Mathematics Test - 28

The given word contains \(9\) letters, namely \(4 A , 2 L , 1 H , 1 B\) and \(1 D\).

\(\therefore\) Required number of ways \(=\frac{9 !}{4 ! 2 ! 1 ! 1 ! 1 !}\)

\(=\frac{9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{4 \times 3 \times 2 \times 1 \times 2}\)

\(=7560\)

Given: \(z^{2}+z+1=0\) means \(z=\omega\) or \(\omega^{2},\) where \(\omega\) represents complex cube root of \(1+i 0\)
From the properties of cube roots of \(1\), we know that: (\(1+\omega+\omega^{2}=0\)) And \(1 \times \omega \times \omega^{2}=1\)
\(\Rightarrow \omega \cdot \omega^{2}=1\) 
\(\Rightarrow\omega^{3}=1\) (\(\because\) roots of \(z^{2}+z+1=0\) are reciprocal of each other)
So, In \(\left(z+\frac{1}{z}\right)^{2}+\left(z^{2}+\frac{1}{z^{2}}\right)^{2}+\left(z^{3}+\frac{1}{z^{3}}\right)^{2}+\ldots .+\left(z^{6}+\frac{1}{z^{6}}\right)^{2}\) 
Putting \(z=\omega\) we get; \(\left(z+\frac{1}{z}\right)^{2}=\left(\omega+\frac{1}{\omega}\right)^{2}=\left(\omega+\omega^{2}\right)^{2}\)\(=(-1)^{2}=1\)
\(\left(z^{2}+\frac{1}{z^{2}}\right)^{2}=\left(\omega^{2}+\omega\right)^{2}\)\(=1\)
\(\left(z^{3}+\frac{1}{z^{3}}\right)^{2}=(1+1)^{2}\)\(=4\)
\(\left(z^{4}+\frac{1}{z^{4}}\right)^{2}=\left(\omega+\omega^{2}\right)^{2}\)\(=(-1)^{2}=1\)
\(\left(z^{3}+\frac{1}{z^{3}}\right)^{2}=\left(\omega^{2}+\omega\right)^{2}\)\(=1\)
\(\left(z^{6}+\frac{1}{z^{6}}\right)^{2}=(1+1)^{2}\)\(=4\)
Thus, adding them all, \(\left(z+\frac{1}{z}\right)^{2}+\left(z^{2}+\frac{1}{z^{2}}\right)^{2}+\left(z^{3}+\frac{1}{z^{3}}\right)^{2}+\ldots .+\left(z^{6}+\frac{1}{z^{6}}\right)^{2}\)\(=1+1+4+1+1+4\)\(=12\)

We know that,

If \(\alpha\) and \(\beta\) are the roots of the quadratic equation, \(a x^{2}+b x+c=0\). Then,

\(\alpha+\beta=-\frac{b}{a}\) and \(\alpha \times \beta=\frac{c}{a}\)

The given equation is,

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

By comparing the given equation with the quadratic equation, \(a x^{2}+b x+c=0 .\) We get, \(a=2, b=-2(k-2)\) and \(c=-(k+1)\)

Let \(\alpha\) and \(\beta\) be the roots of the equation \(2 x^{2}-2(k-2) x-(k+1)=0\)

\(\Rightarrow \alpha+\beta=-\frac{-2(k-2)}{2}=k-2\)

\(\Rightarrow \alpha \times \beta=\frac{-(k+1)}{2}\)

As we know that, \(\alpha^{2}+\beta^{2}=(\alpha+\beta)^{2}-2 \alpha \beta\)

\(\Rightarrow a^{2}+\beta^{2}=(k-2)^{2}+(1+k)\)

\(\Rightarrow a^{2}+\beta^{2}=(k-\frac{3}{2})^{2}+ \frac{11}{4}\)

So, the minimum value of \(a^{2}+\beta^{2}\) is found when \(k-\frac{3}{2}=0 \Rightarrow k= \frac{3}{2}\).

The enclosed sides of a parallelogram are \(\vec{a}\) and\(\vec{b}\) then its area is obtained by \(|\vec{a} \times \vec{b}|\).

Now, \(\ \vec{a} \times \vec{b}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 3 & 1 & 4 \\ 1 & -1 & 1\end{array}\right|\)

\(=\hat{i}(1+4)-\hat{j}(3-4)+\widehat{k}(-3-1)\)

\(=5 \hat{i}+\hat{j}-4 \hat{k}\)

Therefore, \(|\vec{a} \times \vec{b}|=\sqrt{25+1+16}=\sqrt{42}\)

Thus the required area is \(\sqrt{42}\).

