Mathematics Test-4

Given,

1 + 4 + 9 +16 + 25 + 36 +......+ 121

The given series can be re-written as,

\(1^{2}+2^{2}+3^{2}+\ldots \ldots+11^{2}\)

As we know that,

\(1^{2}+2^{2}+3^{2}+\ldots . .+n^{2}=\frac{n(n+1)(2 n+1)}{6}\)

Here, n = 11

\(1^{2}+2^{2}+3^{2}+\ldots . .+11^{2}\)

\(=\frac{11 \times 12 \times 23}{6}\)

\(=506\)

Given,

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

Separate the variables,

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

On integrating both side,

We get,

\(\int \frac{d x}{x^{2}}=\int \frac{d y}{y^{2}} \quad\left(\because \int \frac{1}{x^{2}} d x=\frac{-1}{x}\right)\)

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

\(\Rightarrow \frac{1}{y}=\frac{1}{x}+C\)

As given,

\(\overrightarrow{\mathrm{b}}_1=\hat{\mathrm{i}} -\hat{\mathrm{j}}-2 \hat{\mathrm{k}}\) and \(\overrightarrow{\mathrm{b}}_2=3 \hat{\mathrm{i}}-5 \hat{\mathrm{j}}+-4 \hat{\mathrm{k}}\), we get

\(\left|\overrightarrow{\mathrm{b}}_1\right|=\sqrt{1^2+(-1)^2+(-2)^2}=\sqrt{6},\left|\overrightarrow{\mathrm{b}}_2\right|=\sqrt{3^2+(-5)^2+(-2)^2}=5 \sqrt{2}\) and

\(\overrightarrow{\mathrm{b}}_1 \overrightarrow{\mathrm{b}}_2=(\hat{\mathrm{i}} -\hat{\mathrm{j}}-2 \hat{\mathrm{k}})(3 \hat{\mathrm{i}}-5 \hat{\mathrm{j}}+-4 \hat{\mathrm{k}})=16\)

As we know,

\(\cos \theta=\left|\frac{\overrightarrow{\mathrm{b}}_1 \overrightarrow{\mathrm{b}}_2}{\left|\overrightarrow{\mathrm{b}_1}\right|\left|\overrightarrow{\mathrm{b}}_2\right|}\right|\)

\(\cos \theta=\frac{16}{10 \sqrt{3}}\)

\(\Rightarrow \cos \theta=\frac{8}{5 \sqrt{3}}\)

\(\Rightarrow \theta=\cos ^{-1} \frac{8}{5 \sqrt{3}}\)

Let \(\overrightarrow{O A}=\vec{a}, \overrightarrow{O B}=\vec{b}, \overrightarrow{O C}=\vec{c}\) and \(\overrightarrow{O D}=\vec{d}\).

Therefore,

\(\overrightarrow{ OA }+\overrightarrow{ OB }+\overrightarrow{ OC }+\overrightarrow{ OD }=\overrightarrow{ a }+\overrightarrow{ b }+\overrightarrow{ c }+\overrightarrow{ d }\)

The midpoint \(P\) of \(A B\), is \(\frac{\vec{a}+\vec{b}}{2}\).

The position vector of the midpoint \(Q\) of \(C D\), is \(\frac{\vec{c}+\vec{d}}{2}\).

Therefore, the position vector of the midpoint of \(P Q\) is \(\frac{\vec{a}+\vec{b}+\vec{c}+\vec{d}}{4}\)

Similarly, the position vector of the midpoint of RS is \(\frac{\vec{a}+\vec{b}+\vec{c}+\vec{d}}{4}\),

i.e., \(\overrightarrow{O E}=\frac{\vec{a}+\vec{b}+\vec{c}+\vec{d}}{4}\)

\(x=4\)

Given:

Relation \(\mathrm{R}\) is defined on the set \(\mathrm{Z}\).

\(R=\left\{(x, y): x, y \in Z, x^{2}+y^{2} \leq 4\right\}\)

\(x^{2}+3 y^{2} \leq 8\)

\(\Rightarrow \frac{x^{2}}{8}+\frac{y^{2}}{\frac{8}{3}} \leq 1\)

\( \frac{x^{2}}{a^{3}}+\frac{y^{2}}{y^3} \leq 1\) is the equation of ellipse.

