Q.
Let f(x)=α(x)β(x)γ(x) for all real x, where α(x),β(x) and γ(x) are differentiable functions of x. If f′(2)=18f(2),α′(2)=3α(2),β′(2)=−4β(2) and γ′(2)=kγ(2) , then the value of k is
We have, f(x)=α(x)β(x)γ(x) , for all real x ⇒f′(x)=α′(x)β(x)γ(x)+α(x)β′(x)γ(x)+α(x)β(x)γ′(x) ⇒f′(2)=α′(2)β(2)γ(2)+α(2)β′(2)g(2)+a(2)β(2)γ′(2) 18f(2)=3α(2)β(2)γ(2)−4α(2)β(2)γ(2)+kα(2)β(2)γ(2)
[∵f′(2)=18f(2),α′(2)=3α(2),β′(2)=−4β(2) and γ′(2)=kγ(2)] ⇒18f(2)=(−1+k)α(2)β(2)γ(2) =(k−1)f(2) [∵f(2)=α(2)β(2)γ(2)] ⇒k−1=18 ∴k=19