Monotonicity JEE Mathematics Worksheets Set 02

Question. Verify Rolle’s theorem for the function \( f(x) = \log_e \left( \frac{x^2 + ab}{x(a + b)} \right) + p \), for \([a, b]\) where \( 0 < a < b \).
Answer: Logarithm function is continuous & differentiable in its domain.
For \( 0 < a < b \), the function \( f(x) \) is continuous and differentiable.
\( f(a) = \log \left( \frac{a^2 + ab}{a(a + b)} \right) + p \)
\( \implies f(a) = \log 1 + p = p \)
\( f(b) = \log \left( \frac{b^2 + ab}{b(a + b)} \right) + p \)
\( \implies f(b) = \log 1 + p = p \)
Hence \( f(a) = f(b) \).
So, Rolle’s theorem is applicable.

Question. Using Rolle’s theorem prove that the equation \( 3x^2 + px - 1 = 0 \) has at least one real root in the interval \( (-1, 1) \).
Answer: Consider \( F(x) = x^3 + \frac{px^2}{2} - x \).
\( F(x) \) is a polynomial function and hence continuous and differentiable in \( [-1, 1] \).
\( F(-1) = (-1)^3 + \frac{p(-1)^2}{2} - (-1) = -1 + \frac{p}{2} + 1 = \frac{p}{2} \)
\( F(1) = (1)^3 + \frac{p(1)^2}{2} - (1) = 1 + \frac{p}{2} - 1 = \frac{p}{2} \)
Since \( F(-1) = F(1) \), by Rolle’s theorem, \( F'(x) = 0 \) for at least one root \( x \in (-1, 1) \).
\( \implies 3x^2 + px - 1 = 0 \) has at least one real root in the interval \( (-1, 1) \).

Question. If the equation \( a_0 x^n + a_1 x^{n - 1} + \dots + a_{n - 1} x = 0 \) has a +ve root \( \alpha \), prove that the equation \( na_0 x^{n - 1} + (n - 1)a_1 x^{n - 2} + \dots + a_{n - 1} = 0 \) also has a positive root smaller than \( \alpha \).
Answer: Let \( f(x) = a_0 x^n + a_1 x^{n - 1} + \dots + a_{n - 1} x \).
\( f(x) \) is a polynomial, so it is continuous and differentiable for all real \( x \).
Given \( f(\alpha) = 0 \) and we see \( f(0) = 0 \).
Applying Rolle’s theorem on the interval \( [0, \alpha] \):
\( \implies f'(c) = 0 \) for some \( c \in (0, \alpha) \).
\( f'(x) = na_0 x^{n - 1} + (n - 1)a_1 x^{n - 2} + \dots + a_{n - 1} \).
Thus, the equation \( na_0 x^{n - 1} + (n - 1)a_1 x^{n - 2} + \dots + a_{n - 1} = 0 \) has a positive root \( c < \alpha \).

Question. Explain the failure of Lagrange's mean value theorem in the interval \([-1, 1]\) for the function \( f(x) = \frac{1}{x} \).
Answer: For the function \( f(x) = \frac{1}{x} \):
\( f(x) \) is continuous in \( [-1, 0) \cup (0, 1] \) and also differentiable in \( [-1, 1] \setminus \{0\} \).
However, the function is not defined at \( x = 0 \), which lies within the interval \( [-1, 1] \).
Thus, \( f(x) \) is not continuous on the closed interval \( [-1, 1] \).
So, Lagrange's mean value theorem is not applicable in \( [-1, 1] \).

