NCERT Exemplar - MCQs

Chapter 5 Class 12 Continuity and Differentiability
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## (D) 1/β(1 + x^2 )

This question is similar to Ex 5.1, 21 - Chapter 5 Class 12 - Continuity and Differentiability

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Question 9 Differential coefficient of sec (γπ‘ππγ^(β1)x) w.r.t. x is (A) π₯/β(1 + π₯^2 ) (B) π₯/(1 + π₯^2 ) (C) x β(1+π₯^2 ) (D) 1/β(1 + π₯^2 ) Let y = sec (γπ‘ππγ^(β1) π₯) Differential coefficient sec (γπ‘ππγ^(β1)x) means ππ/ππ Finding ππ/ππ ππ/ππ = π(secβ‘(γπ‘ππγ^(β1) π₯))/ππ₯ = sec (γπ‘ππγ^(β1) π₯) tan (γπππγ^(βπ) π) Γ (π(γπ‘ππγ^(β1) π₯))/ππ₯ = sec (γπ‘ππγ^(β1) π₯) . π . 1/(1 + π₯^2 ) = sec(γπππγ^(βπ) π) Γ π/(π + π^π ) Now, to find ππ/ππ , we need to find value of sec(γπππγ^(βπ) π) Finding value of sec(γπππγ^(βπ) π) Let π½ = γπππγ^(βπ) π tanβ‘π=π₯ Now, let ABC be a triangle tanβ‘π=π΄π΅/π΅πΆ π₯=π΄π΅/π΅πΆ π΄π΅/π΅πΆ=π₯/1 β΄ AB = π & BC = 1 So, AC = β(γπ΄π΅γ^2+γπ΅πΆγ^2 ) = β(π₯^2+1^2 ) = β(π^π+π) Now, cosβ‘γπ=π΅πΆ/π΄πΆγ =1/β(π₯^2 + 1) β΄ π¬ππβ‘π½=1/cosβ‘π =1/(1/β(π₯^2+1)) =β(π^π+π) Putting π = γπππγ^(βπ) π sec (γπππγ^(βπ) π) =β(π₯^2+1) Putting value of sec (γπ‘ππγ^(β1) π₯) in (1) ππ/ππ = sec(γπ‘ππγ^(β1) π₯) Γ π₯/(1+ π₯^2 ) = β(π^π+π) Γ π₯/(1 + π₯^2 ) = β(π₯^2+1) Γ π₯/(β(π₯^2 + 1))^2 = π/β(π^π + π) So, the correct answer is (A)