The number of points, at which the function
f(x) = | 2x + 1 | - 3| x + 2 | + | x² + x - 2 |, x∈R is not differentiable, is __________.

Solution

Maths sir · 5 steps

List the three suspect points, but check x = −2 carefully: two kinks there cancel each other out.

1. Find suspect points. A modulus can create a corner only where its inside is zero. 2x+1=0 gives x=-1/2, x+2=0 gives x=-2, and x²+x-2=(x+2)(x-1)=0 gives x=-2 and x=1. So the suspects are -2, -1/2 and 1.
2. Check x = −2. Two terms vanish here, so combine them. Near x=-2 we have x-1<0, so |x²+x-2| = |x+2|(1-x). The other terms add up as shown below.

   -3|x+2| + |x+2|(1-x) = |x+2|(-2-x) = -(x+2) |x+2|

3. No corner there. With t=x+2, the function -t|t| has derivative 0 from both sides at t=0. |2x+1| is smooth near -2, so f is differentiable at x=-2.
4. The other two. At x=-1/2 only |2x+1| is zero, and its slope jumps from -2 to 2, so this is a corner. At x=1 only |x²+x-2|=(x+2)|x-1| is zero, and its slope jumps from -3 to 3, so this is also a corner.
5. So. f is not differentiable only at x=-1/2 and x=1, which is 2 points.

The common slip: 3. Counting every zero of every modulus, that is -2, -1/2 and 1, gives 3. The corners at x=-2 cancel, so that point does not count.

The idea to keep. When two modulus terms vanish at the same point, do not count it straight away. Factor out the common modulus and see whether the corners cancel.

Answer from the NTA final answer key. Solution by PrepWiser.