A functionf:R1,∞ satisfies the equation f(xy)=f(x)f(y)−f(x)−f(y)+2. If differentiable on R−{0}andf(2)=5,fprime(x)=xf(x)−1λ˙thenλ= 2primef(1) b. 3fprime(1) c. 21fprime(1) d. fprime(1)

Click here to view the solution.

Get 2 FREE Instant-Explanations on Filo with code FILOAPP

A function{f}:{R}{{{1},}} satisfies the equation {f{{({x}{y})}

Question

A function $f : R 1, \infty$ satisfies the equation $f (x y) = f (x) f (y) - f (x) - f (y) + 2 .$ If differentiable on $R - {0} a n d f (2) = 5, f^{p r i m e} (x) = \frac{f ( x ) - 1}{x} \dot{λ} t h e n λ =$ $2^{p r i m e} f (1)$ b. $3 f^{p r i m e} (1)$ c. $\frac{1}{2} f^{p r i m e} (1)$ d. $f^{p r i m e} (1)$

Views: 681 students

Found 5 tutors discussing this question

Samuel Discussed

A function

f : R 1, \infty

satisfies the equation

f (x y) = f (x) f (y) - f (x) - f (y) + 2 .

If differentiable on

R - {0} a n d f (2) = 5, f^{p r i m e} (x) = \frac{f ( x ) - 1}{x} \dot{λ} t h e n λ =

2^{p r i m e} f (1)

3 f^{p r i m e} (1)

\frac{1}{2} f^{p r i m e} (1)

f^{p r i m e} (1)

8 mins ago

Discuss this question LIVE

8 mins ago

Question Text

A function

f : R 1, \infty

satisfies the equation

f (x y) = f (x) f (y) - f (x) - f (y) + 2 .

If differentiable on

R - {0} a n d f (2) = 5, f^{p r i m e} (x) = \frac{f ( x ) - 1}{x} \dot{λ} t h e n λ =

2^{p r i m e} f (1)

3 f^{p r i m e} (1)

\frac{1}{2} f^{p r i m e} (1)

f^{p r i m e} (1)