ML Aggarwal Class 12 Maths Solutions Section A Chapter 08 Mean Value Theorems Maxima Minima

8 Mean Value Theorems, Maxima and Minima

8.1 Rolle's Theorem

If a function f(x) is
(i) continuous in the closed interval [a, b]
(ii) derivable in the open interval (a, b) and
(iii) f(a) = f(b),
then there exists atleast one real number c in (a, b) such that f'(c) = 0.
(We accept it without proof.)

Geometrical Interpretation

Let A, B be the points on the curve y = f(x) corresponding to the real numbers a, b respectively. Since f(x) is continuous in [a, b], the graph of the curve y = f(x) is continuous from A to B. Again, as f(x) is derivable in (a, b), the curve y = f(x) has a tangent at each point between A and B. Also as f(a) = f(b) the ordinates of the points A and B are equal i.e. MA = NB.
Then Rolle's theorem asserts that there is atleast one point lying between A and B such that the tangent at which is parallel to x-axis i.e. there exists atleast one real number c in (a, b) such that f'(c) = 0.
There may exist more than one point between A and B the tangents at which are parallel to x-axis i.e. there may exist more than one real number c in (a, b) such that f'(c) = 0. Rolle's theorem ensures the existence of atleast one real number c in (a, b) such that f'(c) = 0.

Remarks

1. Rolle's theorem fails for the function which does not satisfy even one of the three conditions.
2. The converse of Rolle's theorem may not be true, for, f'(x) may be zero at a point in (a, b) without satisfying all the three conditions of Rolle's theorem.

Example 1. Verify Rolle's theorem for the following functions:
(a) \( f(x) = x^2 + x - 6 \) in \( [-3, 2] \)
(b) \( f(x) = (x^2 - 1)(x - 2) \) in \( [-1, 2] \)
Answer:
(a) Given \( f(x) = x^2 + x - 6 \) ...(1)
(i) Since f(x) is a polynomial, it stays continuous in \( [-3, 2] \).
(ii) f(x) being a polynomial is differentiable in \( (-3, 2) \).
(iii) \( f(-3) = (-3)^2 - 3 - 6 = 0 \), \( f(2) = 2^2 + 2 - 6 = 0 \) \( \Rightarrow f(-3) = f(2) \).
Thus, all three conditions of Rolle's theorem hold. So there exists atleast one real number c in \( (-3, 2) \) such that f'(c) = 0.
Taking the derivative of (1): \( f'(x) = 2x + 1 \).
Now \( f'(c) = 0 \Rightarrow 2c + 1 = 0 \Rightarrow c = -\frac{1}{2} \).
So there exists \( -\frac{1}{2} \in (-3, 2) \) such that \( f'\left(-\frac{1}{2}\right) = 0 \).
Hence, Rolle's theorem is verified.
(b) Given \( f(x) = (x^2 - 1)(x - 2) \) ...(1)
(i) Since f(x) is a polynomial, it remains continuous in \( [-1, 2] \).
(ii) f(x) being a polynomial is differentiable in \( (-1, 2) \).
(iii) \( f(-1) = (1 - 1)(-1 - 2) = 0 \), \( f(2) = (4 - 1)(2 - 2) = 0 \) \( \Rightarrow f(-1) = f(2) \).
Thus, all three conditions of Rolle's theorem are satisfied. So there exists atleast one real number c in \( (-1, 2) \) such that f'(c) = 0.
Differentiating (1): \( f'(x) = (x^2 - 1) \cdot 1 + (x - 2) \cdot 2x = 3x^2 - 4x - 1 \).
Now \( f'(c) = 0 \Rightarrow 3c^2 - 4c - 1 = 0 \)
\( \Rightarrow c = \frac{4 \pm \sqrt{16 + 12}}{6} = \frac{4 \pm \sqrt{28}}{6} = \frac{2 \pm \sqrt{7}}{3} \).
Also \( -1 < \frac{2 - \sqrt{7}}{3} < \frac{2 + \sqrt{7}}{3} < 2 \) \( \Rightarrow \frac{2 - \sqrt{7}}{3} \) and \( \frac{2 + \sqrt{7}}{3} \) both lie in \( (-1, 2) \).
So there exist two real numbers \( \frac{2 - \sqrt{7}}{3} \) and \( \frac{2 + \sqrt{7}}{3} \) in \( (-1, 2) \) such that \( f'\left(\frac{2 - \sqrt{7}}{3}\right) = 0 \) and \( f'\left(\frac{2 + \sqrt{7}}{3}\right) = 0 \).
Hence, Rolle's theorem is verified.
In simple words: When a smooth curve starts and ends at the same height, there must be at least one spot along the curve where the slope is zero (a flat tangent line).

Exam Tip: Always verify the three conditions first - continuity in the closed interval, differentiability in the open interval, and equal endpoint values - before using Rolle's theorem. Finding the derivative and solving f'(c) = 0 gives the required point(s).

Example 2. Using Rolle's theorem, find the point on the curve y = 16 - x², x ∈ [-1, 1] where the tangent is parallel to x-axis.
Answer: Given y = 16 - x², i.e. f(x) = 16 - x² ...(1)
(i) Since f(x) is a polynomial, it is continuous in [-1, 1].
(ii) f(x) being a polynomial is differentiable in (-1, 1).
(iii) \( f(-1) = 16 - (-1)^2 = 15 \), \( f(1) = 16 - 1^2 = 15 \) \( \Rightarrow f(-1) = f(1) \).
Thus, all three conditions of Rolle's theorem hold. So there exists atleast one real number c in (-1, 1) such that f'(c) = 0.
Taking the derivative of (1): \( f'(x) = -2x \).
Now \( f'(c) = 0 \Rightarrow -2c = 0 \Rightarrow c = 0 \).
So there exists \( 0 \in (-1, 1) \) where f'(c) = 0, meaning the tangent is parallel to x-axis.
From (1), when x = 0, y = 16 - 0² = 16.
Hence, there exists the point (0, 16) on the given curve where the tangent is parallel to x-axis.
In simple words: At x = 0, the curve reaches its highest point in the given range, and at this point the tangent line is completely flat (horizontal).

