Mathematics Test-22

Given,

\(a_{n}=(2 n-1)^{2}\)

As we know that,

\(S_{n}=\sum a_{n}\)

\(\Rightarrow S_{n}=\sum_{k=1}^{n} a_{k}=\sum_{k=1}^{n}\left[4 k^{2}-4 k+1\right]\)

\(\Rightarrow S_{n}=-4 \sum_{k=1}^{n} k+4 \sum_{k=1}^{n} k^{2}+\sum_{k=1}^{n} 1\)

As we know,

\(\sum \mathrm{n}^{2}\) = \(\frac{n(n+1)(2 n+1)}{8}\)

\(\sum \mathrm{n}\) = \(\frac{n (n+1)}{2}\)

\(\sum 1\) =n

\(\Rightarrow S_{n}=\frac{n \cdot(2 n+1) \cdot(2 n-1)}{3}\)

Given determinant is \(\left|\begin{array}{ccc}\mathrm{i} & \mathrm{i}^{2} & \mathrm{i}^{3} \\ \mathrm{i}^{4} & \mathrm{i}^{6} & \mathrm{i}^{8} \\ \mathrm{i}^{9} & \mathrm{i}^{12} & \mathrm{i}^{15}\end{array}\right|\)

Since, we have,

\(\mathrm{i}=\sqrt{-1}\)

\(\mathrm{i}^{2}=-1, \mathrm{i}^{3}=-\mathrm{i}, \mathrm{i}^{4}=1, \mathrm{i}^{6}=-1, \mathrm{i}^{8}=1, \mathrm{i}^{9}=\mathrm{i}, \mathrm{i}^{12}=1\), and \(\mathrm{i}^{15}=-\mathrm{i}\)

\(=\left|\begin{array}{ccc}\mathrm{i} & -1 & -\mathrm{i} \\ 1 & -1 & 1 \\ \mathrm{i} & 1 & -\mathrm{i}\end{array}\right|\)

\(=\mathrm{i}(\mathrm{i}-1)+1(-\mathrm{i}-\mathrm{i})-\mathrm{i}(1+\mathrm{i})\)

\(=i^{2}-i-2 i-i-i^{2}\)

\(=-4 i\)

If \(\alpha ,\beta \) are the roots of quadratic equation \(a x^{2}+b x+c=0\), then

Sum of roots \(=\alpha+\beta=\frac{-b}{a}\)

Product of roots \(=\alpha \times \beta =\frac{c}{a}\)

Given,

\(3 a x^{2}+2 b x+c=0\)

Comparing with standard quadratic equation \(a x^{2}+b x+c=0\), we get

\(a=3 a, b=2 b\)

Roots are in the ratio \(3: 2\).

Let, roots are \(3 \alpha\) and \(2 \alpha\).

Sum of roots \(=3 \alpha+2 \alpha=\frac{-2 b }{3 a}\)

\(\Rightarrow 5 \alpha=\frac{-2 b }{ 3 a}\)

\(\Rightarrow \alpha=\frac{-2 b }{ 15 a}\)

\(\Rightarrow \alpha^{2}=\frac{4 b^{2}}{15 \times 15 a^{2}}\)

Now, product of roots \(=3 \alpha \times 2 \alpha=\frac{c }{ 3 a}\)

\(\therefore 6 \alpha^{2}=\frac{c }{ 3 a} \)

\(\Rightarrow 6\left(\frac{4 b^{2}}{15 \times 15 a^{2}}\right)=\frac{c}{3 a} \)

\(\Rightarrow 6 \times \frac{4 b^{2}\times 3a}{15 \times 5 a }=c \)

\(\Rightarrow 2 \times 4 b^{2}=25 ac\)

\(\Rightarrow 8 b ^{2}=25 ac\)

Since two coins are tossed, therefore total number of cases \(= {(H,H) (H,T) (T,T) (T,H)}\)

\( =4\)

Probability \(= \frac{Number\ of\ favourable\ outcomes}{Total\ number\ of\ outcomes}\)

probability of getting heads in both coins is:

\(\therefore P(\text { head })=\frac{1}{4}\)

Given curve is \(x^2=4 y\) \(\Rightarrow x=2 \sqrt{y}\)

Area of ABCD \(=\int_2^4 x d y\)

\(=\int_2^4 2 \sqrt{y}~ d y \)

\(=\frac{32-8 \sqrt{2}}{3} \) sq. unit 

Given:

