Mathematics Test-19

As we know,

In Binomial expansion, the independent term in

\(\left[a x^{p}+\left(\frac{b}{x}\right)^{a}\right]^{n}\) is,

\(T_{r+1}={ }^{n} C_{r} \cdot a^{(n-r)} \cdot b^{r} \quad \ldots(1)\)

Given:

\(\left(3 x+\frac{1}{x}\right)^{6}\)

Using equation (1)

\(\mathrm{T}_{\mathrm{r}+1}={ }^{6} \mathrm{C}_{\mathrm{r}} \cdot(3 {x})^{6-\mathrm{r}} \cdot\left(\frac{1}{{x}}\right)^{\mathrm{r}}\)

\(={ }^{6} \mathrm{C}_{\mathrm{r}}(3)^{6-\mathrm{r}} {x}^{6-\mathrm{r}} \cdot {x}^{-\mathrm{r}}\)

\(={ }^{6} \mathrm{C}_{\mathrm{r}}(3)^{6-\mathrm{r}} \cdot {x}^{6-2 \mathrm{r}} \quad \ldots(2)\)

Coefficient of \(x^{2}\)

\(6-2 \mathrm{r}=2\)

\(\Rightarrow 2 \mathrm{r}=4\)

\(\Rightarrow \mathrm{r}=2\)

The value of \(\mathrm{r}\) is putting in equation \((2)\)

\(={ }^{6} \mathrm{C}_{2}(3)^{6-2} \cdot {x}^{2}\)

\(=\frac{6 \times 5}{2 \times 1} \cdot 3^{4}\)

\(=15 \times 81\)

\(=1215\)

We know that,

\(\Rightarrow \sec ^{2} \theta=1+\tan ^{2} \theta\)

We have,

\(\Rightarrow\left(\sec ^{2} \theta\right)^{2}-\sec ^{2} \theta=3 \)

\(\Rightarrow\left(1+\tan ^{2} \theta\right)^{2}-\left(1+\tan ^{2} \theta\right)=3\)

\(\Rightarrow\left(1+\tan ^{4} \theta+2 \tan ^{2} \theta\right)-\left(1+\tan ^{2} \theta\right)=3\)

\(\Rightarrow 1+\tan ^{4} \theta+2 \tan ^{2} \theta-1-\tan ^{2} \theta=3 \)

\(\Rightarrow \tan ^{4} \theta+\tan ^{2} \theta=3\)

Number of horizontal lines in chess board \(=9\)

Number of vertical lines in chess board \(=9\)

2 horizontal lines and 2 vertical lines will for 1 rectangle box

Number of rectangle \(={ }^{9} C_{2} \times{ }^{9} C_{2}\)

\(\Rightarrow 36 \times 36=1296\)

Given:

\(\left(1+4 x+4 x^{2}\right)^{5} \)

As we know,

\((a+b)^{2}=(a^{2}+b^{2}+2ab)\)

\((1+4x+4x^{2})=\left[(1+2 x)^{2}\right]^{5}\)

\(=(1+2 x)^{2\times5}\)

\(=(1+2 x)^{10}\)

For Middle term: \(r=\frac{10}{2}=5\)

As we know,

\({}^{n}C_{r}(p)^{n}(q)^{n-r}\)

After putting the value,

\(p=1, q=2x, r=5\)

Middle Term \(={ }^{10} C_{5}(1)^{10}(2 x)^{5} \)

As we know,\({ }^{n} C_{r}=\)\(\frac{n !}{r !(n-r) !}\)

\(=\frac{10 !}{5 ! 5 !} \times 32 x^{5} \)

\(= \frac{10 \times 9 \times 8 \times 7 \times 6 \times 5!}{5!\times5!}\)

\(= \frac{10 \times 9 \times 8 \times 7 \times 6}{5\times 4 \times 3 \times 2!}\)

\(=36\times7\times32x^{5}\)

\(=252\times32x^{5}\)

\(=8064 x^{5}\)

\(\therefore\) The coefficient of the middle term in the expansion of \(\left(1+4 x+4 x^{2}\right)^{5}\) is \(8064\).

The standard equation of the circle is

\(( x - h )^2+( y - k )^2= r ^2\)

Where centre is \(( h , k\) ) and radius is \(r\).

Now,

The family of circle having centre \((0,0)\) and radius \(r\) is

\(x^2+y^2=r^2\)

\(\because\) There is only one constant \(r\).

Differentiating w.r.t. \(x\)

\(\Rightarrow 2 x+2 y \frac{d y}{d x}=0 \)

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

Given:

\(X+Y=\left[\begin{array}{ll}7 & 0 \\ 2 & 5\end{array}\right]\)...........(1)

\(X-Y=\left[\begin{array}{ll}3 & 0 \\ 0 & 3\end{array}\right]\)...........(2)

Adding equations (1) and (2), we get,

\(2 \mathrm{X}=\left[\begin{array}{ll}7 & 0 \\ 2 & 5\end{array}\right]+\left[\begin{array}{ll}3 & 0 \\ 0 & 3\end{array}\right]\)

\(\Rightarrow 2 \mathrm{X}=\left[\begin{array}{ll}10 & 0 \\ 2 & 8\end{array}\right]\)

\(\Rightarrow \mathrm{X}=\frac{1}{2}\left[\begin{array}{cc}10 & 0 \\ 2 & 8\end{array}\right]\)

\(\therefore X=\left[\begin{array}{ll}5 & 0 \\ 1 & 4\end{array}\right]\)

Given,

\(y=x^{2}, x=1, x=2\) and \(x\)-axis

Area required \(=\) Area \(A B C D\)

\(=\int_{1}^{2} y d x\)

\(y \rightarrow\) Equation of parabola

\(y=x^{2}\)

Thus,

Area required \(=\int_{1}^{2} y d x\)

\(=\int_{1}^{2} x^{2} d x\)

\(\left[\because\int x^{n} d x=\left(\frac{(x^{n+1})}{(n+1)}\right)\right]\)

\(=\left[\frac{x^{3}}{3}\right]_{1}^{2}\)

\(=\left(\frac{2^{3}}{3}-\frac{1}{3}\right)\)

\(=\left(\frac{8}{3}-\frac{1}{3}\right)\)

\(=\left(\frac{7}{3}\right)\)sq. units