Mathematics Test - 58

Given, \(\int_{-1}^{1} 5 x^{4} \sqrt{x^{5}+1}~ dx\)

Put \(t=x^{5}+1\)\(\quad \dots (i)\)

Differentiating with respect to \(x\),

\(\frac{dt}{dx}=5 x^{4}\)

Then, \(d t=5 x^{4} d x\)

New limit to \(eq^{n}(i)\)

if \(x=1\), \(t=1^{5}+1\)

\(t=2\)

Then \(x=-1\), \(t=-1^{5}+1=0\)

Therefore,

\(\int_{-1}^{1} 5 x^{4} \sqrt{x^{5}+1}~ d x\) \(=\int_{0}^{2} \sqrt{t}~ d t\)

\(=[\frac{2}{3} t^{\frac{3}{2}}]_{0}^{2}\)

So,

Put the value of limit,

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

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

\(=\frac{4 \sqrt{2}}{3}\)

\( \lim _{x \rightarrow 0}\left[1+\frac{f(x)}{x^2}\right]=3 \\\)

\( \Rightarrow \lim _{x \rightarrow 0} \frac{f(x)}{x^2}=2\)

So, \(f(x)\) contain terms in \(x^2, x^3\) and \(x^4\).

Let \(f(x)=a_1 x^2+a_2 x^3+a_3 x^4\)

Since \(\lim _{x \rightarrow 0} \frac{f(x)}{x^2}=2 \Rightarrow a_1=2\)

Hence, \(f(x)=2 x^2+a_2 x^3+a_3 x^4\)

\(f^{\prime}(x)=4 x+3 a_2 x^2+4 a_3 x^3\)

As given : \(\mathrm{f}^{\prime}(1)=0\) adn \(\mathrm{f}^{\prime}(2)=0\)

Hence, \(4+3 a_2+4 a_3=0\)...(i)

and \(8+12 a_2+32 a_3=0\)...(ii)

By \(4 \mathrm{x}(\mathrm{i})-(\mathrm{ii})\), we get

\(16+12 a_2+16 a_3-\left(8+12 a_2+32 a_3\right)=0 \\\)

\(\Rightarrow 8-16 a_3=0 \Rightarrow a_3=1 / 2\)

and by eqn. (i), \(4+3 a_2+4 / 2=0 \Rightarrow a_2=-2\)

\(\Rightarrow \mathrm{f}(\mathrm{x})=2 \mathrm{x}^2-2 \mathrm{x}^3+\frac{1}{2} \mathrm{x}^4\)

\(f(2)=2 \times 4-2 \times 8+\frac{1}{2} \times 16=0\)

Let's check the given relation for its type one by one.

Reflexive: Every line is parallel to itself. It means that \(\mathrm{lRl}\) for all \(\mathrm{l \in L}\). Therefore, \(\mathrm{R}\) is reflexive.

Symmetric: If a line \(\mathrm{l}\) is parallel to \(\mathrm{m}\), then \(\mathrm{m}\) is parallel to l, i.e., if \(\mathrm{lRm}\)

\(\mathrm{mRl}, \forall \mathrm{~l}, \mathrm{m} \in \mathrm{L}\).

Therefore, \(\mathrm{R}\) is symmetric.

Transitive: If a line \(\mathrm{l}\) is parallel to \(\mathrm{m}\) and \(\mathrm{m}\) is parallel to \(\mathrm{k}\), then \(\mathrm{l}\) is also parallel to k. i. e. if \(I R m\) and \(\mathrm{mRk}\)

\( \Rightarrow I R k, \forall I, \mathrm{~m}, \mathrm{k} \in \mathrm{L}\).

Therefore, \(\mathrm{R}\) is transitive.

Since, the relation \(\mathrm{R}\) is reflexive, symmetric and transitive as well, it is an equivalence relation.

Given:

\(\underset{{{{x} \rightarrow 0}}}{\lim} \frac{(1-\cos 2 {x})^{3}}{{x}^{6}}\)

\(=\underset{{{x \rightarrow 0}}}{\lim} \frac{\left(2 \sin ^{2} x\right)^{3}}{x^{6}} \quad\left(\because 1-\cos 2 \theta=2 \sin ^{2} \theta\right)\)

\(=\underset{{{x \rightarrow 0}}}{\lim} \frac{8 \sin ^{6} x}{x^{6}}\)

\(=\underset{{{x \rightarrow 0}}}{\lim} 8 \times\left(\frac{\sin x}{x}\right)^{6}\)

