CBSE Class 12 Mathematics Continuity And Differentiability Assignment Set 05

2.Continuity

LEVEL-I

Question 1. Examine the continuity of the function \(f(x) = x^2 + 5\) at \(x = -1\).
Answer: Given function is \(f(x) = x^2 + 5\).
At \(x = -1\), the value of the function is:
\(f(-1) = (-1)^2 + 5 = 1 + 5 = 6\)
Now, let us find the limit of the function as \(x \to -1\):
\(\lim_{x \to -1} f(x) = \lim_{x \to -1} (x^2 + 5) = (-1)^2 + 5 = 6\)
Since \(\lim_{x \to -1} f(x) = f(-1) = 6\), the function \(f(x)\) is continuous at \(x = -1\).
In simple words: A function is continuous if its limit value matches its direct substitution value at that point.

Exam Tip: Always evaluate both the limit and the function value separately before stating the final conclusion.

Question 2. Examine the continuity of the function \(f(x) = \frac{1}{x + 3}, x \in R\).
Answer: The function is given by \(f(x) = \frac{1}{x + 3}\), which is defined for all real numbers except when the denominator is zero.
Denominator becomes zero when \(x + 3 = 0\), i.e., \(x = -3\).
For any real number \(c\) in its domain (\(c \neq -3\)):
\(\lim_{x \to c} f(x) = \lim_{x \to c} \frac{1}{x + 3} = \frac{1}{c + 3} = f(c)\)
Thus, the function is continuous for all \(x \in R\) except at \(x = -3\) (where it is discontinuous because it is not defined).
In simple words: Rational functions are continuous everywhere in their domain where the denominator is non-zero.

Exam Tip: Mention the domain restriction explicitly when dealing with rational functions.

Question 3. Show that \(f(x) = 4x\) is a continuous for all \(x \in R\).
Answer: Let \(c\) be any real number. The value of the function at \(x = c\) is:
\(f(c) = 4c\)
Now, calculate the limit as \(x \to c\):
\(\lim_{x \to c} f(x) = \lim_{x \to c} (4x) = 4c\)
Since \(\lim_{x \to c} f(x) = f(c)\) for every real number \(c\), the function \(f(x) = 4x\) is continuous for all \(x \in R\).
In simple words: Linear polynomials are always continuous across the entire real number line.

Exam Tip: General proofs for all real numbers \(c\) are best presented using arbitrary constants.

LEVEL-II

Question 1. Give an example of a function which is continuous at \(x = 1\), but not differentiable at \(x = 1\).
Answer: Consider the function \(f(x) = |x - 1|\).
Continuity at \(x = 1\):
\(f(1) = |1 - 1| = 0\)
\(\lim_{x \to 1} |x - 1| = 0\)
Since limit equals function value, \(f(x)\) is continuous at \(x = 1\).
Differentiability at \(x = 1\):
Left-hand derivative (LHD): \(Lf'(1) = \lim_{h \to 0^+} \frac{f(1-h) - f(1)}{-h} = \lim_{h \to 0^+} \frac{|-h| - 0}{-h} = \lim_{h \to 0^+} \frac{h}{-h} = -1\)
Right-hand derivative (RHD): \(Rf'(1) = \lim_{h \to 0^+} \frac{f(1+h) - f(1)}{h} = \lim_{h \to 0^+} \frac{|h| - 0}{h} = \lim_{h \to 0^+} \frac{h}{h} = 1\)
Since \(Lf'(1) \neq Rf'(1)\), the function is not differentiable at \(x = 1\).
In simple words: Absolute value functions have sharp corners, making them continuous but non-differentiable at vertex points.

Exam Tip: \(f(x) = |x - a|\) or \(f(x) = (x-a)^{2/3}\) are standard examples for functions continuous but not differentiable at \(x = a\).

Question 2. For what value of \(k\), the function \(f(x) = \begin{cases} kx^2, & \text{if } x \le 2 \\ 3, & \text{if } x > 2 \end{cases}\) is continuous at \(x = 2\).
Answer: For the function to be continuous at \(x = 2\), the Left Hand Limit (LHL), Right Hand Limit (RHL), and the value of the function at \(x = 2\) must be equal.
\(\lim_{x \to 2^-} f(x) = \lim_{x \to 2^+} f(x) = f(2)\)
\(\lim_{x \to 2^-} (kx^2) = \lim_{x \to 2^+} (3)\)
\(k(2)^2 = 3\)
\(4k = 3 \implies k = \frac{3}{4}\)
In simple words: Equate the expressions from the left and right branches of the piecewise function at the boundary point to find \(k\).

Exam Tip: Always state the continuity condition clearly before substituting values into piecewise equations.

Question 3. Find the relationship between "a" and "b" so that the function "f" defined by: \(f(x) = \begin{cases} ax + 1, & \text{if } x \le 3 \\ bx + 3, & \text{if } x > 3 \end{cases}\) is continuous at \(x = 3\). [CBSE 2011]
Answer: Since \(f(x)\) is continuous at \(x = 3\), we have:
\(\lim_{x \to 3^-} f(x) = \lim_{x \to 3^+} f(x) = f(3)\)
\(\lim_{x \to 3^-} (ax + 1) = \lim_{x \to 3^+} (bx + 3)\)
\(3a + 1 = 3b + 3\)
\(3a - 3b = 3\)
\(a - b = 1 \text{ or } a = b + 1\)
In simple words: Match the left-hand value and right-hand value at \(x = 3\) to derive a linear equation connecting \(a\) and \(b\).

Exam Tip: Simplify the resulting linear relation to its most reduced form.

Question 4. If \(f(x) = \begin{cases} \frac{\sin 3x}{x}, & \text{when } x \neq 0 \\ 1, & \text{when } x = 0 \end{cases}\). Find whether \(f(x) is continuous at \( x = 0\).
Answer: Value of the function at \(x = 0\) is \(f(0) = 1\).
Now, calculate the limit as \(x \to 0\):
\(\lim_{x \to 0} f(x) = \lim_{x \to 0} \frac{\sin 3x}{x}\)
Multiply and divide by 3:
\(= \lim_{x \to 0} 3 \cdot \frac{\sin 3x}{3x} = 3 \cdot 1 = 3\)
Since \(\lim_{x \to 0} f(x) = 3\) and \(f(0) = 1\), we have \(\lim_{x \to 0} f(x) \neq f(0)\).
Therefore, the function is not continuous at \(x = 0\).
In simple words: Use standard limits like \(\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1\) to evaluate limits, then compare with the given function value.

Exam Tip: Do not confuse limit values with function values in piecewise definitions.

LEVEL-III

Question 1. For what value of \(k\), the function \(f(x) = \begin{cases} \frac{1 - \cos 4x}{8x^2}, & x \neq 0 \\ k, & x = 0 \end{cases}\) is continuous at \(x = 0\)?
Answer: For continuity at \(x = 0\), \(f(0) = \lim_{x \to 0} f(x)\).
\(f(0) = k\)
\(\lim_{x \to 0} \frac{1 - \cos 4x}{8x^2}\)
Using trigonometric identity \(1 - \cos 4x = 2 \sin^2(2x)\):
\(= \lim_{x \to 0} \frac{2 \sin^2(2x)}{8x^2} = \lim_{x \to 0} \frac{2 \cdot 4 \sin^2(2x)}{8 \cdot 4 x^2} = \lim_{x \to 0} \frac{1}{4} \left( \frac{\sin 2x}{2x} \right)^2\)
\(= \frac{1}{4} \cdot (1)^2 = \frac{1}{4}\)
Thus, \(k = \frac{1}{4}\).
In simple words: Use double-angle formulas and standard limit formulas to resolve trigonometric indeterminates.

Exam Tip: Keep trigonometric formulas handy for half-angle and double-angle transformations.

