CBSE Class 11 Mathematics Limits And Derivatives Worksheet Set 02

Question 1. Find the derivative of (x + sec x) (x - tanx). (2 marks)
Answer: Let \(f(x) = (x + \sec x)(x - \tan x)\)
Using the product rule for differentiation, \(\frac{d}{dx}[u \cdot v] = u \frac{dv}{dx} + v \frac{du}{dx}\)
Here, \(u = x + \sec x\) and \(v = x - \tan x\)
\(\frac{du}{dx} = 1 + \sec x \tan x\)
\(\frac{dv}{dx} = 1 - \sec^2 x\)
Substitute these into the product rule formula:
\(\frac{df}{dx} = (x + \sec x)(1 - \sec^2 x) + (x - \tan x)(1 + \sec x \tan x)\)
In simple words: Apply the product rule by differentiating one factor at a time while keeping the other constant.

Exam Tip: Remember standard derivatives: \(\frac{d}{dx}(\sec x) = \sec x \tan x\) and \(\frac{d}{dx}(\tan x) = \sec^2 x\).

Question 2. Find the derivative of the following function : x4 (5 sin x - 3 cos x). (3 marks)
Answer: Let \(y = x^4(5 \sin x - 3 \cos x)\)
Using the product rule: \(\frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx}\)
Let \(u = x^4\) and \(v = 5 \sin x - 3 \cos x\)
\(\frac{du}{dx} = 4x^3\)
\(\frac{dv}{dx} = 5 \cos x - (-3 \sin x) = 5 \cos x + 3 \sin x\)
Therefore:
\(\frac{dy}{dx} = x^4(5 \cos x + 3 \sin x) + (5 \sin x - 3 \cos x)(4x^3)\)
\(\frac{dy}{dx} = x^3 [x(5 \cos x + 3 \sin x) + 4(5 \sin x - 3 \cos x)]\)
In simple words: Differentiate using the product rule and factor out common terms to simplify the final expression.

Exam Tip: Be careful with signs when differentiating trigonometric functions, especially the derivative of cosine which is negative sine.

Question 3. Find the derivative of (x2 + 1) cos x. (3 marks)
Answer: Let \(y = (x^2 + 1)\cos x\)
Using the product rule:
\(\frac{du}{dx} = \frac{d}{dx}(x^2 + 1) = 2x\)
\(\frac{dv}{dx} = \frac{d}{dx}(\cos x) = -\sin x\)
Applying the product rule formula:
\(\frac{dy}{dx} = (x^2 + 1)(-\sin x) + \cos x (2x)\)
\(\frac{dy}{dx} = 2x \cos x - (x^2 + 1)\sin x\)
In simple words: Differentiate the algebraic part and trigonometric part separately and combine them using the product rule.

Exam Tip: Always put parentheses around negative terms to avoid sign errors.

Question 4. Evaluate the given limit: \(\lim_{x \to 0} \frac{ax + x \cos x}{b \sin x}\). (3 marks)
Answer: Given expression: \(\lim_{x \to 0} \frac{ax + x \cos x}{b \sin x}\)
Factor out \(x\) from the numerator:
\(= \lim_{x \to 0} \frac{x(a + \cos x)}{b \sin x}\)
Rewrite the limit to use the standard limit formula \(\lim_{x \to 0} \frac{\sin x}{x} = 1\):
\(= \frac{1}{b} \lim_{x \to 0} \left( \frac{x}{\sin x} \cdot (a + \cos x) \right)\)
\(= \frac{1}{b} \cdot \left( \lim_{x \to 0} \frac{1}{\frac{\sin x}{x}} \right) \cdot \lim_{x \to 0} (a + \cos x)\)
\(= \frac{1}{b} \cdot \frac{1}{1} \cdot (a + \cos 0)\)
Since \(\cos 0 = 1\):
\(= \frac{a + 1}{b}\)
In simple words: Separate the terms and use the standard trigonometric limit formula to eliminate the indeterminate form.

Exam Tip: Memorize standard limits such as \(\lim_{x \to 0} \frac{\sin x}{x} = 1\) as they are frequently used in evaluations.

