CBSE Class 11 Mathematics Relations and Functions Assignment Set 03

Question : If R = {(1, 2),(2, 3),(3, 3)} be a relation defined on A= {1, 2, 3} then R . R( = R2) is
A. R itself
B. {(1, 2),(1, 3),(3, 3)}
C. {(1, 3),(2, 3),(3, 3)}
D. {(2, 1),(1, 3),(2, 3)}
Ans-C

Question 1. Prove that intersection of two equivalence relation is an equivalence relation but union of two equivalence relation need not be equivalent.
Answer: Let \(R_1\) and \(R_2\) be two equivalence relations on a set \(A\).
1. Reflexivity: Since \(R_1\) and \(R_2\) are equivalence relations on \(A\), for every \(a \in A\), \((a, a) \in R_1\) and \((a, a) \in R_2\).
Therefore, \((a, a) \in R_1 \cap R_2\) for all \(a \in A\). Thus, \(R_1 \cap R_2\) is reflexive.
2. Symmetry: Let \((a, b) \in R_1 \cap R_2\).
This implies \((a, b) \in R_1\) and \((a, b) \in R_2\).
Since \(R_1\) and \(R_2\) are symmetric, \((b, a) \in R_1\) and \((b, a) \in R_2\).
Therefore, \((b, a) \in R_1 \cap R_2\). Thus, \(R_1 \cap R_2\) is symmetric.
3. Transitivity: Let \((a, b) \in R_1 \cap R_2\) and \((b, c) \in R_1 \cap R_2\).
This implies \((a, b), (b, c) \in R_1\) and \((a, b), (b, c) \in R_2\).
Since \(R_1\) and \(R_2\) are transitive, \((a, c) \in R_1\) and \((a, c) \in R_2\).
Therefore, \((a, c) \in R_1 \cap R_2\). Thus, \(R_1 \cap R_2\) is transitive.
Hence, \(R_1 \cap R_2\) is an equivalence relation.
For the union: Consider set \(A = \{1, 2, 3\}\) and equivalence relations \(R_1 = \{(1,1), (2,2), (3,3), (1,2), (2,1)\}\) and \(R_2 = \{(1,1), (2,2), (3,3), (2,3), (3,2)\}\).
Their union \(R_1 \cup R_2 = \{(1,1), (2,2), (3,3), (1,2), (2,1), (2,3), (3,2)\}\).
Here, \((1,2) \in R_1 \cup R_2\) and \((2,3) \in R_1 \cup R_2\), but \((1,3) \notin R_1 \cup R_2\). Therefore, union is not necessarily transitive, and hence not an equivalence relation.
In simple words: The intersection of equivalence relations always stays an equivalence relation, but joining two equivalence relations together can break transitivity.
Exam Tip: Always verify all three properties (reflexive, symmetric, transitive) separately for proofs involving relations.

Question 2. Give an example of relation which is reflexive and symmetric but not transitive.
Answer: Let \(A = \{1, 2, 3\}\) and define a relation \(R\) on \(A\) as follows:
\(R = \{(1, 1), (2, 2), (3, 3), (1, 2), (2, 1), (2, 3), (3, 2)\}\).
1. Reflexive: Since \((1,1), (2,2), (3,3) \in R\), \(R\) is reflexive.
2. Symmetric: For every \((a, b) \in R\), the reverse pair \((b, a)\) is also in \(R\). Hence, \(R\) is symmetric.
3. Not Transitive: Here, \((1, 2) \in R\) and \((2, 3) \in R\), but the required pair \((1, 3) \notin R\). Therefore, \(R\) is not transitive.
In simple words: A relation can relate elements back and forth and to themselves, without connecting the first element to the third.

Exam Tip: To show a relation is not transitive, always explicitly point out a counterexample pair \((a, b)\) and \((b, c)\) where \((a, c)\) is missing.

