Home
Class 12
MATHS
If the equation |2 - x| - |x + 1|= k has...

If the equation `|2 - x| - |x + 1|= k` has exactly one solution, then number of integral values of `k` is (A) 7 (B) 5 (C) 4 (D) 3

Promotional Banner

Similar Questions

Explore conceptually related problems

If the equation ln (x^(2) +5x ) -ln (x+a +3)=0 has exactly one solution for x, then possible integral value of a is:

If the equation ln (x^(2) +5x ) -ln (x+a +3)=0 has exactly one solution for x, then possible integral value of a is:

If the equation ln (x^(2) +5x ) -ln (x+a +3)=0 has exactly one solution for x, then possible integral value of a is:

The equation |x+2|-|x+1|+|x1|=K,x in R has one solution,then find value of K .

Let f(x)=x+2|x+1|+x-1| . If f(x)=k has exactly one real solution, then the value of k is (a) 3 (b) 0 (c) 1 (d) 2

Let f(x)=x+2|x+1|+x-1|dotIff(x)=k has exactly one real solution, then the value of k is 3 (b) 0 (c) 1 (d) 2

Let f(x)=x+2|x+1|+2|x-1|dot If f(x)=k has exactly one real solution, then the value of k is (a) 3 (b) 0 (c) 1 (d) 2

Let f(x)=x+2|x+1|+2|x-1|dot If f(x)=k has exactly one real solution, then the value of k is (a) 3 (b) 0 (c) 1 (d) 2

Let f(x)=x+2|x+1|+2|x-1|dot If f(x)=k has exactly one real solution, then the value of k is (a) 3 (b) 0 (c) 1 (d) 2

Let f(x)=x+2|x+1|+2|x-1| .If f(x)=k has exactly one real solution, then the value of k is (a) 3 (b)0 (c) 1 (d) 2