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QUESTION: 1

Solution:

QUESTION: 2

For a real number y, let [y] denotes the greatest integer less than or equal to y : Then the function

Solution:

By def. [x – π] is an integer whatever be the value of x.

And so p[x – π] is an integral multiple of p.

Consequently tan

And since 1 + [x]2 ≠ 0 for any x, we conclude that f (x) = 0.

Thus f (x) is constant function and so, it is continuous and differentiable any no. of times, that is f '(x), f ''(x), f '''(x),... all exist for every x, their value being 0 at every pt. x. Hence, out of all the alternatives only (d) is correct.

QUESTION: 3

There exist a function f (x), satisfying f (0) = 1, f '(0) = –1, f (x) > 0 for all x, and

Solution:

f (x) = e^{–x} is one such function.

QUESTION: 4

has the value

Solution:

QUESTION: 5

If f (a) = 2, f ' (a) = 1 , g (a) = -1 , g ' (a) = 2 , then the value of

Solution:

QUESTION: 6

The function is not defined at x = 0. The value which should be assigned to f at x = 0 so that it is continuous at x = 0, is

Solution:

For f (x) to be continuous at x = 0

= a + b

QUESTION: 7

Solution:

QUESTION: 8

Where [x] denotes the greatest integer less than or equal to x. then

Solution:

The given function can be restated as

QUESTION: 9

Let f : R → R be a differentiable function and f (1) = 4. Then the value of

Solution:

We have f : R → R, a differentiable function and f (1) = 4

NOTE THIS STEP

QUESTION: 10

Let [.] den ote th e greatest integer function and f (x) = [tan ^{2}x], then:

Solution:

We have f (x) = [tan^{2} x]

tan x is an increasing function for

QUESTION: 11

The function denotes the greatest integer function, is discontinuous at

Solution:

When x is not an integer, both the functions [x] and cos are continuous.

∴f (x) is continuous on all non integral points.

∴ f is continuous at all integral pts as well.

Thus, f is continuous everywhere.

QUESTION: 12

Solution:

QUESTION: 13

The function f(x) = [x]^{2} – [x^{2}] (where [y] is the greatest integer less than or equal to y), is discontinuous at

Solution:

We have f (x) = [x]^{2} – [x^{2}]

At x = 0,

f (1) = [1]^{2} – [1^{2}] = 1 – 1 = 0

∴ L.H.L. = R.H.L. = f (1)

∴ f (x) is continuous at x = 1.

Clearly at other integral pts f (x) is not continuous.

QUESTION: 14

The function f (x) = (x^{2} - 1) | x^{2} - 3x + 2 |+cos (|x|) is NOT differentiable at

Solution:

∴ Given function can be written as

This function is differentiable at all points except possibly at x = 1 and x = 2.

Lf (2) ≠ Rf '(2)

∴ f is not differentiable at x = 2.

QUESTION: 15

Solution:

QUESTION: 16

Solution:

For x ∈ R

QUESTION: 17

Solution:

QUESTION: 18

The left-hand derivative of f(x) = [x] sin(p x) at x = k, k an integer, is

Solution:

QUESTION: 19

Let f : R → R be a function defined by f (x) = max {x, x^{3}}. The set of all points where f (x) is NOT differentiable is

Solution:

**KEY CONCEPT**

A continuous function f (x) is not differentiable at x = a if graphically it takes a sharp turn at x = a.

Graph of f (x) = max {x, x^{3}} is as shown with solid lines.

From graph of f (x) at x = – 1, 0, 1, we have sharp turns.

∴ f (x) is not differentiable at x = – 1, 0, 1.

QUESTION: 20

Which of the following functions is differentiable at x = 0?

Solution:

Let us test each of four options.

QUESTION: 21

The domain of the derivative of the function

Solution:

The given function is

∴ f (x) is discontinuous at x = – 1

Also we can prove in the same way, that f (x) is discontinuous at x = 1

∴ f ' (x) can not be found for x = ± 1 or domain of f ' (x) = R – {– 1, 1}

QUESTION: 22

The integer n for which is a finite non-zero number is

Solution:

Given that,

For this limit to be finite, n – 3 = 0 ⇒ n = 3

QUESTION: 23

Let f : R → R be such that f (1) = 3 and f '(1) = 6. Then

Solution:

Given that f ; R → R such that

f (1) = 3 and f ' (1) = 6

QUESTION: 24

where n is nonzero real number, then a is equal to

Solution:

We are given that

QUESTION: 25

given that f ' (2) = 6 and f '(1) = 4

Solution:

∴ Applying L Hospital's rule, we get

QUESTION: 26

If (x) is differentiable and strictly increasing function, then the value of

Solution:

Using L.H. Rule, we get

QUESTION: 27

The function given by y = ||x| – 1| is differentiable for all real numbers except th e points

Solution:

Graph of y = | | x | – 1| is as follows :

The graph has sharp turnings at x = – 1, 0 and 1; and hence not differentiable at x = – 1, 0, 1.

QUESTION: 28

If f (x) is continuous an d differ en tiable function and

Solution:

Given that f (x) is a contin uous and differ entiable function and

QUESTION: 29

The value of

Solution:

QUESTION: 30

Let f (x) be differentiable on the interval (0, ∞) such that for each x > 0. Then f (x) is

Solution:

Given that f (x) is differentiable on (0, ∞) with

[Linear differential equation]

Integrating factor

NOTE THIS STEP

QUESTION: 31

Solution:

KEY CONCEPT

On applying L' Hospital's rule, we get

QUESTION: 32

m and n are integers, m ≠ 0,n > 0 , and let p be the left hand derivative of |x – 1| at x = 1.

Solution:

As per question,

p = left hand derivative of |x –1| at x = 1 ⇒ p = –1

QUESTION: 33

then the value of θ is

Solution:

QUESTION: 34

Solution:

For this limit to be finite 1 - a = 0 ⇒ a = 1 then given limit reduces to

QUESTION: 35

Solution:

i s not differ entiable at x = 2.

QUESTION: 36

Let α(a) and β (a) be the roots of the equation where

Solution:

The given equation is

Dividing both sides by y – 1, we get

Taking limit as y → 1 i.e. a → 0 on both sides we get

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