Before we dive into this question, let's try and understand the properties of the modulus function
1. |x+a| has a domain (-infinity, infinity) and range[0, infinity).
2. |x+a| can be rewritten as |x-(-a)| and is nothing but the distance of x from (-a).
2. |x+a| divides the x axis into 3 regions:
a. x < -a
b. x = -a
c. x > -a
Therefore, the range of the function |x+5|+ |x-3| can be divided into 3 segments: 1. x<-5 2. -5≤x≤3 3. x>3
When x<-5:
1. One approach would be taking 2 random values of x (x1 and x2) such that x<-5 and checking the values of f(x) = |x+5|+ |x-3|. f(x1) and f(x2) produce different values. Therefore, value of f(x) can not be determined in this range.
2. Second approach would be to check out the graphs of the two functions, |x+5| and |x-3| in this range. We would see ascendent parallel lines with a negative slope. Therefore, f(x) decreases for decreasing x.
When x>3:
1. Here too, approach #1, would be taking 2 random values of x (x1 and x2) such that x>3 and checking the values of f(x) = |x+5|+ |x-3|. f(x1) and f(x2) produce different values. Therefore, value of f(x) can not be determined in this range.
2. Second approach would be to check out the graphs of the two functions, |x+5| and |x-3| in this range. We would see ascendent parallel lines with a positive slope. Therefore, f(x) increases for increasing values of x.
When -5≤x≤3:
The graphs of two functions are straight lines here that converge, intersect and diverge, with increasing x starting at -5 through 3, in this region with equal but opposite slopes. Therefore, for all x in this region, f(x) = 8.
The same can be verified by trying random values of x.
Therefore, let's now have a look at the statements to see whether they can help us determine whether x lies in our favorable range or not.
Statement 1: x^2<16
=> |x|<4 => -4<x<4. Since x is an integer, this translates to -3≤x≤3. This lies in the favorable range. Therefore this is sufficient.
Statement 2: (x+1)^2 <25
=> |x+1|<25 => -5<x+1<5 => -6<x<4. Since x is an integer, this translates to -5≤x≤3. This lies in the favorable range. Therefore this is sufficient.
Therefore, the answer is (D).
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1. |x+a| has a domain (-infinity, infinity) and range[0, infinity).
2. |x+a| can be rewritten as |x-(-a)| and is nothing but the distance of x from (-a).
2. |x+a| divides the x axis into 3 regions:
a. x < -a
b. x = -a
c. x > -a
Therefore, the range of the function |x+5|+ |x-3| can be divided into 3 segments: 1. x<-5 2. -5≤x≤3 3. x>3
When x<-5:
1. One approach would be taking 2 random values of x (x1 and x2) such that x<-5 and checking the values of f(x) = |x+5|+ |x-3|. f(x1) and f(x2) produce different values. Therefore, value of f(x) can not be determined in this range.
2. Second approach would be to check out the graphs of the two functions, |x+5| and |x-3| in this range. We would see ascendent parallel lines with a negative slope. Therefore, f(x) decreases for decreasing x.
When x>3:
1. Here too, approach #1, would be taking 2 random values of x (x1 and x2) such that x>3 and checking the values of f(x) = |x+5|+ |x-3|. f(x1) and f(x2) produce different values. Therefore, value of f(x) can not be determined in this range.
2. Second approach would be to check out the graphs of the two functions, |x+5| and |x-3| in this range. We would see ascendent parallel lines with a positive slope. Therefore, f(x) increases for increasing values of x.
When -5≤x≤3:
The graphs of two functions are straight lines here that converge, intersect and diverge, with increasing x starting at -5 through 3, in this region with equal but opposite slopes. Therefore, for all x in this region, f(x) = 8.
The same can be verified by trying random values of x.
Therefore, let's now have a look at the statements to see whether they can help us determine whether x lies in our favorable range or not.
Statement 1: x^2<16
=> |x|<4 => -4<x<4. Since x is an integer, this translates to -3≤x≤3. This lies in the favorable range. Therefore this is sufficient.
Statement 2: (x+1)^2 <25
=> |x+1|<25 => -5<x+1<5 => -6<x<4. Since x is an integer, this translates to -5≤x≤3. This lies in the favorable range. Therefore this is sufficient.
Therefore, the answer is (D).
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