Question 14 Marks
Find the intervals in which the following functions are increasing or decreasing.
f(x) = x3 - 6x2 + 9x + 15
Answerf(x) = x3 - 6x2 + 9x + 15
f'(x) = 3x2 - 12x + 9
= 3(x2 - 4x + 3)
= 3(x - 1)(x - 3)
For f(x) to be increasing, we must have
f'(x) > 0
⇒ 3(x - 1)(x - 3) > 0
⇒ (x - 1)(x - 3) > 0
[Since, 3 > 0, 3(x - 1)(x - 3) > 0 ⇒ (x - 1)(x - 3) > 0]
⇒ x < 1 or x > 3
$\Rightarrow\text{x}\in(-\infty,1)\cup(3,\infty)$
So, f(x) is increasing on $\text{x}\in(-\infty,1)\cup(3,\infty).$
For f(x) to be decreasing, we must have,
f'(x) < 0
⇒ 3(x - 1)(x - 3) < 0
⇒ (x - 1)(x - 3) < 0
[Since, 3 > 0, 3(x - 1)(x - 3) < 0 ⇒ (x - 1)(x - 3) < 0]
⇒ 1 < x > 3
$\Rightarrow\text{x}\in(1,3)$
So, f(x) is decreasing on $\text{x}\in(1,3).$
View full question & answer→Question 24 Marks
Find the intervals in which the following functions are increasing or decreasing.
$\text{f}(\text{x})=5\text{x}^{\frac{3}{2}}-3\text{x}^{\frac{5}{2}},\text{x}>0$
Answer$\text{f}(\text{x})=5\text{x}^{\frac{3}{2}}-3\text{x}^{\frac{5}{2}},\text{x}>0$
$\text{f}'(\text{x})=\frac{15}{2}\text{x}^{\frac{1}{2}}-\frac{15}{2}\text{x}^{\frac{3}{2}}$
$=\frac{15}{2}\text{x}^{\frac{1}{2}}(1-\text{x})$
Here, 0, 1 are the roots.
The possible intervals are $(-\infty,0),(0,1),(1,2)$ and $(1,\infty)\ ....(1)$
For f(x) to be increasing, we must have
$\text{f}'(\text{x})>0$
$\Rightarrow\frac{15}{2}\text{x}^{\frac{1}{2}}(1-\text{x})<0$
$\Rightarrow\text{x}\in(0,1)$
So, f(x) is increasing on (0, 1).
For f(x) to be decreasing, we must have,
$\text{f}'(\text{x})<0$
$\Rightarrow\frac{15}{2}\text{x}^{\frac{1}{2}}(1-\text{x})<0$
$\Rightarrow\text{x}\in(1,\infty)$
So, f(x) is decreasing on $\text{x}\in(1,\infty).$
View full question & answer→Question 34 Marks
Find the intervals in which the following functions are increasing or decreasing.
f(x) = 6 - 9x - x2
AnswerWe have,
f(x) = 6 - 9x - x2
f'(x) = -2x - 9
For f(x) to be increasing, we must have
f'(x) > 0
⇒ -2x - 9 > 0
⇒ -2x > -9
$\Rightarrow\text{x}<\frac{-9}{2}$
$\Rightarrow\text{x}\in\Big(-\infty,\frac{-9}{2}\Big)$
So, f(x) is increasing on $\Big(-\infty,\frac{-9}{2}\Big).$
For f(x) to be decreasing, we must have
f'(x) < 0
⇒ -2x - 9 < 0
⇒ -2x < -9
$\Rightarrow\text{x}>\frac{-9}{2}$
$\Rightarrow\text{x}\in\Big(\frac{-9}{2},\infty\Big)$
So, f(x) is decreasing on $\Big(\frac{-9}{2},\infty\Big).$
View full question & answer→Question 44 Marks
Find the intervals in which the following functions are increasing or decreasing.
f(x) = x4 - 4x
Answerf(x) = x4 - 4x f'(x) = 4x3 - 4
= 4(x3 - 1)
For f(x) to be increasing, we must have
f'(x) > 0
⇒ 4(x3 - 1) > 0
⇒ x3 - 1 > 0
⇒ x3 > 1
⇒ x > 1
$\Rightarrow\text{x}\in(1,\infty)$
So, f(x) is increasing on $(1,\infty).$
For f(x) to be decreasing, we must have
f'(x) < 0
⇒ 4(x3 - 1) < 0
⇒ x3 - 1 < 0
⇒ x3 < 1
⇒ x < 1
$\Rightarrow\text{x}\in(-\infty,1)$
So, f(x) is decreasing on $\text{x}\in(-\infty,1).$
View full question & answer→Question 54 Marks
Find the intervals in which the following functions are increasing or decreasing.
