Mathematics Test - 1

It is given that, \(m\) and \(n~(m

Therefore, from the given equation,

\(x^{2}-16 x+39=0\)

\(\Rightarrow x^{2}-13 x-3 x+39=0\)

\(\Rightarrow x(x-13)-3(x-13)=(x-13) \times(x-3)=0\)

\(\Rightarrow x=13\) or \(3\)

\(\therefore m=3\) and \(n=13\)

Now, it is given that, four terms \(p, q, r\) and \(s\) are inserted between \(m\) and \(n\) to form an AP.

We know that for an AP, if the first term is \(a\) and common difference is \(d\), then,

Second term = \(a+d\)

Third term = \(a+2d\) & so on.

Therefore, according to the question,

\( p+q+r+s=(m+d)+(m+2 d)+(m+3 d)+(m+4 d)\)

\(\Rightarrow p+q+r+s=4 m+10 d\)

\(\Rightarrow p+q+r+s=2 \times(2 m+5 d)\)

\(\Rightarrow p+q+r+s=2 \times[m+(m+5 d)]\)

\(\Rightarrow p+q+r+s=2 \times(m+n) \quad [\because(n=m+5 d)]\)

\(\Rightarrow p+q+r+s=32\quad [\because(m+n=16)]\)

It is given that,

\(\tan A-\tan B=x\)    ---- (1)

and \(\cot B-\cot A=y\)    ---- (2)

We know that, \(\cot x=\frac{1}{\tan x}\)

Taking \(\cot B-\cot A=y\)

\(\therefore \cot B-\cot A=\frac{1}{\tan B}-\frac{1}{\tan A}\)

\(\Rightarrow  \cot B-\cot A=\frac{\tan A-\tan B}{\tan B \times \tan A}\)

\(\Rightarrow y=\frac{x}{\tan B \times \tan A}\)    [From equation (1) and (2)]

\(\Rightarrow \frac{x}{y}=\tan A \times \tan B\)

\(\therefore \frac{y}{x}=\cot A \times \cot B\)

As we know that, \(\cot (A-B)=\frac{\cot A \times \cot B+1}{\cot B-\cot A}\)

Putting the values,

\(\cot (A-B)=\frac{\frac{y}{x}+1}{y}\)

\(=\frac{1}{x}+\frac{1}{y}\)

v

Hence , options ( D ) is the correct answer.

Required number of ways = 8!

This is a permutation problem because the order of the students is important. Eight students can be arranged in 8! different ways, which is 40,320. If the question had been for all possible combinations of the eight students, the answer would be only 1.