# Important Questions Class 11 Maths Chapter 13

## Important Questions for CBSE Class 11 Mathematics Chapter 13 – Limits and Derivatives

Important Questions For Class 11 Mathematics Chapter 13 can be found here. The important questions will aid you in your preparation for the annual exam. Examine all of the questions to help you clear your doubts while solving the problems.

Limits and Derivatives are extremely important concepts in Mathematics, but they also have applications in other subjects, such as physics. The set of Mathematics Class 11 Chapter 13 Important Questions includes all the concepts of limits and derivatives along with their properties and formulas.

This Chapter 13 Class 11 Mathematics Important Questions also includes important points to remember as well as a detailed explanation of key concepts and derivations for better understanding.

CBSE Class 11 Mathematics Chapter-13 Important Questions

Study Important Questions for Class 11 Mathematics Chapter 13 – Limits and Derivatives

Q.1 What is the result of the derivative of 2x with respect to x?

Ans: Let us assume the given expression as,

y=2x

Now, differentiating on both sides with respect to x then we get,

dydx

= dydx(2x)

= 2x log2

Q.2 What is the derivative of 2xx with respect to x.

Ans: By using uv the formula of differentiating for the given expression, we get,

ddx 2xx

= xddx 2x- 2x ddx (x)x2

= x × 2x ln 2 – ( 2x × 1)x2

= 2x xln2−1×2

Q.3 Find the value of f′(x) at x=100 if f(x)=99x

Ans: By differentiating the given function with respect to x we get,

f′(x)=99

Now, by substituting x=100 in the above derivative we get,

f′(100)=99

Q.4 Find the derivative of the expression 1+x+x2+x3+…+x50 at x=1.

Ans: By using the power rule of differentiation, we get the derivative of the given function as,

f1(x)=1+2x+3×2+…+50×49

Now, by substituting x=100 in the above derivative, we get,

f′(1)

=1+2+3+…+50

= 50(50+1)2

=25×51

=1275

Q1. Let f(x)=  { a+bx, x>1; 4, x=1; b-ax, x>1  if limx→1f(x)=f(1). What is the possible value of a and b

Ans: We know that the limit of a function exists when the left-hand limit and right-hand limit are both equal to the value of the function at that point that is,

limx→a+f(x)=limx→a−f(x)

Now, by using the above equation for the given function at x=1 then we get,

limx→1+f(x)=f(1)=4…….. (1)

limx→1−f(x)=f(1)=4…….. (2)

For  x>1

we have the given function as,

f(x)=a+bx

Now, by using the equation (1) we get,

limx→1+f(x)=limx→1+(a+bx)

⇒4=a+b…….. (3)

Now, for x<1 we have the given function as,

f(x)=b−ax

Now, by using the equation (2) we get,

limx→1−f(x)=limx→1−(b−ax)

⇒4=b−a………. (4)

Now, by adding both equation (3) and equation (4) then we get,

4+4=(a+b)+(b−a)

⇒8=2b

⇒b=4

Similarly, we get the value of another variable as,

⇒a=0

Q.2 Find the derivative of f(x)=1+x+x2+x3+…+x50  at x=1 .

Ans: We are required to find the derivative of f(x)=1+x+x2+x3+…+x50

at the point x=1.

Differentiating the function, we get,

f′(x)=ddx(1+x+x2+x3+…+x50)

=0+1+2x+3×2+…+50×49

=1+2x+3×2+…+50×49

Now, evaluating the derivative at x=1, and using the formula for the sum of first n natural numbers,

[ 1+2+3+ …. + n = n(n+1)2 ]

f′(1)=1+2+3+⋯+50

= 5025(50+1)2

=25×51

=1305

### 6 Marks Answers and Questions

Q.1 Differentiate tanx from the first principle.