$\text{f}(\text{x})=\log(2+\text{x})-\frac{2\text{x}}{2+\text{x}},\text{x}\in\text{R}$
Answer$\text{f}(\text{x})=\log(2+\text{x})-\frac{2\text{x}}{2+\text{x}},\text{x}\in\text{R}$ $\text{f}'(\text{x})=\frac{1}{(2+\text{x})}-\frac{[(2+\text{x})2-2\text{x}]}{(2+\text{x})^2}$ $=\frac{(2+\text{x})-[4+2\text{x}-2\text{x}]}{(2+\text{x})^2}$ $=\frac{2+\text{x}-4}{(2+\text{x})^2}$ $=\frac{(\text{x}-2)}{(2+\text{x})^2},\text{x}\neq-2$ Here, x = 2 is the critical point. The possible intervals are $(-\infty,2)$ and $(2,\infty)\ ....(1)$ For f(x) to be increasing, we must have, $\text{f}'(\text{x})>0$ $\Rightarrow\frac{(\text{x}-2)}{(2+\text{x})^2}>0$ $\Rightarrow\text{x}-2>0,\text{x}\neq-2$ $\Rightarrow\text{x}>2$ $\Rightarrow\text{x}\in(2,\infty)$ [From eq. (1)] So, f(x) is increasing on $\text{x}\in(2,\infty).$ For f(x) to be decreasing, we must have, $\text{f}'(\text{x})<0$ $\Rightarrow\frac{(\text{x}-2)}{(2+\text{x})^2}<0$ $\Rightarrow\text{x}-2<0,\text{x}\neq-2$ $\Rightarrow\text{x}<2$ $\Rightarrow\text{x}\in(-\infty,2)$ [From eq. (1)] So, f(x) is decreasing on $\text{x}\in(-\infty,2).$
View full question & answer→Question 64 Marks
Determine the values of x for which the function f(x) = x2 - 6x + 9 is increasing or decreasing. Also, find the coordinates of the point on the curve y = x2 - 6x + 9 where the normal is parallel to the line y = x + 5.
AnswerWe have, f(x) = x2 - 6x + 9 $\therefore$ f'(x) = 2x - 6 Critical points f'(x) = 0 ⇒ 2(x - 3) = 0 ⇒ x = 3 Clearly, f'(x) > 0 if x > 3 f'(x) < 0 if x < 3 Thus, f(x) is increases in $(3,\infty),$ decreases in $(-\infty,3)$ II part:
The given equation of curves y = x2 - 6x + 9 ....(i) y = x + 5 ....(ii) Slope of (i) $\text{m}_1=\frac{\text{dy}}{\text{dx}}=2\text{x}-6$ Slope of (ii) $\text{m}_2=1$ Given that slope of normal to (i) is parallel to (ii) $\therefore\ \frac{-1}{2\text{x}-6}=1$ $\Rightarrow2\text{x}-6=-1$ $\Rightarrow\text{x}=\frac{5}{2}$ From (i) $\text{y}=\frac{25}{4}-15+9$ $=\frac{25}{4}-6$ $=\frac{1}{4}$ Thus, the required point is $\Big(\frac{5}{2},\frac{1}{4}\Big).$ View full question & answer→Question 74 Marks
Find the intervals in which the following functions are increasing or decreasing.
f(x) = x3 - 12x2 + 36x + 17
Answerf(x) = x3 - 12x2 + 36x + 17
f'(x) = 3x2 - 24x + 36
= 3(x2 - 8x + 12)
= 3(x - 2)(x - 6)
For f(x) to be increasing, we must have
f'(x) > 0
⇒ 3(x - 2)(x - 6) > 0
⇒ (x - 2)(x - 6) > 0
[Since, 3 > 0, 3(x - 2)(x - 6) > 0 ⇒ (x - 2)(x - 6) > 0]
⇒ x < 2 or x > 6
$\Rightarrow\text{x}\in(-\infty,2)\cup(6,\infty)$
So, f(x) is increasing on $\text{x}\in(-\infty,2)\cup(6,\infty).$
For f(x) to be decreasing, we must have,
f'(x) < 0
⇒ 3(x - 2)(x - 6) < 0
⇒ (x - 2)(x - 6) < 0
[Since, 3 > 0, 3(x - 2)(x - 6) < 0 ⇒ (x - 2)(x - 6) < 0]
⇒ 2 < x < 2
$\Rightarrow\text{x}\in(2,6)$
So, f(x) is decreasing on $\text{x}\in(2,6).$
View full question & answer→Question 84 Marks
Show that $\text{f}(\text{x})=\sin\text{x}-\cos\text{x}$ is an increasing function on $\Big(-\frac{\pi}{4},\frac{\pi}{4}\Big).$
Answer$\text{f}(\text{x})=\sin\text{x}-\cos\text{x}$
$\therefore\ \text{f}'(\text{x})=\cos\text{x}+\sin\text{x}$
$=\sqrt2\Big(\frac{1}{\sqrt2}\cos\text{x}+\frac{1}{\sqrt2}\sin\text{x}\Big)$
$=\sqrt2\Big(\frac{\sin\pi}{4}\cos\text{x}+\frac{\cos\pi}{4}\sin\text{x}\Big)$
$=\sqrt2\sin\Big(\frac{\pi}{4}+\text{x}\Big)$
Now,
$\text{x}\in\Big(-\frac{\pi}{4},\frac{\pi}{4}\Big)$
$\Rightarrow-\frac{\pi}{4}<\text{x}<\frac{\pi}{4}$
$\Rightarrow0<\frac{\pi}{4}<\text{x}<\frac{\pi}{2}$
$\Rightarrow\sin0^{\circ}<\sin\Big(\frac{\pi}{4}+\text{x}\Big)<\sin\frac{\pi}{4}$
$\Rightarrow0<\sin\Big(\frac{\pi}{4}+\text{x}\Big)<1$
$\Rightarrow\sqrt2\sin\Big(\frac{\pi}{4}+\text{x}\Big)>0$
$\Rightarrow\text{f}'(\text{x})>0$
Hence, f(x) is increasing function on $\Big(-\frac{\pi}{4},\frac{\pi}{4}\Big).$
View full question & answer→Question 94 Marks
Find the intervals in which the following functions are increasing or decreasing.