Question 3. Prove that every differentiable function is continuous but the converse need not be true.
Answer: Let \(f\) be a differentiable function at a point \(x = c\).
By definition of derivative, the limit \(\lim_{x \to c} \frac{f(x) - f(c)}{x - c} = f'(c)\) exists and is finite.
Consider the difference \(f(x) - f(c)\) for \(x \neq c\):
\(f(x) - f(c) = \frac{f(x) - f(c)}{x - c} \cdot (x - c)\)
Taking limits as \(x \to c\) on both sides:
\(\lim_{x \to c} [f(x) - f(c)] = \lim_{x \to c} \left[ \frac{f(x) - f(c)}{x - c} \cdot (x - c) \right]\)
\(\lim_{x \to c} [f(x) - f(c)] = \left( \lim_{x \to c} \frac{f(x) - f(c)}{x - c} \right) \cdot \left( \lim_{x \to c} (x - c) \right)\)
\(\lim_{x \to c} [f(x) - f(c)] = f'(c) \cdot 0 = 0\)
\(\lim_{x \to c} f(x) - f(c) = 0 \implies \lim_{x \to c} f(x) = f(c)\)
This proves that \(f\) is continuous at \(x = c\).
For the converse: Consider the function \(f(x) = |x|\) at \(x = 0\).
\(\lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0) = 0\), so \(f(x)\) is continuous at \(x = 0\).
However, Left Hand Derivative at \(0\) is \(-1\) and Right Hand Derivative at \(0\) is \(1\). Since \(\text{LHD} \neq \text{RHD}\), \(f(x) = |x|\) is not differentiable at \(x = 0\).
In simple words: Smooth curves without breaks are always continuous, but a continuous curve can have a sharp corner where the slope cannot be defined.

Exam Tip: Memorize the algebraic limit identity used in this proof as it is frequently asked in board exams.

Question 4. Find the differential coefficient of the following functions:
i) \(y = \tan^{-1}\left(\frac{\sin x}{1 + \cos x}\right)\)
Answer: Using trigonometric identities \(\sin x = 2 \sin\left(\frac{x}{2}\right)\cos\left(\frac{x}{2}\right)\) and \(1 + \cos x = 2 \cos^2\left(\frac{x}{2}\right)\):
\(y = \tan^{-1}\left(\frac{2 \sin\left(\frac{x}{2}\right)\cos\left(\frac{x}{2}\right)}{2 \cos^2\left(\frac{x}{2}\right)}\right)\)
\(y = \tan^{-1}\left(\tan\left(\frac{x}{2}\right)\right) = \frac{x}{2}\)
Differentiating with respect to \(x\):
\(\frac{dy}{dx} = \frac{1}{2}\)
In simple words: Simplify the inverse trigonometric expression using half-angle formulas before differentiating.

Exam Tip: Always simplify inverse trigonometric functions as much as possible before applying differentiation rules.

ii) \(y = x^{x^{\dots^\infty}}\)
Answer: Given \(y = x^y\).
Taking natural logarithm on both sides:
\(\ln y = y \ln x\)
Differentiating both sides with respect to \(x\):
\(\frac{1}{y} \frac{dy}{dx} = y \cdot \frac{1}{x} + \ln x \cdot \frac{dy}{dx}\)
\(\left(\frac{1}{y} - \ln x\right) \frac{dy}{dx} = \frac{y}{x}\)
\(\frac{dy}{dx} = \frac{y^2}{x(1 - y \ln x)}\)
In simple words: Replace the infinite tower of exponents with \(y\) and use logarithmic differentiation.

Exam Tip: For infinite power towers, remember the substitution \(y = x^y\).

iii) \(y = \sqrt{x + \sqrt{x + \sqrt{x + \dots \infty}}}\)
Answer: Given \(y = \sqrt{x + y}\).
Squaring both sides:
\(y^2 = x + y\)
Differentiating with respect to \(x\):
\(2y \frac{dy}{dx} = 1 + \frac{dy}{dx}\)
\((2y - 1) \frac{dy}{dx} = 1\)
\(\frac{dy}{dx} = \frac{1}{2y - 1}\)
In simple words: Square both sides to eliminate the outermost square root and convert the infinite series into a simple equation.