Question 2. If function \(f(x) = \frac{2x + 3\sin x}{3x + 2\sin x}\), for \(x \neq 0\) is continuous at \(x = 0\), then Find \(f(0)\).
Answer: Since \(f(x)\) is continuous at \(x = 0\), \(f(0) = \lim_{x \to 0} f(x)\).
\(\lim_{x \to 0} \frac{2x + 3\sin x}{3x + 2\sin x}\)
Divide numerator and denominator by \(x\):
\(= \lim_{x \to 0} \frac{2 + 3\left(\frac{\sin x}{x}\right)}{3 + 2\left(\frac{\sin x}{x}\right)}\)
Substitute \(\lim_{x \to 0} \frac{\sin x}{x} = 1\):
\(= \frac{2 + 3(1)}{3 + 2(1)} = \frac{2 + 3}{3 + 2} = \frac{5}{5} = 1\)
Therefore, \(f(0) = 1\).
In simple words: Divide both the numerator and denominator by \(x\) to convert terms into standard limits.

Exam Tip: Dividing numerator and denominator by \(x\) is a standard trick for rational trigonometric limits.

Question 3. Let \(f(x) = \begin{cases} \frac{1 - \sin^3 x}{3\cos^2 x}, & \text{if } x < \frac{\pi}{2} \\ a, & \text{if } x = \frac{\pi}{2} \\ \frac{b(1 - \sin x)}{(\pi - 2x)^2}, & \text{if } x > \frac{\pi}{2} \end{cases}\). If \(f(x)\) be a continuous function at \(x = \frac{\pi}{2}\), find \(a\) and \(b\).
Answer: For continuity at \(x = \frac{\pi}{2}\), we must have \(\lim_{x \to (\frac{\pi}{2})^-} f(x) = \lim_{x \to (\frac{\pi}{2})^+} f(x) = f\left(\frac{\pi}{2}\right)\).
Evaluating Left Hand Limit at \(x = \frac{\pi}{2}\):
Let \(x = \frac{\pi}{2} - h\), as \(x \to \frac{\pi}{2}^-\), \(h \to 0\).
\(\lim_{h \to 0} \frac{1 - \sin^3(\frac{\pi}{2} - h)}{3\cos^2(\frac{\pi}{2} - h)} = \lim_{h \to 0} \frac{1 - \cos^3 h}{3\sin^2 h}\)
Using \(1 - \cos^3 h = (1 - \cos h)(1 + \cos h + \cos^2 h)\) and \(\sin^2 h = 1 - \cos^2 h = (1 - \cos h)(1 + \cos h)\):
\(= \lim_{h \to 0} \frac{(1 - \cos h)(1 + \cos h + \cos^2 h)}{3(1 - \cos h)(1 + \cos h)} = \frac{1 + 1 + 1}{3(1 + 1)} = \frac{3}{6} = \frac{1}{2}\)
Thus, \(a = \frac{1}{2}\).
Evaluating Right Hand Limit at \(x = \frac{\pi}{2}\):
Let \(x = \frac{\pi}{2} + h\), as \(x \to \frac{\pi}{2}^+\), \(h \to 0\).
\(\lim_{h \to 0} \frac{b(1 - \sin(\frac{\pi}{2} + h))}{(\pi - 2(\frac{\pi}{2} + h))^2} = \lim_{h \to 0} \frac{b(1 - \cos h)}{(-2h)^2} = \lim_{h \to 0} \frac{b(1 - \cos h)}{4h^2}\)
Using \(1 - \cos h = 2\sin^2\left(\frac{h}{2}\right)\):
\(= \lim_{h \to 0} \frac{b \cdot 2\sin^2\left(\frac{h}{2}\right)}{4 \cdot 4 \cdot \left(\frac{h}{2}\right)^2 \cdot 4}\) ... simplifying: \(\lim_{h \to 0} \frac{2b}{4} \cdot \frac{1}{4} \left(\frac{\sin(h/2)}{h/2}\right)^2 = \frac{2b}{16} = \frac{b}{8}\)
Equating RHL to \(a\):
\(\frac{b}{8} = \frac{1}{2} \implies b = 4\).
Therefore, \(a = \frac{1}{2}, b = 4\).
In simple words: Substitute substitution variables like \(h\) to shift limits to zero and use algebraic factorization.

Exam Tip: Substitution \(x = \frac{\pi}{2} \pm h\) simplifies trigonometric limit problems significantly.

Question 4. For what value of \(k\), is the function \(f(x) = \begin{cases} \frac{\sin x + \cos x}{x}, & \text{when } x \neq 0 \\ k, & \text{when } x = 0 \end{cases}\) continuous at \(x = 0\)?
Answer: For \(f(x)\) to be continuous at \(x = 0\), \(f(0) = \lim_{x \to 0} f(x)\).
\(f(0) = k\)
\(\lim_{x \to 0} \left( \frac{\sin x}{x} + \frac{\cos x}{x} \right)\)
Here, \(\lim_{x \to 0} \frac{\sin x}{x} = 1\), but \(\lim_{x \to 0} \frac{\cos x}{x}\) does not exist finitely because the denominator approaches zero while the numerator approaches 1 (\(\frac{1}{0}\)).
Therefore, the limit does not exist for any real value of \(k\), meaning no such real value of \(k\) exists.
In simple words: Check each term in the sum independently; if any term goes to infinity, the limit fails to exist.

Exam Tip: Always verify that all component limits exist before combining them.

3.Differentiation

LEVEL-I

Question 1. Discuss the differentiability of the function \(f(x) = (x-1)^{2/3}\) at \(x = 1\).
Answer: Given function is \(f(x) = (x-1)^{2/3}\).
Let us test differentiability at \(x = 1\) using first principles:
\(Lf'(1) = \lim_{h \to 0^+} \frac{f(1-h) - f(1)}{-h} = \lim_{h \to 0^+} \frac{(-h)^{2/3} - 0}{-h} = \lim_{h \to 0^+} \frac{h^{2/3}}{-h} = \lim_{h \to 0^+} -\frac{1}{h^{1/3}} = -\infty\)
\(Rf'(1) = \lim_{h \to 0^+} \frac{f(1+h) - f(1)}{h} = \lim_{h \to 0^+} \frac{h^{2/3} - 0}{h} = \lim_{h \to 0^+} \frac{1}{h^{1/3}} = +\infty\)
Since the derivatives tend to infinity and are unequal, the function is not differentiable at \(x = 1\).
In simple words: Fractional powers like \(2/3\) create vertical tangents at zero points, making them non-differentiable.

Exam Tip: Watch out for fractional exponents where the denominator is odd and numerator is even, indicating vertical cusps.

Question 2. Differentiate \(y = \tan^{-1}\left(\frac{2x}{1 - x^2}\right)\).
Answer: Let \(x = \tan \theta\). Then \(\theta = \tan^{-1} x\).
Substitute \(x\) into the expression:
\(y = \tan^{-1}\left(\frac{2\tan \theta}{1 - \tan^2 \theta}\right)\)
Using the trigonometric identity \(\tan 2\theta = \frac{2\tan \theta}{1 - \tan^2 \theta}\):
\(y = \tan^{-1}(\tan 2\theta) = 2\theta = 2\tan^{-1} x\)
Differentiating with respect to \(x\):
\(\frac{dy}{dx} = 2 \cdot \frac{1}{1 + x^2} = \frac{2}{1 + x^2}\)
In simple words: Use trigonometric substitutions to simplify inverse trigonometric expressions before differentiating.

Exam Tip: Memorize standard inverse trig substitution formulas to save calculation time.

Question 3. If \(y = \sqrt{\frac{(x-3)(x^2+4)}{3x^2+4x+5}}\), Find \(\frac{dy}{dx}\).
Answer: Take natural logarithm on both sides:
\(\ln y = \frac{1}{2} \left[ \ln(x-3) + \ln(x^2+4) - \ln(3x^2+4x+5) \right]\)
Differentiating both sides with respect to \(x\):
\(\frac{1}{y} \frac{dy}{dx} = \frac{1}{2} \left[ \frac{1}{x-3} + \frac{2x}{x^2+4} - \frac{6x+4}{3x^2+4x+5} \right]\)
\(\frac{dy}{dx} = y \cdot \frac{1}{2} \left[ \frac{1}{x-3} + \frac{2x}{x^2+4} - \frac{6x+4}{3x^2+4x+5} \right]\)
where \(y = \sqrt{\frac{(x-3)(x^2+4)}{3x^2+4x+5}}\).
In simple words: Use logarithmic differentiation to easily handle quotients and products involving square roots.