f(x) = 2x3 - 15x2 + 36x + 1
Answerf(x) = 2x3 - 15x2 + 36x + 1
f'(x) = 6x2 - 30x + 36
= 6(x2 - 5x + 6)
= 6(x - 2)(x - 3)
For f(x) to be increasing, we must have
f'(x) > 0
⇒ 6(x - 2)(x - 3) > 0
⇒ (x - 2)(x - 3) > 0
[Since, 6 > 0, 6 (x - 2)(x - 3) > 0 ⇒ (x - 2)(x - 3) > 0]
⇒ x < 2 or x > 3
$\Rightarrow\text{x}\in(-\infty,2)\cup(3,\infty)$
So, f(x) is increasing on $\text{x}\in(-\infty,2)\cup(3,\infty).$
For f(x) to be decreasing, we must have,
f'(x) < 0
⇒ 6(x - 2)(x - 3) < 0
⇒ (x - 2)(x - 3) < 0
[Since, 6 > 0, 6 (x - 2)(x - 3) < 0 ⇒ (x - 2)(x - 3) < 0]
⇒ 2 < x < 3
$\Rightarrow\text{x}\in(2,3)$
So, f(x) is decreasing on $\text{x}\in(2,3).$
View full question & answer→Question 104 Marks
Show that $\text{f}(\text{x})=\cos\text{x}$ is a decreasing function on $(0,\pi),$ increasing in $(-\pi,0)$ and neither increasing nor decreasing in $(-\pi,\pi).$
Answer$\text{f}(\text{x})=\cos\text{x}$
Domain of $\cos\text{x}$ is $(-\pi,\pi).$
$\Rightarrow\text{f}'(\text{x})=-\sin\text{x}$
For $\text{x}\in(-\pi,0),\sin\text{x}<0$
$[\because$ Sine function is negative in third and fourth quadrant$]$
$\Rightarrow-\sin\text{x}>0$
$\Rightarrow\text{f}'(\text{x})>0$
So, $\cos\text{x}$ is increasing in $(-\pi,0).$
For $\text{x}\in(0,\pi),\sin\text{x}>0$
$[\because$ Sine function is positive in first and second quadrant$]$
$\Rightarrow-\sin\text{x}<0$
$\Rightarrow\text{f}'(\text{x})<0$
So, f(x) is decreasing on $(0,\pi).$
Thus, f(x) is neither increasing nor decreasing in $(-\pi,\pi).$
View full question & answer→Question 114 Marks
Show that the function $\text{f}(\text{x})=\cot^{-1}(\sin\text{x}+\cos\text{x})$ is decreasing on $\Big(0,\frac{\pi}{4}\Big)$ and increasing on $\Big(\frac{\pi}{4},\frac{\pi}{2}\Big).$
Answer$\text{f}(\text{x})=\cot^{-1}(\sin\text{x}+\cos\text{x})$
$\text{f}'(\text{x})=\frac{1}{1+(\sin\text{x}+\cos\text{x})^2}\times(\cos\text{x}-\sin\text{x})$
$\text{f}'(\text{x})=-\frac{(\cos\text{x}-\sin\text{x})}{2+2\sin\text{x}\cos\text{x}}$
For f(x) to be increasing, we must have f'(x) > 0
$\Rightarrow-\frac{(\cos\text{x}-\sin\text{x})}{2+2\sin\text{x}\cos\text{x}}>0$
$\Rightarrow-(\cos\text{x}-\sin\text{x})>0$
$\Rightarrow\sin\text{x}>\cos\text{x}$
$\Rightarrow\text{x}\in\Big(\frac{\pi}{4},\frac{\pi}{2}\Big)$
For f(x) to be decreasing, we must have f'(x) > 0
$\Rightarrow-\frac{(\cos\text{x}-\sin\text{x})}{2+2\sin\text{x}\cos\text{x}}<0$
$\Rightarrow-(\cos\text{x}-\sin\text{x})<0$
$\Rightarrow\sin\text{x}<\cos\text{x}$
$\Rightarrow\text{x}\in\Big(0,\frac{\pi}{4}\Big)$
View full question & answer→Question 124 Marks
Find the intervals in which the following functions are increasing or decreasing.
f(x) = 2x2 - 24x + 7
AnswerWe have
f(x) = 2x2 - 24x + 7
f'(x) = 6x2 - 24
Critical points
f'(x) = 0
⇒ 6x2 - 24 = 0
⇒ 6x2 = 24
⇒ x2 = 4
⇒ x = 2, -2
Clearly, f'(x) > 0 if x > -1 and x < -2
f'(x) < 0 if $-2\leq\text{x}\leq2$
Thus, f(x) increasing in $(-\infty,-2)\cup(2,\infty),$ decreasing in (-2, 2).
View full question & answer→Question 134 Marks
Find the intervals in which $\text{f}(\text{x})=\sin\text{x}-\cos\text{x},$ where $0<\text{x}<2\pi$ is increasing or decreasing.
Answer$\text{f}(\text{x})=\sin\text{x}-\cos\text{x},\text{x}\in(0,2\pi)$ $\text{f}'(\text{x})=\cos\text{x}+\sin\text{x}$ For f(x) to be increasing, we must have $\text{f}'(\text{x})>0$ $\Rightarrow\cos\text{x}+\sin\text{x}>0$ $\Rightarrow\sin\text{x}>-\cos\text{x}$ $\Rightarrow\tan\text{x}>-1$ $\Rightarrow\text{x}\in\Big(0,\frac{3\pi}{4}\Big)\cup\Big(\frac{7\pi}{4},2\pi\Big)$ So, f(x) is increasing on $\Big(0,\frac{3\pi}{4}\Big)\cup\Big(\frac{7\pi}{4},2\pi\Big).$
For f(x) to be decreasing, we must have $\text{f}'(\text{x})<0$ $\Rightarrow\cos\text{x}+\sin\text{x}<0$ $\Rightarrow\sin\text{x}<-\cos\text{x}$ $\Rightarrow\tan\text{x}<-1$ $\Rightarrow\text{x}\in\Big(\frac{3\pi}{4},\frac{7\pi}{4}\Big)$ So, f(x) is decreasing on $\Big(\frac{3\pi}{4},\frac{7\pi}{4}\Big).$ View full question & answer→Question 144 Marks
Find the intervals in which the following functions are increasing or decreasing.
f(x) = x4 - 4x3 - 4x2 + 15
Answerf(x) = x4 - 4x3 - 4x2 + 15
f'(x) = 4x3 - 12x2 + 8x
= 4x(x2 - 3x + 2)
= 4x(x - 1)(x - 2)
Here, 0, 1 and 2 are the critical points.