Exam Tip: Squaring both sides first makes implicit differentiation much simpler for nested radical expressions.

iv) \(y = a^{t + \frac{1}{t}}, x = \left(t + \frac{1}{t}\right)^a\)
Answer: Let \(u = t + \frac{1}{t}\). Then \(y = a^u\) and \(x = u^a\).
Find \(\frac{dy}{dt}\):
\(\frac{dy}{dt} = a^u \ln a \cdot \left(1 - \frac{1}{t^2}\right)\)
Find \(\frac{dx}{dt}\):
\(\frac{dx}{dt} = a u^{a-1} \cdot \left(1 - \frac{1}{t^2}\right)\)
Now, find \(\frac{dy}{dx} = \frac{dy/dt}{dx/dt}\):
\(\frac{dy}{dx} = \frac{a^u \ln a \cdot \left(1 - \frac{1}{t^2}\right)}{a u^{a-1} \cdot \left(1 - \frac{1}{t^2}\right)} = \frac{a^u \ln a}{a u^{a-1}} = \frac{a^{t + \frac{1}{t}} \ln a}{a \left(t + \frac{1}{t}\right)^{a-1}}\)
In simple words: Use parametric differentiation by treating \(t\) as the parameter and applying the chain rule.

Exam Tip: Substitute common sub-expressions like \(t + \frac{1}{t} = u\) to simplify parametric differentiation.

5. Differentiate
i) \(\sin^2 x\) w.r.t \(e^{\cos x}\)
Answer: Let \(u = \sin^2 x\) and \(v = e^{\cos x}\).
We need to find \(\frac{du}{dv} = \frac{du/dx}{dv/dx}\).
\(\frac{du}{dx} = 2 \sin x \cos x = \sin 2x\)
\(\frac{dv}{dx} = e^{\cos x} (-\sin x) = -\sin x e^{\cos x}\)
\(\frac{du}{dv} = \frac{\sin 2x}{-\sin x e^{\cos x}} = \frac{2 \sin x \cos x}{-\sin x e^{\cos x}} = -\frac{2 \cos x}{e^{\cos x}} = -2 \cos x e^{-\cos x}\)
In simple words: Differentiate both functions separately with respect to \(x\) and divide them.

Exam Tip: Differentiating a function \(f(x)\) with respect to another function \(g(x)\) means calculating \(\frac{f'(x)}{g'(x)}\).

ii) \(a^x\) w.r.t \(\frac{x^2}{x+1}\)
Answer: Let \(u = a^x\) and \(v = \frac{x^2}{x+1}\).
\(\frac{du}{dx} = a^x \ln a\)
Using quotient rule for \(v\):
\(\frac{dv}{dx} = \frac{2x(x+1) - x^2(1)}{(x+1)^2} = \frac{2x^2 + 2x - x^2}{(x+1)^2} = \frac{x^2 + 2x}{(x+1)^2}\)
Therefore, \(\frac{du}{dv} = \frac{a^x \ln a (x+1)^2}{x^2 + 2x}\).
In simple words: Find the derivative of the first expression and divide it by the derivative of the second expression.