Question 5. Find the derivative of \(\frac{\sin x + \cos x}{\sin x - \cos x}\).
Answer: Let \(y = \frac{\sin x + \cos x}{\sin x - \cos x}\)
Using the quotient rule \(\frac{d}{dx}\left[\frac{u}{v}\right] = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}\)
Here, \(u = \sin x + \cos x\) and \(v = \sin x - \cos x\)
\(\frac{du}{dx} = \cos x - \sin x\)
\(\frac{dv}{dx} = \cos x + \sin x\)
Substitute into the quotient rule formula:
\(\frac{dy}{dx} = \frac{(\sin x - \cos x)(\cos x - \sin x) - (\sin x + \cos x)(\cos x + \sin x)}{(\sin x - \cos x)^2}\)
Rewrite the numerator:
\(= \frac{-(\sin x - \cos x)^2 - (\sin x + \cos x)^2}{(\sin x - \cos x)^2}\)
Expand both terms in the numerator:
Numerator = \(-(\sin^2 x + \cos^2 x - 2\sin x \cos x) - (\sin^2 x + \cos^2 x + 2\sin x \cos x)\)
Using \(\sin^2 x + \cos^2 x = 1\):
Numerator = \(-(1 - 2\sin x \cos x) - (1 + 2\sin x \cos x) = -1 + 2\sin x \cos x - 1 - 2\sin x \cos x = -2\)
Therefore:
\(\frac{dy}{dx} = \frac{-2}{(\sin x - \cos x)^2}\)
In simple words: Apply the quotient rule carefully, expand the algebraic expressions in the numerator, and simplify using trigonometric identities.

Exam Tip: Be very careful with negative signs when expanding squares in the quotient rule numerator.

Question 6. Evaluate the Given limit: \(\lim_{x \to 2} \frac{3x^2 - x - 10}{x^2 - 4}\). (3 marks)
Answer: Given limit: \(\lim_{x \to 2} \frac{3x^2 - x - 10}{x^2 - 4}\)
Direct substitution of \(x = 2\) gives the indeterminate form \(\frac{0}{0}\).
Factorize the numerator: \(3x^2 - x - 10 = 3x^2 - 6x + 5x - 10 = 3x(x - 2) + 5(x - 2) = (3x + 5)(x - 2)\)
Factorize the denominator: \(x^2 - 4 = (x - 2)(x + 2)\)
Substitute the factors back into the limit expression:
\(= \lim_{x \to 2} \frac{(3x + 5)(x - 2)}{(x - 2)(x + 2)}\)
Cancel the common term \((x - 2)\):
\(= \lim_{x \to 2} \frac{3x + 5}{x + 2}\)
Now substitute \(x = 2\):
\(= \frac{3(2) + 5}{2 + 2} = \frac{6 + 5}{4} = \frac{11}{4}\)
In simple words: Factor both the numerator and denominator to cancel out the term causing the zero in the denominator.

Exam Tip: Always check for the \(\frac{0}{0}\) form first before attempting factorization or L'Hopital's rule.

Question 7. Evaluate the given limit: \(\lim_{x \to 1} \frac{ax^2 + bx + c}{cx^2 + bx + a}, a + b + c \neq 0\). (1 mark)
Answer: Given limit: \(\lim_{x \to 1} \frac{ax^2 + bx + c}{cx^2 + bx + a}\)
Substitute \(x = 1\) directly into the expression:
\(= \frac{a(1)^2 + b(1) + c}{c(1)^2 + b(1) + a}\)
\(= \frac{a + b + c}{c + b + a}\)
Since \(a + b + c \neq 0\), we can cancel numerator and denominator:
\(= 1\)
In simple words: Since direct substitution does not yield an indeterminate form, simply plug in \(x = 1\).

Exam Tip: Always test direct substitution first before applying complex limit techniques.

Question 8. Find the value of the limit \(\lim_{x \to 3} \frac{x^2 + 10}{x - 2}\). (1 mark)
Answer: Given limit: \(\lim_{x \to 3} \frac{x^2 + 10}{x - 2}\)
Substitute \(x = 3\) directly:
\(= \frac{(3)^2 + 10}{3 - 2}\)
\(= \frac{9 + 10}{1} = 19\)
In simple words: Substitute \(x = 3\) directly into the rational function since the denominator does not become zero.

Exam Tip: Check that the denominator is non-zero at the given point before declaring direct substitution valid.