The possible intervals are $(-\infty,0),(0,1),(1,2)$ and $(2,\infty)\ ....(1)$
For f(x) to be increasing, we must have
f'(x) > 0
⇒ 4x(x - 1)(x - 2) > 0
[Since, 4 > 0, 4x(x - 1)(x - 2) > 0 ⇒ x(x - 1)(x - 2) > 0]
⇒ x(x - 1)(x - 2) > 0
$\Rightarrow\text{x}\in(0,1)\cup(2,\infty)$ [From eq. (1)]
So, f(x) is increasing on $\text{x}\in(0,1)\cup(2,\infty).$
For f(x) to be decreasing, we must have,
f'(x) < 0
⇒ 4x(x - 1)(x - 2) < 0
[Since, 4 > 0, 4x(x - 1)(x - 2) < 0 ⇒ x(x - 1)(x - 2) < 0]
⇒ x(x - 1)(x - 2) < 0
$\Rightarrow\text{x}\in(-\infty,0)\cup(1,2)$ [From eq. (1)]
So, f(x) is decreasing on $\text{x}\in(-\infty,0)\cup(1,2).$
View full question & answer→Question 154 Marks
Find the intervals in which the following functions are increasing or decreasing.
$\text{f}(\text{x})=\frac{3}{2}\text{x}^4-4\text{x}^3-45\text{x}^2+51$
Answer$\text{f}(\text{x})=\frac{3}{2}\text{x}^4-4\text{x}^3-45\text{x}^2+51$ f'(x) = 6x3 - 12x2 - 90x = 6x(x2 - 2x - 15) = 6x(x - 5)(x + 3) Here, x = -3, x = 0 and x = 5 are the critical points. The possible intervals are $(-\infty,-3),(-3,0),(0,5)$ and $(5,\infty)\ ....(1)$ For f(x) to be increasing, we must have f'(x) > 0 ⇒ 6x(x - 5)(x + 3) > 0 [Since, 6 > 0, 6x(x - 5)(x + 3) > 0 ⇒ x(x - 5)(x + 3) > 0] ⇒ x(x - 5)(x + 3) > 0 $\Rightarrow\text{x}\in(-3,0)\cup(5,\infty)$ [From eq. 1] So, f(x) is increasing on $\text{x}\in(-3,0)\cup(5,\infty).$ For f(x) to be decreasing, we must have, f'(x) < 0 ⇒ 6x(x - 5)(x + 3) < 0 [Since, 6 > 0, 6x(x - 5)(x + 3) < 0 ⇒ x(x - 5)(x + 3) < 0] ⇒ x(x - 5)(x + 3) < 0 $\Rightarrow\text{x}\in(-\infty,-3)\cup(0,5)$ [From eq. 1] So, f(x) is decreasing on $\text{x}\in(-\infty,-3)\cup(0,5).$
View full question & answer→Question 164 Marks
Find the intervals in which the following functions are increasing or decreasing.
f(x) = 8 + 36x + 3x2 -2x3
Answerf(x) = 8 + 36x + 3x2 -2x3 f'(x) = 36 + 6x - 6x2 = -6(x2 - x - 6) = -6(x - 3)(x + 2) For f(x) to be increasing, we must have f'(x) > 0 ⇒ -6(x - 3)(x + 2) > 0 ⇒ (x - 3)(x + 2) < 0 [Since, -6 > 0, -6(x - 3)(x + 2) > 0 ⇒ (x - 3)(x + 2) < 0] ⇒ -2 < x < 3 $\Rightarrow\text{x}\in(-2,3)$ So, f(x) is increasing on (-2, 3)
. 
For f(x) to be decreasing, we must have f'(x) < 0 ⇒ -6(x - 3)(x + 2) < 0 ⇒ (x - 3)(x + 2) > 0 [Since, -6 < 0, -6(x - 3)(x + 2) < 0 ⇒ (x - 3)(x + 2) > 0] ⇒ x < -2 or x > 3 $\Rightarrow\text{x}\in(-\infty,-2)\cup(3,\infty)$ So, f(x) is decreasing on
$(-\infty,-2)\cup(3,\infty).$ 
View full question & answer→Question 174 Marks
Find the intervals in which $\text{f}(\text{x})=\log(1+\text{x})-\frac{\text{x}}{1+\text{x}}$ is increasing or decreasing.
Answer$\text{f}(\text{x})=\log(1+\text{x})-\frac{\text{x}}{1+\text{x}}$
$\text{f}'(\text{x})=\frac{1}{1+\text{x}}-\Big(\frac{(1+\text{x})-\text{x}}{(1+\text{x})^2}\Big)$
$=\frac{1}{1+\text{x}}-\frac{1}{(1+\text{x})^2}$
$=\frac{\text{x}}{(1+\text{x})^2}$
Critical points,
$\text{f}'(\text{x})=0$
$\Rightarrow\frac{\text{x}}{(1+\text{x})^2}=0$
$\Rightarrow\text{x}=0,-1 $
Clearly, f'(x) > 0 if x > 0 and f'(x) < 0 if -1 < x < 0 or x < -1
Hence, f(x) increases in $(0,\infty),$ decreases in $(-\infty,-1)\cup(-1,0).$
View full question & answer→Question 184 Marks
Write the set of values of 'a' for which $\text{f}(\text{x})=\log_\text{a}\text{x}$ is decreasing in its domain.
