

Important Questions for Class 11 Maths Chapter 13: Chapter 13 of Class 11 Mathematics, Limits and Derivatives, introduces foundational calculus concepts. It covers the idea of a limit, including evaluating limits of algebraic, trigonometric, and rational functions, and basic rules for calculating them.
The derivative is introduced as the rate of change or slope of a function, along with methods for finding derivatives of polynomials and simple functions. Important questions include solving problems on limit evaluation using formulas, derivatives of basic functions using first principles, and applying derivatives to real-world scenarios. Mastery of this chapter is crucial for understanding advanced calculus in higher classes.Important Questions for Class 11 Maths Chapter 13 PDF
Question 1:
Find the derivative of the function x 2 cos x.Solution:
Given function is x 2 cos x Let y = x 2 cos x Differentiate with respect to x on both sides. Then, we get: dy/dx = (d/dx)x 2 cos x Now, using the formula, we can write the above form as: dy/dx = x 2 (d/dx) cos x + cos x (d/dx)x 2 Now, differentiate the function: dy/dx = x 2 (-sin x) + cos x (2x) Now, rearrange the terms, we will get: dy/dx = 2x cos x – x 2 sin xQuestion 2:
Find the positive integer “n” so that lim x → 3 [(x n – 3 n )/(x – 3)] = 108.Solution:
Given limit: lim x → 3 [(x n – 3 n )/(x – 3)] = 108 Now, we have: lim x → 3 [(x n – 3 n )/(x-3)] = n(3) n-1 n(3) n-1 = 108 Now, this can be written as: n(3) n-1 = 4 (27) = 4(3) 4-1 Therefore, by comparing the exponents in the above equation, we get: n = 4 Therefore, the value of positive integer “n” is 4.Question 3:
Find the derivative of f(x) = x 3 using the first principle.Solution:
By definition, f’(x) = lim h→ 0 [f(x+h)-f(x)]/h Now, substitute f(x)=x 3 in the above equation: f’(x) = lim h→ 0 [(x+h) 3 -x 3 ]/h f’(x) = lim h→ 0 (x 3 +h 3 +3xh(x+h)-x 3 )/h f’(x) = lim h→ 0 (h 2 +3x(x+h)) Substitute h = 0, we get: f’(x) = 3x 2 Therefore, the derivative of the function f’(x) = x 3 is 3x 2 .Question 4:
Determine the derivative of cosx/(1+sin x).Solution:
Given function: cosx/(1+sin x) Let y = cosx/(1+sin x) Now, differentiate the function with respect to “x”, we get dy/dx = (d/dx) (cos x/(1+sin x)) Now, use the u/v formula in the above form, we get dy/dx = [(1+sin x)(-sin x) – (cos x)(cos x)]/(1+sin x) 2 dy/dx = (-sin x – sin 2 x-cos 2 x)/(1+sin x) 2 Now, take (-) outside from the numerator, we get: dy/dx = -(sin x + sin 2. x + cos 2 x)/(1+sin x) 2 We know that sin 2. x + cos 2 x = 1 By substituting this, we can get: dy/dx = -(1+sin x)/(1+sin x) 2 Cancel out (1+sin x) from both numerator and denominator, we get: dy/dx = -1/(1+sin x) Therefore, the derivative of cosx/(1+sin x) is -1/(1+sin x).Question 5:
lim x→ 0 |x|/x is equal to: (a)1 (b)-1 (c)0 (d)does not existsSolution:
A correct answer is an option (d)Explanation:
The limit mentioned here is x→0 It has two possibilities: Case 1: x→0 + Now, substitute the limit in the given function: lim x→ 0+ |x|/x = x/x = 1 Case 2: x→0 – Now, substitute the limit in the given function: lim x→ 0- |x|/x = -x/x = -1 Hence, the result for both cases varies, the solution is an option (D)Question 6:
Evaluate the derivative of f(x) = sin 2 x using Leibnitz product rule.Solution:
Given function: f(x) = sin 2 x Let y= sin 2 x Now, by using Leibnitz product rule, we can write it as: dy/dx = (d/dx) sin 2 x Sin 2 x can be written as (sin x)(sin x) Now, it becomes: dy/dx = (d/dx) (sin x)(sin x) dy/dx = (sin x)’(sin x) + (sin x)(sin x)’ dy/dx = cos x sin x + sin x cos x dy/dx = 2 sin x cos x dy/dx = sin 2x Therefore, the derivative of the function sin 2 x is sin 2x.Strong Conceptual Understanding
Helps in grasping the foundational concepts of limits and derivatives, which are essential for calculus in higher classes.Improved Problem-Solving Skills
Enhances the ability to solve problems methodically and apply formulas correctly.Better Exam Preparation
Familiarizes students with the types of questions likely to appear in exams, boosting confidence.Clarity on Applications
Demonstrates real-life applications of limits and derivatives, making the subject more relatable.Time Management
Regular practice improves speed and accuracy, which is crucial during exams.Foundation for Higher Studies
Builds a strong base for advanced topics in Class 12 and competitive exams.