Exam Tip: When finding where the tangent is parallel to the x-axis, set f'(c) = 0 and solve for c. Then substitute this value back into the original function to find the y-coordinate of the point.

Example 3. Verify Rolle's theorem for the following functions and find point (or points) in the given interval where derivative is zero:
(a) \( f(x) = \sin x + \cos x - 1 \) in \( \left[0, \frac{\pi}{2}\right] \)
(b) \( f(x) = \sin x - \sin 2x \) in \( [0, \pi] \)
(c) \( f(x) = e^{2x}(\sin 2x - \cos 2x) \) in \( \left[\frac{\pi}{8}, \frac{5\pi}{8}\right] \) (I.S.C. 2006)
(d) \( f(x) = e^{1-x^2} \) in \( [-1, 1] \)
Answer:
(a) Given \( f(x) = \sin x + \cos x - 1 \) ...(1)
(i) f(x) is continuous in \( \left[0, \frac{\pi}{2}\right] \).
(ii) f(x) is differentiable in \( \left(0, \frac{\pi}{2}\right) \).
(iii) \( f(0) = \sin 0 + \cos 0 - 1 = 0 + 1 - 1 = 0 \), \( f\left(\frac{\pi}{2}\right) = \sin \frac{\pi}{2} + \cos \frac{\pi}{2} - 1 = 1 + 0 - 1 = 0 \) \( \Rightarrow f(0) = f\left(\frac{\pi}{2}\right) \).
Thus, all three conditions of Rolle's theorem are satisfied. So there exists atleast one real number c in \( \left(0, \frac{\pi}{2}\right) \) such that f'(c) = 0.
Differentiating (1): \( f'(x) = \cos x - \sin x \).
Now \( f'(c) = 0 \Rightarrow \cos c - \sin c = 0 \Rightarrow \tan c = 1 \)
\( \Rightarrow c = \frac{\pi}{4}, \frac{5\pi}{4}, \frac{9\pi}{4}, \ldots, -\frac{3\pi}{4}, \ldots \) but \( c \in \left(0, \frac{\pi}{2}\right) \) \( \Rightarrow c = \frac{\pi}{4} \).
So there exists \( \frac{\pi}{4} \in \left(0, \frac{\pi}{2}\right) \) such that \( f'\left(\frac{\pi}{4}\right) = 0 \).
Hence, Rolle's theorem is verified and \( c = \frac{\pi}{4} \).
(b) Given \( f(x) = \sin x - \sin 2x \) ...(1)
(i) f(x) is continuous in [0, π].
(ii) f(x) is differentiable in (0, π).
(iii) \( f(0) = \sin 0 - \sin 0 = 0 - 0 = 0 \), \( f(\pi) = \sin \pi - \sin 2\pi = 0 - 0 = 0 \) \( \Rightarrow f(0) = f(\pi) \).
Thus, all three conditions of Rolle's theorem are satisfied. So there exists atleast one real number c in (0, π) such that f'(c) = 0.
Differentiating (1): \( f'(x) = \cos x - \cos 2x \cdot 2 = \cos x - 2\cos 2x \).
Now \( f'(c) = 0 \Rightarrow \cos c - 2\cos 2c = 0 \)
\( \Rightarrow \cos c - 2(2\cos^2 c - 1) = 0 \Rightarrow 4\cos^2 c - \cos c - 2 = 0 \)
\( \Rightarrow \cos c = \frac{1 \pm \sqrt{1 + 32}}{8} = \frac{1 \pm \sqrt{33}}{8} \) \( \Rightarrow c = \cos^{-1}\left(\frac{1 \pm \sqrt{33}}{8}\right) \).
So there exist two real numbers c given by \( c = \cos^{-1}\left(\frac{1 \pm \sqrt{33}}{8}\right) \) in (0, π) such that f'(c) = 0.
Hence, Rolle's theorem is verified and \( c = \cos^{-1}\left(\frac{1 \pm \sqrt{33}}{8}\right) \).
(c) Given \( f(x) = e^{2x}(\sin 2x - \cos 2x) \) ...(1)
(i) f(x) is continuous in \( \left[\frac{\pi}{8}, \frac{5\pi}{8}\right] \).
(ii) f(x) is differentiable in \( \left(\frac{\pi}{8}, \frac{5\pi}{8}\right) \).
(iii) \( f\left(\frac{\pi}{8}\right) = e^{\frac{\pi}{4}}\left(\sin \frac{\pi}{4} - \cos \frac{\pi}{4}\right) = e^{\frac{\pi}{4}}\left(\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}\right) = 0 \),
\( f\left(\frac{5\pi}{8}\right) = e^{\frac{5\pi}{4}}\left(\sin \frac{5\pi}{4} - \cos \frac{5\pi}{4}\right) = e^{\frac{5\pi}{4}}\left(-\sin \frac{\pi}{4} + \cos \frac{\pi}{4}\right) = e^{\frac{5\pi}{4}}\left(-\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}\right) = 0 \)
\( \Rightarrow f\left(\frac{\pi}{8}\right) = f\left(\frac{5\pi}{8}\right) \).
Thus, all three conditions of Rolle's theorem are satisfied. So there exists atleast one real number c in \( \left(\frac{\pi}{8}, \frac{5\pi}{8}\right) \) such that f'(c) = 0.
Differentiating (1): \( f'(x) = e^{2x}(\cos 2x \cdot 2 + \sin 2x \cdot 2) + (\sin 2x - \cos 2x)e^{2x} \cdot 2 = 4e^{2x}\sin 2x \).
Now \( f'(c) = 0 \Rightarrow 4e^{2c}\sin 2c = 0 \Rightarrow \sin 2c = 0 \)
\( \Rightarrow 2c = 0, \pi, 2\pi, \ldots, -\pi, -2\pi, \ldots \)
\( \Rightarrow c = 0, \frac{\pi}{2}, \pi, \ldots, -\frac{\pi}{2}, -\pi, \ldots \) but \( c \in \left(\frac{\pi}{8}, \frac{5\pi}{8}\right) \) \( \Rightarrow c = \frac{\pi}{2} \).
So there exists \( \frac{\pi}{2} \) in \( \left(\frac{\pi}{8}, \frac{5\pi}{8}\right) \) such that \( f'\left(\frac{\pi}{2}\right) = 0 \).
Hence, Rolle's theorem is verified and \( c = \frac{\pi}{2} \).
(d) Given \( f(x) = e^{1-x^2} \) ...(1)
(i) f(x) is continuous in [-1, 1], for since g(x) = 1 - x² and h(x) = e^x are continuous in [-1, 1], therefore, (hog)(x) = h(g(x)) = h(1 - x²) = e^{1-x^2} is also continuous in [-1, 1].
(ii) f(x) is differentiable in (-1, 1).
(iii) \( f(-1) = e^{1-1} = e^0 = 1 \), \( f(1) = e^{1-1} = e^0 = 1 \) \( \Rightarrow f(-1) = f(1) \).
Thus, all three conditions of Rolle's theorem are satisfied. So there exists atleast one real number c in (-1, 1) such that f'(c) = 0.
Differentiating (1): \( f'(x) = e^{1-x^2}(-2x) = -2xe^{1-x^2} \).
Now \( f'(c) = 0 \Rightarrow -2ce^{1-c^2} = 0 \Rightarrow c = 0 \).
So there exists 0 in (-1, 1) such that f'(0) = 0.
Hence, Rolle's theorem is verified and c = 0.
In simple words: Each function keeps its same value at both endpoints of its interval and has a smooth slope everywhere in between. This guarantees that somewhere inside, the slope must be exactly zero (a flat spot).

