Mathematics Test - 26

Let \(U=\) No. of students in higher secondary group, \(M=\) No. of higher secondary students who qualified NDA exam and \(J=\) No. of higher secondary students who qualified JEE exam.
Given: \(n(U)=100, n(M)=50, n(J)=40\) and \(n(M \cap J)=30\)
As we know that, if \(A\) and \(B\) are two finite sets then, \(n(A \cup B)=n \)
\(n(A)+n(B)-n(A \cap B) \)
\(\Rightarrow n(M \cup J)=n(M)+n(J)-n(M \cap J)=50+40-30 \) \( =60\)
\(\therefore\) There are 60 higher secondary students who have qualified either NDA or JEE exam.
As, we know that the number of higher secondary students who have neither qualified NDA nor JEE exam is given by: \(n[(M \cup J)]\) \(\Rightarrow n[(M \cup J)]=n(U)-n(M \cup J)=100-60=40\)

Given:

\(\int \frac{d x}{2^{x}-1}\)

Let \(2^{x}=t\)

On solving, we get-

\(2^{x} \log _{e} 2 d x=d t\)

\(d x=\frac{d t}{t \log _{e} 2}\)

\(\int \frac{d t}{t \log _{e} 2(t-1)}\)

\(\frac{1}{\log _{e} 2} \int \frac{d t}{t(t-1)}=\frac{1}{\log _{e} 2}\left[\int \frac{-1 d t}{t}+\int \frac{1 d t}{t-1}\right]\)

\(=\frac{1}{\log _{e} 2}[-\log t+\log (t-1)]+C\)

\(=\frac{1}{\log _{e} 2}\left[\log t^{-1}+\log (t-1)\right]+C\)

\(=\frac{1}{\log _{e} 2}\left[\log t^{-1}(t-1)\right]+C\)

\(=\frac{1}{\log _{e} 2}\left[\log \left(1-t^{-1}\right)\right]+C\)

\(=\frac{1}{\log _{e} 2}\left[\log \left(1-2^{-x}\right)\right]+C\)

\(=\frac{\log \left(1-2^{-x}\right)}{\ln 2}+C\)

Given:

\(\alpha\) is any vector

Let \(\alpha=a_{1} \hat{i}+a_{2} \hat{j}+a_{3} \hat{k}\)

As we know that, if \(\vec{a}=a_{1} \hat{i}+a_{2} \hat{j}+a_{3} \hat{k}\) and \(\vec{b}=b_{1} \hat{i}+b_{2} \hat{j}+b_{3} \hat{k}\) then

\(\vec{a} . \vec{b}=a_{1} b_{1}+a_{2} b_{2}+a_{3} b_{3}\)

\((\alpha. \hat{i}) \hat{i}+(\alpha.\hat{j}) \hat{j}+(\alpha . \hat{k}) \hat{k}=a_{1} \hat{i}+a_{2} \hat{j}+a_{3} \hat{k}\)

\((\alpha.\hat{i}) \hat{i}+(\alpha. \hat{j}) \hat{j}+(\alpha. \hat{k}) \hat{k}=\alpha\)

Given \(5\) boys and \(4\) girls are in total

We can select \(3\) boys from \(5\) boys in \({ }^{5} C_{3}\) ways

Similarly, we can select \(3\) boys from \(54\) girls in \({ }^{4} C_{3}\) ways

\(\therefore\) Number of ways a team of \(3\) boys and \(3\) girls can be selected is \({ }^{5} C_{3} \times{ }^{4} C_{3}\)

\({ }^{5} \mathrm{C}_{3} \times{ }^{4} \mathrm{C}_{3}=\frac{5 !}{3 ! 2 !} \times \frac{4 !}{3 ! 1 !}\)

\(=\frac{5 \times 4 \times 3 !}{3 ! \times 2} \times \frac{4 \times 3 !}{3 !}\)

\(=10 \times 4\)

\(=40\)

\(\therefore\) Number of ways a team of \(3\) boys and \(3\) girls can be selected is \({ }^{5} C_{3} \times{ }^{4} C_{3}=40\) ways

Let any point on the intersecting line is:

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

\(\Rightarrow \mathrm{x}=-3 \lambda-1\)

\(\Rightarrow \mathrm{y}=2 \lambda+3\)

\(\Rightarrow z=-\lambda+2\)

So, the coordinates of point of intersection of two lines will be \((-3 \lambda-1,2 \lambda+3,-\lambda+2)\) for some \(\lambda \in \mathrm{R}\).

Let the point \(A=(-3 \lambda-1,2 \lambda+3,-\lambda+2)\)

\(B=(-4,3,1)\)

Then,

\(\mathrm{AB}=\mathrm{OB}-\mathrm{OA}\)

\(\Rightarrow \mathrm{AB}=(-4 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+1 \hat{\mathrm{k}})-\{(-3 \lambda-1) \hat{\mathrm{i}}+(2 \lambda+3) \hat{\mathrm{j}}+(-\lambda+2) \hat{\mathrm{k}}\}\)

\(\Rightarrow \mathrm{AB}=(-4 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+1 \hat{\mathrm{k}})-(-3 \lambda-1) \hat{\mathrm{i}}-(2 \lambda+3) \hat{\mathrm{j}}-(-\lambda+2) \hat{\mathrm{k}}\)

\(\Rightarrow \mathrm{AB}=(-4 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+1 \hat{\mathrm{k}})+(3 \lambda \hat{\mathrm{i}}+1 \hat{\mathrm{i}})+(-2 \lambda \hat{\mathrm{j}}-3 \hat{\mathrm{j}})+(\lambda \hat{\mathrm{k}}-2 \hat{\mathrm{k}})\)

\(\Rightarrow \mathrm{AB}=(-4 \hat{\mathrm{i}}+3 \lambda \hat{\mathrm{i}}+1 \hat{\mathrm{i}})+(3 \hat{\mathrm{j}}-2 \lambda \hat{\mathrm{j}}-3 \hat{\mathrm{j}})+(1 \hat{\mathrm{k}}+\lambda \hat{\mathrm{k}}-2 \hat{\mathrm{k}})\)

\(\Rightarrow \mathrm{AB}=(3 \lambda \hat{\mathrm{i}}-3 \hat{\mathrm{i}})+(-2 \lambda \hat{\mathrm{j}})+(\lambda \hat{\mathrm{k}}-1 \hat{\mathrm{k}})\)

\(\Rightarrow \mathrm{AB}=(3 \lambda-3) \hat{\mathrm{i}}-(2 \lambda) \hat{\mathrm{j}}+(\lambda-1) \hat{\mathrm{k}}\)

Now as the line is parallel to the given plane, therefore AB will be parallel to the given plane and so AB will be perpendicular to the normal of plane.

∴ AB.n = 0

From the plane equation (x + 2y – z – 5 = 0), the normal to the plane is.

\(\mathrm{n}=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}-\hat{\mathrm{k}}\)

Multiplying Equation (1) and (2) we get

\(\Rightarrow \mathrm{AB}=((3 \lambda-3) \hat{\mathrm{i}}-2 \lambda \hat{\mathrm{j}}+(\lambda-1) \hat{\mathrm{k}}) \cdot(\hat{\mathrm{i}}+2 \hat{\mathrm{j}}-\hat{\mathrm{k}})\)

\(\Rightarrow \mathrm{AB}=3(\lambda-1)-4 \lambda+(-1)(\lambda-1)\)

[f \(\mathrm{a}=\mathrm{a}_{1} \hat{\mathrm{i}}+\mathrm{a}_{2} \hat{\mathrm{j}}+\mathrm{a}_{3} \hat{\mathrm{k}}\) and \(\mathrm{b}=\mathrm{b}_{1} \hat{\mathrm{i}}+\mathrm{b}_{2} \hat{\mathrm{j}}+\mathrm{b}_{3} \hat{\mathrm{k}}\)

then \(\left[\mathrm{ab}=\mathrm{a}_{1} \mathrm{~b}_{1}+\mathrm{a}_{2} \mathrm{~b}_{2}+\mathrm{a}_{3} \mathrm{~b}_{3}\right]\)

\(\Rightarrow \mathrm{AB}=3 \lambda-3-4 \lambda-1 \lambda+1\)

\(\Rightarrow \mathrm{AB}=-2 \lambda-2\)

\(\Rightarrow-2 \lambda=2\)

\(\therefore \lambda=-1\)

When we substitute \(\lambda=-1\) in point \(A\),

\(\Rightarrow \mathrm{A}=(-3 \lambda-1,2 \lambda+3,-\lambda+2)\)

\(\Rightarrow \mathrm{A}=(3-1,-2+3,1+2)\)

\(\therefore \mathrm{A}=(2,1,3)\)

Now the required equation of line joining \(A(2,1,3)\) and \(B(-4,3,1)\), which is

\(\Rightarrow \frac{x-(-4)}{2-(-4)}=\frac{y-3}{1-3}=\frac{z-1}{3-1}\)

Equation of line joining \(\left(x_{1}, y_{1}, z_{1}\right)\) and \(\left(x_{2}, y_{2}, z_{2}\right)\) is

\(\left[\frac{\mathrm{x}-\mathrm{x}_{1}}{\mathrm{x}_{2}-\mathrm{x}_{1}}=\frac{\mathrm{y}-\mathrm{y}_{1}}{\mathrm{y}_{2}-\mathrm{y}_{1}}=\frac{\mathrm{z}-\mathrm{z}_{1}}{\mathrm{z}_{2}-\mathrm{z}_{1}}\right]\)

\(\Rightarrow \frac{\mathrm{x}+4}{6}=\frac{\mathrm{y}-3}{-2}=\frac{\mathrm{z}-1}{2}\)

Now, multiplying by 2 ,

\(\Rightarrow \frac{\mathrm{x}+4}{3}=\frac{\mathrm{y}-3}{-1}=\frac{\mathrm{z}-1}{1}\)

Given:

\(\vec{\alpha}=\mathrm{k}\) and \(\vec{\gamma}=2 \mathrm{i}+3 \mathrm{j}+4 \mathrm{k}\)

\(\vec{\beta}\) is perpendicular to both \(\vec{\alpha}\) and \(\vec{\gamma}\)

\(\therefore \vec{\alpha} \times \vec{\gamma}=\vec{\beta}\)

\(\vec{\alpha} \times \vec{\gamma}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 0 & 0 & 1 \\ 2 & 3 & 4\end{array}\right|\)

\(=\hat{i}(0-3)-\hat{j}(0-2)+\hat{k}(0)\)

\(=-3 \hat{i}+2 \hat{j}\)

On solving the given equation-

\(\frac{d^{2} x}{d t^{2}}+\mu x=0\)

\(\Rightarrow\left(D^{2}+\mu\right) x=0\)

\(\Rightarrow m^{2}+\mu=0\)

\(m^{2}=-\mu\)

\(\Rightarrow \mathrm{m}=\pm \sqrt{u} i\)

\(\Rightarrow x(t)=c_{1} \cos \sqrt{u} t+c_{2} \sin \sqrt{u} t\)

The functions \(\sin \theta \& \cos \theta\) are always periodic

\(\therefore \mathrm{x}(\mathrm{t})\) is periodic function.

Given as:

\((1+x)^{43}\)

As we know, the coefficient of the \(r\) term in the expansion of \((1+x)^{n}\) is \({ }^{n} C_{r-1}\).

Therefore, the coefficients of the \((2 r+1)\) and \((r+2)\) terms in the given expansion are \(^{43} C_{2 r+1-1}\) and \({ }^{43} C_{r+2-1}\)

For these coefficients to be equal, we must have,

\({ }^{43} C_{2 r+1-1}={ }^{43} C_{r+2-1}\)

\({ }^{43} C_{2 r}={ }^{43} C_{r+1}\)

\(2 r=r+1\) (or) \(2 r+r+1=43\) [Since, \({ }^{n} C_{r}={ }^{n} C_{s} \Rightarrow r=s\) (or) \(r+s=n]\)

\(2 r-r=1\) (or) \( 3 r+1=43\)

\(r=1\) (or) \( 3 r=43-1\)

\(r=1\) (or) \( 3 r=42\)

\(r=1\) (or) \( r=\frac{42}{3}\)

\(r=1\) (or) \( r=14\)

\(\therefore r=14\) [since, value '\(1\)' gives the same term]

Solution given below:

Let \(z=(x-i y)(3+5 i)\)

\(z=3 x+5 x i-3 y i-5 y i^{2}\)

\(=3 x+5 x i-3 y i+5 y\)

\(=(3 x+5 y)+i(5 x-3 y)\)

\(\therefore \bar{z}=(3 x+5 y)-i(5 x-3 y)\)

It is given that, \(\bar{z}=-6-24 i\)

\(\therefore(3 x+5 y)-i(5 x-3 y)=-6-24 i\)

Equating real and imaginary parts, we obtain

\(3 x+5 y=-6\quad\quad\)....(i)

\(5 x-3 y=24\quad\quad\)....(ii)

Multiplying equation (i) by 3 and equation (ii) by 5 and then adding them, we obtain

\(\begin{array}{r}9 x+15 y=-18 \\25 x-15 y=120 \\\hline 34 x=102\end{array}\)

\(\therefore x=\frac{102}{34}=3\)

Putting the value of \(x\) in equation (i), we obtain

\(3(3)+5 y=-6\)

\(\Rightarrow 5 y=-6-9=-15\)

\(\Rightarrow y=-3\)

Thus, the values of \(x\) and \(y\) are 3 and -3 respectively.

Given differential equation\(\frac{d y}{d x}=e^{x-y}+x^{2} e^{-y}\)

On solving, we get-

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

Integrationon both sides

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

\(e^{y}=e^{x}+\frac{x^{3}}{3}+c\)

\(3\left(e^{y}-e^{x}\right)=x^{3}+c^{\prime}\)