Mathematics Test - 7

Let \(m =2 k +1, n =2 k -1( k \in N )\)

\(\therefore m ^2- n ^2\)

\(=4 k ^2+1+4 k -4 k ^2+4 k -1=8 k\)

Thus, all the numbers of the form \(m^2-n^2\) are always divisible by 8.

If,

f (x) = $\left|x+1\right|$

Now,

$f\left(x^2\right)=\left|x^2+1\right|$

${\left|f\left(x\right)\right|}^2={\left|x+1\right|}^2={\left|x\right|}^2+1^2+2\left|x\right|$

$\Rightarrow f\left(x^2\right)\ne {\left|f\left(x\right)\right|}^2$

$f\left|x\right|=\left|\left|x\right|+1\right|$

$\left|f\left(x\right)\right|=\left|x+1\right|$

$\Rightarrow f\left|x\right|\ne \left|f\left(x\right)\right|$

$f\left(x\right)+f\left(y\right)=\left|x+1\right|+\left|y+1\right|$

$f\left(x+y\right)\ne f\left(x\right)+f\left(y\right)$

Given, 

\(\mathrm{X}=8^{\mathrm{n}}-7 \mathrm{n}-1\)

\(=(1+7)^{\mathrm{n}}-7 n-1\)

\(=1+7 \mathrm{n}+\frac{\mathrm{n}(\mathrm{n}-1)}{2} 7^{2}+\ldots+7^{\mathrm{n}}-7 \mathrm{n}-1\)

\(=\frac{\mathrm{n}(\mathrm{n}-1)}{2} 7^{2}+\ldots+7^{\mathrm{n}}\)

\(=49\left[\frac{\mathrm{n}(\mathrm{n}-1)}{2}+\ldots+7^{\mathrm{n}-2}\right]\)

So, the set \(\mathrm{X}\) will be some specific multiples of \(49\).

\(\Rightarrow\)\(\mathrm{Y}=49(\mathrm{n}-1)\)

Therefore, the set \(Y\) will be all multiples of \(49 .\) So, it will contain the elements of \(\mathrm{X}\) too.

So, \(\mathrm{X} \subset \mathrm{Y}\)

Here, we have to form a 4 digit number lying between 3000 and 5000 using digits 1, 2, 3, 4, 7, 9 such that repetition of digits is allowed.

Unit’s digit can be filled by any one of the given numbers

⇒ No. of ways to fill unit’s digit = 6

Ten’s digit can be filled by any one of the given numbers ⇒ No. of ways to fill ten’s digit = 6

Hundred’s digit can be filled by any one of the given numbers

⇒ No. of ways to fill hundredth digit = 6

Thousand’s digit can be filled by 3 or 4 only

⇒ No. of ways to fill thousand’s digit = 2

∴ The no. of ways to form a 4 digit number lying between 3000 and 5000 using digits 1, 2, 3, 4, 7, 9 such that repetition of digits is allowed = 6 × 6 × 6 × 2 = 432

The eqution of line is given as, \(\frac{x+2}{-3}=\frac{y-2}{4}=\frac{z+2}{1}=\lambda\)

The coordinate of point B on the given line are,

B (-3λ- 2, 4λ+ 2, λ -2)

and the given coordinate of point A  is  ( 2, 3, 7). 

\(\overrightarrow{\mathrm{AB}}=(-3 \lambda-4) \hat{\mathrm{i}}+(4 \lambda-1) \hat{\mathrm{j}}+(\lambda+5) \hat{\mathrm{k}}\)

\(\mathrm{AB}\) is perpendicular to the line, \(\frac{\mathrm{x}+2}{-3}=\frac{\mathrm{y}-2}{4}=\frac{\mathrm{z}+2}{1},\)

We know that, If two lines are Perpendicular, then  a₁a₂+b₁b₂+c₁c₂ = 0,

⇒ (-3λ- 4 ) × (-3) + (4λ-1 )× 4 + (λ + 5 )× 1 = 0

⇒ 26λ + 13 = 0

\(\Rightarrow \lambda=-\frac{1}{2}\)

\(\overrightarrow{\mathrm{AB}}=-\frac{5}{2} \hat{\mathrm{i}}-3 \hat{\mathrm{j}}+\frac{9}{2} \hat{\mathrm{k}}\)

\(\therefore|\overrightarrow{\mathrm{AB}}|=\sqrt{\left(\frac{5}{2}\right)^{2}+3^{2}+\left(\frac{9}{2}\right)^{2}}\)