Exam Tip: For trigonometric and exponential functions, always compute f(a) and f(b) carefully, check that they are equal, and then differentiate using chain rule and product rule where needed. Solve f'(c) = 0 by recognizing when sine, cosine, or exponential terms equal zero.

Example 4. Verify Rolle's theorem for the following functions and find point (or points) in the given interval where derivative is zero:
(a) \( f(x) = (x - a)^m(x - b)^n \) in [a, b], m, n ∈ ℕ
(b) \( f(x) = \log\left(\frac{x^2 + ab}{(a + b)x}\right) \) in [a, b], a > 0 (I.S.C. 2012)
Answer:
(a) Given \( f(x) = (x - a)^m(x - b)^n \), m, n ∈ ℕ ...(1)
Since m, n ∈ ℕ, f(x) is a polynomial in x.
(i) f(x) is continuous in [a, b].
(ii) f(x) is differentiable in (a, b).
(iii) f(a) = 0, f(b) = 0 \( \Rightarrow f(a) = f(b) \).
Thus, all three conditions of Rolle's theorem are satisfied. So there exists atleast one real number c in (a, b) such that f'(c) = 0.
Differentiating (1): \( f'(x) = (x - a)^m \cdot n(x - b)^{n-1} + (x - b)^n \cdot m(x - a)^{m-1} = (x - a)^{m-1}(x - b)^{n-1}(n(x - a) + m(x - b)) = (x - a)^{m-1}(x - b)^{n-1}((m + n)x - (na + mb)) \).
Now \( f'(c) = 0 \Rightarrow (c - a)^{m-1}(c - b)^{n-1}((m + n)c - (na + mb)) = 0 \).
But \( c \neq a, c \neq b \Rightarrow (m + n)c - (na + mb) = 0 \)
\( \Rightarrow c = \frac{mb + na}{m + n} \), which is a point in (a, b), for it divides [a, b] in the ratio m : n internally.
Thus, there exists a real number \( c = \frac{mb + na}{m + n} \) in (a, b) such that f'(c) = 0.
Hence, Rolle's theorem is verified and \( c = \frac{mb + na}{m + n} \).
(b) Given \( f(x) = \log\left(\frac{x^2 + ab}{(a+b)x}\right) = \log(x^2 + ab) - \log(a + b) - \log x \) ...(1)
(i) Since a > 0 and log x is continuous for all x > 0, therefore, f(x) is continuous in [a, b].
(ii) f(x) is differentiable in (a, b).
(iii) \( f(a) = \log\left(\frac{a^2 + ab}{(a+b)a}\right) = \log 1 = 0 \), \( f(b) = \log\left(\frac{b^2 + ab}{(a+b)b}\right) = \log 1 = 0 \) \( \Rightarrow f(a) = f(b) \).
Thus, all three conditions of Rolle's theorem are satisfied. So there exists atleast one real number c in (a, b) such that f'(c) = 0.
Differentiating (1): \( f'(x) = \frac{1}{x^2 + ab} \cdot 2x - 0 - \frac{1}{x} = \frac{2x}{x^2 + ab} - \frac{1}{x} = \frac{2x^2 - (x^2 + ab)}{x(x^2 + ab)} = \frac{x^2 - ab}{x(x^2 + ab)} \).
Now \( f'(c) = 0 \Rightarrow \frac{c^2 - ab}{c(c^2 + ab)} = 0 \Rightarrow c^2 - ab = 0 \Rightarrow c = \pm\sqrt{ab} \).
But \( c \in (a, b) \Rightarrow c = \sqrt{ab} \) (since the geometric mean lies between them).
So there exists a real number \( c = \sqrt{ab} \) in (a, b) such that f'(c) = 0.
Hence, Rolle's theorem is verified and \( c = \sqrt{ab} \).
In simple words: When a polynomial or logarithmic function starts and ends with zero value in a closed interval, there is always at least one interior point where the rate of change is zero. For the polynomial form, this point divides the interval in a specific ratio.

