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NEET-UG Biology

Derivatives of order up to two — practice questions

18 questions in the bank on this idea. Below are 10 of them, exactly as they appear in a test.

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  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), xRx \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)=sinxcosxsinx+cosx+1272827111,xR, 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:RRf: \mathbf{R} \rightarrow \mathbf{R} be a twice differentiable function such that (sinxcosy)(f(2x+2y)f(2x2y))=(cosxsiny)(f(2x+2y)+f(2x2y))(\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,yRx, 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),x00,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)=24π22πf^{\prime \prime}\left(\frac{2}{\pi}\right)=\frac{24-\pi^2}{2 \pi}
    • D. f(2π)=12π22πf^{\prime \prime}\left(\frac{2}{\pi}\right)=\frac{12-\pi^2}{2 \pi}
  5. Question 5 · difficulty L3 · understanding

    If y(θ)=2cosθ+cos2θcos3θ+4cos2θ+5cosθ+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)=2cos4x2sin4x3+sin22x3+2cos4x2sin4xsin22x2cos4x3+2sin4xsin22x,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+tan1xf(x) = 2x + {\tan ^{ - 1}}x and g(x)=loge(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. minf(x)=1+maxg(x)\min f'(x) = 1 + \max g'(x)
    • D. maxf(x)>maxg(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)<61<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 yyy' - 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)=x3x2f(1)+xf(2)f(3),xRf(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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