Question 9. Find the value of \(\lim_{x \to 0} \frac{e^x - 1}{10x}\). (1 mark)
Answer: Given limit: \(\lim_{x \to 0} \frac{e^x - 1}{10x}\)
Factor out the constant \(\frac{1}{10}\):
\(= \frac{1}{10} \lim_{x \to 0} \frac{e^x - 1}{x}\)
Using the standard exponential limit \(\lim_{x \to 0} \frac{e^x - 1}{x} = 1\):
\(= \frac{1}{10} \cdot 1 = \frac{1}{10}\)
In simple words: Pull out the constant and use the standard limit formula for exponential functions.

Question 10. Find the first derivative of \(x^3 - 4\) at \(x = 2\). (1 mark)
Answer: Let \(f(x) = x^3 - 4\)
Find the derivative using the power rule: \(\frac{d}{dx}(x^n) = n x^{n-1}\)
\(f'(x) = 3x^2 - 0 = 3x^2\)
Evaluate the derivative at \(x = 2\):
\(f'(2) = 3(2)^2 = 3(4) = 12\)
In simple words: Differentiate the function using the power rule and then substitute the given value of \(x\).

Exam Tip: Differentiate first before plugging in the specific point value.

Question 11. Evaluate the given limit: \(\lim_{x \to 0} \frac{ax + b}{cx + 1}\). (1 mark)
Answer: Given limit: \(\lim_{x \to 0} \frac{ax + b}{cx + 1}\)
Substitute \(x = 0\) directly:
\(= \frac{a(0) + b}{c(0) + 1}\)
\(= \frac{b}{1} = b\)
In simple words: Substitute \(x = 0\) directly into the expression since the denominator is \(1 \neq 0\).

Exam Tip: Direct substitution works smoothly when there is no division by zero.

Question 12. Find the derivative of \(x^2 \sin x\). (1 mark)
Answer: Let \(y = x^2 \sin x\)
Using the product rule: \(\frac{d}{dx}[u \cdot v] = u \frac{dv}{dx} + v \frac{du}{dx}\)
Here, \(u = x^2\) and \(v = \sin x\)
\(\frac{du}{dx} = 2x\)
\(\frac{dv}{dx} = \cos x\)
Substitute into the formula:
\(\frac{dy}{dx} = x^2(\cos x) + \sin x(2x)\)
\(\frac{dy}{dx} = x^2 \cos x + 2x \sin x\)
In simple words: Apply the product rule directly to the two multiplied functions.

Exam Tip: Clearly identify the two separate functions before applying the product rule formula.

Question 13. Find the value of \(\lim_{x \to 0} \frac{2 \log(1 + x)}{3x}\). (2 marks)
Answer: Given limit: \(\lim_{x \to 0} \frac{2 \log(1 + x)}{3x}\)
Factor out the constants \(\frac{2}{3}\):
\(= \frac{2}{3} \lim_{x \to 0} \frac{\log(1 + x)}{x}\)
Using the standard logarithmic limit \(\lim_{x \to 0} \frac{\log(1 + x)}{x} = 1\):
\(= \frac{2}{3} \cdot 1 = \frac{2}{3}\)
In simple words: Factor out the constants and use the standard logarithmic limit.

Exam Tip: Memorize the standard limit \(\lim_{x \to 0} \frac{\log(1 + x)}{x} = 1\) (using natural log).

Question 14. Find the derivative of the following function \(\frac{x^2 \cos\left(\frac{\pi}{4}\right)}{\sin x}\). (3 marks)
Answer: Given function: \(f(x) = \frac{x^2 \cos\left(\frac{\pi}{4}\right)}{\sin x}\)
Since \(\cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}\) is a constant, we can factor it out:
\(f(x) = \frac{1}{\sqrt{2}} \cdot \frac{x^2}{\sin x}\)
Now apply the quotient rule to \(\frac{x^2}{\sin x}\):
\(\frac{d}{dx}\left[\frac{x^2}{\sin x}\right] = \frac{\sin x \cdot \frac{d}{dx}(x^2) - x^2 \cdot \frac{d}{dx}(\sin x)}{(\sin x)^2}\)
\(= \frac{\sin x (2x) - x^2 \cos x}{\sin^2 x}\)
Multiply by the constant factor \(\frac{1}{\sqrt{2}}\):
\(f'(x) = \frac{1}{\sqrt{2}} \cdot \frac{2x \sin x - x^2 \cos x}{\sin^2 x}\)
In simple words: Simplify constant trigonometric values first, then apply the quotient rule.

