Mix Questions - Worksheet 1

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Chapter 6 Class 12 - Application of Derivatives - Mix Questions Worksheet 1 by teachoo Chapter: Chapter 6 Class 12 - Application of Derivatives Name: _____________________________ School: _____________________________ Roll Number: _____________________________ 2-Mark Questions The volume of a sphere is increasing at a rate of 12π〖" " cm〗^3/s. How fast is the radius increasing when the radius is 3 cm ? State the conditions under which a point c is considered a "critical point" of a function f(x). The profit P(x) from selling x items is given by P(x)=100x-0.1x^2. Find the marginal profit when x=50. What does this value tell you? A function f(x) has a derivative f^' (x)=(x-1)(x-4)^2. Find the intervals where the function is increasing. Explain why the second derivative test is inconclusive when f^'' (c)=0. The side of a square is increasing at a rate of 2" " cm/s. At what rate is the area of the square increasing when the side length is 10 cm ? Find the two positive numbers whose sum is 16 and whose product is a maximum. A function has a local maximum at x=2. What can you say about the sign of the derivative f^' (x) for values of slightly less than 2 and slightly greater than 2 ? 3-Mark Question A rectangular box with a square base and open top must have a volume of 32,000〖" " cm〗^3. Find the dimensions of the box that minimize the amount of material used. Important links Answer of this worksheet - https://www.teachoo.com/25582/5357/Mix-Questions---Worksheet-1/category/Teachoo-Questions---Mix/ Full Chapter with Explanation, Activity, Worksheets and more – https://www.teachoo.com/subjects/cbse-maths/class-12th/ch6-12th-application-of-derivatives/ Maths Class 12 - https://www.teachoo.com/subjects/cbse-maths/class-12th/ For more worksheets, ad-free videos and Sample Papers – subscribe to Teachoo Black here - https://www.teachoo.com/black/   Answer Key to Mix Questions Worksheet 1 V=4/3 πr^3⇒dV/dt=4πr^2 (dr/dt). Given dV/dt=12π and r=3, we have 12π=4π(3)^2 (dr/dt)⇒ 12π=36π(dr/dt)⇒dr/dt=1/3" " cm/s. A point c is a critical point if it is in the domain of the function and either f^' (c)=0 or f^' (c) is undefined. P^' (x)=100-0.2x. At x=50,P^' (50)=100-0.2(50)=90. This means that when 50 items have been sold, the approximate profit from selling the 51st item is $90. f^' (x)>0 when (x-1)>0 (since (x-4)^2 is always non-negative). So, the function is increasing for x>1. The interval is (1,∞). If f^'' (c)=0, the test provides no information about the concavity at that point. The point could be a maximum, a minimum, or a point of inflection, so the first derivative test must be used. A=s^2⇒dA/dt=2s(ds/dt). Given ds/dt=2 and s=10, we get dA/dt=2(10)(2)=40〖" " cm〗^2/s. Let the numbers be x and 16-x. Product P(x)=x(16-x)=16x-x^2⋅P^' (x)=16-2x=0⇒x=8. P^'' (x)=-2<0, so it's a maximum. The numbers are 8 and 8 . For a local maximum at x=2, the function must be increasing just before x=2 and decreasing just after. Therefore, f^' (x) is positive for x slightly less than 2 and negative for x slightly greater than 2 . Let the base be x by x and height be h . Volume V=x^2 h=32000⇒h=32000/x^2. Surface Area A= x^2+4xh=x^2+4x(32000/x^2 )=x^2+128000/x.A^' (x)=2x-128000/x^2=0⇒2x^3=128000⇒ x^3=64000⇒x=40. Then h=32000/(40^2 )=20. The dimensions are 40" " cm×40" " cm×20" " cm.