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Limits Continuity and Differentiability question

2024 · 1 Feb · Shift 2 · Q50
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Limits Continuity and Differentiability question

2024 · 1 Feb · Shift 2 · Q50

JEE MainMathematicsLimits Continuity and DifferentiabilityMCQ+4 / −1
Let f(x)={x−1,x is even, 2x,x is odd, x∈Nf(x)=\left\{\begin{array}{l}x-1, x \text { is even, } \\ 2 x, \quad x \text { is odd, }\end{array} x \in \mathbf{N}\right.f(x)={x−1,x is even, 2x,x is odd, ​x∈N. If for some a∈N,f(f(f(a)))=21\mathrm{a} \in \mathbf{N}, f(f(f(\mathrm{a})))=21a∈N,f(f(f(a)))=21, then lim⁡x→a−{∣x∣3a−[xa]}\lim\limits_{x \rightarrow \mathrm{a}^{-}}\left\{\frac{|x|^3}{\mathrm{a}}-\left[\frac{x}{\mathrm{a}}\right]\right\}x→a−lim​{a∣x∣3​−[ax​]}, where [t][t][t] denotes the greatest integer less than or equal to ttt, is equal to :
  1. A
    169
  2. B
    121
  3. C
    225
  4. D
    144
View written solutionFree

Correct answer: D

  1. Understand the function fff

    The function is defined on natural numbers as

    f(x)={x−1,x is even2x,x is oddf(x)= \begin{cases} x-1, & x \text{ is even} \\ 2x, & x \text{ is odd} \end{cases}f(x)={x−1,2x,​x is evenx is odd​

    We are given f(f(f(a)))=21f(f(f(a)))=21f(f(f(a)))=21 for some a∈Na\in \mathbb Na∈N.

  2. Work backwards from 212121

    We need all possible preimages of 212121 under fff.

    For f(n)=21f(n)=21f(n)=21:

    • If nnn is even, then f(n)=n−1=21⇒n=22f(n)=n-1=21 \Rightarrow n=22f(n)=n−1=21⇒n=22.
    • If nnn is odd, then f(n)=2n=21f(n)=2n=21f(n)=2n=21, impossible.

    So the only number mapping to 212121 is 222222.

    Thus, f(f(a))=22.f(f(a))=22.f(f(a))=22.

  3. Now solve f(n)=22f(n)=22f(n)=22

    • If nnn is even, then n−1=22⇒n=23n-1=22 \Rightarrow n=23n−1=22⇒n=23, but 232323 is not even, so impossible.
    • If nnn is odd, then 2n=22⇒n=112n=22 \Rightarrow n=112n=22⇒n=11, and 111111 is odd, valid.

    Hence, f(a)=11.f(a)=11.f(a)=11.

  4. Now solve f(a)=11f(a)=11f(a)=11

    • If aaa is even, then a−1=11⇒a=12a-1=11 \Rightarrow a=12a−1=11⇒a=12, valid since 121212 is even.
    • If aaa is odd, then 2a=112a=112a=11, impossible.

    Therefore, a=12.a=12.a=12.

  5. Evaluate the limit

    We need

    lim⁡x→a−{∣x∣3a−[xa]}\lim_{x\to a^-}\left\{\frac{|x|^3}{a}-\left[\frac xa\right]\right\}x→a−lim​{a∣x∣3​−[ax​]}

    with a=12a=12a=12.

    Since x→12−x\to 12^-x→12− and 12>012>012>0, for xxx near 121212 we have ∣x∣=x|x|=x∣x∣=x.

    Also, when x→12−x\to 12^-x→12−,

    x12→1−.\frac{x}{12} \to 1^-.12x​→1−.

    Therefore,

    [x12]=0\left[\frac{x}{12}\right]=0[12x​]=0

    for all x<12x<12x<12 sufficiently close to 121212.

    Hence the limit becomes

    lim⁡x→12−x312=12312=122=144.\lim_{x\to 12^-} \frac{x^3}{12} = \frac{12^3}{12}=12^2=144.x→12−lim​12x3​=12123​=122=144.
  6. Match with options

    144144144 corresponds to Option D.

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