Exam Tip: Use the product rule carefully when differentiating products of powers. Factor out common terms to simplify the derivative. Remember that the geometric mean \( \sqrt{ab} \) always lies between two positive numbers a and b.

Example 5. It is given that for the function f(x) = x³ + bx² + ax + 5 on [1, 3] Rolle's theorem holds with c = 2 + \(\frac{1}{\sqrt{3}}\). Find the values of a and b.
Answer: Given \( f(x) = x^3 + bx^2 + ax + 5 \) ...(1)
We note that f(x) is continuous in [1, 3] and derivable in (1, 3) for all values of a and b.
Differentiating (1): \( f'(x) = 3x^2 + 2bx + a \) ...(2)
Since Rolle's theorem holds for f(x) in [1, 3] with \( c = 2 + \frac{1}{\sqrt{3}} \), therefore, we must have \( f(1) = f(3) \) and \( f'\left(2 + \frac{1}{\sqrt{3}}\right) = 0 \).
\( \Rightarrow 1 + b + a + 5 = 27 + 9b + 3a + 5 \Rightarrow 8b + 2a + 26 = 0 \Rightarrow a + 4b + 13 = 0 \) ...(3)
and \( 3\left(2 + \frac{1}{\sqrt{3}}\right)^2 + 2b\left(2 + \frac{1}{\sqrt{3}}\right) + a = 0 \)
\( \Rightarrow 3\left(4 + \frac{4}{\sqrt{3}} + \frac{1}{3}\right) + 4b + \frac{2b}{\sqrt{3}} + a = 0 \)
\( \Rightarrow (a + 4b + 13) + \frac{12}{\sqrt{3}} + \frac{2b}{\sqrt{3}} = 0 \Rightarrow \frac{12}{\sqrt{3}} + \frac{2b}{\sqrt{3}} = 0 \) (using (3))
\( \Rightarrow 2b + 12 = 0 \Rightarrow b = -6 \).
From (3), \( a - 24 + 13 = 0 \Rightarrow a = 11 \).
Hence, a = 11, b = -6.
In simple words: Use the two conditions that must hold for Rolle's theorem - the function values match at the endpoints, and the derivative is zero at the given critical point - to set up two equations and solve for the unknown coefficients.

Exam Tip: When given that Rolle's theorem applies at a specific point, set up the equation f'(c) = 0 with that exact point. Combine this with f(a) = f(b) to create two independent equations, then solve the system algebraically.

Example 6. Discuss the applicability of Rolle's theorem for the function f(x) = |x| in [-2, 2].
Answer: Given \( f(x) = |x|, x \in [-2, 2] \) ...(1)
(i) f(x) is continuous in [-2, 2].
(ii) Differentiating (1): \( f'(x) = \frac{x}{|x|}, x \neq 0 \)
\( \Rightarrow \) the derivative of f(x) does not exist at x = 0
\( \Rightarrow \) f(x) is not derivable in (-2, 2).
Thus, condition (ii) of Rolle's theorem is not satisfied. Therefore, Rolle's theorem is not applicable to the function f(x) = |x| in [-2, 2].
Moreover, f(-2) = |-2| = 2 and f(2) = |2| = 2 \( \Rightarrow f(-2) = f(2) \), so condition (iii) of Rolle's theorem is satisfied.
Further, it is clear from the graph that there is no point of the curve y = |x| in (-2, 2) at which the tangent is parallel to x-axis.
In simple words: Even though the absolute value function starts and ends at the same height, it has a sharp corner at x = 0, so the derivative does not exist there. Rolle's theorem cannot be used because one key condition fails.

Exam Tip: Always check all three conditions carefully. If even one condition fails, Rolle's theorem does not apply. For functions with absolute values, check if the derivative exists at sharp points (corners or cusps).

8.2 Lagrange's Mean Value Theorem

If a function f(x) is
(i) continuous in the closed interval [a, b] and
(ii) derivable in the open interval (a, b),
then there exists atleast one real number c in (a, b) such that \( f'(c) = \frac{f(b) - f(a)}{b - a} \).

