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Let \(m\) and \(n(m
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It is given that, \(m\) and \(n~(mTherefore, 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,\( p+q+r+s=(m+d)+(m+2 d)+(m+3 d)+(m+4 d)\)\(\Rightarrow p+q+r+s=4 m+10 d\)\(\Rightarrow p+q+r+s=2 \times(2 m+5 d)\)\(\Rightarrow p+q+r+s=2 \times[m+(m+5 d)]\)\(\Rightarrow p+q+r+s=2 \times(m+n) \quad [\because(n=m+5 d)]\)\(\Rightarrow p+q+r+s=32\quad [\because(m+n=16)]\)
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,
\( p+q+r+s=(m+d)+(m+2 d)+(m+3 d)+(m+4 d)\)
\(\Rightarrow p+q+r+s=4 m+10 d\)
\(\Rightarrow p+q+r+s=2 \times(2 m+5 d)\)
\(\Rightarrow p+q+r+s=2 \times[m+(m+5 d)]\)
\(\Rightarrow p+q+r+s=2 \times(m+n) \quad [\because(n=m+5 d)]\)
\(\Rightarrow p+q+r+s=32\quad [\because(m+n=16)]\)
Let f : R be a differentiable function and f(1) = 4. Find the value of
A missile fired from the ground level rises x meters vertically upwards in t seconds, where x = 100t - 25/2 t2. The maximum height reached is
If \(\tan {A}-\tan {B}={x}\) and \(\cot {B}-\cot {A}={y}\), then what is the value of \(\cot ({A}-{B}) ?\)
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,
\(\cot (A-B)=\frac{\frac{y}{x}+1}{y}\)
\(=\frac{1}{x}+\frac{1}{y}\)
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v
Hence , options ( D ) is the correct answer.
If the two pairs of lines x2 – 2mxy – y2 = 0 and x2 – 2nxy – y2 = 0 are such that one of them represents the bisector of the angles between the other, then
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In how many ways can 8 students be arranged in a row?
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.
If f(x) = | x |3, then f′(0) =
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