GIven:

(1 + tan α tan β)²+ (tan α – tan β)²– sec²α sec²β

⇒ (1 + tan α tan β)² = 1 + tan² α tan² β + 2 tan α tan β ...(1)

⇒ (tan α – tan β)² = tan² α + tan² β – 2 tan α tan β ...(2)

Adding (1) and (2), we get

⇒ (1 + tan α tan β)² + (tan α – tan β)²

= 1 + tan² α tan² β + 2 tan α tan β + tan² α + tan² β – 2 tan α tan β

= 1 + tan² α tan² β + tan² α + tan² β

= 1 + tan² α + tan² α tan² β + tan² β

= (1 + tan² α) + tan² β (1 + tan² α)

= (1 + tan² α)(1 + tan² β)

= sec² α sec² β

⇒ (1 + tan α tan β)² + (tan α – tan β)² – sec² α sec² β = 0

The word "RUMOUR" has two repetitive \(R\) and \(U\).

Therefore, number of arrangements \(=\frac{6 !}{2 ! \times 2 !}\)

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

\(=180\)

Let \(f(x)=a x^{2}+b x+c>0\) for all \(x \in R\)

\(\Rightarrow a>0\) and \(b^{2}-4 a c<0\quad\quad\)....(i)

\(\therefore g(x)=f(x)+f^{\prime}(x)+f^{\prime \prime}(x)\)

\(\Rightarrow g(x)=a x^{2}+b x+c+2 a x+b+2 a\)

\(\Rightarrow g(x)=a x^{2}+x(b+2 a)+(c+b+2 a)\)

Whole discriminant:

\(b^{2}-4 a c<0\quad\quad\)

\(=(b+2 a)^{2}-4 a(c+b+2 a) \)

\(=b^{2}+4 a^{2}+4 a b-4 a c-4 a b-8 a^{2} \)

\(=b^{2}-4 a^{2}-4 a c \)

\(=\left(b^{2}-4 a c\right)-4 a^{2}<0 \quad[\text { From Eq. }(\mathrm{i})]\)

\(\therefore g(x)>0\) for all \(x\), as \(a>0\) and discriminant \(<0\).

Thus, \(g(x)>0\) for all \(x \in R\).

Given:\(\mathrm{A}(1,1,1), \mathrm{B}(1,2,3)\) and \(\mathrm{C}(2,3,1)\)

According to the question,

\(A=\hat{i}+\hat{j}+\widehat{k}\)

\(B=\hat{i}+2 \hat{j}+3 \widehat{k}\)

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

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

\(=\hat{j}+2 \hat{k}\)

And \(\overrightarrow{\mathrm{AC}}\)\(=(2 \hat{i}+3 \hat{j}+\hat{k})\) \(-(\hat{i}+\hat{j}+\widehat{k})\)

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

The area of a given triangle is \(\frac{1}{2}|\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}}|\)

Now, \(\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}}=\left|\begin{array}{lll}\hat{i} & \hat{j} & \hat{k} \\ 0 & 1 & 2 \\ 1 & 2 & 0\end{array}\right|\)

\(=\hat{\imath}[(1 \times 0)-(2 \times 2)]-\hat{\jmath}[(0 \times 0)-(1 \times 2)]+\hat{k}[(0 \times 2)-(1 \times 1)]\)

\(=-4 \hat{\imath}+2 \hat{\jmath}-1 \widehat{k}\)

So, \(\ |\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}}|=\sqrt{16+4+1}=\sqrt{21}\)

The area is \(\frac{1}{2} \sqrt{21}\).

Given, \(f(x)=\frac{\log (1+a x)-\log (1-b x)}{x}\)

\(\mathrm{f}(\mathrm{x})\) is continuous at \(\mathrm{x}=\mathrm{k}\) and \(\mathrm{f}(0)=\mathrm{k}\)

\(\therefore \lim _{x \rightarrow 0} \mathrm{f}(\mathrm{x})=\lim _{x \rightarrow 0} \frac{\log (1+\operatorname{ax})-\log (1-\mathrm{bx})}{\mathrm{x}}\left(\frac{0}{0}\right.\) form \()\)

\(=\lim _{x \rightarrow 0}\left(\frac{1}{1+\operatorname{ax}} a+\frac{b}{1-b x}\right)\)

\(=\mathrm{a}+\mathrm{b}\)

\(\therefore \quad \mathrm{a}+\mathrm{b}=\mathrm{f}(0)=\mathrm{k}\)