Mathematics Test - 25

\(|A-x I|=0\)

Roots are -1 and 3

Sum of roots \(=\operatorname{tr}(\mathrm{A})=2\)

Product of roots \(=|\mathrm{A}|=-3\)

Let \(A=\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]\)

We have \(a+d=2\)

\(a d-b c=-3\)

\(A^2=\left[\begin{array}{ll} a & b \\ c & d \end{array}\right] \times\left[\begin{array}{ll} a & b \\ c & d \end{array}\right]=\left[\begin{array}{ll} a^2+b c & a b+b d \\ a c+c d & b c+d^2 \end{array}\right]\)

We need \(a^2+b c+b c+d^2\)

\(=a^2+2 b c+d^2 \)

\( =(a+d)^2-2 a d+2 b c \)

\(=4-2(a d-b c) \)

\(=4-2(-3) \)

\(=4+6 \)

\(=10\)

Given: Mean \(=n p=1\)

Variance \(=n p q=\frac{3}{4}\)

\(\Rightarrow p=\frac{1}{4}, q=\frac{3}{4}, n=4\)

Binomial distribution \(P(X=r)={ }^n C_r p^r q^{n-r}\)

\(P(X=3)={ }^4 C_3\left(\frac{1}{4}\right)^3\left(\frac{3}{4}\right)^{4-3}\Rightarrow\frac{4 \times 3 \times 2 \times 1}{3 \times 2 \times 1} \times \frac{1}{64} \times \frac{3}{4}\Rightarrow4 \times \frac{1}{64} \times \frac{3}{4} \)

\(=\frac{3}{64}\)

Given that, \(2 l-m+2 m=0\) ...(i)

and \(l m+m n+n l=0\)......(ii)

From equation (i), \(m=2(l+n)\) put in equation (ii).

\(2 l(l+n)+2 n(l+n)+n l=0\)

\(\Rightarrow 2 l^{2}(l+n)+2 n(l+n)+n l=0\)

\(\Rightarrow 2 l^{2}+5 n l+2 n^{2}=0\)

\(\Rightarrow 2 l^{2}+4 n l+n l+2 n^{2}=0\)

\(\Rightarrow 2 l(l+2 n)+n(l+2 n)=0\)

\(\Rightarrow (l+2 n)(n+2 l)=0\)

\(\Rightarrow l=-2 n\) and \(n=-2 l\)

If \(l=-2 n\), then \(m=2(-2 n+n)=-2 n\)

and if \(n=-2 l\), then \(m=2(l-2 l)=-2 l\)

The Direction Ratio's are \(1,-2,-2\) and \(-2,-2,1\)

Now, \(1(-2)-(2)(-2)-2(1)\)

\(=-2+4-2=0\)

Thus, lines are perpendicular, so angle between then is \(\frac{\pi}{2}\).

CONCEPT:

If \(\mathrm{z}=\mathrm{x}+\) i\(\mathrm{y}\) then conjugate of \(\mathrm{z}\) is given by \(\mathrm{\bar{z}=x}\)-i \(\mathrm{y}\)

If \(\mathrm{z}=\mathrm{x}+\) i\(\mathrm{y}\) then \(|\mathrm{z}|=\sqrt{x^{2}+y^{2}}\)

\(\mathrm{z=x}+\) i\(\mathrm{y}\) then multiplicative inverse of \(\mathrm{z}\) is given by \(\mathrm{z^{-1}=\frac{\bar{z}}{|z|^{2}}}\)

CALCULATION:

Let \(\mathrm{z}=4-3\)i

In order to find out the multiplicative of \(z\) first we need to find out \(\bar{\mathrm{z}}\) and \(\mathrm{|z|}\)

As we know that,

If \(\mathrm{z=x}+\) i\(\mathrm{y}\) then \(\mathrm{\bar{z}=x}\)-i \(\mathrm{y}\) and \(\mathrm{|z|=\sqrt{x^{2}+y^{2}}}\)

\(\Rightarrow \mathrm{\bar{z}}=4+3\)i and \(\mathrm{|z|}^{2}=25\)

As we know that, \(\mathrm{z}^{-1}=\mathrm{\frac{\bar{z}}{|z|^{2}}}\)

\(\Rightarrow \mathrm{z}^{-1}=\frac{4+3 i}{25}=\frac{4}{25}\) + i\(\frac{3}{25}\)

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.