AnswerGiven: $\text{f}(\text{x})=\log_\text{a}\text{x}$ Domain of the given function is $(0,\infty).$ Let, $\text{x}_1,\text{x}_2\in(0,\infty)$ such that x1 < x2. Since given function is logorithmic, either a > 1 or 0 < a < 1. Case I:
Let a > 1 Here, $\text{x}_1<\text{x}_2$ $\Rightarrow\log_\text{a}\text{x}_1<\log_\text{a}\text{x}_2$ $\Rightarrow\text{f}(\text{x}_1)<\text{f}(\text{x}_2)$ $\therefore\ \text{x}_1< \text{x}_2$ $\Rightarrow\text{f}(\text{x}_1)<\text{f}(\text{x}_2),\forall\ \text{x}_1,\text{x}_2\in(0,\infty)$ So, for a > 1, f(x) is increasing on $(0,\infty).$ Case II:
Let 0 < a < 1 Here, $\text{x}_1<\text{x}_2$ $\Rightarrow\log_\text{a}\text{x}_1>\log_\text{a}\text{x}_2$ $\Rightarrow\text{f}(\text{x}_1)>\text{f}(\text{x}_2)$ $\therefore\ \text{x}_1< \text{x}_2$ $\Rightarrow\text{f}(\text{x}_1)>\text{f}(\text{x}_2),\forall\ \text{x}_1,\text{x}_2\in(0,\infty)$ So, f(x) is decreasing on $(0,\infty)$ Thus, for 0 < a < 1, f(x) is decreasing in its domain. View full question & answer→Question 194 Marks
Find the intervals in which the following functions are increasing or decreasing.
f(x) = {x(x - 2)}2
Answerf(x) = {x(x - 2)}2
= (x2 - 2x)2
= x4 + 4x2 - 4x3
f'(x) = 4x3 + 8x - 12x2
= 4x(x2 - 3x + 2)
= 4x(x - 1)(x - 2)
Here, 0, 1 and 2 are the critical points.
The possible intervals are $(-\infty,0),(0,1),(1,2)$ and $(2,\infty).$
For f(x) to be increasing, we must have,
f'(x) > 0
⇒ 4x(x - 1)(x - 2) > 0
⇒ x(x - 1)(x - 2) > 0
$\Rightarrow\text{x}\in(0,1)\cup(2,\infty)$
So, f(x) is increasing on $\text{x}\in(0,1)\cup(2,\infty).$
For f(x) to be decreasing, we must have,
f'(x) < 0
⇒ 4x(x - 1)(x - 2) < 0
⇒ x(x - 1)(x - 2) < 0
$\Rightarrow\text{x}\in(-\infty,0)\cup(1,2)$
So, f(x) is decreasing on $\text{x}\in(-\infty,0)\cup(1,2).$
View full question & answer→Question 204 Marks
Show that $\text{f}(\text{x})=\tan^{-1}(\sin\text{x}+\cos\text{x})$ is a decreasing function on the interval $\Big(\frac{\pi}{4},\frac{\pi}{2}\Big).$
Answer$\text{f}(\text{x})=\tan^{-1}(\sin\text{x}+\cos\text{x})$ $\therefore\ \text{f}'(\text{x})=\frac{1}{1+(\sin\text{x}+\cos\text{x})^2}(\cos\text{x}-\sin\text{x})$ $=\frac{1}{1+1+2\sin\text{x}\cos\text{x}}(\cos\text{x}-\sin\text{x})$ $=\frac{(\cos\text{x}-\sin\text{x})}{2+\sin2\text{x}}$ Here, $\frac{\pi}{4}<\text{x}<\frac{\pi}{2}$ $\Rightarrow\frac{\pi}{2}<2\text{x}<\pi$ $\Rightarrow\sin2\text{x}>0$ $\Rightarrow2+\sin2\text{x}>0\ ...(1)$ Also,
$\frac{\pi}{4}<\text{x}<\frac{\pi}{2}$ $\cos\text{x}<\sin\text{x}$ $\Rightarrow\cos\text{x}-\sin\text{x}<0\ ....(2)$ $\text{f}'(\text{x})=\frac{(\cos\text{x}-\sin\text{x})}{2+\sin2\text{x}}<0,\ \forall\ \text{x}\in\Big(\frac{\pi}{4},\frac{\pi}{2}\Big).$ [From eqs. (1) and (2)] Hence, f(x) is decreasing on $\Big(\frac{\pi}{4},\frac{\pi}{2}\Big).$ View full question & answer→Question 214 Marks
Find the intervals in which the following functions are increasing or decreasing.
f(x) = 2x3 - 12x2 + 18x + 15
Answerf(x) = 2x3 - 12x2 + 18x + 15
f'(x) = 6x2 - 24x + 18
= 6(x2 - 4x + 3)
= 6(x - 1)(x - 3)
For f(x) to be increasing, we must have
f'(x) > 0
⇒ 6(x - 1)(x - 3) > 0
⇒ (x - 1)(x - 3) > 0
[Since, 6 > 0, 6(x - 1)(x - 3) > 0 ⇒ (x - 1)(x - 3) > 0]
⇒ x < 1 or x > 3
$\Rightarrow\text{x}\in(-\infty,0)\cup(3,\infty).$
So, f(x) is increasing on $(-\infty,0)\cup(3,\infty).$
For f(x) to be decreasing, we must have
f'(x) < 0
⇒ 6(x - 1)(x - 3) < 0
⇒ (x - 1)(x - 3) < 0
[Since, 6 > 0, 6(x - 1)(x - 3) < 0 ⇒ (x - 1)(x - 3) < 0]
⇒ x < 1 or x < 3
$\Rightarrow\text{x}\in(1,3)$
So, f(x) is decreasing on (1, 3).
