Test: Application Of Derivatives - 2 - JEE MCQ

# Test: Application Of Derivatives - 2 - JEE MCQ

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## 25 Questions MCQ Test Mathematics (Maths) for JEE Main & Advanced - Test: Application Of Derivatives - 2

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Test: Application Of Derivatives - 2 - Question 1

### The function f (x) = x2 – 2 x is increasing in the interval

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Test: Application Of Derivatives - 2 - Question 2

### The function f (x) = a x + b is strict increasing for all x ∈ R iff

Detailed Solution for Test: Application Of Derivatives - 2 - Question 2

Since f ‘ (x) = a , therefore , f (x) is strict increasing on R iff a > 0.

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Test: Application Of Derivatives - 2 - Question 3

### Tangents to the curve x2+y2 = 2at the points (1, 1) and (– 1, 1)

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therefore , slope of tangent at (1,1) = - 1 and the slope of tangent at (- 1 ,1) = 1. Hence , the two tangents in reference are at right angles.

Test: Application Of Derivatives - 2 - Question 4

If x be real, the minimum value of x2−8x+17 is

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Test: Application Of Derivatives - 2 - Question 5

If a differentiable function f (x) has a relative minimum at x = 0, then the function y = f (x) + a x + b has a relative minimum at x = 0 for

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And y has a relative minimum at x = 0 if f ‘(0)+a = 0 . a =0.

Test: Application Of Derivatives - 2 - Question 6

The function f (x) = x2, for all real x, is

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Since f ‘(x) = 2x > 0 for x > 0,and f ‘ (x) = 2x < 0 for x < 0 ,therefore on R , f is neither increasing nor decreasing. Infact , f is strict increasing on [0 , ∞) and strict decreasing on (- ∞,0].

Test: Application Of Derivatives - 2 - Question 7

The function f (x) = a x + b is strict increasing for all x ∈ R if

Detailed Solution for Test: Application Of Derivatives - 2 - Question 7

Since f‘(x) = a , therefore , f (x) is strict increasing on R if f a > 0.

Test: Application Of Derivatives - 2 - Question 8

Equation of the tangent to the curve  at the point (a, b) is

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Hence ,slope of tangent at (a , b) = -b/a . Therefore , the equation of tangent at (a , b ) is ;

Test: Application Of Derivatives - 2 - Question 9

a log | x | + bx2 + x has its extreme values at x = – 1 and x = 2, then

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= 0, at x = -1 and x = 2 i.e a + 2b = 1 and 4b + a/2 = -1. ⇒ a = 2 , b = -1/2

Test: Application Of Derivatives - 2 - Question 10

Let f (x) be differentiable in (0, 4) and f (2) = f (3) and S = {c : 2 < c < 3, f’ (c) = 0} then

Detailed Solution for Test: Application Of Derivatives - 2 - Question 10

Conditions of Rolle’s Theorem are satisfied by f(x) in [2,3].Hence there exist atleast one real c in (2, 3) s.t. f ‘(c) = 0 . Therefore , the set S contains atleast one element

Test: Application Of Derivatives - 2 - Question 11

The function f (x) = m x + c where m, c are constants, is a strict decreasing function for all x ∈ R if

Detailed Solution for Test: Application Of Derivatives - 2 - Question 11

f (x) = mx + c is strict decreasing on R
if f ‘ (x) < 0 i.e. if m < 0 .

Test: Application Of Derivatives - 2 - Question 12

The function f (x) = x2−2x is strict decreasing in the interval

Detailed Solution for Test: Application Of Derivatives - 2 - Question 12

f ‘ (x) = 2x – 2 = 2 (x - 1) < 0 if x < 1 i.e. x x∈ (−∞,1) x∈ (−∞,1). Hence f is strict decreasing in left decreasing in (−∞,1)

Test: Application Of Derivatives - 2 - Question 13

The points on the curve 4 y = |x2−4| at which tangents are parallel to x – axis, are

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Only when x = 0 and when x = 0 , y = 1. So, only at (0,1) tangent is parallel to x- axis.

Test: Application Of Derivatives - 2 - Question 14

The maximum value of

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i.e. x = e. Note that , f ‘(x) changes sign from positive to negative as we move from left to right through e. So, f (e) is maximum i.e. maximum value of f (x) is f (e)

Test: Application Of Derivatives - 2 - Question 15

Every continuous function is

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Obviously, every differentiable function is continuous but every continuous function isn't differentiable.

Test: Application Of Derivatives - 2 - Question 16

The function f(x) = tan−1x is

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Therefore , f is strictly increasing on R.

Test: Application Of Derivatives - 2 - Question 17

For the curve x = t2−1,y = t2−t tangent is parallel to X – axis where

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Test: Application Of Derivatives - 2 - Question 18

The slope of the normal to the curvex = a (cos θ + θ sin θ),y = a (sin θ – θ cos θ) at any point ‘θ’ is

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Test: Application Of Derivatives - 2 - Question 19

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Test: Application Of Derivatives - 2 - Question 20

Rolle’s Theorem is not applicable to the function f(x) = | x | for −2⩽x⩽2 because

Detailed Solution for Test: Application Of Derivatives - 2 - Question 20

which does not exist at x = 0 ∈ (-2 , 2). So , Rolle’s theorem is not applicable.

Test: Application Of Derivatives - 2 - Question 21

Let f (x) = x3−6x2+9x+8, then f (x) is decreasing in

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Test: Application Of Derivatives - 2 - Question 22

The equation of the normal to the curve y = sinx at (0, 0) is

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therefore , slope of tangent at (0 , 0) = cos 0 = 1 and hence slope of normal at (0 , 0) is - 1 .

Test: Application Of Derivatives - 2 - Question 23

The curve y = x1/5 has at (0, 0)

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so,at (0 ,0) , the curve y = x1/5 has a vertical tangent.

Test: Application Of Derivatives - 2 - Question 24

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So, f has a local minima at 2 and a local maxima at - 2 .

Test: Application Of Derivatives - 2 - Question 25

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## Mathematics (Maths) for JEE Main & Advanced

209 videos|443 docs|143 tests