psoting some doubts of functions here...plzz help...
1. range of f(x) = 12-sin 3x, m getting [1/2,1]...but the answer is [1/3,1].
2. If f(x) is an odd peridic function with period 2 then f(4) equals...??
3. f:R→R...then f(x) = x^3 + x^2 f'(1) + xf"(2) + f"'(3) then f(2) - f(1)...???
4. the value of b for which the function f(x) = sin x - bx + c is decreasing in the interval (-infinit, infinity) is?
Can i use the first derivative metho in the fourth q???
1q1.
sin3x-[-1,1]
so range of the given fn is-[1/3,1]
Q4.Differentiate,
so cosx-b<0,as the fn is decreasing.
3.-6
23-1\leq sin3x\leq 1
\Rightarrow 1\geq - sin3x\geq -1
add 2
\Rightarrow 3\geq 2- sin3x\geq 1
take reciprcal
\Rightarrow \frac{1}{3}\leq \frac{1}{2- sin3x }\leq 1
23Q3)
let, f'(1)= a ,f''(2)=b,f'''(3)=c
f(x)=x^{3}+ax^{2}+bx+c
\Rightarrow f'(x)=3x^{2}+2ax+b...............(1)
\Rightarrow f''(x)=6x+2a...............(2)
\Rightarrow f'''(x)=6...............(3)
\Rightarrow f'''(3)=6=c
also , from 2 , and from 1 ,
\Rightarrow f''(2)=b=12+2a.........(4)
f'(1)=a=3+2a+b
\Rightarrow a+b=-3......(5)
from 4,5
b=2, a= - 5
so
f(x)=x^{3}-5x^{2}+2x+6
f(2)-f(1) = (8 - 20 + 10 ) - ( 1 - 5 + 2 + 6 )
=-6
23Q2 ->
not sure but there are 2 methods ,
odd \;function \Rightarrow f(x)+f(-x)=0
also , f(x)=f(x+2).......(1)
so,f(4)+f(-4)=0
f(4)+f(-4+2)=0 ......(from1)
f(4)+f(-4+2+2)=0 ......(from1)
f(4)+f(-4+2+2+2)=0 ......(from1)
f(4)+f(-4+2+2+2+2)=0 ......(from1)
\Rightarrow ,f(4)+f(4)=0
\Rightarrow ,f(4)=0
OR,
since no other specific condition is attached to the function , we can take it as any standard odd function with period = 2
eg: f(x)=sin(\pi x)
is an odd function wid period 2 , defined for all x ,
and clearly f(4) = 0
23f(x) = sinx - bx+ c
f'(x) = cosx - b
for function to be always decreasing
f'(x) < 0
this cosx < b
thus b must be gr8er than the gr8est value of cosx
hence b > 1