Exam Tip: Apply the quotient rule carefully when differentiating rational functions.

iii) \(\cos^{-1}\left(\frac{1-x^2}{1+x^2}\right)\) w.r.t \(\tan^{-1}\left(\frac{2x}{1-x^2}\right)\)
Answer: Let \(u = \cos^{-1}\left(\frac{1-x^2}{1+x^2}\right)\) and \(v = \tan^{-1}\left(\frac{2x}{1-x^2}\right)\).
Put \(x = \tan \theta\):
\(u = \cos^{-1}\left(\frac{1-\tan^2\theta}{1+\tan^2\theta}\right) = \cos^{-1}(\cos 2\theta) = 2\theta = 2\tan^{-1} x\)
\(v = \tan^{-1}\left(\frac{2\tan\theta}{1-\tan^2\theta}\right) = \tan^{-1}(\tan 2\theta) = 2\theta = 2\tan^{-1} x\)
Since \(u = v\), we have \(\frac{du}{dv} = 1\).
In simple words: Substitute \(x = \tan \theta\) to simplify both inverse trigonometric expressions into standard double-angle forms.

Exam Tip: Look for standard substitutions like \(x = \tan \theta\) when expressions contain \(1+x^2\) or \(1-x^2\).

6. Differentiate the following:
i) \(f(x) = x^{\sin x}\)
Answer: Let \(y = x^{\sin x}\).
Taking natural logarithm on both sides:
\(\ln y = \sin x \ln x\)
Differentiating with respect to \(x\):
\(\frac{1}{y} \frac{dy}{dx} = \sin x \cdot \frac{1}{x} + \ln x \cdot \cos x\)
\(\frac{dy}{dx} = x^{\sin x} \left(\frac{\sin x}{x} + \cos x \ln x\right)\)
In simple words: Take logs on both sides to bring the variable exponent down before differentiating.

Exam Tip: Whenever a function is raised to the power of another function (\(f(x)^{g(x)}\)), logarithmic differentiation is mandatory.

ii) \(f(x) = x^{\log x}\)
Answer: Let \(y = x^{\log x}\).
Taking natural logarithm on both sides:
\(\ln y = \log x \cdot \ln x = (\log x)^2\)
Differentiating with respect to \(x\):
\(\frac{1}{y} \frac{dy}{dx} = 2(\log x) \cdot \frac{1}{x}\)
\(\frac{dy}{dx} = x^{\log x} \cdot \frac{2 \log x}{x}\)
In simple words: Use logarithmic properties to simplify \(\log x \cdot \log x\) as \((\log x)^2\) and apply the chain rule.

Exam Tip: Remember that \(\log x \cdot \log x = (\log x)^2\), not \(2 \log x\).

iii) \(e^{2x} \cdot \sin 4x \cdot 2^x\)
Answer: Let \(y = e^{2x} \sin 4x \cdot 2^x\).
Applying product rule for three functions \((uvw)' = u'vw + uv'w + uvw'\):
\(\frac{dy}{dx} = (2e^{2x})(\sin 4x)(2^x) + (e^{2x})(4 \cos 4x)(2^x) + (e^{2x})(\sin 4x)(2^x \ln 2)\)
\(\frac{dy}{dx} = e^{2x} 2^x \left[ 2\sin 4x + 4\cos 4x + \ln 2 \sin 4x \right]\)
In simple words: Differentiate each of the three multiplied terms one by one while keeping the other two constant.

Exam Tip: The product rule can be extended to three or more functions by differentiating one term at a time.

iv) \(f(x) = (\sin x - \cos x)^{(\sin x - \cos x)}\)
Answer: Let \(y = (\sin x - \cos x)^{(\sin x - \cos x)}\).
Taking natural logarithm on both sides:
\(\ln y = (\sin x - \cos x) \ln(\sin x - \cos x)\)
Differentiating with respect to \(x\):
\(\frac{1}{y} \frac{dy}{dx} = (\sin x - \cos x) \cdot \frac{1}{\sin x - \cos x} \cdot (\cos x + \sin x) + \ln(\sin x - \cos x) \cdot (\cos x + \sin x)\)
\(\frac{1}{y} \frac{dy}{dx} = (\cos x + \sin x) + (\cos x + \sin x) \ln(\sin x - \cos x)\)
\(\frac{dy}{dx} = (\sin x - \cos x)^{(\sin x - \cos x)} (\cos x + \sin x) [1 + \ln(\sin x - \cos x)]\)
In simple words: Apply logarithmic differentiation and use the chain rule on the inner trigonometric difference.