Proof. Consider a function g defined by \( g(x) = f(x) + kx \) ...(1)
where k is a constant (real number) to be chosen in such a way that \( g(a) = g(b) \)
\( \Rightarrow f(a) + ka = f(b) + kb \Rightarrow k(a - b) = f(b) - f(a) \)
\( \Rightarrow k = -\frac{f(b) - f(a)}{b - a} \) ...(2)
Now (i) g(x) is continuous in [a, b] (since f(x) is given to be continuous in [a, b] and kx being polynomial is continuous for all x ∈ ℝ - their sum is continuous in [a, b])
(ii) g(x) is derivable in (a, b) (since f(x) is given to be derivable in (a, b) and kx being polynomial is derivable for all x ∈ ℝ - their sum is derivable in (a, b))
(iii) \( g(a) = g(b) \).
Thus, all three conditions of Rolle's theorem are satisfied by the function g in [a, b]. Therefore, there exists atleast one real number c in (a, b) such that g'(c) = 0.
Differentiating (1): \( g'(x) = f'(x) + k \cdot 1 \)
Now \( g'(c) = 0 \Rightarrow f'(c) + k = 0 \Rightarrow k = -f'(c) \) ...(3)
From (2) and (3), we get \( -f'(c) = -\frac{f(b) - f(a)}{b - a} \Rightarrow f'(c) = \frac{f(b) - f(a)}{b - a} \).

Geometrical Interpretation

Let A, B be the points on the curve y = f(x) corresponding to the real numbers a, b respectively. Since f(x) is continuous in [a, b], the graph of the curve y = f(x) is continuous from A to B. Again, as f(x) is derivable in (a, b) the curve y = f(x) has a tangent at each point between A and B. Also as \( a \neq b \), the slope of the chord AB exists and the slope of the chord AB = \( \frac{f(b) - f(a)}{b - a} \).
Then Lagrange's Mean Value Theorem asserts that there is atleast one point lying between A and B such that the tangent at which is parallel to the chord AB. There may exist more than one point between A and B the tangents at which are parallel to the chord AB. Lagrange's mean value theorem ensures the existence of atleast one real number c in (a, b) such that \( f'(c) = \frac{f(b) - f(a)}{b - a} \).

Remarks

1. Lagrange's mean value theorem fails for the function which does not satisfy even one of the two conditions.
2. The converse of Lagrange's mean value theorem may not be true, for, \( f'(c) \) may be equal to \( \frac{f(b) - f(a)}{b - a} \) at a point c in (a, b) without satisfying both the conditions of Lagrange's mean value theorem.

Example 1. Verify Lagrange's mean value theorem for the following functions in the given interval and find 'c' of this theorem.
(a) \( f(x) = 3x^2 - 5x + 1 \) in [2, 5] (I.S.C. 2007)
(b) \( f(x) = (x - 1)(x - 2)(x - 3) \) in [0, 4]
Answer:
(a) Given \( f(x) = 3x^2 - 5x + 1, x \in [2, 5] \) ...(1)
(i) f(x) being a polynomial is continuous in [2, 5].
(ii) f(x) being a polynomial is derivable in (2, 5).
Thus, both conditions of Lagrange's mean value theorem are satisfied. Therefore, there exists atleast one real number c in (2, 5) such that \( f'(c) = \frac{f(5) - f(2)}{5 - 2} \).
\( f(5) = 3 \cdot 5^2 - 5 \cdot 5 + 1 = 51 \), \( f(2) = 3 \cdot 2^2 - 5 \cdot 2 + 1 = 3 \).
Differentiating (1): \( f'(x) = 6x - 5 \Rightarrow f'(c) = 6c - 5 \).
\( \therefore f'(c) = \frac{f(5) - f(2)}{5 - 2} \Rightarrow 6c - 5 = \frac{51 - 3}{3} \Rightarrow 6c - 5 = 16 \)
\( \Rightarrow 6c = 21 \Rightarrow c = \frac{7}{2} \).
Thus, there exists \( c = \frac{7}{2} \) in (2, 5) such that \( f'\left(\frac{7}{2}\right) = \frac{f(5) - f(2)}{5 - 2} \).
Hence, Lagrange's mean value theorem is verified and \( c = \frac{7}{2} \).
(b) Given \( f(x) = (x - 1)(x - 2)(x - 3) = x^3 - 6x^2 + 11x - 6 \) ...(1)
(i) f(x) being a polynomial is continuous in [0, 4].
(ii) f(x) being a polynomial is derivable in (0, 4).
Thus, both conditions of Lagrange's mean value theorem are satisfied. Therefore, there exists atleast one real number c in (0, 4) such that \( f'(c) = \frac{f(4) - f(0)}{4 - 0} \).
\( f(4) = 64 - 96 + 44 - 6 = 6 \), \( f(0) = 0 - 0 + 0 - 6 = -6 \).
Differentiating (1): \( f'(x) = 3x^2 - 12x + 11 \Rightarrow f'(c) = 3c^2 - 12c + 11 \).
\( \therefore f'(c) = \frac{f(4) - f(0)}{4 - 0} \Rightarrow 3c^2 - 12c + 11 = \frac{6 - (-6)}{4 - 0} \)
\( \Rightarrow 3c^2 - 12c + 11 = 3 \Rightarrow 3c^2 - 12c + 8 = 0 \)
\( \Rightarrow c = \frac{12 \pm \sqrt{144 - 96}}{6} = \frac{12 \pm \sqrt{48}}{6} = \frac{12 \pm 4\sqrt{3}}{6} = 2 \pm \frac{2\sqrt{3}}{3} = 2 \pm \frac{2}{\sqrt{3}} \).
As \( 0 < 2 - \frac{2}{\sqrt{3}} < 2 + \frac{2}{\sqrt{3}} < 4 \Rightarrow 2 - \frac{2}{\sqrt{3}} \) and \( 2 + \frac{2}{\sqrt{3}} \) both lie in (0, 4).
Thus, there exist \( c = 2 \pm \frac{2}{\sqrt{3}} \) in (0, 4) such that \( f'(c) = \frac{f(4) - f(0)}{4 - 0} \).
Hence, Lagrange's mean value theorem is verified and \( c = 2 \pm \frac{2}{\sqrt{3}} \).
In simple words: There is always at least one point inside an interval where the slope of the tangent line equals the slope of the line joining the endpoints of the curve.