Question. If a, b are two real numbers with \( a < b \) show that a real number 'c' can be found between a and b such that \( 3c^2 = b^2 + ab + a^2 \).
Answer: Let \( f(x) = x^3 \).
\( f(x) \) is continuous on \( [a, b] \) and differentiable on \( (a, b) \).
So using Lagrange’s Mean Value Theorem in \( (a, b) \):
\( f'(c) = \frac{f(b) - f(a)}{b - a} \)
\( \implies 3c^2 = \frac{b^3 - a^3}{b - a} \)
\( \implies 3c^2 = \frac{(b - a)(b^2 + ab + a^2)}{b - a} \)
\( \implies 3c^2 = a^2 + ab + b^2 \)

Question. If \( a > b > 0 \), with the aid of Lagrange’s formula, prove validity of the inequality \( nb^{n - 1} (a - b) < a^n - b^n < na^{n - 1} (a - b) \), if \( n > 1 \). Also prove that the inequalities are in opposite sense if \( 0 < n < 1 \).
Answer: Let \( f(x) = x^n \).
\( f(x) \) is continuous and differentiable in \( [b, a] \).
So using Lagrange’s Mean Value Theorem:
\( f'(c) = \frac{a^n - b^n}{a - b} \) ......(i)
Since \( b < c < a \):
If \( n > 1 \):
\( \implies b^{n - 1} < c^{n - 1} < a^{n - 1} \)
\( \implies nb^{n - 1} < nc^{n - 1} < na^{n - 1} \)
From (i), substitute \( nc^{n - 1} \):
\( \implies nb^{n - 1} < \frac{a^n - b^n}{a - b} < na^{n - 1} \)
\( \implies nb^{n - 1} (a - b) < a^n - b^n < na^{n - 1} (a - b) \).
If \( 0 < n < 1 \), the power function \( x^{n-1} \) is decreasing, so the inequalities will be in the opposite sense.

Question. Using Rolle’s theorem show that the derivative of the function \( f(x) = \begin{cases} x \sin \frac{\pi}{x} & \text{for } x > 0 \\ 0 & \text{for } x = 0 \end{cases} \) vanishes at an infinite set of points of the interval (0, 1).
Answer: \( f(x) \) is continuous on \( [0, 1] \) and differentiable on \( (0, 1) \).
\( f(0) = 0 \) and \( f(1) = 1 \cdot \sin \pi = 0 \).
Also, \( f(x) = 0 \) whenever \( \sin(\pi/x) = 0 \), which occurs at \( x = 1/k \) for \( k \in \mathbb{N} \).
Consider the intervals \( [1/(k+1), 1/k] \) for \( k = 1, 2, 3, \dots \).
Within each such interval, \( f(1/(k+1)) = f(1/k) = 0 \).
By Rolle’s theorem, there exists some \( x_k \in (1/(k+1), 1/k) \) such that \( f'(x_k) = 0 \).
As there are infinitely many such intervals within \( (0, 1) \), \( f'(x) \) vanishes at an infinite set of points.

Question. A function f is differentiable in the interval \( 0 \le x \le 5 \) such that \( f(0) = 4 \) & \( f(5) = -1 \). If \( g(x) = \frac{f(x)}{x + 1} \), then prove that there exists some \( c \in (0, 5) \) such that \( g'(c) = -\frac{5}{6} \).
Answer: \( f(x) \) is continuous and differentiable in \( [0, 5] \).
The term \( (x + 1) \) is also continuous and non-zero on \( [0, 5] \).
So \( g(x) = \frac{f(x)}{x + 1} \) is continuous in \( [0, 5] \) and differentiable in \( (0, 5) \).
By Lagrange’s Mean Value Theorem, there exists some \( c \in (0, 5) \) such that:
\( g'(c) = \frac{g(5) - g(0)}{5 - 0} \)
\( g(5) = \frac{f(5)}{5 + 1} = \frac{-1}{6} \)
\( g(0) = \frac{f(0)}{0 + 1} = 4 \)
\( \implies g'(c) = \frac{-1/6 - 4}{5} = \frac{-25/6}{5} \)
\( \implies g'(c) = -\frac{5}{6} \)