View full question & answer→Question 224 Marks
Show that $\text{f}(\text{x})=\sin\text{x}$ is increasing on $\Big(0,\frac{\pi}{2}\Big)$ and decreasing on $\Big(\frac{\pi}{2},\pi\Big)$ and neither increasing nor decreasing in $(0,\pi).$
Answer$\text{f}(\text{x})=\sin\text{x}$ Domain of $\sin\text{x}$ is $(0,\pi).$ $\text{f}'(\text{x})=\cos\text{x}$ For, $\text{x}\in\Big(0,\frac{\pi}{2}\Big),\cos\text{x}>0$ $[\because\ \cos\text{x}$ is positive in first quadrant$]$ $\text{f}'(\text{x})>0$
So, f(x) is increasing for $\Big(0,\frac{\pi}{2}\Big).$ For $\text{x}\in\Big(\frac{\pi}{2},\pi\Big),\cos\text{x}<0$ $[\because\ \cos\text{x}$ is negative in second quadrant$]$ So, f(x) is decreasing for $\Big(\frac{\pi}{2},\pi\Big).$ Since, f(x) is increasing on $\Big(0,\frac{\pi}{2}\Big)$ and decreasing on $\Big(\frac{\pi}{2},\pi\Big),\text{f}(\text{x})$ is neither increasing nor decreasing in $(0,\pi).$ View full question & answer→Question 234 Marks
What are the values of 'a' for which f(x) = ax is increasing on R?
Answer$\text{f}(\text{x})=\text{a}^\text{x}$ $\text{f}'(\text{x})=\text{a}^{\text{x}}\log\text{a}$ Given: f(x) is increasing on R. $\Rightarrow\ \text{f}'(\text{x})>0$ $\Rightarrow\text{a}^{\text{x}}\log\text{a}>0$ Logarithmic function is defined for positive values of a. $\Rightarrow\text{a}>0$ $\Rightarrow\text{a}^\text{x}>0$ We know,
$\text{a}^{\text{x}}\log\text{a}>0$ It can be possible when $\text{a}^\text{x}>0$ and $\log\text{a}>0$ or $\text{a}^\text{x}<0$ and $\log\text{a}<0.$ $\Rightarrow\log\text{a}>0$ $\Rightarrow\text{a}>1$ So, f(x) is increasing when a > 1. View full question & answer→Question 244 Marks
Find the intervals in which the following functions are increasing or decreasing.
f(x) = x2 + 2x - 5
AnswerWe have,
f(x) = x2 + 2x - 5
$\therefore$ f'(x) = 2x + 2
Now.
f'(x) = 0 ⇒ x = -1
Point x = -1 divides the real line into two disjoints intervals i.e., $(-\infty,-1)$ and $(-1,\infty).$
In interval $(-\infty,-1),$ f'(x) = 2x + 2 < 0.
$\therefore$ f is strictly decreasing in interval $(-\infty,-1).$
Thus, f is strictly decreasing for x < -1.
In interval $(-1,\infty),$ f'(x) = 2x + 2 > 0.
$\therefore$ f is strictly increasing in interval $(-1,\infty).$
Thus, f is strictly increasing for x > -1.
View full question & answer→Question 254 Marks
Write the set of values of 'a' for which $\text{f}(\text{x})=\log_\text{a}\text{x}$ is increasing in its domain.
Answer$\text{f}(\text{x})=\log_\text{a}\text{x}$ Let, $\text{x}_1,\text{x}_2\in(0,\infty)$ such that x1 < x2. Since given function is logorithmic, either a > 1 or 0 < a < 1. Case I:
Let a > 1 Here, $\text{x}_1<\text{x}_2$ $\Rightarrow\log_\text{a}\text{x}_1<\log_\text{a}\text{x}_2$ $\Rightarrow\text{f}(\text{x}_1)<\text{f}(\text{x}_2)$ $\therefore\ \text{x}_1< \text{x}_2$ $\Rightarrow\text{f}(\text{x}_1)<\text{f}(\text{x}_2),\forall\ \text{x}_1,\text{x}_2\in(0,\infty)$ Thus, for a > 1, f(x) is increasing on $(0,\infty).$ Case II:
Let 0 < a < 1 Here, $\text{x}_1<\text{x}_2$ $\Rightarrow\log_\text{a}\text{x}_1>\log_\text{a}\text{x}_2$ $\Rightarrow\text{f}(\text{x}_1)>\text{f}(\text{x}_2)$ $\therefore\ \text{x}_1< \text{x}_2$ $\Rightarrow\text{f}(\text{x}_1)>\text{f}(\text{x}_2),\forall\ \text{x}_1,\text{x}_2\in(0,\infty)$ Thus, for a > 1, f(x) is increasing in its domain. View full question & answer→Question 264 Marks
Show that $\text{f}(\text{x})=\frac{1}{1+\text{x}^2}$ is neither increasing nor decreasing on R.