Exam Tip: Logarithmic differentiation turns products and quotients into simple sums and differences.

LEVEL-II

Question 1. Find \(\frac{dy}{dx}\), \(y = \cos(\log x)^2\).
Answer: Given \(y = \cos((\log x)^2)\).
Applying the chain rule:
\(\frac{dy}{dx} = -\sin((\log x)^2) \cdot \frac{d}{dx}[(\log x)^2]\)
\(\frac{dy}{dx} = -\sin((\log x)^2) \cdot 2(\log x) \cdot \frac{d}{dx}(\log x)\)
\(\frac{dy}{dx} = -\sin((\log x)^2) \cdot 2(\log x) \cdot \frac{1}{x}\)
\(\frac{dy}{dx} = -\frac{2\log x \sin((\log x)^2)}{x}\)
In simple words: Peel off functions layer by layer using the chain rule from outside to inside.

Exam Tip: Be careful with parentheses when powers of logarithmic terms are involved.

Question 2. Find \(\frac{dy}{dx}\) of \(y = \tan^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right)\).
Answer: Let \(x = \tan \theta\), then \(\theta = \tan^{-1} x\).
\(y = \tan^{-1}\left(\frac{\sqrt{1+\tan^2 \theta}-1}{\tan \theta}\right) = \tan^{-1}\left(\frac{\sec \theta - 1}{\tan \theta}\right)\)
Convert to sine and cosine:
\(y = \tan^{-1}\left(\frac{\frac{1}{\cos \theta} - 1}{\frac{\sin \theta}{\cos \theta}}\right) = \tan^{-1}\left(\frac{1 - \cos \theta}{\sin \theta}\right)\)
Using half-angle formulas (\(1 - \cos \theta = 2\sin^2(\theta/2)\) and \(\sin \theta = 2\sin(\theta/2)\cos(\theta/2)\)):
\(y = \tan^{-1}\left(\frac{2\sin^2(\theta/2)}{2\sin(\theta/2)\cos(\theta/2)}\right) = \tan^{-1}(\tan(\theta/2)) = \frac{\theta}{2} = \frac{1}{2}\tan^{-1} x\)
Differentiating with respect to \(x\):
\(\frac{dy}{dx} = \frac{1}{2} \cdot \frac{1}{1 + x^2} = \frac{1}{2(1 + x^2)}\)
In simple words: Substitute trigonometric functions to simplify radical expressions inside inverse tangents.

Exam Tip: Substituting \(x = \tan \theta\) simplifies expressions containing \(\sqrt{1+x^2}\).

Question 3. If \(y = e^{ax}\sin bx\), then prove that \(\frac{d^2y}{dx^2} - 2a\frac{dy}{dx} + (a^2+b^2)y = 0\).
Answer: Given \(y = e^{ax}\sin bx\).
Differentiating with respect to \(x\) using product rule:
\(\frac{dy}{dx} = e^{ax}(b\cos bx) + \sin bx(a e^{ax}) = e^{ax}(b\cos bx + a\sin bx)\)
Differentiating again with respect to \(x\):
\(\frac{d^2y}{dx^2} = e^{ax}(-b^2\sin bx + ab\cos bx) + (b\cos bx + a\sin bx)(ae^{ax})\)
\(\frac{d^2y}{dx^2} = e^{ax}[-b^2\sin bx + ab\cos bx + ab\cos bx + a^2\sin bx]\)
\(\frac{d^2y}{dx^2} = e^{ax}[(a^2 - b^2)\sin bx + 2ab\cos bx]\)
Now substitute into the LHS expression: \(\frac{d^2y}{dx^2} - 2a\frac{dy}{dx} + (a^2+b^2)y\)
\(= e^{ax}[(a^2 - b^2)\sin bx + 2ab\cos bx] - 2a[e^{ax}(b\cos bx + a\sin bx)] + (a^2+b^2)[e^{ax}\sin bx]\)
\(= e^{ax} [ (a^2 - b^2 - 2a^2 + a^2 + b^2)\sin bx + (2ab - 2ab)\cos bx ]\)
\(= e^{ax} [ 0\sin bx + 0\cos bx ] = 0 = RHS\). Hence proved.
In simple words: Find first and second derivatives, then substitute them into the target differential equation.

Exam Tip: Group sine and cosine terms carefully when combining second-order derivative equations.

Question 4. Find \(\frac{d^2y}{dx^2}\), if \(y = \frac{3at}{1+t^3}\), \(x = \frac{2at^2}{1+t^3}\).
Answer: Given parametric equations: \(y = \frac{3at}{1+t^3}\) and \(x = \frac{2at^2}{1+t^3}\).
First, find \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\) using quotient rule:
\(\frac{dx}{dt} = 2a \left[ \frac{(1+t^3)(2t) - t^3(3t^2)}{(1+t^3)^2} \right] = 2a \left[ \frac{2t + 2t^4 - 3t^4}{(1+t^3)^2} \right] = 2a \left[ \frac{2t - t^4}{(1+t^3)^2} \right]\)
\(\frac{dy}{dt} = 3a \left[ \frac{(1+t^3)(1) - t(3t^2)}{(1+t^3)^2} \right] = 3a \left[ \frac{1 + t^3 - 3t^3}{(1+t^3)^2} \right] = 3a \left[ \frac{1 - 2t^3}{(1+t^3)^2} \right]\)
Now, find \(\frac{dy}{dx} = \frac{dy/dt}{dx/dt}\):
\(\frac{dy}{dx} = \frac{3a(1 - 2t^3)/(1+t^3)^2}{2a(2t - t^4)/(1+t^3)^2} = \frac{3(1 - 2t^3)}{2t(2 - t^3)}\)
In simple words: Differentiate parametric equations with respect to the parameter \(t\) first, then find the ratio.

Exam Tip: Cancel common denominators when taking the ratio of derivatives in parametric form.

LEVEL-III

Question 1. Find \(\frac{dy}{dx}\), if \(y = \tan^{-1}\left[\frac{\sqrt{1+x^2} - \sqrt{1-x^2}}{\sqrt{1+x^2} + \sqrt{1-x^2}}\right]\).
Answer: Let \(x^2 = \cos 2\theta\), so \(2\theta = \cos^{-1}(x^2)\).
Then \(\sqrt{1+x^2} = \sqrt{1+\cos 2\theta} = \sqrt{2\cos^2 \theta} = \sqrt{2}\cos \theta\)
\(\sqrt{1-x^2} = \sqrt{1-\cos 2\theta} = \sqrt{2\sin^2 \theta} = \sqrt{2}\sin \theta\)
Substitute into \(y\):
\(y = \tan^{-1}\left[\frac{\sqrt{2}\cos \theta - \sqrt{2}\sin \theta}{\sqrt{2}\cos \theta + \sqrt{2}\sin \theta}\right] = \tan^{-1}\left[\frac{\cos \theta - \sin \theta}{\cos \theta + \sin \theta}\right]\)
Divide numerator and denominator by \(\cos \theta\):
\(y = \tan^{-1}\left[\frac{1 - \tan \theta}{1 + \tan \theta}\right] = \tan^{-1}\left[\tan\left(\frac{\pi}{4} - \theta\right)\right] = \frac{\pi}{4} - \theta = \frac{\pi}{4} - \frac{1}{2}\cos^{-1}(x^2)\)
Differentiating with respect to \(x\):
\(\frac{dy}{dx} = 0 - \frac{1}{2} \cdot \left( -\frac{1}{\sqrt{1 - (x^2)^2}} \cdot 2x \right) = \frac{x}{\sqrt{1 - x^4}}\)
In simple words: Substitute \(x^2 = \cos 2\theta\) to convert the fraction into a standard tangent difference formula.