If \(a_{1} x+b_{1} y+c_{1} z+d_{1}=0\) and \(a_{2} x+b_{2} y+c_{2} z+d_{2}=0\) are a plane equations, then angle between planes can be found using the following formula:

\(\cos \theta=\frac{ a _{1} a _{2}+ b _{1} b _{2}+ c _{1} c _{2}}{\sqrt{ a _{1}^{2}+ b _{1}^{2}+ c _{1}^{2}} \sqrt{ a _{2}^{2}+ b _{2}^{2}+ c _{2}^{2}}}\)

Here,

\(a_{1}=\sqrt{3}-1\)

\(a_{2}=-\sqrt{3}-1\)

\(b_{1}=-\sqrt{3}-1\)

\(b_{2}=\sqrt{3}-1\)

\(c_{1}=4\)

\(c_{2}=4\)

Therefore, required angle will be given by:

\(\cos \theta=\frac{ a _{1} a _{2}+ b _{1} b _{2}+ c _{1} c _{2}}{\sqrt{ a _{1}^{2}+ b _{1}^{2}+ c _{1}^{2}} \sqrt{a_{2}^{2}+ b _{2}^{2}+ c _{2}^{2}}}\)

\(\cos \theta=\frac{-2-2+16}{\sqrt{24} \sqrt{24}}\)

\(=\frac{12}{24}\)

\(=\frac{1}{2}\)

\(\cos \theta = \cos \frac{\pi}{3}\)

\(\Rightarrow \theta=\frac{\pi}{3}\)

\(x^2+y^2-4 x-6 y-12=0 \ldots(1)\)

Centre, \(\mathrm{C}_1=(2,3)\)

Radius, \(r_1=5\) units

\(x^2+y^2+6 x+18 y+26=0 \ldots(02)\)

Centre, \(\mathrm{C}_2=(-3,-9)\)

Radius, \(\mathrm{r}_2=8\) units

\(\mathrm{C}_1 \mathrm{C}_2=\sqrt{(2+3)^2+(3+9)^2}=13\) units

\(\mathrm{r}_1+\mathrm{r}_2=5+8=13\)

\(\therefore \mathrm{C}_1 \mathrm{C}_2=\mathrm{r}_1+\mathrm{r}_2\)

Therefore there are three common tangents.

\(f(x)=2 x-\tan ^{-1}x -\log \left\{x+\sqrt{x^{2}+1}\right\}\)

\(\Rightarrow f^{\prime}(x)=2-\frac{1}{1+x^{2}}-\frac{1}{x+\sqrt{x^{2}+1}}\left(1+\frac{2 x}{2 \sqrt{x^{2}+1}}\right)\)

\(\Rightarrow f^{\prime}(x)=2-\frac{1}{1+x^{2}}-\frac{1}{\sqrt{x^{2}+1}}\)

\(\Rightarrow f^{\prime}(x)=\frac{1+2 x^{2}}{1+x^{2}}-\frac{1}{\sqrt{x^{2}+1}}\)

\(\Rightarrow f^{\prime}(x)=\frac{1+2 x^{2}-\sqrt{x^{2}+1}}{1+x^{2}}\)

Function is increasing monotonically.

\(\Rightarrow \frac{1+2 x^{2}-\sqrt{{x}^{2}+1}}{1+{x}^{2}}>0\)

\(\Rightarrow 1+2 x^{2}-\sqrt{x^{2}+1}>0\)

\(\Rightarrow 1+2 x^{2}>\sqrt{x^{2}+1}\)

Squaring on both sides,

\(\Rightarrow\left(1+2 x^{2}\right)^{2}>x^{2}+1\)

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

For all \(x \in R\)

Given:

The binomial coefficients of \((3 r-1)\) and \((r+2)\) th terms are equal.

3rd term in the expansion of \((1+x)^{2 n}\)

\(={ }^{2 \mathrm{n}} \mathrm{C}_{3 \mathrm{r}-1} \mathrm{x}^{3 \mathrm{r}-1}\)

\((r+2)\) th term in the expansion of \((1+x)\)

\(={ }^{2 \mathrm{n}} \mathrm{C}_{\mathrm{r}+1} \mathrm{x}^{\mathrm{r}+1}\)

According to question:

\({ }^{2 \mathrm{n}} \mathrm{C}_{3 \mathrm{r}-1}={ }^{2 \mathrm{n}} \mathrm{C}_{\mathrm{r}+1}\)

\(\Rightarrow 3 \mathrm{r}-1=\mathrm{r}+1\)

So, \(2 \mathrm{n}=(3 \mathrm{r}-1)+(\mathrm{r}+1)\)

\( 2 \mathrm{r}=2\) or \(2 \mathrm{n}=4 \mathrm{r}\)

\(\Rightarrow \mathrm{r}=1\) or \(\mathrm{n}=2 \mathrm{r}\)

But \(\mathrm{r}>1\)

Therefore, \(\mathrm{n}=2 \mathrm{r}\)

We have,

\(e^{x}=(1-x)\left(B_{0}+B_{1} x+B_{2} x^{2}+\ldots+B_{n-1} x^{n-1}+B_{n} x^{n}+\ldots\right)\)

By the expansion of e^x, we get,

\(1+\frac{x}{1 !}+\frac{x^{2}}{2 !}+\ldots+\frac{x^{n}}{n !}+\ldots\)

\(=(1-x)\left(B_{0}+B_{1} x+B_{2} x^{2}+\ldots+B_{n-1} x^{n-1}+B_{n} x^{n}+\ldots\right)\)

Equating the coefficient of xⁿ on both sides, we get,

\(B_{n}-B_{n-1}=\frac{1}{n !}\)