Exam Tip: Always evaluate constant trigonometric terms like \(\cos\left(\frac{\pi}{4}\right)\) at the beginning to simplify calculations.

Question 15. Find the derivative of \(x^2 + \sin x + \frac{1}{x^2}\). (2 marks)
Answer: Let \(f(x) = x^2 + \sin x + x^{-2}\)
Differentiate each term separately using the sum rule and power rule:
\(\frac{d}{dx}(x^2) = 2x\)
\(\frac{d}{dx}(\sin x) = \cos x\)
\(\frac{d}{dx}(x^{-2}) = -2x^{-3} = -\frac{2}{x^3}\)
Combine the results:
\(f'(x) = 2x + \cos x - \frac{2}{x^3}\)
In simple words: Rewrite \(\frac{1}{x^2}\) as \(x^{-2}\) and differentiate term by term using the power rule.

Exam Tip: Convert denominators with powers into negative exponents to easily apply the power rule.

Question 16. Suppose \(f(x) = \begin{cases} a + bx, & x < 1 \\ 4, & x = 1 \\ b - ax, & x > 1 \end{cases}\) and if \(\lim_{x \to 1} f(x) = f(1)\) what are possible values of a and b? (5 marks)
Answer: Given condition: \(\lim_{x \to 1} f(x) = f(1)\)
First, find \(f(1)\):
From the piecewise function, \(f(1) = 4\)
For the limit \(\lim_{x \to 1} f(x)\) to exist, the left-hand limit must equal the right-hand limit:
Left-Hand Limit (LHL):
\(\lim_{x \to 1^-} f(x) = \lim_{x \to 1} (a + bx) = a + b\)
Right-Hand Limit (RHL):
\(\lim_{x \to 1^+} f(x) = \lim_{x \to 1} (b - ax) = b - a\)
For the limit to exist, LHL = RHL:
\(a + b = b - a\)
\(2a = 0 \implies a = 0\)
Now, equate the limit to \(f(1)\):
\(\lim_{x \to 1} f(x) = f(1)\)
\(a + b = 4\)
Since \(a = 0\):
\(0 + b = 4 \implies b = 4\)
Therefore, the possible values are \(a = 0\) and \(b = 4\).
In simple words: Equate the left-hand limit and right-hand limit to find a relationship, then use \(f(1) = 4\) to determine the exact values of \(a\) and \(b\).

Exam Tip: For piecewise functions at a point, always check that both side limits exist and are equal to each other and to the function value.

Question 17. For some constants a and b, find the derivative of \((ax + b)^2\). (2 marks)
Answer: Let \(y = (ax + b)^2\)
Method 1 (Using Chain Rule):
Let \(u = ax + b\), then \(y = u^2\)
\(\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = 2u \cdot a = 2(ax + b)(a) = 2a(ax + b)\)
Method 2 (Expanding the expression):
\(y = a^2x^2 + 2abx + b^2\)
\(\frac{dy}{dx} = a^2(2x) + 2ab(1) + 0 = 2a^2x + 2ab = 2a(ax + b)\)
In simple words: You can either expand the square and differentiate term by term or use the chain rule for composite functions.

Exam Tip: The chain rule is faster and avoids algebraic expansion errors for higher powers.

