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JEE Main Mathematics · Class 12

Derivatives of order up to two — NEET & JEE Questions

Derivatives of order up to two is a topic in Limit, Continuity and Differentiability, a Class 12 Mathematics chapter of the JEE Main syllabus. TestBudy holds 18 published JEE Main questions on Derivatives of order up to two, each with the marking scheme of the real paper.

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Derivatives of order up to two MCQs with solutions

10 of the 18 published questions on this topic, exactly as they appear in a test. Answers and worked solutions come with your result rather than on this page.

  1. Question 1 · difficulty L3 · understanding

    Let f(x)=x3+x2f′(1)+2xf′′(2)+f′′′(3)f(x) = x^3 + x^2 f'(1) + 2x f''(2) + f'''(3), x∈Rx \in \mathbb{R}. Then the value of f′(5)f'(5) is :

    • A. 6575\dfrac{657}{5}
    • B. 1175\dfrac{117}{5}
    • C. 25\dfrac{2}{5}
    • D. 625\dfrac{62}{5}
  2. Question 2 · difficulty L3 · understanding

     If y(x)=∣sin⁡xcos⁡xsin⁡x+cos⁡x+1272827111∣,x∈R, then d2ydx2+y is equal to  \text { If } y(x)=\left|\begin{array}{ccc} \sin x & \cos x & \sin x+\cos x+1 \\ 27 & 28 & 27 \\ 1 & 1 & 1 \end{array}\right|, x \in \mathbb{R} \text {, then } \frac{d^2 y}{d x^2}+y \text { is equal to }

    • A. 28
    • B. 27
    • C. -1
    • D. 1
  3. Question 3 · difficulty L3 · understanding

    Let f:R→Rf: \mathbf{R} \rightarrow \mathbf{R} be a twice differentiable function such that (sin⁡xcos⁡y)(f(2x+2y)−f(2x−2y))=(cos⁡xsin⁡y)(f(2x+2y)+f(2x−2y))(\sin x \cos y)(f(2 x+2 y)-f(2 x-2 y))=(\cos x \sin y)(f(2 x+2 y)+f(2 x-2 y)), for all x,y∈Rx, y \in \mathbf{R}. If f′(0)=12f^{\prime}(0)=\frac{1}{2}, then the value of 24f′′(5π3)24 f^{\prime \prime}\left(\frac{5 \pi}{3}\right) is :

    • A. 2
    • B. 3
    • C. −-3
    • D. −-2
  4. Question 4 · difficulty L3 · understanding

     If f(x)={x3sin⁡(1x),x≠00,x=0, then \text { If } f(x)=\left\{\begin{array}{ll} x^3 \sin \left(\frac{1}{x}\right), & x \neq 0 \\ 0 & , x=0 \end{array}\right. \text {, then }

    • A. f′′(0)=0f^{\prime \prime}(0)=0
    • B. f′′(0)=1f^{\prime \prime}(0)=1
    • C. f′′(2π)=24−π22πf^{\prime \prime}\left(\frac{2}{\pi}\right)=\frac{24-\pi^2}{2 \pi}
    • D. f′′(2π)=12−π22πf^{\prime \prime}\left(\frac{2}{\pi}\right)=\frac{12-\pi^2}{2 \pi}
  5. Question 5 · difficulty L3 · understanding

    If y(θ)=2cos⁡θ+cos⁡2θcos⁡3θ+4cos⁡2θ+5cos⁡θ+2y(\theta)=\frac{2 \cos \theta+\cos 2 \theta}{\cos 3 \theta+4 \cos 2 \theta+5 \cos \theta+2}, then at θ=π2,y′′+y′+y\theta=\frac{\pi}{2}, y^{\prime \prime}+y^{\prime}+y is equal to :

    • A. 12\frac{1}{2}
    • B. 1
    • C. 32\frac{3}{2}
    • D. 2
  6. Question 6 · difficulty L3 · understanding

    If f(x)=∣2cos⁡4x2sin⁡4x3+sin⁡22x3+2cos⁡4x2sin⁡4xsin⁡22x2cos⁡4x3+2sin⁡4xsin⁡22x∣,f(x)=\left|\begin{array}{ccc} 2 \cos ^4 x & 2 \sin ^4 x & 3+\sin ^2 2 x \\ 3+2 \cos ^4 x & 2 \sin ^4 x & \sin ^2 2 x \\ 2 \cos ^4 x & 3+2 \sin ^4 x & \sin ^2 2 x \end{array}\right|, then 15f′(0)=\frac{1}{5} f^{\prime}(0)= is equal to :

    • A. 2
    • B. 1
    • C. 0
    • D. 6
  7. Question 7 · difficulty L3 · understanding

    Let f(x)=2x+tan⁡−1xf(x) = 2x + {\tan ^{ - 1}}x and g(x)=log⁡e(1+x2+x),x∈[0,3]g(x) = {\log _e}(\sqrt {1 + {x^2}} + x),x \in [0,3]. Then

    • A. there exists x^∈[0,3]\widehat x \in [0,3] such that f′(x^)<g′(x^)f'(\widehat x) < g'(\widehat x)
    • B. there exist 0<x1<x2<30 < {x_1} < {x_2} < 3 such that f(x)<g(x),∀x∈(x1,x2)f(x) < g(x),\forall x \in ({x_1},{x_2})
    • C. min⁡f′(x)=1+max⁡g′(x)\min f'(x) = 1 + \max g'(x)
    • D. max⁡f(x)>max⁡g(x)\max f(x) > \max g(x)
  8. Question 8 · difficulty L3 · understanding

    Let ff and gg be the twice differentiable functions on R\mathbb{R} such that f′′(x)=g′′(x)+6xf''(x)=g''(x)+6x f′(1)=4g′(1)−3=9f'(1)=4g'(1)-3=9 f(2)=3g(2)=12f(2)=3g(2)=12. Then which of the following is NOT true?

    • A. g(−2)−f(−2)=20g(-2)-f(-2)=20
    • B. There exists x0∈(1,3/2)x_0\in(1,3/2) such that f(x0)=g(x0)f(x_0)=g(x_0)
    • C. ∣f′(x)−g′(x)∣<6⇒−1<x<1|f'(x)-g'(x)| < 6\Rightarrow -1 < x < 1
    • D. If −1<x<2-1 < x < 2, then ∣f(x)−g(x)∣<8|f(x)-g(x)| < 8
  9. Question 9 · difficulty L3 · understanding

    Let y(x)=(1+x)(1+x2)(1+x4)(1+x8)(1+x16)y(x) = (1 + x)(1 + {x^2})(1 + {x^4})(1 + {x^8})(1 + {x^{16}}). Then y′−y′′y' - y'' at x=−1x = - 1 is equal to

    • A. 496
    • B. 976
    • C. 464
    • D. 944
  10. Question 10 · difficulty L3 · understanding

    If f(x)=x3−x2f′(1)+xf′′(2)−f′′′(3),x∈Rf(x) = {x^3} - {x^2}f'(1) + xf''(2) - f'''(3),x \in \mathbb{R}, then

    • A. 2f(0)−f(1)+f(3)=f(2)2f(0) - f(1) + f(3) = f(2)
    • B. f(1)+f(2)+f(3)=f(0)f(1) + f(2) + f(3) = f(0)
    • C. f(3)−f(2)=f(1)f(3) - f(2) = f(1)
    • D. 3f(1)+f(2)=f(3)3f(1) + f(2) = f(3)

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