Exam Tip: Calculate f(a), f(b), and the slope of the chord first. Then differentiate f(x) and set f'(c) equal to the chord slope. Solve for c algebraically. If multiple values result, check that they all lie within the open interval (a, b).

Example 2. Verify Lagrange's mean value theorem for the following functions in the given intervals:
(a) \( f(x) = \tan^{-1} x \) in [0, 1]
(b) \( f(x) = \sqrt{x^2 - x} \) in [1, 4] (I.S.C. 2013)
Answer:
(a) Given \( f(x) = \tan^{-1} x, x \in [0, 1] \) ...(1)
(i) f(x) is continuous in [0, 1].
(ii) f(x) is derivable in (0, 1).
Thus, both conditions of Lagrange's mean value theorem are satisfied. Therefore, there exists atleast one real number c in (0, 1) such that \( f'(c) = \frac{f(1) - f(0)}{1 - 0} \).
\( f(1) = \tan^{-1}(1) = \frac{\pi}{4} \), \( f(0) = \tan^{-1}(0) = 0 \).
Differentiating (1): \( f'(x) = \frac{1}{1 + x^2} \Rightarrow f'(c) = \frac{1}{1 + c^2} \).
\( \therefore f'(c) = \frac{f(1) - f(0)}{1 - 0} \Rightarrow \frac{1}{1 + c^2} = \frac{\frac{\pi}{4} - 0}{1} \Rightarrow \frac{1}{1 + c^2} = \frac{\pi}{4} \)
\( \Rightarrow 1 + c^2 = \frac{4}{\pi} \Rightarrow c^2 = \frac{4}{\pi} - 1 \Rightarrow c = \pm\sqrt{\frac{4 - \pi}{\pi}} \).
But \( c \in (0, 1) \Rightarrow c = \sqrt{\frac{4 - \pi}{\pi}} \).
Thus, there exists \( c = \sqrt{\frac{4 - \pi}{\pi}} \) in (0, 1) such that \( f'(c) = \frac{f(1) - f(0)}{1 - 0} \).
Hence, Lagrange's mean value theorem is verified.
(b) Given \( f(x) = \sqrt{x^2 - x}, x \in [1, 4] \) ...(1)
(i) Since g(x) = x² - x is continuous on ℝ and h(x) = \(\sqrt{x}\) is continuous in [0, ∞), therefore, (hog)(x) = h(g(x)) = h(x² - x) = \(\sqrt{x^2 - x}\) is continuous for all x such that x² - x ≥ 0 \( \Rightarrow \sqrt{x^2 - x} \) is continuous in (-∞, 0] ∪ [1, ∞) \( \Rightarrow \sqrt{x^2 - x} \) is continuous in [1, 4].
(ii) Differentiating (1): \( f'(x) = \frac{1}{2}(x^2 - x)^{-1/2}(2x - 1) = \frac{2x - 1}{2\sqrt{x^2 - x}} \)
which exists for all x such that x² - x > 0 i.e. for x in (-∞, 0) ∪ (1, ∞) \( \Rightarrow \) f(x) is derivable for all x in (1, 4).
Thus, both conditions of Lagrange's mean value theorem are satisfied. Therefore, there exists atleast one real number c in (1, 4) such that \( f'(c) = \frac{f(4) - f(1)}{4 - 1} \).
\( \Rightarrow \frac{2c - 1}{2\sqrt{c^2 - c}} = \frac{\sqrt{16 - 4} - \sqrt{1 - 1}}{4 - 1} = \frac{\sqrt{12} - 0}{3} = \frac{2\sqrt{3}}{3} \)
\( \Rightarrow 3(2c - 1)^2 = 16(c^2 - c) \Rightarrow 4c^2 - 4c - 3 = 0 \)
\( \Rightarrow (2c - 3)(2c + 1) = 0 \Rightarrow c = \frac{3}{2}, -\frac{1}{2} \).
Thus, there exists \( c = \frac{3}{2} \in (1, 4) \) such that \( f'(c) = \frac{f(4) - f(1)}{4 - 1} \).
Hence, Lagrange's mean value theorem is verified.
In simple words: For both the inverse tangent and square root functions, the conditions of the theorem hold in the given intervals. There is always a point where the rate of change of the function matches the average rate of change over the interval.

Exam Tip: For inverse trigonometric and radical functions, carefully verify both conditions by checking domain restrictions. When solving f'(c) = [chord slope], simplify algebra carefully and verify that any solution found actually lies in the open interval (a, b).

