Continuity & DifferentiabilitymediumFree

For Which m Is x^m cos(x - 1/x) Continuous or Differentiable at 0 | JEE

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Question
If
f(x)={xmcos⁡(x−1x)x>00x=0f(x) = \begin{cases} x^{m}\cos\left(x-\dfrac1x\right) & x > 0 \\ 0 & x = 0 \end{cases}
 then
Afor all m∈(−∞,0]m \in (-\infty, 0], f(x)f(x) is discontinuous at x=0x = 0correct
Bfor all m∈(0,∞)m \in (0, \infty), f(x)f(x) is differentiable at x=0x = 0
Cfor all m∈(1,∞)m \in (1, \infty), f(x)f(x) is differentiable at x=0x = 0correct
Dfor all m∈(0,1]m \in (0, 1], f(x)f(x) is continuous but not differentiable at x=0x = 0correct
Solution
Step 1: The only concept wet needed the cosine factor is bounded.
∣cos⁡(x−1x)∣≤1for every x>0,\left|\cos\left(x-\frac1x\right)\right| \le 1 \quad \text{for every } x > 0 ,
 though it oscillates wildly as x→0+x \to 0^{+} and has no limit there. Step 2: Test continuity at 00. We need
lim⁡x→0+xmcos⁡(x−1x)=f(0)=0.\lim_{x\to0^{+}}x^{m}\cos\left(x-\frac1x\right) = f(0) = 0 .
 Step 3: By the squeeze theorem this holds precisely when xm→0x^{m} \to 0, i.e. when
m>0.m > 0 .
 If m=0m = 0 the expression is the non-convergent cos⁡(x−1x)\cos\left(x-\frac1x\right); if m<0m < 0 it is unbounded. Either way the limit fails. Step 4: So for m∈(−∞,0]m \in (-\infty, 0] the function is discontinuous at 00 - option (A) is true. Step 5: Test differentiability at 00 from the definition.
f′(0)=lim⁡h→0+hmcos⁡(h−1h)−0h=lim⁡h→0+hm−1cos⁡(h−1h).f'(0) = \lim_{h\to0^{+}}\frac{h^{m}\cos\left(h-\frac1h\right)-0}{h} = \lim_{h\to0^{+}}h^{m-1}\cos\left(h-\frac1h\right).
 Step 6: By the same squeeze argument, this limit exists (and equals 00) precisely when
m−1>0  ⟹  m>1.m-1 > 0 \implies m > 1 .
 Step 7: So option (C) is true, and option (B) is false - on (0,1](0,1] the function is continuous but the difference quotient does not settle down. Step 8: That last observation is exactly option (D): for m∈(0,1]m \in (0,1], ff is continuous (Step 3) but not differentiable (Step 6). True. Answer: (1), (3) and (4).
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