\(\mathrm{I}=\int_{-a}^{a}\left(x^{2}+\sin x\right) d x\)

\(=\int_{-a}^{a} x^{2} d x+\int_{-a}^{a} \sin x d x\)

\(=\mathrm{I}_{1}+\mathrm{I}_{2}\)

Now,

\(\mathrm{I}_{1}=\int_{-{a}}^{{a}} {x}^{2} {dx}\)

Here \(f(x)=x^{2}\)

Replace \({x}\) by \(-{x}\), we get

\(\Rightarrow f(-x)=(-x)^{2}=x^{2}\)

\(\Rightarrow {f}(-{x})={f}({x})\)

So, \({f}({x})\) is even function.

As we know, If \({f}({x})\) even function then,

\(\int_{-{a}}^{{a}} {f}({x}) {dx}=2 \int_{0}^{{a}} {f}({x}) {dx}\)

Therefore, \(\mathrm{I}_{1}=2 \int_{0}^{{a}} {x}^{2} {dx}\)

\(=2 \times\left[\frac{{x}^{3}}{3}\right]_{0}^{{a}}\)

\(=2 \times\left[\frac{a^{3}}{3}-0\right]=\frac{2 a^{3}}{3}\)

Now,

\(\mathrm{I}_{2}=\int_{-{a}}^{{a}} \sin {x} {dx}\)

Here \(f(x)=\sin x\)

Replace \(x\) by \(-x\), we get

\(\Rightarrow {f}(-{x})=\sin (-{x})=-\sin {x} \quad(\because \sin (-\theta)=-\sin \theta)\)

\(\Rightarrow {f}(-{x})=-{f}({x})\)

So, \(f(x)\) is odd function.

As we know, If \(f(x)\) even function then \(\int_{-a}^{a} f(x) d x=0\)

\(\mathrm{I}=\mathrm{I}_{1}+\mathrm{I}_{2}\)

\(=\frac{2 {a}^{3}}{3}+0\)

\(=\frac{2 {a}^{3}}{3}\)

Given that:

\(d y=\left(1+y^{2}\right) d x\)

\(\Rightarrow \frac{\mathrm{dy}}{1+\mathrm{y}^{2}}=\mathrm{dx}\)

Integrating both sides, we get

\(\Rightarrow \int \frac{\mathrm{dy}}{1+\mathrm{y}^{2}}=\int \mathrm{dx}\)

\(\Rightarrow \tan ^{-1} \mathrm{y}=\mathrm{x}+\mathrm{c}\)

\(\therefore y=\tan (x+c)\)

Concept:

Arrangements of n different objects around a circle, then the number of arrangements is (n -1)!

Calculation:

Given: All boys are to sit together

So, All boys can be considered as a single group.

\(\therefore\) Total no of students \(=4\) Girls \(+ 1\)Group \(=5\)

The number of ways of arranging 5 students in a round table is \((5-1) !=4!\)

Now, no of the ways of arranging 5 boys is \(5 !\)

Therefore, The total number of ways \(=4 ! 5 !\)

The given equation is, \(x^{2}-5 x+4=0\)

We need to find out the roots of the given equation. So, for that, we can write the equation as,

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

\(\Rightarrow x(x-4)-(x-4)=0\)

\(\Rightarrow(x-4)(x-1)=0\)

\(\therefore \mathrm{x}=1,4\)

\(\cot \alpha\) and \(\cot \beta\) are the roots of the equation \(x^{2}-5 x+4=0\)

\(\cot \alpha=1\) and \(\cot \beta=4\)

\(\cot (\alpha+\beta)=\frac{(\cot \alpha \cot \beta-1)}{(\cot \alpha+\cot \beta)}\)

Putting the value of the roots we get,

\(\cot (\alpha+\beta)=\frac{(1 \times 4-1)}{(1+4)}\)

\(\therefore \cot (\alpha+\beta)=\frac{3}{5}\)

Given, \(\angle A O B=\angle P O Q=30^{\circ}\)

\(\angle X O Y=\angle Z O W=40^{\circ}\)

\(XY=4\) units and \(AB= 3.5\) units

As we know that, opposite side of equal angles of two different traingles are always same. Therefore, \(AB=PQ= 3.5\) and \(XY=ZW=4\)

\(\therefore AB+ ZW+ PQ+ XY= 3.5+4+3.5+4 \)

\(\Rightarrow 15\) units