Sign in
12thPass logo
New chatPYQ LibraryDoubtsRank report
Sign in to see Recents

Your guest activity stays on this device

Sign in to save progress →
Sign in

Application of Derivatives question

2021 · 17 Mar · Shift 2 · Q30
Guest · filters and generic practice availableBrowsing as a guest · PYQ filters and generic practice are available. Sign in only for personalised features and saved progress.
  1. PYQ Library
  2. /JEE Main
  3. /Mathematics
  4. /Application of Derivatives
  5. /2021 · 17 Mar · Shift 2 · Q30

Application of Derivatives question

2021 · 17 Mar · Shift 2 · Q30

JEE MainMathematicsApplication of DerivativesMCQ+4 / −1
Consider the function f:R→Rf : \mathbb{R} \to \mathbb{R}f:R→R defined by f(x)={(2−sin⁡(1x))∣x∣x≠00x=0f(x) = \left\{ \begin{array}{ll} \left(2 - \sin\left(\dfrac{1}{x}\right)\right)|x| & x \ne 0 \\ 0 & x = 0 \end{array} \right.f(x)=⎩⎨⎧​(2−sin(x1​))∣x∣0​x=0x=0​ Then fff is :
  1. A
    not monotonic on (−∞-\infty−∞, 0) and (0, ∞\infty∞)
  2. B
    monotonic on (0, ∞\infty∞) only
  3. C
    monotonic on (−∞-\infty−∞, 0) only
  4. D
    monotonic on (−∞-\infty−∞, 0) ∪\cup∪(0, ∞\infty∞)
View written solutionFree

Correct answer: A

  1. Given function

    \begin{cases} \left(2-\sin\left(\dfrac1x\right)\right)|x|, & x\neq 0,\\[4pt] 0,& x=0 \end{cases}$$ We need to check whether $f$ is monotonic on: - $(-\infty,0)$ - $(0,\infty)$

  2. Write separately on the two intervals

    Since ∣x∣=x|x|=x∣x∣=x for x>0x>0x>0 and ∣x∣=−x|x|=-x∣x∣=−x for x<0x<0x<0:

    • For x>0x>0x>0, f(x)=x(2−sin⁡(1x))f(x)=x\left(2-\sin\left(\frac1x\right)\right)f(x)=x(2−sin(x1​))
    • For x<0x<0x<0, f(x)=(−x)(2−sin⁡(1x))f(x)=(-x)\left(2-\sin\left(\frac1x\right)\right)f(x)=(−x)(2−sin(x1​))

  3. Check monotonicity on (0,∞)(0,\infty)(0,∞)

    For x>0x>0x>0, differentiate:

    f(x)=2x−xsin⁡(1x)f(x)=2x-x\sin\left(\frac1x\right)f(x)=2x−xsin(x1​)

    Now,

    =\sin\left(\frac1x\right)+x\cos\left(\frac1x\right)\left(-\frac1{x^2}\right) =\sin\left(\frac1x\right)-\frac{\cos(1/x)}{x}$$ Hence, $$f'(x)=2-\left(\sin\left(\frac1x\right)-\frac{\cos(1/x)}{x}\right) =2-\sin\left(\frac1x\right)+\frac{\cos(1/x)}{x}$$ So, $$f'(x)=2-\sin\left(\frac1x\right)+\frac{\cos(1/x)}{x}$$ The term $\dfrac{\cos(1/x)}{x}$ oscillates and becomes arbitrarily large positive and negative as $x\to 0^+$. Therefore $f'(x)$ changes sign infinitely often near $0$. Let us verify using specific sequences: - Take $$x_n=\frac{1}{2n\pi}$$ then $$\cos\left(\frac1{x_n}\right)=\cos(2n\pi)=1, \quad \sin\left(\frac1{x_n}\right)=0$$ so $$f'(x_n)=2+\frac1{x_n}>0$$ - Take $$y_n=\frac{1}{(2n+1)\pi}$$ then $$\cos\left(\frac1{y_n}\right)=\cos((2n+1)\pi)=-1, \quad \sin\left(\frac1{y_n}\right)=0$$ so $$f'(y_n)=2-\frac1{y_n}<0$$ for large $n$. Thus $f'(x)$ takes both positive and negative values on $(0,\infty)$, so $f$ is **not monotonic** on $(0,\infty)$.

