Differentiation-Quiz-7

 

Differentiation Quiz

Dear Readers,

Important topics for Maths has been designed in such a way that it offers very practical and application-based learning to further make it easier for students to understand every concept or topic by correlating it with the day-to-day experiences.

Q1. 

 

•  y
•  y+x^n/n!
•  y-x^n/n!
•  y-1-x^n/n!

Solution

 

Q2.The derivative of tan^(-1)⁡((√(1+x^2)-1)/x) with respect to tan^(-1)⁡((2x√(1-x^2))/(1-2x^2)) at x=0 is

•  1/8
•  1/4
•  1/2
•  1

Solution

 

 

Q3.  If y=|cos⁡x|+|sin⁡x|, then dy/dx at x=2π/3 is

•   (1-√3)/2
•  0
•  1/2(√3-1)
•  None of these

Solution
(c) In neighbourhood of x=2π/3,|cos⁡x |=-cos⁡x and |sin⁡x | =sin⁡x ⇒y=-cos⁡x + sin⁡x ⇒dy/dx=sin⁡x + cos⁡x ⇒ At x=2π/3,dy/dx=sin⁡2π/3 + cos⁡2π/3=(√3-1)/2

Q4. Let f(x) be a quadratic expression which is positive for all the real values of x. If g(x)=f(x)+f'(x)+f''(x), then for any real x,

•  g(x)<0
•  g(x)>0
•  g(x)=0
•  g(x)≥0

Solution

Q5.Suppose f(x)=e^ax+e^bx, where a≠b, and that f"(x)-2f'(x)-15f(x)=0 for all x. Then the product ab is

•  25
•  9
•  -15
•  -9

Solution
 (c) (a^2-2a-15) e^ax+(b^2-2b-15) d^bx=0 ⇒(a^2-2a-15)=0 and b^2-2b-15=0 ⇒(a-5)(a+3)=0 and (b-5)(b+3)=0 ⇒a=5 or -3 and b=5 or-3 ∴a≠b hence a=5 and b=-3 Or a=-3 and b=5 ⇒ab=-15

Q6. If x=∅(t),y=Ψ(t), then (d^2y)/(dx^2) is

•  (ϕ'Ψ"-Ψ'ϕ")/(ϕ')^2
•  (ϕ'Ψ"-Ψ'ϕ")/(ϕ')^3
• ϕ"/Ψ"
•  Ψ"/ϕ"

Solution

Q7.If y=(x+√(x^2+a^2))^n, then dy/dx is

•  ny/√(x^2+a^2)
•  -ny/√(x^2+a^2)
•  nx/√(x^2+a^2)
•  -nx/√(x^2+a^2)

Solution

Q8.Let g(x)=log⁡f(x), where f(x) is a twice differentiable positive function on (0,∞) such that f(x+1)=xf(x).Then,for N=1,2,3,…,g'' (N+1/2)-g''(1/2) is equal to

•  -4 {1+1/9+1/25+⋯+1/(2N-1)^2}
•  4 {1+1/9+1/25+⋯+1/(2N-1)^2}
•  -4 {1+1/9+1/25+⋯+1/(2N+1)^2}
•  4 {1+1/9+1/25+⋯+1/(2N+1)^2}

Solution

Q9.

 

•  0
•  2/(1+cos⁡x)
•  4/(1+cos⁡x)
•  (-4)/(1+cos⁡x)^2

Solution
(a) y=2 In(1+cos⁡x) dy/dx=(-2 sin⁡x)/(1+cos⁡x ) (d^2 y)/(dx^2)=-2[((1+cos⁡x)cos⁡x-sin⁡x(-sin⁡x))/(1+cos⁡x )^2 ] =-2[cos⁡x+1/(1+cos⁡x )^2 ]=(-2)/((1+cos⁡x ) ) Now 2e^(-y/2)=2∙e^(-(In (1+cos⁡x )^2)/2)=2/(1+cos⁡x) ∴(d^2 y)/(dx^2 )+2/e^(y/2) =0

Q10. 

 

•  1
•  0
•  7
• -7

Solution
(a) u(x)=7v(x)⇒u' (x)=7v' (x)⇒p=7 (given) Again (u(x))/(v(x))=7⇒((u(x))/(v(x)))=0⇒q=0 Now (p+q)/(p-q)=(7+0)/(7-0)=1

Written by: AUTHORNAME

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