Ans: Let the given function be f(x)=tanx. From this we get,

f(x+h)=tan(x+h)

By definition, we have,

f′(x)=lim h→0 f(x+h)−f(x)h

Substituting the values of the functions, we get,

f′(x)=limh→0 = tan(x+h)−tanxh

=limh→0 sin(x+h)cos(x+h) – sinxcos xh

Simplifying, we get

f’ x = limh→0

sin(x+h−x)hcos(x+h)cosx

= limh→0 sin hhcos(x+h)cosx

= limh→0sinhlimh→0cos(x+h)cosx

Applying the formula  limh→0 sin hh we get,

f′(x)= 1cos x cos x

= 1cos2 x

= sec2 x

Q.2 Differentiate (x+4)5 from the first principle.

Ans: Let the given function be f(x)=(x+4)5 . From this, we get,

f(x+h)=(x+h+4)5

By definition, we have,

f(x+h)=(x+h+4)5

Substituting the values of the functions, and applying the formula

limx→a  xn – anx-a=nan−1 , we get,

f′(x)=limh→0 ( x+ h+ 4)5 ( x+4)5h

Limh→0 (x+h+4)5−(x+4)5(x+h+4)−(x+4)

= 6(x+4)(6-1)

= 6 (x+4)5

Meaning of Limit and Derivative

Calculus is primarily concerned with the study of how the value of a function changes as the points in the domain change. When we need to find the value of a function that is close to some value, we use the limit. If the right and left limits coincide, we refer to that as the limit of f(x) at x = a and denote it by limxaf(x). The derivative is used to calculate the function’s instantaneous rate of change, as opposed to its average rate of change. The derivative can also be defined as the maximum rate of change in the function as the length of the interval on which the average is computed approaches zero.

Limits and Derivatives Introduction

The fundamentals of differentiation and calculus serve as the foundation for advanced mathematics, modern physics, and a variety of other modern sciences and engineering fields. Derivatives and limits Calculus is introduced to CBSE students in Class 11.

Limits of a Function

A function’s limit f(x) is defined as a value at which the function reaches the limit at some value. Limits are used to define other topics such as integration, integral calculus, and function continuity.

Limit Formula

Consider f(y) is a function, then the limit of the function can be represented as;

lim

y→b

Properties of Limits

Let p and q be two functions and a be a value such that limx→ap(x) and limx→aq(x) exists:

limx→a[p(x)+q(x)]=limx→ap(x)+limx→aq(x)

limx→a[p(x)−q(x)]=limx→ap(x)−limx→aq(x)

For every real number k

limx→a[kp(x)]=klimx→ap(x)

limx→a[p(x)q(x)]=limx→ap(x)×limx→aq(x)

limx→ap(x)q(x)=limx→ap(x)limx→aq(x)

Derivatives of a Function

The “derivative” is the instantaneous rate of change of one quantity with respect to another.The derivative of a function is represented by the following formula:

Derivative Formula

lim     f(x+h) -f (x)h

h→0

Properties of Derivatives

Algebra of the derivative of the function is given below:

Consider f and g be two functions such that their derivatives are defined in a common domain.

(i) The derivative of the sum of two functions is the sum of the derivatives of the functions.

dd(x) [p(x)+q (x)]= dd(x)(p(x)) + dd(x)(q(x))

(ii) The derivative of the difference of two functions is the difference of the derivatives of the functions

dd(x) [p(x)-q (x)]= dd(x)(p(x)) – dd(x)(q(x))

(iii) The derivative of product of two functions is given by the following product rule

dd(x) [p(x)-q (x)]= dd(x) [p(x)] q(x) + p(x) dd(x) [q(x)]

(iv) The derivative of the quotient of two functions is given by the following quotient rule (whenever the denominator is non–zero).

dd(x) [p(x)q(x)]= dd(x) [p(x)] q(x) – p(x) dd(x) [q(x)](g(x))2

Steps to Find the Derivative

1. Change x by the smallest possible value and let that be ‘h’ and so the function becomes f(x+h).
2. Get the change in value of function that is : f(x + h) – f(x)
3. The rate of change in function f(x) on changing from ‘x’ to ‘x+h’ will be

dydx= f (x+h) – f(x)h

We can ignore d(x) because it is considered to be too small.

Types of Derivative

Derivatives are classified according to their order, such as first- and second-order derivatives. These are defined further below.

