1. Class 12
2. Important Question for exams Class 12

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Ex7.9, 22 Choose the correct answer β«_0^(2/3)βππ₯/(4 +9π₯^2 ) equals (A) π/6 (B) π/12 (C) π/24 (D) π/4 Step 1 :- Let F(π₯)=β«1βππ₯/(4 + 9π₯^2 ) Divide and multiply the integrate by 4 =β«1ββ(ππ₯@4)/((4 + 9π₯^2)/4) =1/4 β«1βππ₯/(1 + 9/4 π₯^2 ) =1/4 β«1βππ₯/(1 + (3/2 π₯)^2 ) Put 3/2 π₯=π‘ Differentiating w.r.t.π₯ π(3/2 π₯)/ππ₯=ππ‘/ππ₯ 3/2=ππ‘/ππ₯ ππ₯=ππ‘/(3/2) ππ₯=2/3 ππ‘ Hence 1/4 β«1βγππ₯/(1+(3/2 π₯)^2 )=1/4 β«1βγ1/(1+π‘^2 ) 2/3γ ππ‘γ =1/4 Γ 2/3 β«1βππ‘/(1+π‘^2 ) =1/6 tan^(β1)β‘π‘ Putting π‘=3/2 π₯ =1/6 tan^(β1)β‘γ3/2 π₯γ Hence F(π₯)=1/6 tan^(β1)β‘γ3/2 π₯γ Step 2 :- β«_0^(2/3)βγππ₯/(4+9π₯^2 )=πΉ(2/3)βπΉ(0) γ =1/6 tan^(β1)β‘γ(3/2 Γ 2/3)β1/6 tan^(β1)β‘(0) γ =1/6 tan^(β1)β‘γ(1)β1/6 tan^(β1)β‘(0) γ =1/6 Γ π/4β0 =π/24 β΅ Option (C) is correct.

Class 12
Important Question for exams Class 12