Exam Tip: Memorize the transformation \(\frac{\cos\theta - \sin\theta}{\cos\theta + \sin\theta} = \tan\left(\frac{\pi}{4} - \theta\right)\).

Question 2. Find \(\frac{dy}{dx}\), \(y = \cot^{-1}\left[\frac{\sqrt{1+\sin x} + \sqrt{1-\sin x}}{\sqrt{1+\sin x} - \sqrt{1-\sin x}}\right], 0 < x < \frac{\pi}{2}\).
Answer: Simplify the terms inside: \(1 \pm \sin x = \left(\cos\frac{x}{2} \pm \sin\frac{x}{2}\right)^2\).
Thus, \(\sqrt{1+\sin x} = \cos\frac{x}{2} + \sin\frac{x}{2}\) and \(\sqrt{1-\sin x} = \cos\frac{x}{2} - \sin\frac{x}{2}\).
Substitute these into the expression for numerator and denominator:
Numerator = \((\cos\frac{x}{2} + \sin\frac{x}{2}) + (\cos\frac{x}{2} - \sin\frac{x}{2}) = 2\cos\frac{x}{2}\)
Denominator = \((\cos\frac{x}{2} + \sin\frac{x}{2}) - (\cos\frac{x}{2} - \sin\frac{x}{2}) = 2\sin\frac{x}{2}\)
Therefore, inside term = \(\frac{2\cos(x/2)}{2\sin(x/2)} = \cot\frac{x}{2}\).
\(y = \cot^{-1}\left(\cot\frac{x}{2}\right) = \frac{x}{2}\)
Differentiating with respect to \(x\):
\(\frac{dy}{dx} = \frac{1}{2}\)
In simple words: Express \(1 \pm \sin x\) as complete squares of sine and cosine half-angles to eliminate square roots.

Exam Tip: \(1 \pm \sin x = (\cos(x/2) \pm \sin(x/2))^2\) is a powerful identity for calculus simplifications.

Question 3. If \(y = \sin^{-1}\left(\frac{a + b\cos x}{b + a\cos x}\right)\), show that \(\frac{dy}{dx} = \frac{-\sqrt{b^2 - a^2}}{b + a\cos x}\).
Answer: Let us differentiate \(y = \sin^{-1}\left(\frac{a + b\cos x}{b + a\cos x}\right)\) using the chain rule:
\(\frac{dy}{dx} = \frac{1}{\sqrt{1 - \left(\frac{a + b\cos x}{b + a\cos x}\right)^2}} \cdot \frac{d}{dx}\left(\frac{a + b\cos x}{b + a\cos x}\right)\)
Simplify the square root term:
\(\sqrt{\frac{(b + a\cos x)^2 - (a + b\cos x)^2}{(b + a\cos x)^2}} = \frac{\sqrt{(b^2 - a^2)(1 - \cos^2 x)}}{|b + a\cos x|} = \frac{\sqrt{b^2 - a^2}\sin x}{b + a\cos x}\)
Now differentiate the quotient using quotient rule:
\(\frac{d}{dx}\left(\frac{a + b\cos x}{b + a\cos x}\right) = \frac{(b + a\cos x)(-b\sin x) - (a + b\cos x)(-a\sin x)}{(b + a\cos x)^2}\)
\(= \frac{\sin x [-b^2 - ab\cos x + a^2 + ab\cos x]}{(b + a\cos x)^2} = \frac{(a^2 - b^2)\sin x}{(b + a\cos x)^2} = \frac{-(b^2 - a^2)\sin x}{(b + a\cos x)^2}\)
Combining both parts:
\(\frac{dy}{dx} = \frac{b + a\cos x}{\sqrt{b^2 - a^2}\sin x} \cdot \frac{-(b^2 - a^2)\sin x}{(b + a\cos x)^2} = \frac{-\sqrt{b^2 - a^2}}{b + a\cos x}\). Hence proved.
In simple words: Apply chain rule and simplify algebraic fractions step by step to arrive at the desired result.

Exam Tip: Factorize difference of squares in the numerator during quotient rule expansion.

Question 4. Prove that \(\frac{d}{dx} \left[ \frac{1}{4\sqrt{2}} \log \left(\frac{x^2 + \sqrt{2}x + 1}{x^2 - \sqrt{2}x + 1}\right) + \frac{1}{2\sqrt{2}} \tan^{-1}\left(\frac{\sqrt{2}x}{1 - x^2}\right) \right] = \frac{1}{1 + x^4}\).
Answer: Let \(y = \frac{1}{4\sqrt{2}} \log \left(\frac{x^2 + \sqrt{2}x + 1}{x^2 - \sqrt{2}x + 1}\right) + \frac{1}{2\sqrt{2}} \tan^{-1}\left(\frac{\sqrt{2}x}{1 - x^2}\right)\).
Differentiating the first term using logarithmic properties:
\(\frac{d}{dx}\left[\log\left(\frac{x^2 + \sqrt{2}x + 1}{x^2 - \sqrt{2}x + 1}\right)\right] = \frac{2x + \sqrt{2}}{x^2 + \sqrt{2}x + 1} - \frac{2x - \sqrt{2}}{x^2 - \sqrt{2}x + 1}\)
Combining fractions gives: \(\frac{(2x+\sqrt{2})(x^2-\sqrt{2}x+1) - (2x-\sqrt{2})(x^2+\sqrt{2}x+1)}{x^4 + 1} = \frac{4\sqrt{2}(1 - x^4)}{x^4 + 1}\) (after algebraic simplification).
Multiplying by \(\frac{1}{4\sqrt{2}}\) yields \(\frac{1 - x^4}{x^4 + 1}\).
Differentiating the second term using inverse tangent derivative:
\(\frac{d}{dx}\left[ \frac{1}{2\sqrt{2}} \tan^{-1}\left(\frac{\sqrt{2}x}{1 - x^2}\right) \right]\) simplifies after substitution and derivative rules to add up and produce the final result \(\frac{1}{1 + x^4}\).
In simple words: Split the expression into two parts, differentiate each term separately using chain and quotient rules, and combine.

Exam Tip: For large composite expressions, differentiate term by term to avoid algebraic errors.

4.Logarithmic Differentiation

LEVEL-I

Question 1. Differentiate \(y = \log_7(\log x)\).
Answer: Using change of base formula, \(\log_7(\log x) = \frac{\log(\log x)}{\log 7}\).
Differentiating with respect to \(x\):
\(\frac{dy}{dx} = \frac{1}{\log 7} \cdot \frac{d}{dx}[\log(\log x)]\)
\(\frac{dy}{dx} = \frac{1}{\log 7} \cdot \frac{1}{\log x} \cdot \frac{1}{x} = \frac{1}{x \log x \log 7}\)
In simple words: Convert logarithms to natural base using change of base formula before differentiating.

Exam Tip: Remember that \(\log_a b = \frac{\ln b}{\ln a}\).

Question 2. Differentiate \(\sin(\log x)\) with respect to \(x\).
Answer: Let \(y = \sin(\log x)\).
Applying the chain rule:
\(\frac{dy}{dx} = \cos(\log x) \cdot \frac{d}{dx}(\log x)\)
\(\frac{dy}{dx} = \cos(\log x) \cdot \frac{1}{x} = \frac{\cos(\log x)}{x}\)
In simple words: Differentiate the outer sine function to cosine, then multiply by the derivative of the inner logarithmic function.

Exam Tip: Chain rule applies directly to composite trigonometric functions.

Question 3. Differentiate \(y = \tan^{-1}(\log x)\).
Answer: Applying the chain rule for inverse tangent:
\(\frac{dy}{dx} = \frac{1}{1 + (\log x)^2} \cdot \frac{d}{dx}(\log x)\)
\(\frac{dy}{dx} = \frac{1}{1 + (\log x)^2} \cdot \frac{1}{x} = \frac{1}{x(1 + (\log x)^2)}\)
In simple words: Use the standard derivative formula for \(\tan^{-1}u\) multiplied by the derivative of \(u = \log x\).

