Mathematics Test - 13

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

We know that: \(\log m+\log n=\log m n\)

Using the above identity, we get:

\(\mathrm{LHS}=\log x+\log 5=\log 5 x\)

We know that: \(\log m-\log n=\log \left(\frac{m}{n}\right)\)

Using the above identity, we get:

RHS \(=\log x^{2}-\log 14=\log \left(\frac{x^{2}}{14}\right)\)

So, we have \(\log 5 x=\log \left(\frac{x^{2}}{14}\right)\)

Since log \(x\) is a one-one function, we have \(5 x=\frac{x^{2}}{14}\)

Thus, we have \(x=0\) or \(x=5 \times 14=70\)

Since log \(x\) is defined for \(x>0\), we have \(x=0\) is rejected.

Thus, we have \(x=70\) is the solution of the given equation.

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

If \(\alpha, \beta\) are roots of equation \(a x^{2}+b x+c=0\) then
\(\alpha+\beta=\frac{-b}{a} ~~\& ~~\alpha \beta=\frac{c}{a}\)
\((1+\alpha)(1+\beta)\) \(=1+\alpha+\boldsymbol{\beta}+\boldsymbol{\alpha} \boldsymbol{\beta}\) \(=1+\left(\frac{-b}{a}\right)+\frac{c}{a}\) \(=1+\left(\frac{c-b}{a}\right)\)

If,

\(f(x)=\left\{\begin{array}{c}x^{2} \ln |x|, x \neq 0 \\ 0, x=0\end{array}=\left\{\begin{array}{ll}x^{2} \ln x & , x>0 \\ x^{2} \ln (-x) & , x<0 \\ 0, & x=0\end{array}\right.\right.\)

\(\mathrm{RHD}\)

\(\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}\)

\(=\lim _{h \rightarrow 0} \frac{(x+h)^{2} \ln (x+h)-x^{2} \ln x}{h}\)

\(=\left(\frac{0}{0}\right)\) form

Applying L' Hospital Rule. \(=\lim _{h \rightarrow 0} \frac{2(x+h) \ln (x+h)+(x+h)^{2} \frac{1}{(x+h)}-2 x \ln x-x^{2} \cdot \frac{1}{x}}{1}\)

\(=2 x \ln x+x-2 x \ln x-x\)

\(=0\)

Similarly, LHD \(=0\)

\(\Rightarrow \mathrm{f}^{\prime}(0)=0\)

\(z=\left(\frac{\sqrt{3}}{2}+\frac{i}{2}\right)^5+\left(\frac{\sqrt{3}}{2}-\frac{i}{2}\right)^5\)
\(\because\) Euler's form of\(\frac{\sqrt{3}}{2}+\frac{i}{2}=\left(\cos \frac{\pi}{6}+i \sin \frac{\pi}{6}\right)=e^{i( \pi / 6)}\)
\(\frac{\sqrt{3}}{2}-\frac{i}{2}=\cos \left(\frac{-\pi}{6}\right)+i \sin \left(-\frac{\pi}{6}\right)=e^{-i \pi / 6}\)
So,\( z=\left(e^{i \pi / 6}\right)^5+\left(e^{-i \pi / 6}\right)^5=e^{i \frac{5 \pi}{6}}+e^{-i \frac{5 \pi}{6}}\)\(=\left(\cos \frac{5 \pi}{6}+i \sin \frac{5 \pi}{6}\right)+\left(\cos \frac{5 \pi}{6}-i \sin \frac{5 \pi}{6}\right)\) \(\left[\because e^{i \theta}=\cos \theta+i \sin \theta\right]\)
\(=2 \cos \frac{5 \pi}{6}\)
\(\therefore I(z)=0\) and \(R(z)=-2 \cos \frac{\pi}{6}=-\sqrt{3}<0\) \(\left[\because \cos \frac{5 \pi}{6}=\cos \left(\pi-\frac{\pi}{6}\right)=-\cos \frac{\pi}{6}\right]\)

The angle \(\theta\) between two vectors \(\vec{a}\) and \(\vec{b}\) is given by:

\(\cos \theta=\frac{\vec{a} \cdot \vec{b}}{|\vec{a} \| b|}\)

Now, \(\vec{a} \cdot \vec{b}=(\hat{i}+\hat{j}-\hat{k}) \cdot(\hat{i}-\hat{j}+\hat{k})=1-1-1=-1\)

\(|\vec{a}|=\sqrt{1^{2}+(1)^{2}+(-1)^{2}}\)\(=\sqrt{3}\)
\(|\vec{b}|=\sqrt{1^{2}+(-1)^{2}+1^{2}}\)\(=\sqrt{3}\)

\(|\vec{a}|.|\vec{b}|\) \(=\sqrt{3} \times \sqrt{3}\) = 3

So, \(\cos \theta=-\frac{1}{3}\)

Thus, required angle\(\theta=\cos ^{-1}\left(-\frac{1}{3}\right)\)

The coefficient matrix for the given system of equations is:

\(\operatorname{det}(A)=1(10-4)+1(16-20)+1(2-4)=6-4-2=0\)

Therefore, either there is no solution (inconsistent) or there are infinitely many solutions (consistent and dependent).

Let, us convert the augmented matrix into the row echelon form to find its solutions. The augmented matrix is:

\(A \mid B =\left[\begin{array}{ccccc}1 & 1 & 1 & \mid & 1 \\ 2 & 1 & 4 & \mid & k \\ 4 & 1 & 10 & \mid & k ^{2}\end{array}\right]\)

\(R _{3} \rightarrow R _{3}-4 R _{1}\)

\(R _{2} \rightarrow R _{2}-2 R _{1}\)

\(A \mid B =\left[\begin{array}{ccccc}1 & 1 & 1 & \mid & 1 \\ 0 & -1 & 2 & \mid & k -2 \\ 0 & -3 & 6 & \mid & k ^{2}-4\end{array}\right]\)

\(R _{3} \rightarrow R _{3}-3 R _{2}\)

\(A \mid B =\left[\begin{array}{ccccc}1 & 1 & 1 & \mid & 1 \\ 0 & -1 & 2 & \mid & k -2 \\ 0 & 0 & 0 & \mid & k ^{2}-3 k +2\end{array}\right]\)

The rank of both the coefficient matrix and the augmented matrix must be equal for the system to be consistent.

\(\therefore k ^{2}-3 k +2=0\)

\(\Rightarrow k ^{2}-2 k - k +2=0\)

\(\Rightarrow k ( k -2)-( k -2)=0\)

\(\Rightarrow( k -2)( k -1)=0\)

\(\Rightarrow k -2=0\) OR \(k -1=0\)

\(\Rightarrow k =2\) or \(k =1\)

Given,

\(A=\left[\begin{array}{lll}3 & 1 & 2 \\ 4 & 2 & 1 \\ 2 & a & 1\end{array}\right]\) is a singular matrix.

Therefore, \(|\mathrm{A}|=0\)

\(|\mathrm{A}|=\left|\begin{array}{lll}3 & 1 & 2 \\ 4 & 2 & 1 \\ 2 & a & 1\end{array}\right|=0\)

\(\Rightarrow|A|=3(2-a)-1(4-2)+2(4 a-4)=0\)

\(\Rightarrow |A|=6-3 a-2+8 a-8=0\)

\(\Rightarrow |A|=5 a-4=0\)

\(\Rightarrow 5 a-4=0\)

\(\Rightarrow {a}=\frac{4}{5}\)