frac{1}{\mathrm{x}}+\mathrm{c}\)

\(\frac{1}{x}+\frac{1}{y}=-c\)

\(\mathrm{x}+\mathrm{y}=-\mathrm{cxy}\)

Here \(c\) is differential constant, take \(-c=\frac{1}{c}\) (Because \(\frac{1}{c}\) is also a constant)

\(\Rightarrow c(x+y)=x y\)

Given:

\(f(x)=x^{3}+3 x^{2}+3 x-7\)

We know that:

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

Derivative of a constant, i.e.,

\(\frac{\mathrm{d}(\text { constant })}{\mathrm{dx}}=0\)

Therefore, 

\(\frac{d f(x)}{d x}=3 x^{2}+6 x+3\)

Putting \(x=2\) in above, we get:

\(\frac{\mathrm{d} \mathrm{f}(\mathrm{x})}{\mathrm{dx}}=3(2)^{2}+6(2)+3\)

\(=3(4)+12+3\)

\(=12+12+3\)

\(\frac{\mathrm{d} \mathrm{f}(\mathrm{x})}{\mathrm{dx}}=27\)

The value of \(\frac{\mathrm{d} \mathrm{f}(x)}{\mathrm{dx}}\) at \(\mathrm{x}=2\) is 27.

Given,

Second degree equation, \(x^{2}+x y+y^{2}+x+y=1\)

Compare with second-degree equation \(a x^{2}+2 h x y+b y^{2}+2 g x+\) \(2 f y+c=0\)

So, \(a=1, b=1, h=0.5, g=0.5, f=0.5\) and \(c=-1\)

Now, \(\Delta=a b c+2 f g h-a f^{2}-b g^{2}-c h^{2}\)

\(\Delta=1 \times 1 \times(-1)+2 \times 0.5 \times 0.5 \times 0.5-1 \times(0.5)^{2}-1 \times(0.5)^{2}-(-1) \times\) \((0.5)^{2}\)

\(=-1-0.25-0.25-0.25+0.25\)

So, \(\Delta=-1.5\)

Now, \(h^{2}-a b=(0.5)^{2}-1 \times 1=-0.75\)

So, \(h^{2}-a b<0\) and \(a=b\)

It represents a circle.

Let,

\(z=\frac{1+i}{1-i}\)

Multiplying by \((1+i)\) in numerator and denominator, we get

\(z=\frac{1+i}{1-i} \times \frac{1+i}{1+i}\)

\(\Rightarrow z=\frac{(1+i)^{2}}{1^{2}-i^{2}} \)

\(\Rightarrow z=\frac{1+i^{2}+2 i}{1-i^{2}}\)

\(\Rightarrow z=\frac{1+(-1)+2 i}{1-(-1)} \quad\left[\because i^{2}=-1\right] \)

\(\Rightarrow z=\frac{0-2 i}{1+1}\)

\(\Rightarrow z=\frac{-2 i}{2} \)

\(\Rightarrow z=-i\)

So, Conjugate of \(z=\bar{z}=i\)

Given,

Quadratic equation \(7 x^{2}-28 x+21\)

\( 7 x^{2}-28 x+21=0 \)

\(\Rightarrow x^{2}-4 x+3=0 \)

\(\Rightarrow x^{2}-x-3 x+3=0 \)

\(\Rightarrow(x-1)(x-3)=0 \)

\(\Rightarrow x=1,3\)

Now,

\( \frac{1}{\alpha}+\frac{1}{\beta}=\frac{k}{\alpha \beta} \)

\(\Rightarrow \frac{1}{1}+\frac{1}{3}=\frac{k}{1 \times 3} \)

\(\Rightarrow \frac{4}{3}=\frac{k}{3}\)

\(\Rightarrow k=4\)

So, the value of \(k\) is \(4\).

We know that,

Probability \(=\frac{\text { Number of favorable outcomes }}{\text { Total outcomes }}\)

Number of all combinations of \(\boldsymbol{n}\) things, taken \(r\) at a time, is given by \({ }^{n} C_{\tau}=\frac{n !}{(r) !(n-r) !}\)

According to question,

Total outcomes \(={ }^{52} C_{2}\)

\(=\frac{52 !}{(2) !(50) !} \)

\(=\frac{52 \times 51}{2}=1326\)

Number of favorable outcomes \(={ }^{4} C_{2}=\frac{4 !}{(2) !(2) !}\) \(=\frac{4 \times 3 \times 2 \times 1}{2 \times 1 \times 2 \times 1}=6\)

\(\therefore\) Probability \(=\frac{6}{1326}=\frac{1}{221}\)

Given,

Mean of a set of observations \(x_{1}, x_{2}, x_{3}, \ldots, x_{10}\) is 50.

Here \(n=10\),

\(\text { Mean }=\frac{\sum \mathrm{x}_{\mathrm{i}}}{\mathrm{n}}=\frac{\mathrm{x}_{1}+\mathrm{x}_{2}+\mathrm{x}_{3}+\ldots+\mathrm{x}_{10}}{10}=50\)

\(\Rightarrow \mathrm{x}_{1}+\mathrm{x}_{2}+\mathrm{x}_{3}+\ldots+\mathrm{x}_{10}=500\)...(1)

Now find out the mean of new observations \(x_{1}+5, x_{2}+10, x_{3}+15, \ldots, x_{10}+50\)

Mean \(=\frac{\sum \mathrm{x}_{\mathrm{i}}}{\mathrm{n}}=\frac{\left(\mathrm{x}_{1}+5\right)+\left(\mathrm{x}_{2}+10\right)+\left(\mathrm{x}_{3}+15\right)+\ldots+\left(\mathrm{x}_{10}+50\right)}{10}\)

\(=\frac{\left(\mathrm{x}_{1}+\mathrm{x}_{2}+\mathrm{x}_{3}+\ldots+\mathrm{x}_{10}\right)+(5+10+15+\ldots+50)}{10}\)

From equation (1),

\(=\frac{500+5(1+2+3+\ldots+10)}{10}\)

\(=50+\frac{1}{2} \times \frac{10 \times 11}{2}\)

\(=50+27.5\)

\(=77.5\)

Here, \(y=\cos ^{2} x^{2}\)

Let, \(x^{2}=t\)

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

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

\(\Rightarrow \frac{d t}{d x}=2 x\)..(1)

\(y=\cos ^{2} t\)

\(=\frac{\cos 2 t+1}{2}=\frac{\cos 2 t}{2}+\frac{1}{2}\)

\(\frac{d y}{d x}=\frac{1}{2} \frac{d}{d t}(\cos 2 t) \frac{d t}{d x}+0\)

\(=\frac{1}{2}(-2 \sin 2 t) \frac{d t}{d x}\) from eq (1)

\(=-\sin 2 x^{2} \times 2 x\)

\(=-4 x \cos x^{2} \sin x^{2}\)