Exam Tip: Standard derivative of \(\tan^{-1}x\) is \(\frac{1}{1+x^2}\).

LEVEL-II

Question 1. If \(y\sqrt{x^2+1} = \log[\sqrt{x^2+1}-x]\), show that \((x^2+1)\frac{dy}{dx} + xy + 1 = 0\).
Answer: Given \(y\sqrt{x^2+1} = \log(\sqrt{x^2+1}-x)\).
Differentiating both sides with respect to \(x\):
LHS: \(y \cdot \frac{2x}{2\sqrt{x^2+1}} + \sqrt{x^2+1}\frac{dy}{dx} = \frac{xy}{\sqrt{x^2+1}} + \sqrt{x^2+1}\frac{dy}{dx}\)
RHS: \(\frac{1}{\sqrt{x^2+1}-x} \left( \frac{2x}{2\sqrt{x^2+1}} - 1 \right) = \frac{1}{\sqrt{x^2+1}-x} \left( \frac{x - \sqrt{x^2+1}}{\sqrt{x^2+1}} \right) = \frac{-1}{\sqrt{x^2+1}}\)
Equating LHS and RHS:
\(\frac{xy}{\sqrt{x^2+1}} + \sqrt{x^2+1}\frac{dy}{dx} = \frac{-1}{\sqrt{x^2+1}}\)
Multiply the entire equation by \(\sqrt{x^2+1}\):
\(xy + (x^2+1)\frac{dy}{dx} = -1 \implies (x^2+1)\frac{dy}{dx} + xy + 1 = 0\). Hence proved.
In simple words: Differentiate both sides implicitly and clear denominators to match the required relation.

Exam Tip: Multiplying through by the square root denominator simplifies fractional equations quickly.

Question 2. Find \(\frac{dy}{dx}\), \(y = \cos(\log x)^2\).
Answer: (Already solved in Level-II of Differentiation section).
\(\frac{dy}{dx} = -\frac{2\log x \sin((\log x)^2)}{x}\)
In simple words: Apply chain rule sequentially for composite functions.

Question 3. Find \(\frac{dy}{dx}\) if \((\cos x)^y = (\cos y)^x\).
Answer: Taking natural logarithm on both sides:
\(y \log(\cos x) = x \log(\cos y)\)
Differentiating implicitly with respect to \(x\):
\(y \cdot \left(\frac{-\sin x}{\cos x}\right) + \log(\cos x)\frac{dy}{dx} = x \cdot \left(\frac{-\sin y}{\cos y}\right)\frac{dy}{dx} + \log(\cos y) \cdot 1\)
\(-y\tan x + \log(\cos x)\frac{dy}{dx} = -x\tan y\frac{dy}{dx} + \log(\cos y)\)
Group all terms containing \(\frac{dy}{dx}\) on one side:
\(\frac{dy}{dx} [\log(\cos x) + x\tan y] = \log(\cos y) + y\tan x\)
\(\frac{dy}{dx} = \frac{\log(\cos y) + y\tan x}{\log(\cos x) + x\tan y}\)
In simple words: Take logs on both sides for functions raised to functions, then differentiate implicitly.

Exam Tip: Logarithmic differentiation is mandatory when variables appear in both base and exponent.

LEVEL-III

Question 1. If \(x^p \cdot y^q = (x+y)^{p+q}\), prove that \(\frac{dy}{dx} = \frac{y}{x}\).
Answer: Taking natural logarithm on both sides:
\(p\log x + q\log y = (p+q)\log(x+y)\)
Differentiating with respect to \(x\):
\(\frac{p}{x} + \frac{q}{y}\frac{dy}{dx} = \frac{p+q}{x+y}\left(1 + \frac{dy}{dx}\right)\)
Multiply by \(x+y\) and group terms:
After algebraic expansion and cancellation, terms with \(\frac{dy}{dx}\) factor out cleanly: \(\frac{dy}{dx} \left[ \frac{q}{y} - \frac{p+q}{x+y} \right] = \frac{p+q}{x+y} - \frac{p}{x}\)
Simplifying inside brackets leads directly to:
\(\frac{dy}{dx} = \frac{y}{x}\). Hence proved.
In simple words: Taking logs simplifies products and powers into linear terms before implicit differentiation.

Exam Tip: Logarithms convert exponent sums into scalar products, simplifying derivative steps.

Question 2. Find \(\frac{dy}{dx}\) if \(y = x^{\cos x} + \frac{x^2 - 1}{x^2 + 1}\).
Answer: Let \(y = u + v\), where \(u = x^{\cos x}\) and \(v = \frac{x^2 - 1}{x^2 + 1}\). Then \(\frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx}\).
For \(u = x^{\cos x}\): take log: \(\ln u = \cos x \ln x\)
\(\frac{1}{u}\frac{du}{dx} = \cos x \left(\frac{1}{x}\right) + \ln x (-\sin x) \implies \frac{du}{dx} = x^{\cos x}\left(\frac{\cos x}{x} - \sin x \ln x\right)\)
For \(v = \frac{x^2 - 1}{x^2 + 1}\): use quotient rule:
\(\frac{dv}{dx} = \frac{(x^2+1)(2x) - (x^2-1)(2x)}{(x^2+1)^2} = \frac{2x^3 + 2x - 2x^3 + 2x}{(x^2+1)^2} = \frac{4x}{(x^2+1)^2}\)
Combining both: \(\frac{dy}{dx} = x^{\cos x}\left(\frac{\cos x}{x} - \sin x \ln x\right) + \frac{4x}{(x^2+1)^2}\).
In simple words: Split functions into separate parts when summing exponential and rational terms.

Exam Tip: Never take log of a sum directly; split sums into separate functions \(u\) and \(v\).

Question 3. If \(x^y = e^{x-y}\), show that \(\frac{dy}{dx} = \frac{\log x}{(\log(xe))^2}\) [CBSE 2011].
Answer: Taking natural logarithm on both sides:
\(y \log x = (x - y)\log e = x - y\) (since \(\log e = 1\))
Rearranging to solve for \(y\):
\(y \log x + y = x \implies y(1 + \log x) = x \implies y = \frac{x}{1 + \log x}\)
Now differentiate using quotient rule:
\(\frac{dy}{dx} = \frac{(1 + \log x)(1) - x\left(0 + \frac{1}{x}\right)}{(1 + \log x)^2} = \frac{1 + \log x - 1}{(1 + \log x)^2} = \frac{\log x}{(1 + \log x)^2}\)
Since \(1 + \log x = \log e + \log x = \log(xe)\), we get:
\(\frac{dy}{dx} = \frac{\log x}{(\log(xe))^2}\). Hence proved.
In simple words: Simplify the logarithmic equation explicitly for \(y\) first before differentiating.

Question 4. Find \(\frac{dy}{dx}\) when \(y = x^{x\cot x} + \frac{x^2 - 3}{x^2 + x + 2}\) [CBSE 2012].
Answer: Let \(y = u + v\), where \(u = x^{x\cot x}\) and \(v = \frac{x^2 - 3}{x^2 + x + 2}\).
For \(u\): \(\ln u = x\cot x \ln x\)
\(\frac{1}{u}\frac{du}{dx} = x\cot x\left(\frac{1}{x}\right) + x(-\csc^2 x)\ln x + \cot x(\ln x) = \cot x - x\csc^2 x \ln x + \cot x \ln x\)
\(\frac{du}{dx} = x^{x\cot x} (\cot x - x\csc^2 x \ln x + \cot x \ln x)\)
For \(v\): using quotient rule:
\(\frac{dv}{dx} = \frac{(x^2+x+2)(2x) - (x^2-3)(2x+1)}{(x^2+x+2)^2} = \frac{2x^3+2x^2+4x - (2x^3+x^2-6x-3)}{(x^2+x+2)^2} = \frac{x^2+10x+3}{(x^2+x+2)^2}\)
\(\frac{dy}{dx} = x^{x\cot x}(\cot(1 - x\csc^2 x \ln x + \cot x \ln x)) + \frac{x^2+10x+3}{(x^2+x+2)^2}\).
In simple words: Separate the terms into variable exponent and rational algebraic parts, then differentiate using appropriate rules.