• First-Order Derivative

First-order derivatives are used to determine the direction of a function, whether it is increasing or decreasing. The first-order derivative can be interpreted as the rate of change in an instant. The slope of the tangent line is used to predict the first-order derivative.

• Second-Order Derivative

The second-order derivatives are used to determine the shape of the given function’s graph. If the value of a function’s second-order derivatives is positive, the graph is upwardly concave. If the value of the second-order derivative is negative, then the graph of the function is downwardly open.

### Conclusion:

Students can get an idea of the type of questions asked in the examination by answering the above Class 11 Mathematics Limits and Derivatives Important Questions. These questions are based on the CBSE Board latest syllabus. There are various types of questions available, such as 1 mark, 2 marks, 4 marks, and 6 marks. Use the Important Questions for Class 11 Mathematics Chapter 13 provided here to improve your final exam score.

Q.1 If f(x) =xx,  x00,   x=0 , the limx0f(x) equals

Marks:1

Ans

LHL = limx0-f(x)=limx0-xx=limx0x-x=-1 RHL = limx0+f(x)=limx0+xx=limx0xx=1 limx0-f(x)  limx0+f(x)So limit does not exist.

Q.2 The value of is

Marks:1

Ans

Q.3 Find the derivative of :(a)3x4x2 (b)(5y42y2+1)(y1)

Marks:4

Ans

(a) f(x) =3x4x23df(x)dx=3x4x2=34‹2x=38x(b)f(y) =5y42y2+1y1 Using the product rule,

df(y)dy =ddy5y42y2+1(y1) =5y42y2+1(y1)+5y42y2+1(y1) =20y34y(y1)+5y42y2+1(1)

=20y420y34y2+4y+5y42y2+1 =25y420y36y2+4y+1

Q.4 If f(x)={mx2+n,x<0nx+m,0x1nx3+m,x>1For what values of m and n dolimx0f (x) and limx1f(x) both exist.

Marks:4

Ans

It is given that fx=mx2+n,x<0nx+m,0x1nx3+m,x>1limx 0f(x)and limx1f(x) both exist.limx 0f(x)=limx0+f(x)and limx1f(x)=limx1+f(x)limh0f(0h)

=limh 0f (0+h)andlimh0f(1h) =limh0f(1+ h)limh0m(h)2+n =limh0{nh + m}andlimh0{n(1h) + m} =limh0n (1 + h)3+ mn=m and n+m=n+m Hence limx 0f

(x) and limx1f (x) both exist for n=m.

Q.5 Differentiate with respect to x:ax2+bx+cx.

Marks:4

Ans

ddxax2+bx+cx =ddxax2x+bxx+cx

=ddxax32+bx12+cx12 =ddxax32+ddxbx12+ddxcx12

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### 1. What is the best way to prepare for Class 11 Maths Chapter 13 Limits and Derivatives?

While Limits and Derivatives questions and answers cover the chapter thoroughly, there are a few steps that can be taken to prepare effectively – (1) Review basic mathematics, such as arithmetic, algebra, trigonometry, and geometry. (2) Understand the Calculus part, whether Integral or Differential. (3) Understand and memorise Calculus formulas; (4) Proceed to Limits and learn about it; (5) Understand Calculus’ fundamental theorem; and (6) Practice Calculus problems. Students who take this approach will find it much easier to prepare for Limits and Derivatives Class 11.

### 2. Are derivatives necessary?

A derivative describes the changing relationship between two variables. The derivative formula can be used to calculate the slope of a line, the slope of a curve, and the difference between two measurements. As a result, derivatives are extremely important, and one must learn them thoroughly as well as practise various types of problems.

### 3. Which concepts are included in Limits and Derivatives Class 11 solutions?

Students must keep in mind that the concepts for Class 11 Maths Chapter 13 lay down the foundation for subsequent advanced topics of differentiation and integration.  The concepts in Limits and Derivatives Class 11 are – (1) Defining derivative of a function, (2) Description of Limits, (3) Limits of trigonometric functions and (4) Derivatives.