Question 3 - NCERT Exemplar MCQ - Chapter 7 Class 12 Integrals (Term 2)

Transcript

Question 3 If ∫1▒〖(3𝑒^𝑥 − 5𝑒^(−𝑥) )/(4𝑒^𝑥 + 5𝑒^(−𝑥) ) 𝑑𝑥=𝑎𝑥+𝑏 log⁡〖|4𝑒^𝑥+5𝑒^(−𝑥) |〗+𝐶〗, then 𝑎=(−1)/8, 𝑏=7/8 (B) 𝑎=1/8, 𝑏=7/8 (C) 𝑎=(−1)/8, 𝑏=(−7)/8 (D) 𝑎=1/8, 𝑏=(−7)/8 To solve this, We write (𝟑𝒆^𝒙 − 𝟓𝒆^(−𝒙) )/(𝟒𝒆^𝒙 + 𝟓𝒆^(−𝒙) ) as (3𝑒^𝑥 − 5𝑒^(−𝑥) )/(4𝑒^𝑥 + 5𝑒^(−𝑥) ) = a + b × (𝐝(𝟒𝒆^𝒙 + 𝟓𝒆^(−𝒙) )/𝒅𝒙)/(𝟒𝒆^𝒙 + 𝟓𝒆^(−𝒙) ) (3𝑒^𝑥 − 5𝑒^(−𝑥) )/(4𝑒^𝑥 + 5𝑒^(−𝑥) ) = a + b × (4𝑒^𝑥 − 5𝑒^(−𝑥))/(4𝑒^𝑥 + 5𝑒^(−𝑥) ) (3𝑒^𝑥 − 5𝑒^(−𝑥) )/(4𝑒^𝑥 + 5𝑒^(−𝑥) ) = (𝑎 × (4𝑒^𝑥 + 5𝑒^(−𝑥) ) + 𝑏 × (4𝑒^𝑥 − 5𝑒^(−𝑥)))/(4𝑒^𝑥 + 5𝑒^(−𝑥) ) 3𝑒^𝑥 − 5𝑒^(−𝑥) = 𝑎 × (4𝑒^𝑥+ 5𝑒^(−𝑥) ) + 𝑏 × (4𝑒^𝑥 − 5𝑒^(−𝑥)) 3𝑒^𝑥 − 5𝑒^(−𝑥) = 4𝑎 𝑒^𝑥+5𝑎 𝑒^(−𝑥)+4𝑏 𝑒^𝑥−5𝑏 𝑒^(−𝑥) 𝟑𝒆^𝒙 − 𝟓𝒆^(−𝒙) = 𝒆^𝒙 (𝟒𝒂+𝟒𝒃)+𝒆^(−𝒙) (𝟓𝒂−𝟓𝒃) 𝟑 = 𝟒𝒂+𝟒𝒃 4𝑎+4𝑏=3 𝒂+𝒃=3/4 −𝟓 = 𝟓𝒂−𝟓𝒃𝒃 5𝑎−5𝑏=−5 𝒂−𝒃=−1 Thus, our equations are 𝒂+𝒃=3/4 …(1) 𝒂−𝒃=−1 …(2) Adding (1) and (2) 𝑎+𝑏+𝑎−b=3/4−1 2a=(−1)/4 𝐚=(−𝟏)/𝟖 Putting value of a in (1) 𝑎+𝑏=3/4 (−𝟏)/𝟖+𝒃=𝟑/𝟒 𝑏=3/4+1/8 𝑏=6/8+1/8 𝒃=7/8 Thus, 𝐚=(−𝟏)/𝟖 and 𝒃=7/8 So, the correct answer is (a)