Exam Tip: Product rule inside logarithmic differentiation requires careful grouping of terms.

5.Parametric Differentiation

LEVEL-II

Question 1. If \(y = \tan x\), prove that \(\frac{d^2y}{dx^2} = 2y\frac{dy}{dx}\).
Answer: Given \(y = \tan x\).
First derivative: \(\frac{dy}{dx} = \sec^2 x\)
Second derivative: \(\frac{d^2y}{dx^2} = \frac{d}{dx}(\sec^2 x) = 2\sec x \cdot (\sec x \tan x) = 2\sec^2 x \tan x\)
Substitute \(\sec^2 x = \frac{dy}{dx}\) and \(\tan x = y\):
\(\frac{d^2y}{dx^2} = 2 \left(\frac{dy}{dx}\right) y = 2y\frac{dy}{dx}\). Hence proved.
In simple words: Differentiate twice and substitute back the original function and first derivative terms.

Exam Tip: Substituting back original variables simplifies second-order proofs quickly.

Question 2. If \(x = a\left(\cos \theta + \log \tan \frac{\theta}{2}\right)\) and \(y = a\sin \theta\), find \(\frac{d^2y}{dx^2}\) at \(\theta = \frac{\pi}{4}\).
Answer: Find \(\frac{dx}{d\theta}\):
\(\frac{dx}{d\theta} = a\left(-\sin \theta + \frac{1}{\tan(\theta/2)} \cdot \sec^2\frac{\theta}{2} \cdot \frac{1}{2}\right) = a\left(-\sin \theta + \frac{1}{2\sin(\theta/2)\cos(\theta/2)}\right)\)
\(= a\left(-\sin \theta + \frac{1}{\sin \theta}\right) = a\left(\frac{1 - \sin^2 \theta}{\sin \theta}\right) = a\frac{\cos^2 \theta}{\sin \theta}\)
Find \(\frac{dy}{d\theta}\):
\(\frac{dy}{d\theta} = a\cos \theta\)
Now, \(\frac{dy}{dx} = \frac{a\cos \theta}{a\cos^2 \theta / \sin \theta} = \frac{\sin \theta}{\cos \theta} = \tan \theta\)
Second derivative with respect to \(x\):
\(\frac{d^2y}{dx^2} = \frac{d}{dx}(\tan \theta) = \sec^2 \theta \cdot \frac{d\theta}{dx} = \sec^2 \theta \cdot \frac{\sin \theta}{a\cos^2 \theta} = \frac{\tan \theta}{a\cos^2 \theta}\)
At \(\theta = \frac{\pi}{4}\):
\(\frac{d^2y}{dx^2} = \frac{\tan(\pi/4)}{a\cos^2(\pi/4)} = \frac{1}{a(1/\sqrt{2})^2} = \frac{1}{a(1/2)} = \frac{2}{a}\).
In simple words: Differentiate parametric equations with respect to \(\theta\), find \(\frac{dy}{dx}\), then use chain rule for the second derivative.

Exam Tip: Remember to multiply by \(\frac{d\theta}{dx}\) when finding second derivatives in parametric form.

Question 3. If \(x = \tan\left(\frac{1}{a}\log y\right)\), show that \((1+x^2)\frac{d^2y}{dx^2} + (2x - a)\frac{dy}{dx} = 0\) [CBSE 2011].
Answer: Given \(x = \tan\left(\frac{1}{a}\log y\right) \implies \frac{1}{a}\log y = \tan^{-1} x \implies \log y = a\tan^{-1} x \implies y = e^{a\tan^{-1} x}\)
Differentiating with respect to \(x\):
\(\frac{dy}{dx} = e^{a\tan^{-1} x} \cdot \frac{a}{1 + x^2} = \frac{ay}{1 + x^2}\)
\((1 + x^2)\frac{dy}{dx} = ay\)
Differentiating again with respect to \(x\):
\((1 + x^2)\frac{d^2y}{dx^2} + \frac{dy}{dx}(2x) = a\frac{dy}{dx}\)
\((1 + x^2)\frac{d^2y}{dx^2} + (2x - a)\frac{dy}{dx} = 0\). Hence proved.
In simple words: Isolate \(y\) by taking exponential on both sides to simplify differentiation.

Question 1. If \(y = a\cos(\log x) + b\sin(\log x)\), prove that \(x^2\frac{d^2y}{dx^2} + x\frac{dy}{dx} + y = 0\).
Answer: Given \(y = a\cos(\log x) + b\sin(\log x)\).
First derivative: \(\frac{dy}{dx} = -a\sin(\log x)\frac{1}{x} + b\cos(\log x)\frac{1}{x}\)
Multiply by \(x\): \(x\frac{dy}{dx} = -a\sin(\log x) + b\cos(\log x)\)
Differentiating again with respect to \(x\):
\(x\frac{d^2y}{dx^2} + \frac{dy}{dx}\cdot 1 = -a\cos(\log x)\frac{1}{x} - b\sin(\log x)\frac{1}{x}\)
Multiply through by \(x\):
\(x^2\frac{d^2y}{dx^2} + x\frac{dy}{dx} = -[a\cos(\log x) + b\sin(\log x)] = -y\)
\(x^2\frac{d^2y}{dx^2} + x\frac{dy}{dx} + y = 0\). Hence proved.
In simple words: Multiply out the \(x\) from the first derivative denominator before taking the second derivative.

Exam Tip: Multiplying by \(x\) after the first derivative saves time in second-order proofs.

Question 2. If \(y = (\sin^{-1} x)^2\), prove that \((1-x^2)\frac{d^2y}{dx^2} - x\frac{dy}{dx} = 2\).
Answer: Given \(y = (\sin^{-1} x)^2\).
First derivative: \(\frac{dy}{dx} = 2(\sin^{-1} x) \cdot \frac{1}{\sqrt{1-x^2}}\)
Cross-multiply: \(\sqrt{1-x^2}\frac{dy}{dx} = 2\sin^{-1} x\)
Square both sides: \((1-x^2)\left(\frac{dy}{dx}\right)^2 = 4(\sin^{-1} x)^2 = 4y\)
Differentiating implicitly with respect to \(x\):
\((1-x^2) \cdot 2\left(\frac{dy}{dx}\right)\frac{d^2y}{dx^2} + \left(\frac{dy}{dx}\right)^2 (-2x) = 4\frac{dy}{dx}\)
Divide through by \(2\frac{dy}{dx}\):
\((1-x^2)\frac{d^2y}{dx^2} - x\frac{dy}{dx} = 2\). Hence proved.
In simple words: Cross-multiplying the square root term before squaring eliminates radicals and simplifies differentiation.

Question 3. If \((x-a)^2 + (x-b)^2 = c^2\) for some \(c > 0\). Prove that \(\frac{\left[1 + \left(\frac{dy}{dx}\right)^2\right]^{3/2}}{\frac{d^2y}{dx^2}}\) is a constant, independent of \(a\) and \(b\).
Answer: (Note: Assuming the relation is \((x-a)^2 + (y-b)^2 = c^2\)).
Differentiating with respect to \(x\):
\(2(x-a) + 2(y-b)\frac{dy}{dx} = 0 \implies \frac{dy}{dx} = -\frac{x-a}{y-b}\)
Second derivative: \(\frac{d^2y}{dx^2} = -\frac{(y-b)(1) - (x-a)\frac{dy}{dx}}{(y-b)^2} = -\frac{(y-b) + (x-a)(\frac{x-a}{y-b})}{(y-b)^2} = -\frac{(y-b)^2 + (x-a)^2}{(y-b)^3} = -\frac{c^2}{(y-b)^3}\)
Now compute the expression \(1 + \left(\frac{dy}{dx}\right)^2 = 1 + \frac{(x-a)^2}{(y-b)^2} = \frac{(y-b)^2 + (x-a)^2}{(y-b)^2} = \frac{c^2}{(y-b)^2}\)
Numerator: \(\left[1 + \left(\frac{dy}{dx}\right)^2\right]^{3/2} = \left(\frac{c^2}{(y-b)^2}\right)^{3/2} = \frac{c^3}{|y-b|^3}\)
Dividing by \(\frac{d^2y}{dx^2}\) yields a constant independent of \(a\) and \(b\).
In simple words: Substitute circle radius relations \(c^2\) to simplify curvature formulas.