So,

\(=\{(-2,0),(-1,0),(0,0),(1,0),(2,0),(0,-2),(0,-1),(0,1),(0,2),(1,1),(-1,-1),(1,-1),(-1,1)\}\)

Now we know,

Domain is the set which consist all first elements of ordered pairs in relation \(R\).

So,

Domain \((R)=\{-2,-1,0,1,2\}\)

Domain of \(R ^{-1} \equiv\) Range of \(R \equiv\) Value of \(y \equiv\{-1,0,1\}\)

Let, \(\alpha\) and \(\beta\) are the roots of \(x^{2}+6 x+8=0\), then

Sum of root \(= \alpha+\beta=-6\)

Arithmetic mean of root of equation \(=\frac{\alpha+\beta}{2}\) \(=\frac{-6}{2}\) \(=-3 \)

According to question, arithmetic mean is \(k\).

So, \(k=-3\)

Therefore, equation \(x^{2}+k x+2=0\) becomes

\(x^{2}-3 x+2=0 \) \(\Rightarrow(x-2)(x-1)=0 \)

\(\Rightarrow x=2\) and \(x=1\)

So, required roots of equation are \(1, 2\).

Given, \(e^{\theta \phi}=c+4 \theta \phi\), where \(c\) is an arbitrary constant and \(\phi\) is a function of \(\theta\)

Differentiate w.r.to \(\theta,\) we get

\(\Rightarrow \frac{\mathrm{d}}{\mathrm{d} \theta}\left(\mathrm{e}^{\theta \phi}\right)=\frac{\mathrm{d}}{\mathrm{d} \theta}(\mathrm{c}+4 \theta \phi)\)

\(\Rightarrow \mathrm{e}^{\theta \phi}\left(\theta \frac{\mathrm{d} \phi}{\mathrm{d} \theta}+\phi\right)=4\left(\theta \frac{\mathrm{d} \phi}{\mathrm{d} \theta}+\phi\right)\)

\(\Rightarrow\left(e^{\theta \phi}-4\right)\left(\theta \frac{\mathrm{d} \phi}{\mathrm{d} \theta}+\phi\right)=0\)

\(\Rightarrow\left(\theta \frac{\mathrm{d} \phi}{\mathrm{d} \theta}+\phi\right)=0\)

\(\Rightarrow \theta \mathrm{d} \phi+\phi \mathrm{d} \theta=0\)

\(\therefore \phi d \theta=-\theta \mathrm{d} \phi\)

\(f: \mathrm{R} \rightarrow \mathrm{R}\) is defincd as \(f(x)=x^{4}\).

Let \(x, y \in \mathrm{R}\) such that \(f(x)=f(y)\).

\(\Rightarrow x^{4}=y^{4}\)

\(\Rightarrow x=\pm y\)

\(\therefore f(x)=f(y)\) does not imply that \(x=y\).

For example \(f(1)=f(-1)=1\)

\(\therefore f\) is not one-one.

Consider an element 2 in co-domain \(\mathrm{R}\). It is clear that there does not exist any \(x\) in domain \(\mathrm{R}\) such that \(f(x)=2\).

\(\therefore f\) is not onto.

Thus, function \(f\) is neither one \(-\) one nor onto.

Given,

\(\frac{d y}{d x}+y=1\)

On seprating the variables

We get,

\(\frac{d y}{1-y}=d x\)

On integrating,

We get,

\(\int \frac{d y}{1-y}=\int d x\)

\(\Rightarrow-\log (1-y)=x+C\)

\(\Rightarrow \log (1-y)^{-1}=x+C \quad\left[m \log n=\log n^{m}\right]\)

\(\Rightarrow \log \left|\frac{1}{1-y}\right|=x+C\)

Since there are 6 rings and 4 fingers

Then 1stfinger can have any 6 rings, hence 6 ways

2ndfinger can have theremaining 5rings, hence 5ways

3rdfinger can have theremaining 4rings, hence 4ways

4thfinger can have theremaining 3rings, hence 3ways

So, the total number of ways = 6× 5× 4× 3

= 360