If f:R→R and g:R→R be two given functions, then 2 min f(x)-g(x),0 equals
f(x)+g(x)-g(x)-f(x)
f(x)+g(x)+g(x)-f(x)
f(x)-g(x)+g(x)-f(x)
f(x)-g(x)-g(x)-f(x)
Explanation for the correct answer:
Step 1: Case 1, when f(x)>g(x)
Let h(x)=2minf(x)-g(x),0
⇒f(x)>g(x)⇒f(x)-g(x)>0⇒2minf(x)-g(x),0=0
Therefore, h(x)=0
Step 2: Case 2, when f(x)<g(x)
⇒f(x)<g(x)⇒f(x)-g(x)<0⇒2minf(x)-g(x),0=2min(f(x)-g(x))
⇒h(x)=2(f(x)-g(x))
Therefore, h(x)=f(x)-g(x)-g(x)-f(x)
Hence, the correct answer is option (D).
If f:R→R is defined by f(x)=2x-2x,∀x∈R,where x is the greatest integer not exceeding x, then the range of f is
Chooseanappropriateoptionandfillintheblanks:
Rs10.1=......paise 101/1010