AnswerHere, $\text{f}(\text{x})=\frac{1}{1+\text{x}^2}$ R can be divided into two intervals $0,\infty$ and $(-\infty,0].0,$ Case 1:
Let $\text{x}_1,\text{x}_2\in (0,\infty)$ such that $\text{x}_1<\text{x}_2.$ Then, $\text{x}_1<\text{x}_2$
$\Rightarrow\text{x}_1^2<\text{x}_2^2$ $\Rightarrow1+\text{x}_1^2<1+\text{x}_2^2$ $\Rightarrow\frac{1}{1+\text{x}_1^2}>\frac{1}{1+\text{x}_2^2}$ $\Rightarrow\text{f}(\text{x}_1)>\text{f}(\text{x}_2)\ \forall\ \text{x}_1,\text{x}_2\in(0,\infty)$ So, f(x) is decreasing on $(0,\infty).$ Case 2:
Let $\text{x}_1,\text{x}_2\in(-\infty,0]$ such that $\text{x}_1<\text{x}_2.$ Then, $\text{x}_1<\text{x}_2$
$\Rightarrow\text{x}_1^2>\text{x}_2^2$ $\Rightarrow1+\text{x}_1^2>1+\text{x}_2^2$ $\Rightarrow\frac{1}{1+\text{x}_1^2}<\frac{1}{1+\text{x}_2^2}$ $\Rightarrow\text{f}(\text{x}_1)<\text{f}(\text{x}_2)$ $\Rightarrow\text{f}(\text{x}_1)<\text{f}(\text{x}_2)\ \forall\ \text{x}_1,\text{x}_2\in(-\infty,0]$ So, f(x) is increasing on $(-\infty,0].$ Here, f(x) is decreasing on $(0,\infty)$ and increasing on $(-\infty,0].$ Thus, f(x) is neither increasing nor decreasing on R. View full question & answer→Question 274 Marks
Show that the function $\text{f}(\text{x})=\sin\Big(2\text{x}+\frac{\pi}{4}\Big)$ is decreasing on $\Big(\frac{3\pi}{8},\frac{5\pi}{8}\Big).$
Answer$\text{f}(\text{x})=\sin\Big(2\text{x}+\frac{\pi}{4}\Big)$
$\text{f}'(\text{x})=2\cos\Big(2\text{x}+\frac{\pi}{4}\Big)$
Here,
$\frac{3\pi}{8}<\text{x}<\frac{5\pi}{8}$
$\Rightarrow\frac{3\pi}{4}<2\text{x}<\frac{5\pi}{4}$
$\Rightarrow\pi<2\text{x}+\frac{\pi}{4}<\frac{3\pi}{2}$
$\Rightarrow\cos\Big(2\text{x}+\frac{\pi}{4}\Big)<0$ $[\because$ Cos function is negative in third quadrent$]$
$\Rightarrow2\cos\Big(2\text{x}+\frac{\pi}{4}\Big)<0$
$\Rightarrow\text{f}'(\text{x})<0,\ \forall\ \text{x}\in\Big(\frac{3\pi}{8},\frac{5\pi}{8}\Big)$
So, f(x) is decreasing on $\Big(\frac{3\pi}{8},\frac{5\pi}{8}\Big).$
View full question & answer→Question 284 Marks
Find the intervals in which the following functions are increasing or decreasing.
f(x) = 5x3 - 15x2 - 120x + 3.
Answerf(x) = 5x3 - 15x2 - 120x + 3 f'(x) = 15x2 - 30x - 120 = 15(x2 - 2x - 8) = 15(x - 4)(x + 2) For f(x) to be increasing, we must have f'(x) > 0 ⇒ 15(x - 4)(x + 2) > 0 ⇒ (x - 4)(x + 2) > 0 [Since, 15 > 0, 15(x - 4)(x + 2) > 0 ⇒ (x - 4)(x + 2) > 0] ⇒ x < -2 or x > 4 $\Rightarrow\text{x}\in(-\infty,-2)\cup(4,\infty)$ So, f(x) is increasing on $\text{x}\in(-\infty,-2)\cap(4,\infty).$ 
For f(x) to be decreasing, we must have, f'(x) < 0 ⇒ 6(x - 4)(x + 2) < 0 ⇒ (x - 4)(x + 2) < 0 [Since, 15 > 0, 15(x - 4)(x + 2) > 0 ⇒ (x - 4)(x + 2) > 0] ⇒ -2 < x < 4 $\Rightarrow\text{x}\in(-2,4)$ So, f(x) is decreasing on $\text{x}\in(-2,4).$ View full question & answer→Question 294 Marks
Show that the function x2 - x + 1 is neither increasing nor decreasing on (0, 1).
Answerf(x) = x2 - x + 1
f'(x) = 2x - 1
For, f(x) to be increasing, we must have
f'(x) > 0
⇒ 2x - 1 > 0
⇒ 2x > 1
$\Rightarrow\text{x}>\frac{1}{2}$
$\Rightarrow\text{x}\in\Big(\frac{1}{2},1\Big)$
So, f(x) is increasing on $\text{x}\in\Big(\frac{1}{2},1\Big).$
For, f(x) to be decreasing, we must have
f'(x) < 0
⇒ 2x - 1 < 0
⇒ 2x < 1
$\Rightarrow\text{x}<\frac{1}{2}$
$\Rightarrow\text{x}\in\Big(0,\frac{1}{2}\Big)$
So, f(x) is increasing on $\Big(0,\frac{1}{2}\Big)$
Since, f(x) is increasing on $\Big(\frac{1}{2},1\Big)$ and decreasing on $\Big(0,\frac{1}{2}\Big),\text{f}(\text{x})$ is neither increasing nor decreasing on (0, 1).