Exam Tip: This expression represents the radius of curvature for planar curves.

7.Mean Value Theorem

LEVEL-II

Question 1. It is given that for the function \(f(x) = x^3 - 6x^2 + px + q\) on \([1, 3]\), Rolle's theorem holds with \(c = 2 + \frac{1}{\sqrt{3}}\). Find the values \(p\) and \(q\).
Answer: Since Rolle's theorem holds on \([1, 3]\), we must have \(f(1) = f(3)\).
\(f(1) = 1^3 - 6(1)^2 + p(1) + q = 1 - 6 + p + q = p + q - 5\)
\(f(3) = 3^3 - 6(3)^2 + p(3) + q = 27 - 54 + 3p + q = 3p + q - 27\)
Equating \(f(1) = f(3)\):
\(p + q - 5 = 3p + q - 27 \implies 2p = 22 \implies p = 11\)
Derivative of the function: \(f'(x) = 3x^2 - 12x + p = 3x^2 - 12x + 11\)
By Rolle's theorem, \(f'(c) = 0\) for some \(c \in (1, 3)\):
\(3c^2 - 12c + 11 = 0\)
Given \(c = 2 + \frac{1}{\sqrt{3}}\), substituting this value satisfies the derivative equation, confirming \(p = 11\). Since \(q\) does not affect the derivative, \(q\) can be any real number (often taken as 0 or evaluated from boundary conditions if specified).
In simple words: Use \(f(a) = f(b)\) to find parameters, and set \(f'(c) = 0\) at the given critical point.

Exam Tip: Constants like \(q\) often vanish during differentiation in Rolle's theorem problems.

Question 2. Verify Rolle's theorem for the function \(f(x) = \sin x\) in \([0, \pi]\). Find \(c\), if verified.
Answer: 1. \(f(x) = \sin x\) is continuous on \([0, \pi]\) and differentiable on \((0, \pi)\).
2. Evaluate boundaries: \(f(0) = \sin 0 = 0\) and \(f(\pi) = \sin \pi = 0\). Thus, \(f(0) = f(\pi)\).
All conditions of Rolle's theorem are satisfied.
Now, find \(c \in (0, \pi)\) such that \(f'(c) = 0\):
\(f'(x) = \cos x \implies f'(c) = \cos c = 0\)
Within interval \((0, \pi)\), \(c = \frac{\pi}{2}\).
Hence, Rolle's theorem is verified with \(c = \frac{\pi}{2}\).
In simple words: Sine functions satisfy Rolle's theorem on intervals between multiples of \(\pi\).

Exam Tip: Always verify continuity and differentiability explicitly before applying Rolle's theorem.

Question 3. Verify Lagrange's Mean Value Theorem for \(f(x) = \sqrt{x-2}\) in the interval \([2, 4]\).
Answer: 1. \(f(x) = \sqrt{x-2}\) is continuous on \([2, 4]\) and differentiable on \((2, 4)\).
2. By Lagrange's Mean Value Theorem, there exists \(c \in (2, 4)\) such that:
\(f'(c) = \frac{f(4) - f(2)}{4 - 2}\)
Evaluate function values: \(f(4) = \sqrt{4-2} = \sqrt{2}\) and \(f(2) = \sqrt{2-2} = 0\).
\(\frac{f(4) - f(2)}{4 - 2} = \frac{\sqrt{2} - 0}{2} = \frac{1}{\sqrt{2}}\)
Derivative: \(f'(x) = \frac{1}{2\sqrt{x-2}} \implies f'(c) = \frac{1}{2\sqrt{c-2}}\)
Equate: \(\frac{1}{2\sqrt{c-2}} = \frac{1}{\sqrt{2}} \implies 2\sqrt{c-2} = \sqrt{2} \implies \sqrt{c-2} = \frac{1}{\sqrt{2}}\)
Square both sides: \(c - 2 = \frac{1}{2} \implies c = \frac{5}{2} = 2.5\), which lies inside \((2, 4)\).
Thus, LMVT is verified.
In simple words: LMVT states that there is a point where the tangent slope equals the secant slope connecting interval endpoints.

Question 1. For what value of \(k\) is the following function continuous at \(x = 2\)?
\(f(x) = \begin{cases} 2x + 1, & x < 2 \\ k, & x = 2 \\ 3x - 1, & x > 2 \end{cases}\)
Answer: For continuity at \(x = 2\), LHL = RHL = \(f(2)\).
\(\lim_{x \to 2^-} (2x + 1) = 2(2) + 1 = 5\)
\(\lim_{x \to 2^+} (3x - 1) = 3(2) - 1 = 5\)
Since LHL = RHL = 5, and \(f(2) = k\), we must have \(k = 5\).
In simple words: Evaluate the limit from both sides at \(x = 2\) and set equal to \(k\).

Exam Tip: Both left and right limits must match for the limit to exist.

Question 2. If \(f(x) = \begin{cases} 3ax + b, & \text{if } x > 1 \\ 11, & \text{if } x = 1 \\ 5ax - 2b, & \text{if } x < 1 \end{cases}\), continuous at \(x = 1\), find the values of \(a\) and \(b\) [CBSE 2012 Comptt.].
Answer: For continuity at \(x = 1\), LHL = RHL = \(f(1)\).
\(\lim_{x \to 1^-} (5ax - 2b) = 5a - 2b\)
\(\lim_{x \to 1^+} (3ax + b) = 3a + b\)
Equating both to \(f(1) = 11\):
1) \(3a + b = 11\)
2) \(5a - 2b = 11\)
Multiply equation (1) by 2: \(6a + 2b = 22\)
Add to equation (2): \(11a = 33 \implies a = 3\)
Substitute \(a = 3\) into (1): \(3(3) + b = 11 \implies 9 + b = 11 \implies b = 2\).
Thus, \(a = 3, b = 2\).
In simple words: Set up a system of two linear equations in \(a\) and \(b\) using left and right limits.

Exam Tip: Elimination or substitution method solves simultaneous linear equations efficiently.

Question 3. Discuss the continuity of \(f(x) = |x - 1| + |x - 2|\) at \(x = 1\) & \(x = 2\).
Answer: Rewrite the function in piecewise form across critical points 1 and 2:
For \(x < 1\): \(f(x) = -(x-1) - (x-2) = -2x + 3\)
For \(1 \le x < 2\): \(f(x) = (x-1) - (x-2) = 1\)
For \(x \ge 2\): \(f(x) = (x-1) + (x-2) = 2x - 3\)
Check continuity at \(x = 1\):
\(\lim_{x \to 1^-} f(x) = -2(1) + 3 = 1\)
\(\lim_{x \to 1^+} f(x) = 1\)
\(f(1) = 1\). Hence continuous at \(x = 1\).
Check continuity at \(x = 2\):
\(\lim_{x \to 2^-} f(x) = 1\)
\(\lim_{x \to 2^+} f(x) = 2(2) - 3 = 1\)
\(f(2) = 1\). Hence continuous at \(x = 2\).
In simple words: Sums of continuous absolute value functions remain continuous everywhere.

Exam Tip: Break absolute value functions into intervals to analyze behavior at vertex points.

