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Limits Continuity and Differentiability question

2021 · 25 Feb · Shift 1 · Q43
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Limits Continuity and Differentiability question

2021 · 25 Feb · Shift 1 · Q43

JEE MainMathematicsLimits Continuity and DifferentiabilityNumerical+4 / −1
The number of points, at which the function f(x) = | 2x + 1 | −-− 3| x + 2 | + | x2 + x −-− 2 |, x ∈\in∈ R is not differentiable, is ‾\underline{\hspace{2cm}}​.
Numerical answer
View written solutionFree

Correct answer: 2

  1. We need the number of points where f(x)=∣2x+1∣−3∣x+2∣+∣x2+x−2∣f(x)=|2x+1|-3|x+2|+|x^2+x-2|f(x)=∣2x+1∣−3∣x+2∣+∣x2+x−2∣ is not differentiable.

  2. A sum of absolute value functions is non-differentiable only at points where the expression inside some modulus becomes zero.

So we check the zeros of:

  • 2x+1=0⇒x=−122x+1=0 \Rightarrow x=-\frac122x+1=0⇒x=−21​
  • x+2=0⇒x=−2x+2=0 \Rightarrow x=-2x+2=0⇒x=−2
  • x2+x−2=0x^2+x-2=0x2+x−2=0

Factor the quadratic: x2+x−2=(x+2)(x−1)x^2+x-2=(x+2)(x-1)x2+x−2=(x+2)(x−1) So its zeros are: x=−2,  1x=-2,\;1x=−2,1

Hence the only candidate points are: x=−2, −12, 1x=-2,\,-\frac12,\,1x=−2,−21​,1

  1. Now test differentiability at each candidate.

At x=−12x=-\frac12x=−21​

Only ∣2x+1∣|2x+1|∣2x+1∣ is problematic there; the other two terms are differentiable since their inside expressions are nonzero at x=−12x=-\frac12x=−21​.

Since ∣2x+1∣|2x+1|∣2x+1∣ has a cusp at 2x+1=02x+1=02x+1=0, f(x)f(x)f(x) is not differentiable at x=−12.x=-\frac12.x=−21​.

At x=1x=1x=1

Only ∣x2+x−2∣|x^2+x-2|∣x2+x−2∣ is problematic there, because x2+x−2=(x−1)(x+2)x^2+x-2=(x-1)(x+2)x2+x−2=(x−1)(x+2) changes sign at x=1x=1x=1 and has a simple root there. Thus ∣x2+x−2∣|x^2+x-2|∣x2+x−2∣ is not differentiable at x=1x=1x=1. So f(x)f(x)f(x) is not differentiable at x=1.x=1.x=1.

At x=−2x=-2x=−2

Here both ∣x+2∣|x+2|∣x+2∣ and ∣x2+x−2∣|x^2+x-2|∣x2+x−2∣ may create trouble. So we simplify carefully.

Since x2+x−2=(x+2)(x−1),x^2+x-2=(x+2)(x-1),x2+x−2=(x+2)(x−1), we get ∣x2+x−2∣=∣x+2∣ ∣x−1∣.|x^2+x-2|=|x+2|\,|x-1|.∣x2+x−2∣=∣x+2∣∣x−1∣. Near x=−2x=-2x=−2, we have x−1<0x-1<0x−1<0, so ∣x−1∣=1−x|x-1|=1-x∣x−1∣=1−x in a neighborhood of −2-2−2. Thus near x=−2x=-2x=−2, ∣x2+x−2∣=(1−x)∣x+2∣.|x^2+x-2|=(1-x)|x+2|.∣x2+x−2∣=(1−x)∣x+2∣.

Therefore near x=−2x=-2x=−2,

=|2x+1|-(x+2)|x+2|.$$ Wait, combine carefully: $$-3|x+2|+(1-x)|x+2| = (-3+1-x)|x+2|=-(x+2)|x+2|.$$ So $$f(x)=|2x+1|-(x+2)|x+2|.$$ Now at $x=-2$, note $2x+1=-3\neq 0$, so $|2x+1|$ is differentiable there. Also, $$g(x)=(x+2)|x+2|$$ **is differentiable** at $x=-2$ (in fact, writing $t=x+2$, $g=t|t|$, whose derivative at $t=0$ exists and equals $0$). Hence $f(x)$ is differentiable at $x=-2$. We can also verify by piecewise form: - For $x>-2$, $|x+2|=x+2$, so $-(x+2)|x+2|=-(x+2)^2$ - For $x<-2$, $|x+2|=-(x+2)$, so $-(x+2)|x+2|=(x+2)^2$ Both side derivatives at $x=-2$ are $0$. So $f(x)$ **is differentiable** at $$x=-2.$$ 4. Therefore the non-differentiable points are exactly: $$x=-\frac12,\;1$$ So the number of such points is $$\boxed{2}.$$ 5. Comparison with stored answer: Stored correct answer = $2$. Our derived answer also = $2$. So they agree.
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