Example 3. Find a point on the graph of y = x³ where the tangent is parallel to the chord joining (1, 1) and (3, 27).
Answer: Let us apply Lagrange's mean value theorem to the function \( f(x) = x^3 \) in the interval [1, 3] ...(1)
(i) f(x) being polynomial is continuous in [1, 3].
(ii) f(x) being polynomial is derivable in (1, 3).
Thus, both conditions of Lagrange's mean value theorem are satisfied by the function f(x) in [1, 3]. Therefore, there exists atleast one real number c in (1, 3) such that \( f'(c) = \frac{f(3) - f(1)}{3 - 1} \).
\( f(3) = 3^3 = 27 \) and \( f(1) = 1^3 = 1 \).
Differentiating (1): \( f'(x) = 3x^2 \Rightarrow f'(c) = 3c^2 \).
Now \( f'(c) = \frac{f(3) - f(1)}{3 - 1} \Rightarrow 3c^2 = \frac{27 - 1}{3 - 1} \Rightarrow 3c^2 = 13 \)
\( \Rightarrow c^2 = \frac{13}{3} \Rightarrow c = \pm\sqrt{\frac{39}{9}} = \pm\frac{\sqrt{39}}{3} \).
But \( c \in (1, 3) \Rightarrow c = \frac{\sqrt{39}}{3} \).
When \( x = \frac{\sqrt{39}}{3} \), from (1), \( y = \left(\frac{\sqrt{39}}{3}\right)^3 = \frac{39\sqrt{39}}{27} = \frac{13\sqrt{39}}{9} \).
Hence, there exists a point \( \left(\frac{\sqrt{39}}{3}, \frac{13\sqrt{39}}{9}\right) \) on the given curve y = x³ where the tangent is parallel to the chord joining the points (1, 1) and (3, 27).
In simple words: Find the slope between the two given points, then use the derivative to find where the curve has that same slope.

Exam Tip: Write both points' coordinates clearly, calculate the chord slope, set up f'(c) equal to this slope, solve for c, and finally substitute back into f(x) to find the y-coordinate of the point on the curve.

Example 4. Does the Lagrange's mean value theorem apply to f(x) = x^{1/3}, -1 ≤ x ≤ 1? What conclusions can be drawn?
Answer: Given \( f(x) = x^{1/3}, x \in [-1, 1] \) ...(1)
(i) f(x) is continuous in [-1, 1].
(ii) Differentiating (1): \( f'(x) = \frac{1}{3}x^{-2/3} = \frac{1}{3x^{2/3}}, x \neq 0 \) ...(2)
\( \Rightarrow \) the derivative of f(x) does not exist at x = 0
\( \Rightarrow \) f(x) is not derivable in (-1, 1).
Thus, condition (ii) of Lagrange's mean value theorem is not satisfied by the function f(x) = x^{1/3} in [-1, 1] and hence Lagrange's mean value theorem is not applicable to the given function f(x) = x^{1/3} in [-1, 1].
Conclusion. However, from (2), \( f'(c) = \frac{1}{3c^{2/3}}, c \neq 0 \)
Also \( f(-1) = (-1)^{1/3} = -1 \), \( f(1) = 1^{1/3} = 1 \) (we have taken only real values)
\( \therefore f'(c) = \frac{f(1) - f(-1)}{1 - (-1)} \Rightarrow \frac{1}{3c^{2/3}} = \frac{1 - (-1)}{1 - (-1)} = \frac{2}{2} = 1 \)
\( \Rightarrow c^{2/3} = \frac{1}{3} \Rightarrow c^2 = \frac{1}{27} \Rightarrow c = \pm\frac{1}{3\sqrt{3}} \).
As \( -1 < -\frac{1}{3\sqrt{3}} < \frac{1}{3\sqrt{3}} < 1 \Rightarrow c = \pm\frac{1}{3\sqrt{3}} \) both lie in (-1, 1).
Thus, we find that there exist two real numbers \( c = \pm\frac{1}{3\sqrt{3}} \) in (-1, 1) such that \( f'(c) = \frac{f(1) - f(-1)}{1 - (-1)} \). It follows that the converse of Lagrange's mean value theorem may not be true.
In simple words: Even though the function is not smooth everywhere (it has a corner at x = 0), there are still points where the slope matches the average slope. This shows that the converse of the theorem is not guaranteed.

Exam Tip: When a function fails the conditions of a theorem, state clearly which condition is violated. Nevertheless, it is possible that the conclusion might still hold by coincidence - examine whether the equation f'(c) = [chord slope] has solutions in the open interval, and note that this does not prove the theorem applies.

8.3 Maxima and Minima

8.3.1 Absolute Maxima and Absolute Minima

Let f be a real valued function defined on D (subset of ℝ), then
(i) f is said to have absolute maxima at x = c (in D) iff f(x) ≤ f(c) for all x ∈ D, and c is called point of absolute maxima and f(c) is called absolute maximum (or greatest) value of f on D.
(ii) f is said to have absolute minima at x = d (in D) iff f(d) ≤ f(x) for all x ∈ D, and d is called point of absolute minima and f(d) is called absolute minimum (or smallest) value of f on D.
Obviously, absolute maximum and absolute minimum values of a function (if they exist) are unique. However, absolute maximum or absolute minimum values of a given function f on D (subset of ℝ) may be obtained at more than one points. Also it is not essential that a given function must have maxima or minima in its domain.

8.3.2 Local Maxima and Local Minima

Let f be a real valued function defined on D (subset of ℝ), then
(i) f is said to have a local (or relative) maxima at x = c (in D) iff there exists a positive real number δ such that f(x) ≤ f(c) for all x in (c - δ, c + δ) i.e. f(x) ≤ f(c) for all x in the immediate neighbourhood of c, and c is called point of local maxima and f(c) is called local maximum value.
(ii) f is said to have local (or relative) minima at x = d (in D) iff there exists some positive real number δ such that f(d) ≤ f(x) for all x ∈ (d - δ, d + δ) i.e. f(d) ≤ f(x) for all x in the immediate neighbourhood of d, and d is called point of local minima and f(d) is called local minimum value.
Geometrically, a point c in the domain of the given function f is a point of local maxima or local minima according as the graph of f has a peak or trough (cavity) at c.
(iii) a point (in D) which is either a point of local maxima or a point of local minima is called an extreme point, and the value of the function at this point is called an extreme value.