Question 4. If \(f(x)\) defined by the following is continuous at \(x = 0\), find the values of \(a, b, c\):
\(f(x) = \begin{cases} \frac{\sin(a+1)x + \sin x}{x}, & x < 0 \\ c, & x = 0 \\ \frac{\sqrt{x + bx^2} - \sqrt{x}}{bx^{3/2}}, & x > 0 \end{cases}\)
Answer: For continuity at \(x = 0\), LHL = RHL = \(f(0) = c\).
Evaluate LHL:
\(\lim_{x \to 0^-} \left( \frac{\sin(a+1)x}{x} + \frac{\sin x}{x} \right) = (a+1) + 1 = a + 2\)
Evaluate RHL:
\(\lim_{x \to 0^+} \frac{\sqrt{x}( \sqrt{1 + bx} - 1 )}{bx \cdot x^{1/2}} = \lim_{x \to 0^+} \frac{\sqrt{1 + bx} - 1}{bx}\)
Multiply numerator and denominator by conjugate \((\sqrt{1+bx} + 1)\):
\(= \lim_{x \to 0^+} \frac{1 + bx - 1}{bx(\sqrt{1+bx} + 1)} = \lim_{x \to 0^+} \frac{bx}{bx(\sqrt{1+bx} + 1)} = \frac{1}{1 + 1} = \frac{1}{2}\)
Equating LHL and RHL to \(c\):
\(c = \frac{1}{2}\) and \(a + 2 = \frac{1}{2} \implies a = -\frac{3}{2}\). Parameter \(b\) can be any non-zero real number since it cancels out in RHL evaluation.
In simple words: Evaluate limits from both sides independently using standard limit formulas and rationalization.

Question 5. If \(x = a\left(\cos \theta + \log \tan \frac{\theta}{2}\right)\) and \(y = a\sin \theta\), find \(\frac{dy}{dx}\) at \(\theta = \frac{\pi}{4}\).
Answer: (Previously derived in Parametric Differentiation Level-II, Question 2).
\(\frac{dy}{dx} = \tan \theta\). At \(\theta = \frac{\pi}{4}\), \(\frac{dy}{dx} = \tan\left(\frac{\pi}{4}\right) = 1\).
In simple words: First find the parametric derivative ratio \(\frac{dy/d\theta}{dx/d\theta}\), then substitute the angle.

Exam Tip: Make sure not to confuse first derivative with second derivative questions.

Question 6. If \(y = (\log x)^{\cos x} + \frac{x^2+1}{x^2-1}\), find \(\frac{dy}{dx}\).
Answer: Let \(y = u + v\), where \(u = (\log x)^{\cos x}\) and \(v = \frac{x^2+1}{x^2-1}\).
For \(u\): take log: \(\ln u = \cos x \ln(\log x)\)
\(\frac{1}{u}\frac{du}{dx} = \cos x \left(\frac{1}{\log x} \cdot \frac{1}{x}\right) + \ln(\log x)(-\sin x)\)
\(\frac{du}{dx} = (\log x)^{\cos x} \left[ \frac{\cos x}{x \log x} - \sin x \log(\log x) \right]\)
For \(v\): quotient rule:
\(\frac{dv}{dx} = \frac{(x^2-1)(2x) - (x^2+1)(2x)}{(x^2-1)^2} = \frac{2x^3 - 2x - 2x^3 - 2x}{(x^2-1)^2} = \frac{-4x}{(x^2-1)^2}\)
Combining: \(\frac{dy}{dx} = (\log x)^{\cos x} \left[ \frac{\cos x}{x \log x} - \sin x \log(\log x) \right] - \frac{4x}{(x^2-1)^2}\).
In simple words: Split functions into sums and differentiate each term using logarithmic and quotient rules.

Exam Tip: Chain rule inside logarithmic differentiation requires careful handling of composite logs.

Question 7. If \(xy + y^2 = \tan x + y\), find \(\frac{dy}{dx}\).
Answer: Differentiating implicitly with respect to \(x\):
\(\left(x\frac{dy}{dx} + y \cdot 1\right) + 2y\frac{dy}{dx} = \sec^2 x + \frac{dy}{dx}\)
Group all terms containing \(\frac{dy}{dx}\) on one side:
\(\frac{dy}{dx} (x + 2y - 1) = \sec^2 x - y\)
\(\frac{dy}{dx} = \frac{\sec^2 x - y}{x + 2y - 1}\)
In simple words: Apply product and chain rules to each term and collect all derivative terms on the left.

Question 8. If \(y = \log\left(\sqrt{\frac{x-1}{x+1}}\right)\), find \(\frac{dy}{dx}\).
Answer: Using logarithmic properties: \(y = \frac{1}{2}[\log(x-1) - \log(x+1)]\)
Differentiating with respect to \(x\):
\(\frac{dy}{dx} = \frac{1}{2} \left[ \frac{1}{x-1} - \frac{1}{x+1} \right] = \frac{1}{2} \left[ \frac{(x+1) - (x-1)}{(x^2-1)} \right] = \frac{1}{2} \left[ \frac{2}{x^2-1} \right] = \frac{1}{x^2-1}\)
In simple words: Simplify logarithmic expressions using exponent and division laws before differentiating.

Exam Tip: Logarithms simplify square roots and quotients instantly.

Question 9. If \(\sqrt{1-x^2} + \sqrt{1-y^2} = a(x - y)\), prove that \(\frac{dy}{dx} = \frac{\sqrt{1-y^2}}{\sqrt{1-x^2}}\).
Answer: Let \(x = \sin \theta\) and \(y = \sin \phi\). Then \(\sqrt{1-x^2} = \cos \theta\) and \(\sqrt{1-y^2} = \cos \phi\).
Substitute into the given equation:
\(\cos \theta + \cos \phi = a(\sin \theta - \sin \phi)\)
Using trigonometric sum-to-product formulas:
\(2\cos\left(\frac{\theta+\phi}{2}\right)\cos\left(\frac{\theta-\phi}{2}\right) = 2a\cos\left(\frac{\theta+\phi}{2}\right)\sin\left(\frac{\theta-\phi}{2}\right)\)
Dividing both sides gives \(\cot\left(\frac{\theta-\phi}{2}\right) = a \implies \frac{\theta-\phi}{2} = \cot^{-1} a \implies \theta - \phi = 2\cot^{-1} a\)
Substitute back \(\theta = \sin^{-1} x\) and \(\phi = \sin^{-1} y\):
\(\sin^{-1} x - \sin^{-1} y = 2\cot^{-1} a\)
Differentiating with respect to \(x\):
\(\frac{1}{\sqrt{1-x^2}} - \frac{1}{\sqrt{1-y^2}}\frac{dy}{dx} = 0 \implies \frac{dy}{dx} = \frac{\sqrt{1-y^2}}{\sqrt{1-x^2}}\). Hence proved.
In simple words: Use trigonometric substitutions to simplify complex radical implicit equations into inverse function differences.

Exam Tip: Trigonometric substitutions like \(x = \sin\theta\) turn radical expressions into clean sum-to-product identities.

Question 10. Find \(\frac{dy}{dx}\) if \((\cos x)^y = (\cos y)^x\).
Answer: (Already solved in Logarithmic Differentiation Level-II, Question 3).
\(\frac{dy}{dx} = \frac{\log(\cos y) + y\tan x}{\log(\cos x) + x\tan y}\)
In simple words: Take natural logarithms on both sides and differentiate implicitly.

Question 11. If \(y = a\cos(\log x) + b\sin(\log x)\), prove that \(x^2\frac{d^2y}{dx^2} + x\frac{dy}{dx} + y = 0\).
Answer: (Already solved in Second Order Derivatives Level-II, Question 1).
Proof verified successfully.
In simple words: Differentiate twice and multiply by \(x^2\) to clear denominators.

Exam Tip: Standard trigonometric-logarithmic combinations yield homogeneous second-order relations.

Question 12. If \(x^p \cdot y^q = (x+y)^{p+q}\), prove that \(\frac{dy}{dx} = \frac{y}{x}\).
Answer: (Already solved in Logarithmic Differentiation Level-III, Question 1).
Proof verified successfully.
In simple words: Apply logarithms to simplify product powers before implicit differentiation.

Exam Tip: Homogeneous logarithmic expansions always simplify to \(\frac{dy}{dx} = \frac{y}{x}\).