Question. Let \( f(x) \) & \( g(x) \) be differentiable function so that \( f(x) g'(x) \neq f'(x) g(x) \). Prove that between any two roots of \( f(x) \) there exist atleast one root of \( g(x) \).
Answer: Let \( h(x) = \frac{f(x)}{g(x)} \).
Let \( \alpha \) and \( \beta \) be two consecutive roots of \( f(x) \).
Then \( f(\alpha) = f(\beta) = 0 \).
This implies \( h(\alpha) = h(\beta) = 0 \).
If \( g(x) \) has no roots in \( [\alpha, \beta] \), then \( h(x) \) is continuous and differentiable on \( [\alpha, \beta] \).
By Rolle’s theorem, there exists some \( c \in (\alpha, \beta) \) such that \( h'(c) = 0 \).
\( h'(x) = \frac{f'(x)g(x) - f(x)g'(x)}{g(x)^2} \)
\( \implies h'(c) = 0 \)
\( \implies f'(c)g(c) - f(c)g'(c) = 0 \)
This contradicts the given condition that \( f(x) g'(x) \neq f'(x) g(x) \) for all \( x \).
Hence, \( g(x) \) must have at least one root between any two roots of \( f(x) \).

Question. f is continuous in \([a, b]\) and differentiable in \((a, b)\) (where \( a > 0 \)) such that \( \frac{f(a)}{a} = \frac{f(b)}{b} \). Prove that there exist \( x_0 \in (a, b) \) such that \( f'(x_0) = \frac{f(x_0)}{x_0} \).
Answer: If \( f \) is continuous in \( [a, b] \) and differentiable in \( (a, b) \), then by Lagrange’s Mean Value Theorem:
\( f'(x_0) = \frac{f(b) - f(a)}{b - a} \) ......(i)
Given \( \frac{f(a)}{a} = \frac{f(b)}{b} \)
\( \implies f(b) = \frac{b \cdot f(a)}{a} \)
Substitute in (i):
\( f'(x_0) = \frac{\frac{b \cdot f(a)}{a} - f(a)}{b - a} = \frac{f(a) \cdot (\frac{b - a}{a})}{b - a} = \frac{f(a)}{a} \)
Since \( \frac{f(a)}{a} = \frac{f(x_0)}{x_0} \) at the point \( x_0 \) where the ratio is constant,
\( \implies f'(x_0) = \frac{f(x_0)}{x_0} \)

Question. Verify Rolles theorem for \( f(x) = (x - a)^m (x - b)^n \) on \([a, b]\); m, n being positive integer.
Answer: As \( f(x) \) is a polynomial, \( f(x) \) is continuous in \( [a, b] \).
Also \( f(x) \) is differentiable in \( (a, b) \).
Checking boundary values:
\( f(a) = (a - a)^m (a - b)^n = 0 \)
\( f(b) = (b - a)^m (b - b)^n = 0 \)
So \( f(a) = f(b) \).
Therefore, Rolle’s theorem is applicable for \( f(x) \).

Question. Let \( f : [a, b] \to R \) be continuous on \([a, b]\) and differentiable on \((a, b)\). If \( f(a) < f(b) \), then show that \( f'(c) > 0 \) for some \( c \in (a, b) \).
Answer: \( f \) is continuous in \( [a, b] \) and differentiable in \( (a, b) \).
By Lagrange’s Mean Value Theorem, there exists some \( c \in (a, b) \) such that:
\( f'(c) = \frac{f(b) - f(a)}{b - a} \) ......(i)
Given \( f(b) - f(a) > 0 \) and \( b - a > 0 \).
\( \implies f'(c) > 0 \) for some \( c \in (a, b) \).

Question. Let \( f(x) = 4x^3 - 3x^2 - 2x + 1 \), use Rolle’s theorem to prove that there exist c, \( 0 < c < 1 \) such that \( f(c) = 0 \).
Answer: Let \( F(x) \) be the integral of \( f(x) \):
\( F(x) = x^4 - x^3 - x^2 + x \).
\( F(x) \) is a polynomial, so it is continuous in \( [0, 1] \) and differentiable in \( (0, 1) \).
\( F(0) = 0 \)
\( F(1) = 1 - 1 - 1 + 1 = 0 \)
Since \( F(0) = F(1) \), by Rolle’s theorem, there exists some \( c \in (0, 1) \) such that \( F'(c) = 0 \).
\( \implies f(c) = 0 \).

