A function f : R − { 0 } → R satisfies
⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ tan ( x f ( x ) ) = x − 1 ∣ f ( x ) ∣ < ∣ ∣ ∣ 2 x π ∣ ∣ ∣
for all nonzero reals x . The average of f ( x ) when x is uniformly chosen from the interval [ 1 , 2 ] is p π q ln r + s for nonnegative rational numbers p , q , r , and s . Find p + q + r + s to 4 decimal places.
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0 isn't a positive rational number.
0 is not positive
Super sol and sums (+1)...But sir give the answer as 12.0000 since u said upto 4 places of decimal and it is accepting only integers..XD!!!!
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Multiply ∣ x ∣ to the inequality to obtain ∣ x f ( x ) ∣ < 2 π .
This means we can rewrite f ( x ) as f ( x ) = x arctan ( x − 1 ) .
The average value of f ( x ) over [ 1 , 2 ] is ∫ 1 2 x arctan ( x − 1 ) d x .
Use integration by parts, differentiating arctan ( x − 1 ) and integrating x 1 .
∫ 1 2 x arctan ( x − 1 ) d x = [ ln x ⋅ arctan ( x − 1 ) ] 1 2 − ∫ 1 2 ( x − 1 ) 2 + 1 ln x d x ( ⟹ x − 1 = tan t ) = 4 π ln 2 − ∫ 0 π / 4 ln ( 1 + tan t ) d t
Now, substitute t = 4 π − u .
∫ 0 π / 4 ln ( 1 + tan t ) d t = ∫ 0 π / 4 ln ( 1 + tan ( 4 π − u ) ) d u = ∫ 0 π / 4 ln ( 1 + 1 + tan u 1 − tan u ) d u = 4 π ln 2 − ∫ 0 π / 4 ln ( 1 + tan u ) d u
Hence, ∫ 0 π / 4 ln ( 1 + tan t ) d t = 8 π ln 2 .
From this we conclude that
∫ 1 2 x arctan ( x − 1 ) d x = 4 π ln 2 − 8 π ln 2 = 8 π ln 2 .
Then, p = 8 1 , q = 1 , r = 2 , s = 0 .
∴ p + q + r + s = 3 . 1 2 5 0 .