\(\Rightarrow|\overrightarrow{\mathrm{AB}}|=\sqrt{\frac{71}{2}}\)

\(y =\cos x , x =0 \text { and } x =\pi \text { is shaded area } \)

\( \text { Area }=\int_0^{\frac{\pi }{ 2}} ydx +\int_{\frac{\pi }{ 2}}^0 ydx \)

\( =\int_0^{\frac{\pi }{ 2}}(\cos x -0) dx +\int_{\frac{\pi }{ 2}}^\pi(0-\cos x ) dx \)

\( =\int_0^{\frac{\pi }{ 2}} \cos x -\int_{\frac{\pi }{ 2}}^\pi \cos xdx \)

\( =|\sin x|_0^{\frac{\pi }{ 2}}-|\sin x|_{\frac{\pi }{ 2}}^\pi \)

\( =[1-0]-[0-1] \)

=2 sq. units 

Given,

\(\frac{(x-3)}{1}=\frac{(y-2)}{-1}=\frac{(z+\lambda)}{-2}=k\)

Let \(P\) be any point on the plane,

\(P=(k+3,-k-2,-2 k-\lambda)\)

\(P\) lie on the plane, \(2 x-4 y+3 z=2,\) Which means \(P\) satisfy the equationn of plane.

So, \(2(k+3)-4(-k-2)+3(-2 k-\lambda)=2\)

Solving above equation, we have \(\lambda=4\)

Now, the shortest distance between the below lines:

\(\frac{(x-3)}{1}=\frac{(y-2)}{-1}=\frac{(z+4)}{-2}\) and \(\frac{(x-1)}{12}=\frac{y}{9}=\frac{z}{4}\) is:

\(\begin{aligned} \mathrm{d}^{2} &=\left|\begin{array}{ccc}x_{1}-x_{2} & y_{1}-y_{2} & z_{1}-z_{2} \\ l_{1} & m_{1} & n_{1} \\ l_{2} & m_{2} & n_{2}\end{array}\right| \\ &=\left|\begin{array}{ccc}3-1 & -2-0 & -4-0 \\ 1 & -1 & -2 \\ 12 & 9 & 4\end{array}\right| \end{aligned}\)

\(=\mid2(14)+2(28)-4(21) \mid\)

\(=0\)

\(d^{2}=0\) or \({d}=0\)

The given equation \(x^{4}-2 x^{2} \sin ^{2} \frac{\pi x}{2}+1=0\)

On solving, we get-

\(\left(x^{2}-\sin ^{2} \frac{\pi x}{2}\right)^{2}+1-\sin ^{4} \frac{\pi x}{2}=0\)

So, \(\left(x^{2}-\sin ^{2} \frac{\pi x}{2}\right)^{2}=0\) and \(1-\sin ^{4} \frac{\pi x}{2}=0\)

So, \(1-\sin ^{4} \frac{\pi x}{2}=0 \Rightarrow \frac{\pi x}{2}=(2 n+1) \frac{\pi}{2}\)

So, \(x=2 n+1\)

\(\left(x^{2}-\sin ^{2} \frac{\pi x}{2}\right)^{2}=0\)

\((2 n+1)^{2}-1=0\)

\(n=0,-1\)

\(x=1,-1\)

We know that.

\(A^{-1}=\frac{\operatorname{adj}(A)}{|A|}\)

Now,

\(|A|=\cos \theta(\cos \theta)-\sin \theta(-\sin \theta)\)

\(=\cos ^{2} \theta+\sin ^{2} \theta\)

\(=1\)

Now,

\(\operatorname{Adj} A=\left(\begin{array}{ccc}\cos \theta & -\sin \theta & 0 \\ \sin \theta & \cos \theta & 0 \\ 0 & 0 & 1\end{array}\right)\)

\(\therefore A^{-1}=\left(\begin{array}{ccc}\cos \theta & -\sin \theta & 0 \\ \sin \theta & \cos \theta & 0 \\ 0 & 0 & 1\end{array}\right)\)

Given,

\(\lim _{x \rightarrow 2} \frac{x^{3}+x^{2}}{x^{2}+3 x+2}\)

\(=\lim _{x \rightarrow 2} \frac{x^{2}(x+1)}{(x+1)(x+2)}\)

\(=\lim _{x \rightarrow 2} \frac{x^{2}}{x+2}\)

\(=\frac{2^{2}}{2+2}\)

\(=\frac{4}{4}\)

\(=1\)