Question. Using LMVT prove that : (a) \( \tan x > x \) in \( (0, \frac{\pi}{2}) \), (b) \( \sin x < x \) for \( x > 0 \)
Answer: (a) Let \( f(x) = \tan x - x \).
\( f'(x) = \sec^2 x - 1 = \tan^2 x > 0 \) for \( x \in (0, \pi/2) \).
So \( f(x) \) is an increasing function.
Thus, for \( x > 0 \), \( f(x) > f(0) \).
\( \implies \tan x - x > 0 \)
\( \implies \tan x > x \).
(b) Let \( f(x) = x - \sin x \).
\( f'(x) = 1 - \cos x \ge 0 \).
\( f(x) \) is an increasing function for \( x > 0 \).
\( \implies f(x) > f(0) \)
\( \implies x - \sin x > 0 \)
\( \implies x > \sin x \).

Question. Prove that if f is differentiable on \([a, b]\) and if \( f(a) = f(b) = 0 \) then for any real \( \alpha \) there is an \( x \in (a, b) \) such that \( \alpha f(x) + f'(x) = 0 \).
Answer: Let \( g(x) = e^{\alpha x} \cdot f(x) \).
Since \( f(a) = f(b) = 0 \), then \( g(a) = g(b) = 0 \).
Also \( g(x) \) is continuous and differentiable in \( [a, b] \).
So Rolle’s theorem is applicable for \( g(x) \).
There exists some \( x \in (a, b) \) such that \( g'(x) = 0 \).
\( \implies \alpha e^{\alpha x} f(x) + e^{\alpha x} f'(x) = 0 \)
Dividing by \( e^{\alpha x} \) (which is never zero):
\( \implies \alpha f(x) + f'(x) = 0 \).

Question. For what value of a, m and b does the function \( f(x) = \begin{cases} 3 & x = 0 \\ -x^2 + 3x + a & 0 < x < 1 \\ mx + b & 1 \le x \le 2 \end{cases} \) satisfy the hypothesis of the mean value theorem for the interval [0, 2].
Answer: \( f(x) \) should be continuous in \( [0, 2] \) and differentiable in \( (0, 2) \).
For continuity at \( x = 0 \):
\( f(0^+) = f(0) \)
\( \implies a = 3 \).
For continuity at \( x = 1 \):
\( f(1^-) = f(1^+) \)
\( \implies -1 + 3 + a = m + b \)
\( \implies 5 = m + b \) ......(i)
For differentiability at \( x = 1 \):
\( f'(1^-) = f'(1^+) \)
\( \implies -2(1) + 3 = m \)
\( \implies m = 1 \).
From (i), \( 5 = 1 + b \)
\( \implies b = 4 \).
Result: \( a = 3, m = 1, b = 4 \).

Question. Suppose that on the interval \([-2, 4]\) the function f is differentiable, \( f(-2) = 1 \) and \( |f'(x)| \le 5 \). Find the bounding functions of f on \([-2, 4]\), using LMVT.
Answer: \( f'(x) = \pm 5 \) (for bounding function).
Integrating, we get:
\( \implies f(x) = \pm (5x + c) \).
Given \( f(-2) = 1 \).
If \( f(x) = 5x + c \):
\( 1 = 5(-2) + c \implies c = 11 \).
If \( f(x) = -(5x + c) \):
\( 1 = -(-10 + c) = 10 - c \implies c = 9 \).
Thus, \( f(x) = \pm (5x + 9) \) are the bounding functions.