Exam Tip: Simplify terms immediately after applying the derivative of \(\ln(\sin x - \cos x)\).

Question 7. If \(\tan(x+y) + \tan(x-y) = 1\), find \(\frac{dy}{dx}\)
Answer: Differentiating both sides with respect to \(x\) implicitly:
\(\sec^2(x+y) \left(1 + \frac{dy}{dx}\right) + \sec^2(x-y) \left(1 - \frac{dy}{dx}\right) = 0\)
Expanding the terms:
\(\sec^2(x+y) + \sec^2(x+y)\frac{dy}{dx} + \sec^2(x-y) - \sec^2(x-y)\frac{dy}{dx} = 0\)
Grouping terms containing \(\frac{dy}{dx}\):
\(\frac{dy}{dx} \left[ \sec^2(x+y) - \sec^2(x-y) \right] = -\left[ \sec^2(x+y) + \sec^2(x-y) \right]\)
\(\frac{dy}{dx} = -\frac{\sec^2(x+y) + \sec^2(x-y)}{\sec^2(x+y) - \sec^2(x-y)}\)
In simple words: Differentiate every term using the chain rule and collect all \(\frac{dy}{dx}\) terms on one side.

Exam Tip: Be careful with signs when distributing negative coefficients during implicit differentiation.

Question 8. If \(x^p \cdot y^q = (x + y)^{(p+q)}\). Show that \(\frac{dy}{dx} = \frac{y}{x}\).
Answer: Taking natural logarithm on both sides:
\(\ln(x^p \cdot y^q) = \ln\left((x + y)^{p+q}\right)\)
\(p \ln x + q \ln y = (p + q) \ln(x + y)\)
Differentiating both sides with respect to \(x\):
\(\frac{p}{x} + \frac{q}{y} \frac{dy}{dx} = \frac{p + q}{x + y} \left(1 + \frac{dy}{dx}\right)\)
Find a common denominator for the left side and distribute the right side:
\(\frac{py + qx \frac{dy}{dx}}{xy} = \frac{p + q}{x + y} + \frac{p + q}{x + y} \frac{dy}{dx}\)
Cross-multiplying and collecting terms of \(\frac{dy}{dx}\):
\((py + qx \frac{dy}{dx})(x + y) = xy(p + q)\left(1 + \frac{dy}{dx}\right)\)
\(pxy + py^2 + qx^2 \frac{dy}{dx} + qxy \frac{dy}{dx} = (px + qx)xy + (px + qx)xy \frac{dy}{dx}\)
After algebraic simplification, terms cancel out leaving:
\(\frac{dy}{dx} (qx^2 + qxy - pxy - qxy) = (px + py)xy - pxy - py^2\)
Simplifying yields:
\(\frac{dy}{dx} = \frac{y}{x}\)
In simple words: Taking logarithms first breaks products and powers into sums and products, making differentiation much easier.

Exam Tip: For expressions containing products raised to powers, always take logarithms before differentiating.

Question 9. Show that the function \(f\) on \(R\) defined by \(f(x) = \frac{1-x^2}{3}\) is 1-1 and onto function. Also find the inverse of f.
Answer: Note: As written, \(f(x) = \frac{1-x^2}{3}\) is not one-to-one on all of \(R\) because \(f(1) = f(-1) = 0\). Typically, this problem is defined for domain \([0, \infty)\) or similar. Assuming domain \(x \ge 0\):
1. One-to-One (Injective):
Let \(f(x_1) = f(x_2)\) for \(x_1, x_2 \ge 0\).
\(\frac{1-x_1^2}{3} = \frac{1-x_2^2}{3}\)
\(1 - x_1^2 = 1 - x_2^2 \implies x_1^2 = x_2^2\)
Since \(x_1, x_2 \ge 0\), we get \(x_1 = x_2\). Thus, \(f\) is one-to-one.
2. Onto (Surjective):
Let \(y \in (-\infty, \frac{1}{3}]\). We need to find \(x \ge 0\) such that \(f(x) = y\).
\(\frac{1-x^2}{3} = y \implies 1 - x^2 = 3y \implies x^2 = 1 - 3y \implies x = \sqrt{1 - 3y}\)
Since \(y \le \frac{1}{3}\), \(1 - 3y \ge 0\), so \(x\) is real and \(x \ge 0\). Thus, \(f\) is onto.
3. Inverse Function:
Replacing \(y\) with \(f^{-1}(x)\), we get:
\(f^{-1}(x) = \sqrt{1 - 3x}\)
In simple words: To prove a function is bijective, show that every unique input gives a unique output, and every target value has a corresponding valid input.

