Question 10 - NCERT Exemplar - MCQs - Chapter 5 Class 12 Continuity and Differentiability (Term 1)
Last updated at Nov. 17, 2021 by Teachoo
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If u = γsinγ^(-1) (2x/(1 + x^2 )) and v = γtanγ^(-1) (2x/(1 - x^2 )), then du/dv is
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Question 10
If u = γπ ππγ^(β1) (2π₯/(1 + π₯^2 )) and v = γπ‘ππγ^(β1) (2π₯/(1 β π₯^2 )), then ππ’/ππ£ is
(A) 1/2 (B) π₯
(C) (1 β π₯^2)/(1 + π₯^2 ) (D) 1
π π/π π=(π πβπ π)/(π πβπ π)
So, we will find ππ’/ππ₯ and ππ£/ππ₯
Simplifying u
u = γπ ππγ^(β1) (2π₯/(1 + π₯^2 ))
Putting π₯=tanβ‘π
u = γπ ππγ^(β1) ((2 πππβ‘π½)/(1 + γπππγ^π π½))
Using sinβ‘γ2π=γ (2 tanβ‘π)/(1 + γπ‘ππγ^2 π)
u = γπ ππγ^(β1) (sinβ‘2π )
u = 2π
Putting π=γπ‘ππγ^(β1) π₯
u = πγπππγ^(βπ) π
Simplifying v
v = γπ‘ππγ^(β1) (2π₯/(1 β π₯^2 ))
Putting π₯=tanβ‘π
v = γπ‘ππγ^(β1) ((2 πππβ‘π½)/(1 β γπππγ^π π½))
Using tanβ‘γ2π=γ (2 tanβ‘π)/(1 β γπ‘ππγ^2 π)
v = γπ‘ππγ^(β1) (tanβ‘2π )
v = 2π
Putting π=γπ‘ππγ^(β1) π₯
v = πγπππγ^(βπ) π
Finding π π/π π
u = πγπππγ^(βπ) π
Differentiating w.r.t. π₯
π π/π π=(πγ(2π‘ππγ^(β1) π₯))/ππ₯
=2 Γ(πγ(π‘ππγ^(β1) π₯))/ππ₯
=2 Γ1/γ1 + π₯γ^2
= π/γπ + πγ^π
Finding π π/π π
v = πγπππγ^(βπ) π
Differentiating w.r.t. π₯
π π/π π=(πγ(2π‘ππγ^(β1) π₯))/ππ₯
=2 Γ(πγ(π‘ππγ^(β1) π₯))/ππ₯
=2 Γ1/γ1 + π₯γ^2
= π/γπ + πγ^π
Now,
π π/π π=(ππ’βππ₯)/(ππ£βππ₯)
=(2/γ1 + π₯γ^2 )/(1/γ1 + π₯γ^2 )
=π
So, the correct answer is (D)
Made by
Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.
