Previous Year Papers

Download Solved Question Papers Free for Offline Practice and view Solutions Online.

Test Series

Take Zigya Full and Sectional Test Series. Time it out for real assessment and get your results instantly.

Test Yourself

Practice and master your preparation for a specific topic or chapter. Check you scores at the end of the test.

Application of Derivatives

Name: Zigya - For The Curious Learner
Rating: 4.4 (740 reviews)

Multiple Choice Questions

131.

The slope of the normal to the curve y = 2 x² + 3 sin x at x = 0 is

3
$1 third$
-3
-3

84 Views

132.

The line y = x + 1 is a tangent to the curve y ² = 4x at the point

(1, 2)
(2, 3)
(1, -2)
(1, -2)

75 Views

133. The normal at the point (1, 1) on the curve 2y + x² = 3 is

x + y = 0
x – y = 0
x + y + 1 = 0
x + y + 1 = 0

114 Views

134. The normal to the curve x² = 4 y passing (1, 2) is

x + y = 3
x – y = 3
x + y = 1
x + y = 1

95 Views

Short Answer Type

135. Show that the function f (x) = 2 x + 3 is a strictly increasing function on R.

85 Views

136. Without using the derivative show that the function f (a) = 7x – 3 is a strictly increasing function on R.

85 Views

137. Show that the function f (x) = x² is an increasing function in (0, ∞).

Let x₁ , x₂ ∊ (0, ∞) and let x₁, < x₂.
Now, $straight x subscript 1 space less than space straight x subscript 2$
$rightwards double arrow space space space space space straight x subscript 1. space straight x subscript 1 space space space less than space straight x subscript 1. space straight x subscript 2$ $open square brackets because space space straight x subscript 1 greater than 0 close square brackets$

rightwards double arrow space space space space straight x subscript 1 squared space space less than space straight x subscript 1. straight x subscript 2

...(1)
Again

straight x subscript 1 less than straight x subscript 2

rightwards double arrow space space space space space straight x subscript 1. space space straight x subscript 2 space space space less than space space straight x subscript 2. space straight x subscript 2 space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space

open square brackets because space space straight x subscript 2 space greater than space 0 close square brackets

rightwards double arrow space space space straight x subscript 1. space end subscript straight x subscript 2 space space less than space straight x subscript 2 squared

...(2)
From (1) and (2), we get,

straight x subscript 1 squared space less than space straight x subscript 2 squared space space space space space space or space space space space space straight f left parenthesis straight x subscript 1 right parenthesis space less than space straight f left parenthesis straight x subscript 2 right parenthesis

therefore space space straight x subscript 1 space less than space straight x subscript 2 space space space space space space space space space space space space space space space space space space space rightwards double arrow space space space space straight f left parenthesis straight x subscript 1 right parenthesis space less than space straight f left parenthesis straight x subscript 2 right parenthesis therefore space space space space straight f space is space an space increasing space function space in space left parenthesis 0 comma space infinity right parenthesis.

81 Views

138.

Show that the function f(x) = x² is a decreasing function in (– ∞ 0).

84 Views

139.

Construct an example of a functions which is strictly increasing but whose derivative vanishes at a point in the domain of definition of the function.

82 Views