\(=8 \times 1\quad\) \(\left(\because\underset{{{x \rightarrow 0} }}{\lim}\frac{\sin x}{x}=1\right)\)

\(=8\)

\(A=\left[\begin{array}{lll}0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0\end{array}\right]\)

\(\Rightarrow A^2=\left[\begin{array}{lll}0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0\end{array}\right] \times\left[\begin{array}{lll}0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0\end{array}\right]=\left[\begin{array}{lll}0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0\end{array}\right]\)

\(\Rightarrow A^3=\left[\begin{array}{lll}0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0\end{array}\right]\left[\begin{array}{lll}0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0\end{array}\right]=\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]=1\)

Now \(B_0=A^{49}+2 A^{98}=\left(A^3\right)^{16} \cdot A+2\left(A^3\right)^{32} \cdot A^2\)

\(B_0=A+2 A^2=\left[\begin{array}{lll}0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0\end{array}\right]+\left[\begin{array}{lll}0 & 0 & 2 \\ 2 & 0 & 0 \\ 0 & 2 & 0\end{array}\right]=\left[\begin{array}{lll}0 & 1 & 2 \\ 2 & 0 & 1 \\ 1 & 2 & 0\end{array}\right]\)

\(\left|B_0\right|=9\)

Since, \(B_n=\operatorname{Adj}\left|B_{n-1}\right| \Rightarrow\left|B_n\right|=\left|B_{n-1}\right|^2\)

Hence \(\left|B_4\right|=\left|B_3\right|^2=\left|B_2\right|^4=\left|B_1\right|^8=\left|B_0\right|^{16}=\left|3^2\right|^{16}=3^{32}\)

Given that:

4 boys and 4 girls can be arranged in a row so that no two girls and no two boys are together.

It means they can sit alternately.

Case-I: 1st person in the row is a boy.

\({B}_{1} {G}_{1} {~B}_{2} {G}_{2} {~B}_{3} {G}_{3} {~B}_{4} {G}_{4}\)

The no. of ways in which the boys can be rearranged among themselves is \(4 !\).

Similarly, the no. of ways in which the girls can be rearranged among themselves is \(4 !\)

So, the total number of ways in this case \(=4 ! \times 4 !=(4 !)^{2}\)

Case-II: 1st person in the row is a girl.

\({G}_{1} {~B}_{1} {G}_{2} {~B}_{2} {G}_{3} {~B}_{3} {G}_{4} {~B}_{4}\)

The no. of ways in which the girls can be rearranged among themselves is \(4 ! .\)

Similarly, the no. of ways in which the boys can be rearranged among themselves is \(4 !\)

So, the total number of ways in this case \(=4 ! \times 4 !=(4 !)^{2}\)

Therefore, Total Number of ways \(=(4 !)^{2}+(4 !)^{2}=2(4 !)^{2}\)

Given,

The area of a triangle is 5 square units.

Equation of line is \(5 x-y=0\)

The equation of a line perpendicular to a given line is

\( x+5 y=\lambda\)....(i)

\(\Rightarrow \frac{x}{\lambda}+\frac{5 y}{\lambda}=1\)

\(\Rightarrow \frac{x}{\lambda}+\frac{y}{\left(\frac{\lambda}{5}\right)}=1\)

Area of triangle \(=5\) square units

Area of triangle = \(\frac{1}{2}×b×h\)

\(\frac{1}{2} \times \lambda \times \frac{\lambda}{5}=5\)

\(\Rightarrow \lambda^{2}=50 \)

\(\Rightarrow\lambda=\pm 5 \sqrt{2}\)

Put the value of \(\lambda\) in equation (i) we get,

 \(x+5 y=\pm 5 \sqrt{2}\)

\(\Rightarrow x+5 y \pm 5 \sqrt{2}=0\)

Given:

\(f(x)=X^{5}-20 X^{3}+240 X\)

Differentiate w.r.t \(x\),

\(f^{\prime}(x)=5 X^{4}-60 X^{2}+240=X^{4}-12 X^{2}+48\)

For checking whether the function is Monotonically increasing or decreasing put \(X_{1}=-1\)

\(f^{\prime}\left(X_{1}\right)=(-1)^{4}-12(-1)^{2}+48=1-12+48=37\)

Now, put \(X_{2}=0\)

\(f^{\prime}\left(X_{2}\right)=0-0+48=48\)

Therefore, \(f^{\prime}\left(X_{1}\right) \leq f^{\prime}\left(X_{2}\right)\)

From the above condition we can assume that given function is Monotonically increasing everywhere.