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Matching list on 1^infinity and infinity^0 limits, roots and |x(x-1)(x-2)f(x)| | JEE Advanced

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Question
Match the following List-I with List-II.
List-I
IIf lim⁡x→0(sin⁡xx)11−cos⁡x\displaystyle\lim_{x\to0}\left(\frac{\sin x}{x}\right)^{\frac{1}{1-\cos x}} is LL then ln⁡(L)\ln\left(L\right) equals kk. The value of 1k2\dfrac{1}{k^2} is equal to
II The value of lim⁡n→∞(n2n−1)tan⁡1n\displaystyle\lim_{n\to\infty}\left(\frac{n^2}{n-1}\right)^{\tan\frac{1}{\sqrt n}} is
IIINumber of solution of the equation 8sin⁡4x+8cos⁡4x=58\sin^4x+8\cos^4x = 5 in the interval 0<x<2π0<x<2\pi is
IVLet f(x)f\left(x\right) be a non-constant polynomial function and g(x)=∣x(x−1)(x−2)f(x)∣g\left(x\right) = \left|x\left(x-1\right)\left(x-2\right)f\left(x\right)\right|. If g(x)g\left(x\right) is differentiable ∀ x∈R\forall\, x \in \mathbb{R}, then minimum number of distinct roots of f(x)=0f\left(x\right) = 0 is
List-II
P00
Q 11
R 33
S 88
T 99
AI →\to Q
BII →\to S
CIII →\to P
DIV →\to Rcorrect
Solution
Step 1: For (I), a 1∞1^\infty form.
k=ln⁡L=lim⁡x→0ln⁡(sin⁡xx)1−cos⁡xk = \ln L = \lim_{x\to0}\frac{\ln\left(\dfrac{\sin x}{x}\right)}{1-\cos x}
 sin⁡xx=1−x26+O(x4)\dfrac{\sin x}{x} = 1-\dfrac{x^2}{6}+O\left(x^4\right), so ln⁡sin⁡xx=−x26+O(x4)\ln\dfrac{\sin x}{x} = -\dfrac{x^2}{6}+O\left(x^4\right), while 1−cos⁡x=x22+O(x4)1-\cos x = \dfrac{x^2}{2}+O\left(x^4\right).
k=lim⁡x→0−x2/6x2/2=−13,1k2=9k = \lim_{x\to0}\frac{-x^2/6}{x^2/2} = -\frac13 ,\qquad \frac{1}{k^2} = 9
(I)→(T)\text{(I)}\to\textbf{(T)}
 Step 2: For (II), an ∞0\infty^0 form. Let LL be this limit.
ln⁡L=lim⁡n→∞tan⁡1n⋅ln⁡n2n−1\ln L = \lim_{n\to\infty}\tan\frac{1}{\sqrt n}\cdot\ln\frac{n^2}{n-1}
 As n→∞n\to\infty, tan⁡1n∼1n\tan\dfrac{1}{\sqrt n}\sim\dfrac{1}{\sqrt n} and ln⁡n2n−1=2ln⁡n−ln⁡(n−1)∼ln⁡n\ln\dfrac{n^2}{n-1} = 2\ln n-\ln\left(n-1\right)\sim\ln n, so the product behaves like ln⁡nn→0\dfrac{\ln n}{\sqrt n}\to0, since ln⁡n\ln n grows slower than any positive power of nn. Therefore
L=e0=1,(II)→(Q)L = e^0 = 1 ,\qquad \text{(II)}\to\textbf{(Q)}
 Step 3: For (III),
sin⁡4x+cos⁡4x=1−sin⁡22x2⟹8−4sin⁡22x=5⟹sin⁡22x=34\sin^4x+\cos^4x = 1-\frac{\sin^2 2x}{2} \quad\Longrightarrow\quad 8-4\sin^2 2x = 5 \quad\Longrightarrow\quad \sin^2 2x = \frac34
sin⁡2x=±32\sin 2x = \pm\frac{\sqrt3}{2}
 0<x<2π0<x<2\pi gives 2x∈(0,4π)2x \in \left(0,4\pi\right), two full periods. In one period sin⁡θ=±32\sin\theta = \pm\dfrac{\sqrt3}{2} has the four solutions (π3,2π3,4π3,5π3)\left(\dfrac\pi3,\dfrac{2\pi}3,\dfrac{4\pi}3,\dfrac{5\pi}3\right), none of them an end point of the interval, so there are 88 values of 2x2x, and x=θ/2x = \theta/2 being one-one, 88 values of xx.
(III)→(S)\text{(III)}\to\textbf{(S)}
 Step 4: For (IV), put h(x)=x(x−1)(x−2)f(x)h\left(x\right) = x\left(x-1\right)\left(x-2\right)f\left(x\right). ∣h∣\left|h\right| fails to be differentiable at a **simple** zero of hh, the graph having a corner there, and is differentiable at a zero of even order, so every root of hh must have even multiplicity. The cubic contributes multiplicity 11 at each of x=0,1,2x = 0,1,2, so ff must vanish at 00, 11 and 22 to an odd order, at least once each. Hence ff has at least the three distinct roots 0,1,20,1,2, and f(x)=x(x−1)(x−2)f\left(x\right) = x\left(x-1\right)\left(x-2\right) achieves exactly three: then h=[x(x−1)(x−2)]2≥0h = \left[x\left(x-1\right)\left(x-2\right)\right]^2 \ge 0 and ∣h∣=h\left|h\right| = h is a polynomial, certainly differentiable.
(IV)→(R)\text{(IV)}\to\textbf{(R)}
 Step 5:
I→T,II→Q,III→S,IV→R\text{I}\to\text{T},\qquad \text{II}\to\text{Q},\qquad \text{III}\to\text{S},\qquad \text{IV}\to\text{R}
 (1) says I →\to Q, (2) says II →\to S and (3) says III →\to P, all three false; (4) says IV →\to R, which is correct. Answer: (4)
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