Applications of Derivatives Test

Result

Applications of Derivatives Test - 5

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Question 1
1 / -0

A function y = f(x) has a second order derivative f″(x) = 6(x – 1). If its graph passes through the point (2, 1) and at that point the tangent to the graph is y = 3x – 5, then the function is

A
(x-1)²

B
(x-1)³

C
(x+1)³

D
(x+1)²

Solution

f″ (x) = 6(x – 1) ⇒ f′ (x) = 3(x – 1)² + c
and f′ (2) = 3 ⇒ c = 0
⇒ f (x) = (x – 1)³ + k and f (2) = 1 ⇒ k = 0
⇒f (x) = (x – 1)³.
Question 2
1 / -0

The point of intersection of the tangents drawn to the curve x²y = 1–y at the points where it is meet by the curve xy = 1– y, is given by

A
(0, –1)

B
(1, 1)

C
(0, 1)

D
None of these

Solution

x²y = 1–y …(1)
xy = 1–y … (2)
Solving (1) and (2), we get
x(xy) = 1 – y
⇒x (1 – y) = (1 – y)
⇒(1 – y) (x – 1) = 0
⇒x = 1, y = 1
It must satisfy equation of the curve
xy = 1 – y
Linked To Above Question ||| No
Difficulty ||| Medium
Time ||| 60
Marking ||| 4, 1
PYSP ||| Engg-Main-Math-New-Questions
Topic ||| Maths || Application of Derivatives || Tangents and Normals
1. 1 = 1– φ ≠ 0
If take x = 1, y = 0
then 1.0 ≠ 1 – 0
If take x = 0 and y = 1
0.1 = 1 –1 = 0
Hence, option [C] is correct answer.