Question. Let f, g be differentiable on R and suppose that \( f(0) = g(0) \) and \( f'(x) \le g'(x) \) for all \( x \ge 0 \). Show that \( f(x) \le g(x) \) for all \( x \ge 0 \).
Answer: Let \( h(x) = g(x) - f(x) \).
\( h'(x) = g'(x) - f'(x) \).
Given \( f'(x) \le g'(x) \), so \( h'(x) \ge 0 \) for all \( x \ge 0 \).
\( h(x) \) is also continuous & differentiable for all \( x \ge 0 \).
Using LMVT: \( h'(c) = \frac{h(x) - h(0)}{x - 0} \).
Since \( h'(x) \ge 0 \), then \( h(x) - h(0) \ge 0 \).
\( \implies h(x) \ge h(0) \).
Since \( g(0) = f(0) \), \( h(0) = 0 \).
\( \implies g(x) - f(x) \ge 0 \)
\( \implies f(x) \le g(x) \) for all \( x \ge 0 \).

Question. Let f be continuous on \([a, b]\) and differentiable on \((a, b)\). If \( f(a) = a \) and \( f(b) = b \) then show that there exist distinct \( c_1, c_2 \) in \( (a, b) \) such that \( f'(c_1) + f'(c_2) = 2 \).
Answer: \( f(x) \) is continuous on \( [a, b] \) and differentiable on \( (a, b) \).
So using LMVT:
\( f'(C) = \frac{f(b) - f(a)}{b - a} \) ......(i)
Given \( f(a) = a \) and \( f(b) = b \), so \( f'(C) = 1 \).
If we find two points \( c_1, c_2 \) in the interval where the derivative is the same as the average, then \( f'(c_1) = 1 \) and \( f'(c_2) = 1 \).
\( \implies f'(c_1) + f'(c_2) = 1 + 1 = 2 \).

Question. Let f defined on [0, 1] be a twice differentiable function such that, \( |f''(x)| \le 1 \) for all \( x \in [0, 1] \). If \( f(0) = f(1) \), then show that, \( |f'(x)| < 1 \) for all \( x \in [0, 1] \).
Answer: \( f(x) \) and \( f'(x) \) are differentiable and continuous in \( [0, 1] \).
For \( f(x) \), there will be some ‘c’ \( \in (0, 1) \) such that \( f'(c) = 0 \) (by Rolle’s theorem).
Case-1: If \( x = c \), then \( f'(x) = f'(c) = 0 \), so \( |f'(x)| = |0| = 0 < 1 \).
Case-2: If \( x > c \), by LMVT in \( [c, x] \):
\( \frac{f'(x) - f'(c)}{x - c} = f''(\alpha) \) for \( c < \alpha < x \).
\( f'(x) = (x - c) f''(\alpha) \) (since \( f'(c) = 0 \)).
\( \implies |f'(x)| = |x - c| \cdot |f''(\alpha)| \).
Since \( x \in [0, 1] \) and \( c \in (0, 1) \), \( \implies |x - c| < 1 \) and \( f''(x) < 1 \) for all \( x \).
So \( |f'(x)| < 1 \) for all \( x \in [0, 1] \).
Case-3: If \( x < c \), then \( \frac{f'(c) - f'(x)}{c - x} = f''(\alpha) \)
\( \implies |-f'(x)| = |c - x| \cdot |f''(\alpha)| \)
\( \implies |f'(x)| < 1 \).
So \( |f'(x)| < 1 \) for all \( x \in [0, 1] \).

Question. \( f(x) \) and \( g(x) \) are differentiable functions for \( 0 \le x \le 2 \) such that \( f(0) = 5, g(0) = 0, f(2) = 8, g(2) = 1 \). Show that there exists a number c satisfying \( 0 < c < 2 \) and \( f'(c) = 3 g'(c) \).
Answer: Let \( h(x) = f(x) - 3g(x) \).
\( h(x) \) is continuous on \( [0, 2] \) and differentiable on \( (0, 2) \).
\( h(0) = 5 - 3(0) = 5 \).
\( h(1) = 8 - 3(1) = 5 \).
So Rolle's theorem is applicable for \( h(x) \) in \( [0, 2] \).
There is some \( c \in (0, 2) \) such that \( h'(c) = 0 \).
\( \implies f'(c) - 3g'(c) = 0 \)
\( \implies f'(c) = 3g'(c) \).

Question. If f, \( \phi, \psi \) are continuous in \([a, b]\) and derivable in \( ]a, b[ \) then show that there is a value of c lying between a & b such that, \( \begin{vmatrix} f(a) & f(b) & f'(c) \\ \phi(a) & \phi(b) & \phi'(c) \\ \psi(a) & \psi(b) & \psi'(c) \end{vmatrix} = 0 \).
Answer: By LMVT:
\( f'(c) = \frac{f(b) - f(a)}{b - a} \), \( \phi'(c) = \frac{\phi(b) - \phi(a)}{b - a} \), \( \psi'(c) = \frac{\psi(b) - \psi(a)}{b - a} \).
By expanding the determinant:
\( \begin{vmatrix} f(a) & f(b) & f'(c) \\ \phi(a) & \phi(b) & \phi'(c) \\ \psi(a) & \psi(b) & \psi'(c) \end{vmatrix} \)
\( \implies f(a) [\phi(b) \psi'(c) - \phi'(c) \psi(b)] - f(b) [\phi(a) \psi'(c) - \psi(a) \phi'(c)] + f'(c) [\phi(a) \psi(b) - \psi(a) \phi(b)] \).
By putting the value of \( f'(c), \phi'(c) \) and \( \psi'(c) \), we will get the value equal to zero.

Question. Show that exactly two real values of x satisfy the equation \( x^2 = x \sin x + \cos x \).
Answer: Let \( f(x) = x^2 - x \sin x - \cos x \).
\( f'(x) = 2x - \sin x - x \cos x + \sin x \)
\( f'(x) = x(2 - \cos x) \)
Setting \( f'(x) = 0 \):
\( \implies x = 0 \).
The slope changes sign at \( x = 0 \).
\( f(0) = -1 \).
As \( x \to \pm \infty \), \( f(x) \to \infty \).
Two real value of \( x \) are possible where \( f(x) = 0 \).

Question. Let \( a > 0 \) and f be continuous in \([-a, a]\). Suppose that \( f'(x) \) exists and \( f'(x) \le 1 \) for all \( x \le (-a, a) \). If \( f(a) = a \) and \( f(-a) = -a \), show that \( f(0) = 0 \).
Answer: Using LMVT on \( [-a, a] \):
\( f'(x) = \frac{f(a) - f(-a)}{a - (-a)} = \frac{a - (-a)}{2a} = 1 \).
So \( f'(x) = 1 \).
Integrating, we get:
\( \implies f(x) = x + c \).
Using \( f(a) = a \), we find \( a = a + c \implies c = 0 \).
So \( f(x) = x \).
Hence, \( f(0) = 0 \).

Question. Prove the inequality \( e^x > (1 + x) \) using LMVT for all \( x \in R_0 \) and use it to determine which of the two numbers \( e^\pi \) and \( \pi^e \) is greater.
Answer: Using LMVT for the function \( e^x \) on \( [0, x] \):
\( f'(c) = \frac{e^x - e^0}{x - 0} = e^c \).
Since \( c > 0 \), \( e^c > 1 \).
\( \implies \frac{e^x - 1}{x} > 1 \)
\( \implies e^x > x + 1 \) for all \( x > 0 \).
To compare \( e^\pi \) and \( \pi^e \), consider the function \( f(x) = x^{1/x} \).
The derivative \( f'(x) = x^{1/x} \left( \frac{1 - \ln x}{x^2} \right) \).
Maximum is at \( x = e \). For \( x > e \), the function is decreasing.
Since \( \pi > e \), then \( f(e) > f(\pi) \).
\( \implies e^{1/e} > \pi^{1/\pi} \).
Raising both sides to the power \( e\pi \):
\( \implies e^\pi > \pi^e \).