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# Find all real numbers A , B such that the function f (x) = A.2^x + B, satisfies the conditions f ' (1) = 2 andintegral of f (x)dx = 7. where upper limit is 3 , lower limit is 0

bharat bajaj IIT Delhi
7 years ago
`Differentiate the function :f`(x) = 2Af`(1) = 2A = 2Hence, A = 1Integrate the function :Ax^2 + Bx : 0 to 3 = 79A + 3B = 79 + 3B = 7B = -2/3ThanksBharat Bajajaskiitians facultyIIT Delhi`
71 Points
7 years ago
Given f(x)=A.x^2+B, integration of f(x)=7 differentiate on both sides f`(x)=2*x*A, But f`(1)=2 Therefore f`(1)=1*2*A 2=2A Hence,A=1 Integrate the function:f(x)=A.x^2+B with limits 0 to 3 Therefore[( A.x^3)/3+Bx] 0 to 3=7 [(x^3)/3+Bx] 0 to 3=7 (27/3+B.3)-(0)=7 9+3B=7 3B=-2 B=-2/3 Therefore A=1 and B=-2/3
chaitnyakishore
26 Points
7 years ago
Given f(x)=A.x^2+B, integration of f(x)=7 differentiate on both sides f`(x)=2*x*A, But f`(1)=2 Therefore f`(1)=1*2*A 2=2A Hence,A=1 Integrate the function:f(x)=A.x^2+B with limits 0 to 3 Therefore {[(A.x^3)/3]+Bx}0 to 3=7 (27/3)+B*3=7 3B=7-9 B=-2/3 Therefore A=1 and B=-2/3