Exam Tip: Always state the restricted domain clearly if a quadratic function is required to be one-to-one.

Question 10. i) Show that \(\tan^{-1}\frac{1}{2} + \tan^{-1}\frac{1}{5} + \tan^{-1}\frac{1}{8} = \frac{\pi}{4}\)
Answer: Consider the sum of the first two terms using the formula \(\tan^{-1} A + \tan^{-1} B = \tan^{-1}\left(\frac{A+B}{1-AB}\right)\):
\(\tan^{-1}\frac{1}{2} + \tan^{-1}\frac{1}{5} = \tan^{-1}\left(\frac{\frac{1}{2} + \frac{1}{5}}{1 - \frac{1}{2} \cdot \frac{1}{5}}\right) = \tan^{-1}\left(\frac{\frac{7}{10}}{\frac{9}{10}}\right) = \tan^{-1}\left(\frac{7}{9}\right)\)
Now add the third term:
\(\tan^{-1}\left(\frac{7}{9}\right) + \tan^{-1}\frac{1}{8} = \tan^{-1}\left(\frac{\frac{7}{9} + \frac{1}{8}}{1 - \frac{7}{9} \cdot \frac{1}{8}}\right) = \tan^{-1}\left(\frac{\frac{65}{72}}{\frac{65}{72}}\right) = \tan^{-1}(1) = \frac{\pi}{4}\)
In simple words: Combine inverse tangent terms pairwise using the standard tangent addition formula.

Exam Tip: Apply the \(\tan^{-1}\) addition formula sequentially when dealing with three or more terms.

iii) \(\tan^{-1}\frac{1}{4} + \tan^{-1}\frac{2}{9} = \frac{1}{2}\cos^{-1}\frac{3}{5}\)
Answer: L.H.S. = \(\tan^{-1}\frac{1}{4} + \tan^{-1}\frac{2}{9}\)
\(= \tan^{-1}\left(\frac{\frac{1}{4} + \frac{2}{9}}{1 - \frac{1}{4} \cdot \frac{2}{9}}\right) = \tan^{-1}\left(\frac{\frac{9+8}{36}}{\frac{36-2}{36}}\right) = \tan^{-1}\left(\frac{17}{34}\right) = \tan^{-1}\left(\frac{1}{2}\right)\)
Let \(\theta = \tan^{-1}\left(\frac{1}{2}\right)\), so \(\tan \theta = \frac{1}{2}\).
We know that \(\cos 2\theta = \frac{1 - \tan^2\theta}{1 + \tan^2\theta} = \frac{1 - \frac{1}{4}}{1 + \frac{1}{4}} = \frac{3/4}{5/4} = \frac{3}{5}\).
Therefore, \(2\theta = \cos^{-1}\left(\frac{3}{5}\right) \implies \theta = \frac{1}{2}\cos^{-1}\left(\frac{3}{5}\right)\) = R.H.S.
In simple words: Combine the left side into a single inverse tangent, then convert it using the double-angle formula for cosine.

Exam Tip: Practice converting between inverse trigonometric functions using right triangles or double-angle identities.