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Limits Continuity and Differentiability question

2024 · 1 Feb · Shift 2 · Q41
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  5. /2024 · 1 Feb · Shift 2 · Q41

Limits Continuity and Differentiability question

2024 · 1 Feb · Shift 2 · Q41

JEE MainMathematicsLimits Continuity and DifferentiabilityMCQ+4 / −1
Let f(x)=∣2x2+5∣x∣−3∣,x∈Rf(x)=\left|2 x^2+5\right| x|-3|, x \in \mathbf{R}f(x)=​2x2+5​x∣−3∣,x∈R. If m\mathrm{m}m and n\mathrm{n}n denote the number of points where fff is not continuous and not differentiable respectively, then m+n\mathrm{m}+\mathrm{n}m+n is equal to :
  1. A
    5
  2. B
    3
  3. C
    2
  4. D
    0
View written solutionFree

Correct answer: B

We interpret the function as f(x)=∣2x2+5∣x∣−3∣,x∈R.f(x)=\left|2x^2+5|x|-3\right|, \qquad x\in \mathbb R.f(x)=​2x2+5∣x∣−3​,x∈R.

We need:

  1. the number of points where fff is not continuous = mmm
  2. the number of points where fff is not differentiable = nnn
  3. then compute m+nm+nm+n.

1. Check continuity

The inner expression is g(x)=2x2+5∣x∣−3.g(x)=2x^2+5|x|-3.g(x)=2x2+5∣x∣−3.

Now:

  • x2x^2x2 is continuous for all xxx
  • ∣x∣|x|∣x∣ is continuous for all xxx
  • hence g(x)g(x)g(x) is continuous for all xxx
  • and f(x)=∣g(x)∣f(x)=|g(x)|f(x)=∣g(x)∣ is also continuous for all xxx

Therefore, there are no points of discontinuity.

So, m=0.m=0.m=0.


2. Check differentiability

A modulus function ∣g(x)∣|g(x)|∣g(x)∣ can fail to be differentiable at points where:

  1. g(x)=0g(x)=0g(x)=0 and g′(x)≠0g'(x)\neq 0g′(x)=0, or
  2. g(x)g(x)g(x) itself is not differentiable.

So first study g(x)=2x2+5∣x∣−3.g(x)=2x^2+5|x|-3.g(x)=2x2+5∣x∣−3.

2.1 Where is g(x)g(x)g(x) not differentiable?

Since ∣x∣|x|∣x∣ is not differentiable at x=0x=0x=0, g(x)g(x)g(x) is also not differentiable at x=0.x=0.x=0.

Check whether this causes non-differentiability of fff: g(0)=2⋅02+5∣0∣−3=−3≠0.g(0)=2\cdot 0^2+5|0|-3=-3\neq 0.g(0)=2⋅02+5∣0∣−3=−3=0.

Near x=0x=0x=0, g(x)g(x)g(x) remains negative (by continuity), so f(x)=∣g(x)∣=−g(x)f(x)=|g(x)|=-g(x)f(x)=∣g(x)∣=−g(x) in a neighborhood of 000. Since ggg is not differentiable at 000, fff is also not differentiable at x=0.x=0.x=0.

So one non-differentiable point is x=0x=0x=0.


2.2 Find points where g(x)=0g(x)=0g(x)=0

Solve 2x2+5∣x∣−3=0.2x^2+5|x|-3=0.2x2+5∣x∣−3=0.

Let t=∣x∣≥0.t=|x| \ge 0.t=∣x∣≥0. Then 2t2+5t−3=0.2t^2+5t-3=0.2t2+5t−3=0.

Factor: 2t2+5t−3=(2t−1)(t+3)=0.2t^2+5t-3=(2t-1)(t+3)=0.2t2+5t−3=(2t−1)(t+3)=0.

So t=12ort=−3.t=\frac12 \quad \text{or} \quad t=-3.t=21​ort=−3.

Since t≥0t\ge 0t≥0, only t=12t=\frac12t=21​ is valid. Hence ∣x∣=12  ⟹  x=±12.|x|=\frac12 \implies x=\pm \frac12.∣x∣=21​⟹x=±21​.


2.3 Differentiability of fff at x=±12x=\pm \frac12x=±21​

For x≠0x\neq 0x=0, ggg is differentiable.

For x>0x>0x>0, g(x)=2x2+5x−3,g′(x)=4x+5.g(x)=2x^2+5x-3, \qquad g'(x)=4x+5.g(x)=2x2+5x−3,g′(x)=4x+5. At x=12x=\frac12x=21​, g′(12)=4⋅12+5=7≠0.g'\left(\frac12\right)=4\cdot \frac12+5=7\neq 0.g′(21​)=4⋅21​+5=7=0. So ∣g(x)∣|g(x)|∣g(x)∣ has a sharp corner at x=12x=\frac12x=21​. Hence fff is not differentiable at x=12x=\frac12x=21​.

For x<0x<0x<0, g(x)=2x2−5x−3,g′(x)=4x−5.g(x)=2x^2-5x-3, \qquad g'(x)=4x-5.g(x)=2x2−5x−3,g′(x)=4x−5. At x=−12x=-\frac12x=−21​, g′(−12)=4⋅(−12)−5=−7≠0.g'\left(-\frac12\right)=4\cdot \left(-\frac12\right)-5=-7\neq 0.g′(−21​)=4⋅(−21​)−5=−7=0. So fff is not differentiable at x=−12x=-\frac12x=−21​.

Thus additional non-differentiable points are x=±12.x=\pm \frac12.x=±21​.

So total number of non-differentiable points is n=3.n=3.n=3.


3. Compute m+nm+nm+n

m+n=0+3=3.m+n=0+3=3.m+n=0+3=3.


4. Compare with stored answer

Derived answer: 3

Stored correct answer: B = 3

They match.

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