Remarks

1. A local maximum (minimum) value may not be the absolute maximum (minimum) value.
2. A local maximum value at some point may be less than a local minimum value of the function at another point.

Stationary (or Turning) Point

Let f be a real valued function defined on D (subset of ℝ), then a point c (in D) is called a stationary (or turning or critical) point of f iff f is differentiable at x = c and f'(c) = 0.
However, it is not essential that an extreme point is a stationary point, and a stationary point is an extreme point.

Exercise 8.2

Question 1.
Answer:
(i) 2
(ii) 2
(iii) \( \frac{9}{2} \)
(iv) \( \frac{-8 + 4\sqrt{13}}{3} \)

Exam Tip: Check each sub-part calculation by verifying intermediate steps; errors in differentiation or substitution accumulate quickly in multi-part problems.

Question 2.
Answer:
(i) \( \pm 1 \)
(ii) \( 1 - \frac{\sqrt{21}}{6} \)
(iii) \( 6 - \frac{\sqrt{39}}{3} \)

Exam Tip: When finding critical points, remember that \( \pm \) solutions often indicate both maximum and minimum candidates - test both in the original function.

Question 3.
Answer:
(i) \( \sqrt{3} \)
(ii) \( \frac{1 + 3\sqrt{5}}{4} \)
(iii) \( \frac{1}{3}(2 + \sqrt{13}) \)
(iv) \( \frac{8}{27} \)

Exam Tip: Simplify radical expressions in your final answer - leave them in exact form unless the question specifies decimal approximation.

Question 4.
Answer:
(i) \( \cos^{-1}\left(\frac{2}{\pi}\right) \)
(ii) \( \pm\frac{\pi}{2} \)
(iii) \( \frac{\pi}{3} \)

Exam Tip: Be careful with inverse trigonometric notation and ensure your answers fall within the principal range of the respective function.

Question 5.
Answer:
(i) \( \log_2 e \)
(ii) \( \sqrt{6} \)
(iii) \( \cos^{-1}\left(\frac{1 \pm \sqrt{33}}{8}\right) \)

Exam Tip: When the answer involves both positive and negative roots, state both values clearly and verify that each lies within the valid domain.

Question 6.
Answer: \( \left(\frac{7}{2}, \frac{1}{4}\right) \)

Exam Tip: For coordinate answers, always present points as ordered pairs in the form (x, y).

Question 7.
Answer: \( \left(\frac{7}{3}, -\frac{2}{3}\sqrt{\frac{7}{3}}\right) \)

Exam Tip: When a point appears in a radical form, verify that the expression cannot be simplified further before finalizing.

Question 8.
Answer: \( (2, -7) \)

Exam Tip: State coordinate pairs in standard notation with the x-coordinate first.

Question 9.
Answer: \( \left(\frac{9}{4}, \frac{1}{2}\right) \)

Exam Tip: Double-check fractional coordinates by substituting back into the original function or constraint equation.

Exercise 8.3

Question 1.
Answer:
(i) Minimum value = 3, no maximum value
(ii) Maximum value = 7, no minimum value

Exam Tip: On unbounded domains, always state "no maximum" or "no minimum" rather than leaving it blank - the examiner needs to see you've considered the endpoint behavior.

Question 2.
Answer:
(i) Minimum value = -2, no maximum value
(ii) Maximum value = 9, no minimum value

Exam Tip: When a function approaches infinity in one direction, explicitly state that no maximum or minimum exists in that direction.

Question 3.
Answer:
(i) Neither maximum nor minimum
(ii) Minimum value = 0, no maximum value

Exam Tip: Recognize that oscillating or monotonic functions may have neither extremum, while functions with boundaries may have only one.

Question 4.
Answer:
(i) Maximum value = 3, no minimum value
(ii) Maximum value = 6, minimum value = 4

Exam Tip: On closed intervals, always check endpoints and critical points to find both extrema.

Question 5.
Answer:
(i) Maximum value = 4, minimum value = 2
(ii) Maximum value = -2, minimum value = -3

Exam Tip: For closed and bounded domains, both a maximum and minimum must exist by the Extreme Value Theorem.

Question 6.
Answer:
(i) Maximum value = 5, minimum value = -1
(ii) Maximum value = 5, minimum value = -5

Exam Tip: Compare function values at all critical points and interval endpoints to identify global extrema.

Question 7.
Answer: -2

Exam Tip: For single-value answers, ensure your working clearly shows how you arrived at that unique result.

Question 8.
Answer: 1

Exam Tip: Verify single numerical answers by substituting back into the function definition.

Question 9.
Answer: \( x = \frac{\pi}{4} \)

Exam Tip: When the answer is an angle, express it in exact form (as a multiple of \( \pi \)) unless otherwise directed.

Question 11. Complete the table for maximum value, minimum value, point of maxima, and point of minima.
Answer:

Exam Tip: For multi-part extrema tables, organize answers clearly in tabular form and always state the actual points where maxima and minima occur - do not omit the x-values.

Question 12.
Answer: 18; -9

Exam Tip: When two separate answers are requested, clearly label or separate them with a semicolon to show they are distinct values.

Question 13.
Answer: \( \frac{\pi}{4}, \frac{5\pi}{4} \)

Exam Tip: For trigonometric answers involving multiple angles in different quadrants or periods, list all solutions that satisfy the domain constraint.

Question 14.
Answer: Maximum value = \( 2\pi \), minimum value = 0

Exam Tip: Always present both maximum and minimum values clearly, even if one is zero - stating zero explicitly shows you've found both extrema on the given interval.