Application of DerivativeshardFree

Piecewise limit of [ln(3 + x^2) - x^(2m) sin(x^2)]/(1 + x^(2m)) | JEE Advanced

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Question
Let f(x)=lim⁡m→∞ln⁡(3+x2)−x2msin⁡(x2)1+x2mf(x) = \displaystyle\lim_{m\to\infty}\frac{\ln\left(3+x^2\right)-x^{2m}\sin\left(x^2\right)}{1+x^{2m}}, then which of the following is/are CORRECT?
Af(x)f(x) is discontinuous at two pointscorrect
BThere exists α\alpha such that f(α)=1f(\alpha) = 1correct
CThe equation f(x)=0f(x) = 0 has at least one root in (1,∞)\left(1,\infty\right)correct
DMinimum value of f(x)f(x) is equal to (−sin⁡1)\left(-\sin1\right)
Solution
Step 1: Everything depends on whether ∣x∣\left|x\right| is below, at, or above 11:
x2m⟶{0,∣x∣<11,∣x∣=1∞,∣x∣>1.x^{2m}\longrightarrow \begin{cases}0, & \left|x\right|<1\\ 1, & \left|x\right| = 1\\ \infty, & \left|x\right|>1 .\end{cases}
 Step 2: ∣x∣<1\left|x\right|<1: numerator →ln⁡(3+x2)\to \ln\left(3+x^2\right), denominator →1\to1, so
f(x)=ln⁡(3+x2)f(x) = \ln\left(3+x^2\right)
 ∣x∣=1\left|x\right| = 1: f(x)=ln⁡4−sin⁡12f(x) = \dfrac{\ln4-\sin1}2. ∣x∣>1\left|x\right|>1: dividing top and bottom by x2mx^{2m},
f(x)=lim⁡m→∞ln⁡(3+x2)x2m−sin⁡(x2)1x2m+1=−sin⁡(x2)f(x) = \lim_{m\to\infty}\frac{\dfrac{\ln\left(3+x^2\right)}{x^{2m}}-\sin\left(x^2\right)} {\dfrac1{x^{2m}}+1} = -\sin\left(x^2\right)
 So
f(x)={−sin⁡(x2),∣x∣>1ln⁡4−sin⁡12,∣x∣=1ln⁡(3+x2),∣x∣<1.f(x) = \begin{cases}-\sin\left(x^2\right), & \left|x\right|>1\\[4pt] \dfrac{\ln4-\sin1}2, & \left|x\right| = 1\\[4pt] \ln\left(3+x^2\right), & \left|x\right|<1 .\end{cases}
 Step 3: (1). At x=1x = 1,
lim⁡x→1−f=ln⁡4=1.3863,lim⁡x→1+f=−sin⁡1=−0.8415,f(1)=ln⁡4−sin⁡12=0.2724\lim_{x\to1^-}f = \ln4 = 1.3863 ,\qquad \lim_{x\to1^+}f = -\sin1 = -0.8415 ,\qquad f(1) = \frac{\ln4-\sin1}2 = 0.2724
 All three differ, so ff is discontinuous at x=1x = 1, and ff being even, also at x=−1x = -1. Everywhere else each branch is continuous. Exactly two points, so (1) is TRUE. Step 4: (2). On ∣x∣>1\left|x\right|>1, f(x)=−sin⁡(x2)f(x) = -\sin\left(x^2\right) takes every value in [−1,1]\left[-1,1\right]; in particular f(x)=1f(x) = 1 when x2=3π2x^2 = \dfrac{3\pi}2, i.e.
α=3π2≈2.1708>1\alpha = \sqrt{\frac{3\pi}2} \approx 2.1708 > 1
 (2) is TRUE. On ∣x∣<1\left|x\right|<1 the values are ln⁡(3+x2)∈[ln⁡3,ln⁡4)\ln\left(3+x^2\right) \in \left[\ln3,\ln4\right), all above 11 anyway. Step 5: (3).
−sin⁡(x2)=0⟺x2=π⟹x=π≈1.7725∈(1,∞)-\sin\left(x^2\right) = 0 \quad\Longleftrightarrow\quad x^2 = \pi \quad\Longrightarrow\quad x = \sqrt\pi \approx 1.7725 \in \left(1,\infty\right)
 (3) is TRUE. Step 6: (4). On ∣x∣>1\left|x\right|>1 the value −sin⁡(x2)=−1-\sin\left(x^2\right) = -1 is attained at x2=π2x^2 = \dfrac\pi2, i.e. x=π/2≈1.2533x = \sqrt{\pi/2} \approx 1.2533, which does satisfy ∣x∣>1\left|x\right|>1. On ∣x∣=1\left|x\right| = 1, f=0.2724f = 0.2724, and on ∣x∣<1\left|x\right|<1, f≥ln⁡3=1.0986f \ge \ln3 = 1.0986, both above −1-1. So the minimum of ff over R\mathbb R is
−1 ≠ −sin⁡1=−0.8415-1 \ \ne\ -\sin1 = -0.8415
 (4) is FALSE. Answer: (1), (2), (3)
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