Sign in
Please select an account to continue using cracku.in
↓ →
If a function f(a) = max (a, 0) then the smallest integer value of x for which the equation f(x — 3) + 2𝑓(x + 1) = 8 holds true is _______________.
Correct Answer: 3
f(a) = max (a, 0). This means that is a>0, then f(a) = a, and if a<0, then f(a) = 0.
We are given - f(x — 3) + 2𝑓(x + 1) = 8
=> [max (x-3, 0)] + 2[max (x+1, 0)] = 8
The above equation will depend on the range of x. Thus, there are three cases possible -
Case-1: $$x\ge3$$
f(x-3) = [max (x-3, 0)] = x-3 (Because x-3 will always be greater than or equal to 0 for this case)
f(x+1) = [max (x+1, 0)] = x+1 (Because x+1 will always be greater than or equal to 0 for this case)
Thus, our equation becomes => (x-3) + 2(x+1) = 8 => 3x-1 = 8 => x = 3.
And, since we assumed that x is greater than or equal to 3, x = 3 satisfies this equation.
Case-2: $$-1\le x<3$$
f(x-3) = [max (x-3, 0)] = 0 (Because x-3 will always be less than 0 for this case, and thus the max value will be 0)
f(x+1) = [max (x+1, 0)] = x+1 (Because x+1 will always be greater than or equal to 0 for this case)
Thus, our equation becomes => 0 + 2(x+1) = 8 => x + 1 = 4 => x = 3.
But, since we assumed that x is less than 3, x = 3 does not satisfy this equation.
Case-3: $$x\le-1$$
f(x-3) = [max (x-3, 0)] = 0 (Because x-3 will always be less than 0 for this case, and thus the max value will be 0)
f(x+1) = [max (x+1, 0)] = 0 (Because x+1 will always be less than 0 for this case)
Thus, our equation becomes => 0 + 0 = 8. But zero is never equal to 8 in any case, thus this does not satisfy the equation.
After all three cases, we have only one value of x, x = 3, which satisfies the equation. Therefore $$x=3$$.
Create a FREE account and get: