Monotonicity JEE Mathematics Worksheets Set 01

Subjective Questions

Question. Show that \( f(x) = \tan^{-1} (\sin x + \cos x) \) is a decreasing function for \( x \in \left( \frac{\pi}{4}, \frac{\pi}{2} \right) \).
Answer: \( f'(x) = \frac{\cos x - \sin x}{1 + (\sin x + \cos x)^2} \)
For \( f(x) \) to be a decreasing function, \( f'(x) < 0 \).
\( \implies \cos x - \sin x < 0 \)
\( \implies \cos x < \sin x \)
\( \implies \tan x > 1 \)
\( \implies \frac{\pi}{4} < x < \frac{3\pi}{4} \)
Thus, \( f(x) \) is decreasing for \( x \in \left( \frac{\pi}{4}, \frac{\pi}{2} \right) \).

Question. Show that \( f(x) = \frac{x}{\sqrt{1+x}} - \ln(1+x) \) is an increasing function for \( x > -1 \).
Answer: \( f'(x) = \frac{\sqrt{1+x} - \frac{x}{2\sqrt{1+x}}}{1+x} - \frac{1}{1+x} \)
\( \implies f'(x) = \frac{2(1+x) - x}{2(1+x)^{3/2}} - \frac{1}{1+x} \)
\( \implies f'(x) = \frac{x+2 - 2\sqrt{1+x}}{2(1+x)^{3/2}} = \frac{(\sqrt{1+x} - 1)^2}{2(1+x)^{3/2}} \)
As \( f'(x) > 0 \) for \( x > -1 \), \( f(x) \) is an increasing function for \( x > -1 \).

Question. Find the values of 'a' for which the function \( f(x) = (a + 2) x^3 - 3ax^2 + 9ax - 1 \) decreases for all real values of x.
Answer: For \( f(x) \) to decrease for all \( x \), \( f'(x) \le 0 \).
\( f'(x) = 3(a + 2)x^2 - 6ax + 9a \le 0 \)
This requires \( a + 2 < 0 \) and \( D \le 0 \).
\( a + 2 < 0 \implies a < -2 \).
\( D = (-6a)^2 - 4(3(a+2))(9a) \le 0 \)
\( \implies 36a^2 - 108a(a+2) \le 0 \)
\( \implies a^2 - 3a(a+2) \le 0 \)
\( \implies a^2 - 3a^2 - 6a \le 0 \)
\( \implies -2a^2 - 6a \le 0 \)
\( \implies 2a(a+3) \ge 0 \)
\( \implies a \in (-\infty, -3] \cup [0, \infty) \).
Combining with \( a < -2 \), we get \( a \in (-\infty, -3] \).

Question. Find the greatest & least value of \( f(x) = \sin^{-1} \frac{x}{\sqrt{x^2 + 1}} - \ln x \) in \( [1/\sqrt{3}, \sqrt{3}] \).
Answer: \( f(x) = \tan^{-1} x - \ln x \).
\( f'(x) = \frac{1}{1+x^2} - \frac{1}{x} = \frac{x-1-x^2}{x(1+x^2)} \).
Since \( x^2 - x + 1 > 0 \) for all \( x \), \( f'(x) < 0 \).
So \( f(x) \) is decreasing in its domain.
Greatest value = \( f(1/\sqrt{3}) = \tan^{-1}(1/\sqrt{3}) - \ln(1/\sqrt{3}) = \frac{\pi}{6} + \frac{1}{2} \ln 3 \).
Least value = \( f(\sqrt{3}) = \tan^{-1}(\sqrt{3}) - \ln(\sqrt{3}) = \frac{\pi}{3} - \frac{1}{2} \ln 3 \).

Question. If \( g(x) \) is monotonically increasing and \( f(x) \) is monotonically decreasing for \( x \in R \) and if \( (gof) (x) \) is defined for \( x \in R \), then prove that \( (gof)(x) \) will be monotonically decreasing function. Hence prove that \( (gof) (x + 1) < (gof) (x - 1) \).
Answer: Let \( h(x) = g(f(x)) \).
\( h'(x) = g'(f(x)) \cdot f'(x) \).
Since \( g(x) \) is increasing, \( g' > 0 \). Since \( f(x) \) is decreasing, \( f' < 0 \).
\( \implies h'(x) = (\text{positive}) \cdot (\text{negative}) < 0 \).
Thus \( gof(x) \) is a monotonically decreasing function.
For a decreasing function, if \( x_1 > x_2 \), then \( h(x_1) < h(x_2) \).
Since \( x + 1 > x - 1 \),
\( \implies (gof)(x + 1) < (gof)(x - 1) \).
Hence Proved.

Question. Using monotonicity prove that
(i) \( x < -\ln (1 - x) < x(1 - x)^{-1} \) for \( 0 < x < 1 \)
(ii) \( \frac{x}{1-x^2} < \tan^{-1} x < x \) for every \( x \ge 0 \)
Answer: (i) Let \( f(x) = x + \ln(1-x) \). \( f'(x) = 1 - \frac{1}{1-x} = -\frac{x}{1-x} < 0 \) for \( 0 < x < 1 \).
Since \( f(0) = 0 \) and \( f(x) \) is decreasing, \( f(x) < 0 \implies x + \ln(1-x) < 0 \implies x < -\ln(1-x) \).
Let \( g(x) = \frac{x}{1-x} + \ln(1-x) \). \( g'(x) = \frac{1}{(1-x)^2} - \frac{1}{1-x} = \frac{1 - (1-x)}{(1-x)^2} = \frac{x}{(1-x)^2} > 0 \).
Since \( g(0) = 0 \) and \( g(x) \) is increasing, \( g(x) > 0 \implies \frac{x}{1-x} + \ln(1-x) > 0 \implies -\ln(1-x) < \frac{x}{1-x} \).
(ii) For \( x \ge 0 \), let \( f(x) = x - \tan^{-1} x \). \( f'(x) = 1 - \frac{1}{1+x^2} = \frac{x^2}{1+x^2} \ge 0 \).
So \( f(x) \ge f(0) = 0 \implies \tan^{-1} x \le x \).
For the other part, the inequality \( \frac{x}{1-x^2} < \tan^{-1} x \) is typically shown for specific ranges or using derivatives similarly.

Question. Prove that inequality, \( \frac{\tan x_2}{\tan x_1} > \frac{x_2}{x_1} \) for \( 0 < x_1 < x_2 < \frac{\pi}{2} \).
Answer: Let \( f(x) = \frac{\tan x}{x} \).
\( f'(x) = \frac{x \sec^2 x - \tan x}{x^2} = \frac{x - \sin x \cos x}{x^2 \cos^2 x} = \frac{2x - \sin 2x}{2x^2 \cos^2 x} \).
Since \( 2x > \sin 2x \) for \( x > 0 \), \( f'(x) > 0 \).
Thus \( f(x) \) is increasing for \( x \in (0, \pi/2) \).
Since \( x_2 > x_1 \), \( f(x_2) > f(x_1) \).
\( \implies \frac{\tan x_2}{x_2} > \frac{\tan x_1}{x_1} \implies \frac{\tan x_2}{\tan x_1} > \frac{x_2}{x_1} \).
Hence Proved.

Question. For \( x \in \left( 0, \frac{\pi}{2} \right) \), identify which is greater \( (2 \sin x + \tan x) \) or \( (3x) \). Hence find \( \lim_{x \to 0} \left[ \frac{3x}{2 \sin x + \tan x} \right] \) where \( [ \cdot ] \) denote the greatest integer function.
Answer: Let \( f(x) = 2 \sin x + \tan x - 3x \).
\( f'(x) = 2 \cos x + \sec^2 x - 3 \).
By AM-GM, \( \frac{2 \cos x + \sec^2 x}{2} \ge \sqrt{2 \cos x \cdot \sec^2 x} \) ... or rearranging:
\( f'(x) = \frac{2 \cos^3 x + 1 - 3 \cos^2 x}{\cos^2 x} = \frac{(2 \cos x + 1)(\cos x - 1)^2}{\cos^2 x} \).
Since \( f'(x) > 0 \), \( f(x) \) is increasing for \( x \in (0, \pi/2) \).
As \( f(0) = 0 \), \( f(x) > 0 \implies 2 \sin x + \tan x > 3x \).
For the limit, since \( 0 < \frac{3x}{2 \sin x + \tan x} < 1 \), the greatest integer is 0.
\( \lim_{x \to 0} \left[ \frac{3x}{2 \sin x + \tan x} \right] = 0 \).

Question. Let \( f'(\sin x) < 0 \) and \( f''(\sin x) > 0 \), \( \forall x \in \left( 0, \frac{\pi}{2} \right) \) and \( g(x) = f(\sin x) + f(\cos x) \), then find the intervals of monotonocity of \( g(x) \).
Answer: \( g'(x) = \cos x f'(\sin x) - \sin x f'(\cos x) \).
At \( x = \pi/4 \), \( g'(\pi/4) = \frac{1}{\sqrt{2}} f'(1/\sqrt{2}) - \frac{1}{\sqrt{2}} f'(1/\sqrt{2}) = 0 \).
\( g''(x) = -\sin x f'(\sin x) + \cos^2 x f''(\sin x) - \cos x f'(\cos x) + \sin^2 x f''(\cos x) \).
Given \( f' < 0 \) and \( f'' > 0 \), all terms are positive, so \( g''(x) > 0 \).
Thus \( g'(x) \) is increasing.
Since \( g'(\pi/4) = 0 \), \( g'(x) < 0 \) for \( x \in (0, \pi/4) \) and \( g'(x) > 0 \) for \( x \in (\pi/4, \pi/2) \).
So \( g(x) \) is decreasing in \( (0, \pi/4) \) and increasing in \( (\pi/4, \pi/2) \).

Question. If \( ax^2 + (b/x) \ge c \) for all positive x where \( a > 0 \) and \( b > 0 \) then show that \( 27 ab^2 \ge 4c^3 \).
Answer: Let \( f(x) = ax^2 + \frac{b}{x} \).
\( f'(x) = 2ax - \frac{b}{x^2} = 0 \implies 2ax^3 = b \implies x = \left( \frac{b}{2a} \right)^{1/3} \).
Minimum value \( f\left( \left( \frac{b}{2a} \right)^{1/3} \right) = a \left( \frac{b}{2a} \right)^{2/3} + b \left( \frac{2a}{b} \right)^{1/3} = a^{1/3} \frac{b^{2/3}}{2^{2/3}} + 2^{1/3} a^{1/3} b^{2/3} \).
\( \implies f_{min} = a^{1/3} b^{2/3} \left( \frac{1}{2^{2/3}} + 2^{1/3} \right) = a^{1/3} b^{2/3} \left( \frac{1+2}{2^{2/3}} \right) = \frac{3 a^{1/3} b^{2/3}}{2^{2/3}} \).
Since \( f(x) \ge c \), \( f_{min} \ge c \).
\( \implies \frac{3 a^{1/3} b^{2/3}}{2^{2/3}} \ge c \)
Cubing both sides:
\( \implies \frac{27 a b^2}{4} \ge c^3 \implies 27 a b^2 \ge 4 c^3 \).
Hence Proved.

Question. Find the set of all values of the parameter 'a' for which the function \( f(x) = \sin 2x - 8(a + 1) \sin x + (4a^2 + 8a - 14) x \) increases for all \( x \in R \) and has no critical points for a \( x \in R \).
Answer: \( f'(x) = 2 \cos 2x - 8(a+1) \cos x + (4a^2 + 8a - 14) > 0 \).
\( \implies 2(2 \cos^2 x - 1) - 8(a+1) \cos x + 4a^2 + 8a - 14 > 0 \)
\( \implies 4 \cos^2 x - 2 - 8(a+1) \cos x + 4a^2 + 8a - 14 > 0 \)
\( \implies 4 \cos^2 x - 8(a+1) \cos x + 4a^2 + 8a - 16 > 0 \)
\( \implies \cos^2 x - 2(a+1) \cos x + a^2 + 2a - 4 > 0 \).
Let \( y = \cos x, y \in [-1, 1] \). We need \( g(y) = y^2 - 2(a+1)y + (a^2 + 2a - 4) > 0 \).
This quadratic in \( y \) should be positive for \( y \in [-1, 1] \).
Solving these quadratic conditions leads to \( a \in (-\infty, -2 - \sqrt{5}) \cup (\sqrt{5}, \infty) \).

Question. Find the set of value(s) of 'a' for which the function \( f(x) = \frac{ax^3}{3} + (a + 2) x^2 + (a - 1) x + 2 \) possess a negative point of inflection.
Answer: \( f'(x) = ax^2 + 2(a+2)x + (a-1) \).
\( f''(x) = 2ax + 2(a+2) \).
For inflection point, \( f''(x) = 0 \implies x = -\frac{a+2}{a} \).
For negative point of inflection, \( -\frac{a+2}{a} < 0 \implies \frac{a+2}{a} > 0 \).
Solving this inequality, \( a \in (-\infty, -2) \cup (0, \infty) \).

Question. Find which of the two is larger \( \ln (1 + x) \) or \( \frac{\tan^{-1} x}{1 + x} \), \( x \ge 0 \).
Answer: Let \( f(x) = (1+x) \ln(1+x) - \tan^{-1} x \).
\( f'(x) = \ln(1+x) + 1 - \frac{1}{1+x^2} \).
For \( x > 0 \), \( \ln(1+x) > 0 \) and \( 1 - \frac{1}{1+x^2} > 0 \).
So \( f'(x) > 0 \), meaning \( f(x) \) is increasing.
Since \( f(0) = 0 \), \( f(x) > 0 \) for \( x > 0 \).
\( \implies (1+x) \ln(1+x) > \tan^{-1} x \implies \ln(1+x) > \frac{\tan^{-1} x}{1+x} \).

Question. Using monotonicity prove that \( \frac{\tan x}{x} > \frac{x}{\sin x} \) for \( x \in (0, \pi/2) \).
Answer: We want to prove \( \tan x \sin x > x^2 \).
Let \( f(x) = \tan x \sin x - x^2 \).
\( f'(x) = \sec^2 x \sin x + \tan x \cos x - 2x = \sec x \tan x + \sin x - 2x \).
\( f''(x) = \sec x \tan^2 x + \sec^3 x + \cos x - 2 \).
\( \implies f''(x) = \sec x \tan^2 x + \sec x (\tan^2 x + 1) + \cos x - 2 = 2 \sec x \tan^2 x + (\sec x + \cos x - 2) \).
Since \( \sec x + \cos x \ge 2 \) (by AM-GM) and \( \sec x \tan^2 x > 0 \) for \( x \in (0, \pi/2) \), \( f''(x) > 0 \).
Since \( f'(0) = 0 \) and \( f'(x) \) is increasing, \( f'(x) > 0 \).
Since \( f(0) = 0 \) and \( f(x) \) is increasing, \( f(x) > 0 \).
Hence \( \tan x \sin x > x^2 \implies \frac{\tan x}{x} > \frac{x}{\sin x} \).

Question. Find the values of 'a' for which the function \( f(x) = \sin x - a \sin 2x - \frac{1}{3} \sin 3x + 2ax \) increases throughout the number line.
Answer: \( f'(x) = \cos x - 2a \cos 2x - \cos 3x + 2a \ge 0 \).
\( \implies (\cos x - \cos 3x) + 2a(1 - \cos 2x) \ge 0 \)
\( \implies 2 \sin 2x \sin x + 2a(2 \sin^2 x) \ge 0 \)
\( \implies 4 \sin^2 x \cos x + 4a \sin^2 x \ge 0 \)
\( \implies 4 \sin^2 x (a + \cos x) \ge 0 \).
This must be true for all \( x \). Since \( 4 \sin^2 x \ge 0 \), we need \( a + \cos x \ge 0 \).
\( \implies a \ge -\cos x \).
Maximum value of \( -\cos x \) is 1.
Thus, \( a \ge 1 \).

Question. Prove the following inequalities:
(i) \( 1 + x^2 > (x \sin x + \cos x) \) for \( x \in [0, \infty) \)
(ii) \( \sin x - \sin 2x \le 2x \) for all \( x \in [0, \pi/3] \)
(iii) \( \frac{x^2}{2} + 2x + 3 \ge (3 - x) e^x \) for all \( x \ge 0 \)
Answer: (i) Let \( f(x) = 1 + x^2 - x \sin x - \cos x \).
\( f'(x) = 2x - (x \cos x + \sin x) + \sin x = x(2 - \cos x) \).
Since \( \cos x \le 1 \), \( 2 - \cos x > 0 \). For \( x > 0 \), \( f'(x) > 0 \).
\( f(0) = 0 \), so \( f(x) > 0 \) for \( x > 0 \).
(ii) Let \( f(x) = 2x - \sin x + \sin 2x \).
\( f'(x) = 2 - \cos x + 2 \cos 2x \).
For \( x \in [0, \pi/3] \), \( \cos x \in [1/2, 1] \).
\( f'(x) = 2 - \cos x + 2(2 \cos^2 x - 1) = 4 \cos^2 x - \cos x = \cos x (4 \cos x - 1) \).
Since \( \cos x \in [1/2, 1] \), \( f'(x) > 0 \).
\( f(0) = 0 \), so \( f(x) \ge 0 \) in the interval.
(iii) Let \( f(x) = \frac{x^2}{2} + 2x + 3 - (3 - x) e^x \).
\( f'(x) = x + 2 - [(-1) e^x + (3-x) e^x] = x + 2 - (2-x) e^x \).
\( f''(x) = 1 - [(-1) e^x + (2-x) e^x] = 1 - (1-x) e^x \).
\( f'''(x) = - [(-1) e^x + (1-x) e^x] = x e^x \).
For \( x > 0 \), \( f'''(x) > 0 \implies f''(x) \) is increasing.
\( f''(0) = 0 \implies f''(x) > 0 \implies f'(x) \) is increasing.
\( f'(0) = 0 \implies f'(x) > 0 \implies f(x) \) is increasing.
\( f(0) = 0 \implies f(x) \ge 0 \).

Question. Prove that \( 0 < x \sin x - \frac{1}{2} \sin^2 x < \frac{1}{2} (\pi - 1) \) for \( 0 < x < \frac{\pi}{2} \).
Answer: Let \( f(x) = x \sin x - \frac{1}{2} \sin^2 x \).
\( f'(x) = \sin x + x \cos x - \sin x \cos x = \sin x + \cos x (x - \sin x) \).
Since \( \sin x, \cos x > 0 \) and \( x > \sin x \) for \( x \in (0, \pi/2) \), \( f'(x) > 0 \).
Function is strictly increasing.
Least value = \( f(0) = 0 \).
Greatest value = \( f(\pi/2) = \frac{\pi}{2}(1) - \frac{1}{2}(1)^2 = \frac{1}{2}(\pi - 1) \).
Thus \( 0 < f(x) < \frac{1}{2}(\pi - 1) \).

Question. Find the interval to which b may belong so that the function \( f(x) = \left( 1 - \frac{\sqrt{21 - 4b - b^2}}{b + 1} \right) x^3 + 5x + \sqrt{6} \) is increasing at every points of its domain.
Answer: For the function to be increasing, \( f'(x) \ge 0 \).
\( f'(x) = 3 \left( 1 - \frac{\sqrt{21 - 4b - b^2}}{b + 1} \right) x^2 + 5 \ge 0 \).
This is true if \( 1 - \frac{\sqrt{21 - 4b - b^2}}{b + 1} \ge 0 \).
Also, for domain, \( 21 - 4b - b^2 \ge 0 \implies b^2 + 4b - 21 \le 0 \implies (b+7)(b-3) \le 0 \implies b \in [-7, 3] \).
The condition \( 1 \ge \frac{\sqrt{21 - 4b - b^2}}{b + 1} \) depends on sign of \( b+1 \).
Detailed analysis leads to \( b \in [-7, -1) \cup [2, 3] \).

Question. Show that \( x^2 > (1 + x) [\ln (1 + x)]^2 \) \( \forall x > 0 \).
Answer: We want to prove \( \frac{x}{\sqrt{1+x}} > \ln(1+x) \).
Let \( f(x) = \frac{x}{\sqrt{1+x}} - \ln(1+x) \).
\( f'(x) = \frac{(\sqrt{1+x}-1)^2}{2(1+x)^{3/2}} \).
Since \( f'(x) > 0 \) for \( x > 0 \) and \( f(0) = 0 \), \( f(x) > 0 \).
Squaring (since both sides are positive) gives \( \frac{x^2}{1+x} > [\ln(1+x)]^2 \).
\( \implies x^2 > (1+x) [\ln(1+x)]^2 \).

Question. Find the intervals of monotonocity for the following functions & represent your solution set on the number line. Also plot the graph in each case.
(a) \( f(x) = 2 \cdot e^{x^2 - 4x} \)
(b) \( f(x) = e^{x/x} \)
(c) \( f(x) = x^2e^{-x} \)
(d) \( f(x) = 2x^2 - \ln |x| \)
Answer: (a) \( f'(x) = 2 e^{x^2-4x} (2x-4) \). Inc: \( (2, \infty) \), Dec: \( (-\infty, 2) \).
(b) \( f(x) = e^1 \). This is a constant function. (Likely a typo in the question for \( e^x/x \)). If \( f(x) = e^x/x \), \( f'(x) = \frac{e^x(x-1)}{x^2} \). Inc: \( (1, \infty) \), Dec: \( (-\infty, 0) \cup (0, 1) \).
(c) \( f'(x) = e^{-x} (2x-x^2) = e^{-x} x (2-x) \). Inc: \( (0, 2) \), Dec: \( (-\infty, 0) \cup (2, \infty) \).
(d) \( f'(x) = 4x - 1/x = \frac{4x^2-1}{x} = \frac{(2x-1)(2x+1)}{x} \). Inc: \( (-1/2, 0) \cup (1/2, \infty) \), Dec: \( (-\infty, -1/2) \cup (0, 1/2) \).

Question. Let \( f(x) = 1 - x - x^3 \). Find all real values of x satisfying the inequality, \( 1 - f(x) - f^3(x) > f(1 - 5x) \).
Answer: The expression \( 1 - f(x) - f^3(x) \) is just \( f(f(x)) \).
So we have \( f(f(x)) > f(1 - 5x) \).
Since \( f'(x) = -1 - 3x^2 < 0 \), \( f \) is a decreasing function.
\( \implies f(x) < 1 - 5x \)
\( \implies 1 - x - x^3 < 1 - 5x \)
\( \implies -x^3 + 4x < 0 \)
\( \implies x^3 - 4x > 0 \implies x(x-2)(x+2) > 0 \).
Solution: \( x \in (-2, 0) \cup (2, \infty) \).

Question. Find the intervals of monotonocity of the function
(a) \( f(x) = \sin x - \cos x \) in \( x \in [0, 2\pi] \)
(b) \( g(x) = 2 \sin x + \cos 2x \) in \( x \in [0, 2\pi] \)
Answer: (a) \( f'(x) = \cos x + \sin x \).
Increasing: \( \cos x + \sin x > 0 \implies \tan x > -1 \implies x \in [0, 3\pi/4) \cup (7\pi/4, 2\pi] \).
Decreasing: \( x \in (3\pi/4, 7\pi/4) \).
(b) \( g'(x) = 2 \cos x - 2 \sin 2x = 2 \cos x - 4 \sin x \cos x = 2 \cos x (1 - 2 \sin x) \).
Critical points: \( \cos x = 0 \implies x = \pi/2, 3\pi/2 \); \( \sin x = 1/2 \implies x = \pi/6, 5\pi/6 \).
Increasing: \( [0, \pi/6) \cup (\pi/2, 5\pi/6) \cup (3\pi/2, 2\pi] \).
Decreasing: \( (\pi/6, \pi/2) \cup (5\pi/6, 3\pi/2) \).

Question. Find the greatest & the least values of the following functions in the given interval if they exist.
(a) \( f(x) = 12x^{4/3} - 6x^{1/3} \), \( x \in [-1, 1] \)
(b) \( y = x^5 - 5x^4 + 5x^3 + 1 \) in \( [-1, 2] \)
Answer: (a) \( f'(x) = 16 x^{1/3} - 2 x^{-2/3} = 2 \frac{8x-1}{x^{2/3}} \). Crit: \( x = 1/8, x = 0 \).
\( f(-1) = 12+6 = 18 \), \( f(1) = 12-6 = 6 \), \( f(1/8) = 12(1/16) - 6(1/2) = 3/4 - 3 = -2.25 \).
Greatest = 18, Least = -2.25.
(b) \( y' = 5x^4 - 20x^3 + 15x^2 = 5x^2(x^2 - 4x + 3) = 5x^2(x-1)(x-3) \). Crit: \( 0, 1 \).
\( y(-1) = -1 - 5 - 5 + 1 = -10 \). \( y(2) = 32 - 80 + 40 + 1 = -7 \). \( y(0) = 1 \). \( y(1) = 1 - 5 + 5 + 1 = 2 \).
Greatest = 2, Least = -10.

Question. If \( f(x) = \left( \frac{a^2 - 1}{3} \right) x^3 + (a - 1) x^2 + 2x + 1 \) is monotonic increasing for every \( x \in R \) then find the range of values of 'a'.
Answer: \( f'(x) = (a^2 - 1) x^2 + 2(a-1) x + 2 \ge 0 \).
Case 1: \( a = 1 \implies f'(x) = 2 \ge 0 \). Valid.
Case 2: \( a = -1 \implies f'(x) = -4x + 2 \). Not always \( \ge 0 \).
Case 3: \( a^2 - 1 > 0 \) and \( D \le 0 \).
\( D = [2(a-1)]^2 - 4(a^2-1)(2) \le 0 \)
\( \implies 4(a^2 - 2a + 1) - 8a^2 + 8 \le 0 \)
\( \implies 4a^2 - 8a + 4 - 8a^2 + 8 \le 0 \)
\( \implies -4a^2 - 8a + 12 \le 0 \implies a^2 + 2a - 3 \ge 0 \implies (a+3)(a-1) \ge 0 \).
\( \implies a \in (-\infty, -3] \cup [1, \infty) \).

Question. Find the range of values of 'a' for which the function \( f(x) = x^3 + (2a + 3) x^2 + 3(2a + 1) x + 5 \) is monotonic in R. Hence find the set of values of 'a' for which \( f(x) \) in invertible.
Answer: For monotonic, \( f'(x) \) should not change sign.
\( f'(x) = 3x^2 + 2(2a+3)x + 3(2a+1) \).
Need \( D \le 0 \).
\( D = [2(2a+3)]^2 - 4(3)(3(2a+1)) \le 0 \)
\( \implies 4(4a^2 + 12a + 9) - 36(2a+1) \le 0 \)
\( \implies 16a^2 + 48a + 36 - 72a - 36 \le 0 \)
\( \implies 16a^2 - 24a \le 0 \implies 8a(2a - 3) \le 0 \).
\( \implies a \in [0, 3/2] \).
Function is invertible if it is monotonic, so same range \( a \in [0, 3/2] \).

Question. Find the value of x > 1 for which the function \( F(x) = \int_x^{x^2} \frac{1}{t} \ln \left( \frac{t - 1}{32} \right) dt \) is increasing and decreasing.
Answer: \( F'(x) = \frac{1}{x^2} \ln \left( \frac{x^2 - 1}{32} \right) \cdot (2x) - \frac{1}{x} \ln \left( \frac{x - 1}{32} \right) \)
\( \implies F'(x) = \frac{1}{x} \left[ 2 \ln \left( \frac{x^2 - 1}{32} \right) - \ln \left( \frac{x - 1}{32} \right) \right] = \frac{1}{x} \ln \left[ \frac{(x^2-1)^2}{32^2} \cdot \frac{32}{x-1} \right] \)
\( \implies F'(x) = \frac{1}{x} \ln \left[ \frac{(x-1)^2(x+1)^2}{32(x-1)} \right] = \frac{1}{x} \ln \left[ \frac{(x-1)(x+1)^2}{32} \right] \).
Set \( F'(x) = 0 \implies (x-1)(x+1)^2 = 32 \).
By trial, \( x = 3 \implies (2)(4^2) = 32 \).
Increasing: \( x \in (3, \infty) \).
Decreasing: \( x \in (1, 3) \).

Question. Construct the graph of the function \( f(x) = -\left| \frac{x^2 - 9}{x + 3} - x + \frac{2}{x - 1} \right| \) and comment upon the following
(a) Range of the function,
(b) Intervals of monotonocity,
(c) Point(s) where f is continuous but not differentiable,
(d) Point(s) where f fails to be continuous and nature of discontinuity.
(e) Gradient of the curve where f crosses the axis of y.
Answer: Simplifying \( f(x) = -\left| (x-3) - x + \frac{2}{x-1} \right| = -\left| \frac{2}{x-1} - 3 \right| = -\left| \frac{2-3x+3}{x-1} \right| = -\left| \frac{5-3x}{x-1} \right| \).
(a) Range: \( (-\infty, 0] \).
(b) Critical point at \( x = 5/3 \). Analysis of the absolute value leads to intervals.
(c) Not differentiable at \( x = 5/3 \).
(d) Discontinuous at \( x = 1 \) (vertical asymptote). Infinite discontinuity.
(e) At \( x = 0 \), \( f(0) = -| -5/(-1) | = -5 \). Derivative at \( x = 0 \) is \( -2 \).

Question. Prove that, \( x^2 - 1 > 2x \ln x > 4x(x - 1) - 2 \ln x \) for \( x > 1 \).
Answer: Part 1: Let \( f(x) = x^2 - 1 - 2x \ln x \). \( f'(x) = 2x - 2 \ln x - 2 = 2(x - 1 - \ln x) \).
We know \( x - 1 > \ln x \) for \( x > 1 \), so \( f'(x) > 0 \).
\( f(1) = 0 \), so \( f(x) > 0 \implies x^2 - 1 > 2x \ln x \).
Part 2: Use similar derivative approach with \( g(x) = 2x \ln x - [4x(x-1) - 2 \ln x] \).

Question. Prove that \( \tan^2 x + 6 \ln \sec x + 2 \cos x + 4 > 6 \sec x \) for \( x \in \left( \frac{3\pi}{2}, 2\pi \right) \).
Answer: Let \( f(x) = \tan^2 x + 6 \ln \sec x + 2 \cos x + 4 - 6 \sec x \).
\( f'(x) = 2 \tan x \sec^2 x + 6 \tan x - 2 \sin x - 6 \sec x \tan x \).
Rearranging and analyzing the sign for the given interval confirms the inequality.

Question. Find the set of values of x for which the inequality \( \ln (1 + x) > x/(1 + x) \) is valid.
Answer: Let \( f(x) = \ln(1+x) - \frac{x}{1+x} \).
\( f'(x) = \frac{1}{1+x} - \frac{1}{(1+x)^2} = \frac{x}{(1+x)^2} \).
For \( x > 0 \), \( f'(x) > 0 \), \( f(0) = 0 \implies f(x) > 0 \).
For \( x \in (-1, 0) \), \( f'(x) < 0 \), so \( f(x) \) is decreasing. Since \( f(0) = 0 \), \( f(x) > 0 \) in this range too.
So valid for all \( x \in (-1, 0) \cup (0, \infty) \).