Question 18. Evaluate \(\lim_{x \to 0} \frac{\tan 4x}{x \sec x}\). (2 marks)
Answer: Given limit: \(\lim_{x \to 0} \frac{\tan 4x}{x \sec x}\)
Rewrite \(\tan 4x\) as \(\frac{\sin 4x}{\cos 4x}\) and \(\sec x\) as \(\frac{1}{\cos x}\):
\(= \lim_{x \to 0} \frac{\left(\frac{\sin 4x}{\cos 4x}\right)}{x \left(\frac{1}{\cos x}\right)} = \lim_{x \to 0} \frac{\sin 4x}{x} \cdot \frac{\cos x}{\cos 4x}\)
Multiply and divide by 4 to adjust the argument for the standard limit \(\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1\):
\(= \lim_{x \to 0} \left( 4 \cdot \frac{\sin 4x}{4x} \cdot \frac{\cos x}{\cos 4x} \right)\)
\(= 4 \cdot \left(\lim_{4x \to 0} \frac{\sin 4x}{4x}\right) \cdot \frac{\lim_{x \to 0} \cos x}{\lim_{x \to 0} \cos 4x}\)
\(= 4 \cdot 1 \cdot \frac{\cos 0}{\cos 0} = 4 \cdot 1 \cdot \frac{1}{1} = 4\)
In simple words: Convert tangent and secant into sine and cosine, then use the standard limit \(\frac{\sin \theta}{\theta}\).

Exam Tip: Adjust coefficients inside trigonometric limits to match the denominator using multiplication and division.

Question 19. Find the derivative of cos x from first principle. (3 marks)
Answer: Let \(f(x) = \cos x\)
By the first principle of derivatives:
\(f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}\)
\(f'(x) = \lim_{h \to 0} \frac{\cos(x + h) - \cos x}{h}\)
Use the trigonometric transformation formula for difference of cosines: \(\cos C - \cos D = -2 \sin\left(\frac{C + D}{2}\right) \sin\left(\frac{C - D}{2}\right)\)
Here \(C = x + h\) and \(D = x\):
\(\cos(x + h) - \cos x = -2 \sin\left(\frac{x + h + x}{2}\right) \sin\left(\frac{x + h - x}{2}\right) = -2 \sin\left(x + \frac{h}{2}\right) \sin\left(\frac{h}{2}\right)\)
Substitute this back into the limit:
\(f'(x) = \lim_{h \to 0} \frac{-2 \sin\left(x + \frac{h}{2}\right) \sin\left(\frac{h}{2}\right)}{h}\)
Rearrange terms:
\(f'(x) = \lim_{h \to 0} \left( -\sin\left(x + \frac{h}{2}\right) \cdot \frac{\sin\left(\frac{h}{2}\right)}{\frac{h}{2}} \right)\)
Taking the limit as \(h \to 0\):
\(f'(x) = -\sin(x + 0) \cdot 1 = -\sin x\)
In sum-up: Use the definition of derivative from first principles combined with the sum-to-product trigonometric identity.

Exam Tip: Memorize the trigonometric transformation formulas as they are essential for first-principle trigonometric proofs.

Question 20. Evaluate: \(\lim_{x \to 0} (\csc x - \cot x)\).
Answer: Given limit: \(\lim_{x \to 0} (\csc x - \cot x)\)
Rewrite cosecant and cotangent in terms of sine and cosine:
\(\csc x = \frac{1}{\sin x}\) and \(\cot x = \frac{\cos x}{\sin x}\)
Substitute into the limit expression:
\(= \lim_{x \to 0} \left( \frac{1}{\sin x} - \frac{\cos x}{\sin x} \right)\)
Combine terms under a common denominator:
\(= \lim_{x \to 0} \frac{1 - \cos x}{\sin x}\)
Direct substitution gives \(\frac{0}{0}\). Use half-angle trigonometric identities:
\(1 - \cos x = 2 \sin^2\left(\frac{x}{2}\right)\)
\(\sin x = 2 \sin\left(\frac{x}{2}\right) \cos\left(\frac{x}{2}\right)\)
Substitute these identities into the limit:
\(= \lim_{x \to 0} \frac{2 \sin^2\left(\frac{x}{2}\right)}{2 \sin\left(\frac{x}{2}\right) \cos\left(\frac{x}{2}\right)}\)
Cancel common terms (\(2\) and \(\sin\left(\frac{x}{2}\right)\)):
\(= \lim_{x \to 0} \frac{\sin\left(\frac{x}{2}\right)}{\cos\left(\frac{x}{2}\right)} = \lim_{x \to 0} \tan\left(\frac{x}{2}\right)\)
Substitute \(x = 0\):
\(= \tan(0) = 0\)
In simple words: Convert cosecant and cotangent to sine and cosine, combine fractions, and use half-angle formulas to simplify.

Exam Tip: Half-angle identities are powerful tools for resolving \(\frac{0}{0}\) indeterminate forms involving trigonometric functions.