Q 9 A function 'f' is defined for all real numbers and satisfies f (2 + x) = - Maths - Relations and Functions

Question

Q 9 A function 'f' is defined for all real numbers and satisfies f
(2 + x) = f (2 - x) and f (7 + x) = f (7 - x) for all real x If x =
0 is a root of f (x) = 0, then find least number of roots of f (x)
= 0 for x ∈ [- 1000, 1000]

Aarushi Mishra · Accepted Answer

f2+x=f2-x _______________1From above equation one can infer that fx is symmetric about x=2 line as x units on either side of 2 f has same value
Also f7+x=f7-xSimilarly x units on either side of 7 f has same value
 Thus it is symmetric about x=7 line too
A fuction is symmetric about line x=2 as wels as x=7
 Now that is only possible if function is periodic
We can proove this toof7+x=f7-xReplace x by x-5 in above equationf7+x-5=f7-x-5f2+x=f12-xFrom equation 1f2-x=f12-xNow replace x by -xf2+x=f12+xNow replace x by x-2fx=fx+10Therefore 10 is a period of function fxf0=f10=f20=
=f1000We need to find number of terms in A
P   10, 20, 30,
,1000a1=10d=10an=a1+n-1d=100010+n-1×10=1000n-1=99n=100Alsof0=f-10=f-20=
=f-1000Here also in -10,-20,-30,
,-1000we have total 100 termsTherefore total number of roots=100+100+1=201At last we have added 1 because we also need to count the root at x=0

Q.9. A function 'f' is defined for all real numbers and satisfies f (2 + x) = f (2 - x) and f (7 + x) = f (7 - x) for all real x. If x = 0 is a root of f (x) = 0, then find least number of roots of f (x) = 0 for x ∈ [- 1000, 1000].

Q.9. A function 'f' is defined for all real numbers and satisfies f (2 + x) = f (2 - x) and f (7 + x) = f (7 - x) for all real x. If x = 0 is a root of f (x) = 0, then find least number of roots of f (x) = 0 for x $\in$ [- 1000, 1000].