Mathematics Test - 14

Given \(\alpha, \beta\) are the roots of \(a x^{2}+b x+c=0,\)
\(\alpha+\beta=\frac{-b}{a},\alpha.\beta=\frac{c}{a}\)
\(\frac{1}{a \alpha+b}+\frac{1}{a \beta+b}\) \(=\frac{1}{a \alpha-a(\alpha+\beta)}+\frac{1}{a \beta-a(\alpha+\beta)}\)
\(\Rightarrow {b}=-{a}({\alpha}+{\beta})\) \(=\frac{1}{a \alpha-a \alpha-a \beta}+\frac{1}{a \beta-a \alpha-a \beta}\) \(=-\frac{1}{a}\left(\frac{1}{\alpha}+\frac{1}{\beta}\right)\) \(=\frac{-1}{a}\left(\frac{-b}{c}\right)\) \(=\frac{{b}}{{a c}}\)

The sum of the given vector,

\(\vec{a}+\vec{b}=\vec{c}\)

\(=(2 \hat{i}+2 \hat{j}-5 \hat{k})\) + \((2 \hat{i}+\hat{j}+3 \hat{k})=\vec{c}\)

Or \(\vec{c}=4 \hat{i}+3 \hat{j}-2 \hat{k}\)

And \(\quad|\vec{c}|=\sqrt{4^{2}+3^{2}+(-2)^{2}}=\sqrt{29}\)

Therefore, the required unit vector,

\(\vec{c}=\frac{1}{|\vec{c}|} \vec{c}=\frac{1}{\sqrt{29}}(4 \hat{i}+3 \hat{j}-2 \hat{k})=\frac{4}{\sqrt{29}} \hat{i}+\frac{3}{\sqrt{29}} \hat{j}-\frac{2}{\sqrt{29}} \hat{k}\)

Given, probability of success \(\frac{2}{3}\) and number of trials is 10.

Consider, the most probable number of successes in 10 trials with probability of success \(\frac{2}{3}\) is n.

Now, \(\mathrm{n}=\frac{2}{3}+\frac{2}{3}+\frac{2}{3}+\ldots\) ( 10 times)

\(\mathrm{n}=\frac{2}{3} \times 10\)

\(\Rightarrow \mathrm{n}=6.67\)

\(\Rightarrow \mathrm{n}=7\)

The most probable number of successes in 10 trials with probability of success \(\frac{2}{3}\) is 7.

Given:

\(3 \tan ^{-1} x+\cot ^{-1} x=\pi\)

We know that,

\(\cot ^{-1} x=\frac{\pi}{2}-\tan ^{-1} x\)

Now,

\(3 \tan ^{-1} x+\left(\frac{\pi}{2}-\tan ^{-1} x\right)=\pi\)

\(\Rightarrow 2 \tan ^{-1} x+\frac{\pi}{2}=\pi\)

\(\Rightarrow 2 \tan ^{-1} x=\frac{\pi}{2}\)

\(\Rightarrow \tan ^{-1} x=\frac{\pi}{4}\)

\(\Rightarrow x=\tan \frac{\pi}{4}\)

\(\Rightarrow x=1\)

\({ }^{n} P_{3}=n \times(n-1) \times(n-2)\)

\({ }^{(n+1)} P_{3}=(n+1) \times n \times(n-1)\)

Now, \(5 \times n \times(n-1) \times(n-2)=4 \times(n+1) \times n \times(n-1)\)

or, \(5(n-2)=4(n+1)\)

or, \(5 n-10=4 n+4\)

or, \(5 n-4 n=4+10\)

So, \(n=14\)

Consider a point (2, 0) on the x-axis.

Substituting \(x=2, y=0\) in \(3 x+12 y=6<400\).

Hence, one constraints is \(3 x+12 y \leq 400\)

Again, substituting \(x=2, y=0\) in \(x-4 y=2-0>0\)

\(\therefore x-4 y \geq 0\) is other constraints and also the third constraint from the figure is \(y \leq 25\).

So, the correct alternative is \(3 x+12 y \leq 400, y \leq 25, x \geq 4 y\).

When, a coin is tossed

Outcome = S = {H, T}

When a die is rolled,

S = {1, 2, 3, 4, 5, 6}

Now, a coin is tossed and a die is rolled, and we need only head on the coin

So, S = {(H, 1), (H, 2), (H, 3), (H, 4), (H, 5), (H, 6)}

n(S) = 6

A matrix is said to be singular if its determinant is zero

i.e. for matrix \(A\) to be singular, \(|A|=0\)

For a singular matrix, the inverse doesn't exist

Given that, matrix \(\left[\begin{array}{ccc}\cos \theta & \sin \theta & 0 \\ \sin \theta & \cos \theta & 0 \\ 0 & 0 & 1\end{array}\right]\) is singular.

Then, \(\left|\begin{array}{ccc}\cos \theta & \sin \theta & 0 \\ \sin \theta & \cos \theta & 0 \\ 0 & 0 & 1\end{array}\right|=0\)

\(\Rightarrow\left|\begin{array}{ll}\cos \theta & \sin \theta \\ \sin \theta & \cos \theta\end{array}\right|=0\)

\(\Rightarrow \cos ^2 \theta-\sin ^2 \theta=0\)

\(\Rightarrow \cos 2 \theta=\cos \frac{\pi}{2}\)

\(\therefore \theta=\frac{\pi}{4}\)

The given equation is,

$3x^2-2x+4=0$

On comparing the above equation with$a{x}^2+bx+c=0$,

$a=3,b=-2,c=4$

Discriminant,

$b^2-4ac=(-2{)}^2-4\times 3\times 4$

$=4-48$

$=-44$(Negative)

So, the roots are imaginary.

If n = 2017, then

\(\frac{1}{\log _{2} n}+\frac{1}{\log _{3} n}+\ldots+\frac{1}{\log _{2017} n} \ldots . .(1)\)

Now we know that \(\frac{1}{\log _{a} b}=\log _{b}\) a

\(\therefore\) we can rewrite Equation 1 as

\(\frac{1}{\log _{2} n}+\frac{1}{\log _{3} n}+\ldots+\frac{1}{\log _{2017} n}\)

\(=\log _{n} 2+\log _{n} 3+\log _{n} 4+\ldots+\log _{n} 2017\)

\(=\log _{n}(2.3 .4 \ldots . .2017)\) as \(\left[\log _{a} b+\log _{a} c+\log _{a}(b \times c)\right]\)

n = (2017)!

\(\therefore \log _{n}(2.3 .4 \ldots .2017)=\log _{2017 !}(2017) !=1\)

So,

\(\frac{1}{\log _{2} n}+\frac{1}{\log _{3} n}+\ldots+\frac{1}{\log _{2017} n}=1\)