  4. Check monotonicity on (−∞,0)(-\infty,0)(−∞,0)

    For x<0x<0x<0, f(x)=−x(2−sin⁡(1x))=−2x+xsin⁡(1x)f(x)=-x\left(2-\sin\left(\frac1x\right)\right)=-2x+x\sin\left(\frac1x\right)f(x)=−x(2−sin(x1​))=−2x+xsin(x1​)

    Differentiate:

    f′(x)=−2+ddx[xsin⁡(1x)]f'(x)=-2+\frac{d}{dx}\left[x\sin\left(\frac1x\right)\right]f′(x)=−2+dxd​[xsin(x1​)]

    Using the derivative already computed, ddx[xsin⁡(1x)]=sin⁡(1x)−cos⁡(1/x)x\frac{d}{dx}\left[x\sin\left(\frac1x\right)\right]=\sin\left(\frac1x\right)-\frac{\cos(1/x)}{x}dxd​[xsin(x1​)]=sin(x1​)−xcos(1/x)​

    Therefore, f′(x)=−2+sin⁡(1x)−cos⁡(1/x)xf'(x)=-2+\sin\left(\frac1x\right)-\frac{\cos(1/x)}{x}f′(x)=−2+sin(x1​)−xcos(1/x)​

    Again, the term −cos⁡(1/x)x-\dfrac{\cos(1/x)}{x}−xcos(1/x)​ oscillates and changes sign near 0−0^-0−. Let us take sequences:

    • Let xn=−12nπx_n=-\frac{1}{2n\pi}xn​=−2nπ1​ then \quad \cos\left(\frac1{x_n}\right)=1$$ so $$f'(x_n)=-2-\frac{1}{x_n}=-2+2n\pi>0$$ for large $n$ because $x_n<0$.
    • Let yn=−1(2n+1)πy_n=-\frac{1}{(2n+1)\pi}yn​=−(2n+1)π1​ then \quad \cos\left(\frac1{y_n}\right)=-1$$ so $$f'(y_n)=-2-\frac{-1}{y_n}=-2+\frac{1}{y_n}<0$$ for large $n$, since $y_n<0$.

    Thus f′(x)f'(x)f′(x) also changes sign on (−∞,0)(-\infty,0)(−∞,0), so fff is not monotonic on (−∞,0)(-\infty,0)(−∞,0).

  5. Conclusion

    The function is not monotonic on either interval (−∞,0)(-\infty,0)(−∞,0) and (0,∞)(0,\infty)(0,∞).

    Therefore the correct option is: A\boxed{\text{A}}A​

PreviousNext

More from Application of Derivatives

  • Let f : [− 1, 1] → R be defined as f(x) = ax2 + bx + c for all x ∈[− 1, 1], where a, b, c ∈ R such that f(− 1) = 2, f'(− 1) = 1 for x ∈(− 1, 1) the maximum value of f ''(x) is 21​. If f(x) $\le…2021 · Numerical
  • Let A=[aij​] be a 3 × 3 matrix, where aij​=⎩⎨⎧​1−x2x+1​,,,​ifi=jif∣i−j∣=1otherwise.​…2021 · MCQ
  • Let 'a' be a real number such that the function f(x) = ax2 + 6x − 15, x ∈ R is increasing in (−∞,43​) and decreasing in (43​,∞). Then the function g(x) = ax2 − 6x…2021 · MCQ
  • The sum of all the local minimum values of the twice differentiable function f : R → R defined by f(x)=x3−3x2−23f′′(2)​x+f′′(1) is :2021 · MCQ
  • Let f : R → R be defined as f(x)={−34​x3+2x2+3x,3xex,​x>0x≤0​. Then f is increasing function in the interval2021 · MCQ
  • The function f(x) = 64x3−3x2​−2sinx+(2x−1)cosx :2021 · MCQ
  • The minimum value of α for which the equation sinx4​+1−sinx1​=α has at least one solution in (0,2π​) is .......2021 · Numerical
  • Let f:R→R be defined as f(x)=⎩⎨⎧​−55x,2x3−3x2−120x,2x3−3x2−36x−336,​ifx<−5if−5≤x≤4ifx>4,​ Let…2021 · MCQ