View full question & answer→Question 304 Marks
Find the intervals in which the following functions are increasing or decreasing.
f(x) = (x - 1)(x - 2)2
Answerf(x) = (x - 1)(x - 2)2
= (x - 1)(x2 - 4x + 4)
= x3 - 5x2 + 8x - 4
f'(x) = 3x2 - 10x + 8
= 3x2 - 6x - 4x + 8
= (x - 2)(3x - 4)
For f(x) to be increasing, we must have
f'(x) > 0
⇒ (x - 2)(3x - 4) > 0
$\Rightarrow\text{x}<\frac{4}{3}\text{ or }\text{x}>2$
$\Rightarrow\text{x}\in\Big(-\infty,-\frac{4}{3}\Big)\cup(2,\infty)$
So, f(x) is increasing on $\text{x}\in\Big(-\infty,-\frac{4}{3}\Big)\cup(2,\infty).$
For f(x) to be decreasing, we must have,
f'(x) < 0
⇒ (x - 2)(3x - 4) < 0
$\Rightarrow\frac{4}{3}<\text{x}<2$
$\Rightarrow\text{x}\in\Big(\frac{4}{3},2\Big)$
So, f(x) is decreasing on $\text{x}\in\Big(\frac{4}{3},2\Big).$
View full question & answer→Question 314 Marks
Prove that the function f given by $\text{f}(\text{x})=\log\cos\text{x}$ is strictly increasing on $\Big(-\frac{\pi}{2},0\Big)$ and strictly decreasing on $\Big(0,\frac{\pi}{2}\Big).$
AnswerWe have,
$\text{f}(\text{x})=\log\cos\text{x}$
$\therefore\ \text{f}'(\text{x})=\frac{1}{\cos\text{x}}(-\sin\text{x})=-\tan\text{x}$
In interval $\Big(0,\frac{\pi}{2}\Big),\tan\text{x}>0\Rightarrow-\tan\text{x}<0.$
$\therefore\text{f}'(\text{x})<0\text{ on }\Big(0,\frac{\pi}{2}\Big)$
$\therefore$ f is strictly decreasing on $\Big(0,\frac{\pi}{2}\Big).$
In interval $\Big(\frac{\pi}{2},\pi\Big),\tan\text{x}<0\Rightarrow-\tan\text{x}>0.$
$\therefore\text{f}'(\text{x})>0\text{ on }\Big(\frac{\pi}{2},\pi\Big)$
$\therefore$ f is strictly increasing on $\Big(-\frac{\pi}{2},0\Big).$
View full question & answer→Question 324 Marks
Find the intervals in which the following functions are increasing or decreasing.
f(x) = 3x4 - 4x3 - 12x2 + 5
AnswerConsider the given function
f(x) = 3x4 - 4x3 - 12x2 + 5
⇒ f'(x) = 12x3 - 12x2 - 24x
⇒ f'(x) = 12x(x2 - x - 2)
⇒ f'(x) = 12x(x + 1)(x- 2)
For f(x) to be increasing, we must have,
f'(x) > 0
⇒ 12x(x + 1)(x- 2) > 0
⇒ x(x + 1)(x- 2) > 0
$\Rightarrow-1<\text{x}<0\text{ or }2<\text{x}<\infty$
$\Rightarrow\text{x}\in(-1,0)\cup(2,\infty)$
So, f(x) is increasing on $(-1,0)\cup(2,\infty).$
For f(x) to be decreasing, we must have,
f'(x) < 0
⇒ 12x(x + 1)(x- 2) < 0
⇒ x(x + 1)(x- 2) < 0
$\Rightarrow-\infty<\text{x}<-1\text{ or }0<\text{x}<2$
$\Rightarrow\text{x}\in(-\infty,-1)\cup(0,2)$
So, f(x) is decreasing in $(-\infty,-1)\cup(0,2).$
View full question & answer→Question 334 Marks
Show that $\text{f}(\text{x})=\frac{1}{1+\text{x}^2}$ is decreases in the interval $[0,\infty)$ and increases in the interval $(-\infty,0].$
AnswerHere, $\text{f}(\text{x})=\frac{1}{1+\text{x}^2}$ Case 1:
Let
$\text{x}_1,\text{x}_2\in (0,\infty)$ such that $\text{x}_1<\text{x}_2.$ Then, $\text{x}_1<\text{x}_2$
$\Rightarrow\text{x}_1^2<\text{x}_2^2$ $\Rightarrow1+\text{x}_1^2<1+\text{x}_2^2$ $\Rightarrow\frac{1}{1+\text{x}_1^2}>\frac{1}{1+\text{x}_2^2}$ $\Rightarrow\text{f}(\text{x}_1)>\text{f}(\text{x}_2)\ \forall\ \text{x}_1,\text{x}_2\in(0,\infty)$ So, f(x) is decreasing on $(0,\infty).$ Case 2:
Let $\text{x}_1,\text{x}_2\in (0,\infty]$ such that $\text{x}_1<\text{x}_2.$ Then, $\text{x}_1<\text{x}_2$ $\Rightarrow\text{x}_1^2>\text{x}_2^2$ $\Rightarrow1+\text{x}_1^2<1+\text{x}_2^2$ $\Rightarrow\frac{1}{1+\text{x}_1^2}<\frac{1}{1+\text{x}_2^2}$ $\Rightarrow\text{f}(\text{x}_1)<\text{f}(\text{x}_2)$ $\Rightarrow\text{f}(\text{x}_1)<\text{f}(\text{x}_2)\ \forall\ \text{x}_1,\text{x}_2\in(0,\infty]$ So, f(x) is increasing on $(0,\infty].$ View full question & answer→Question 344 Marks
What are the values of 'a' for which f(x) = ax is decreasing on R?
Answer$\text{f}(\text{x})=\text{a}^\text{x}$
$\text{f}'(\text{x})=\text{a}^{\text{x}}\log\text{a}$
Given: f(x) is decreasing on R.
$\Rightarrow\ \text{f}'(\text{x})<0,\forall\ \text{x}\in\text{R}$
$\Rightarrow\text{a}^{\text{x}}\log\text{a}<0,\forall\ \text{x}\in\text{R}$
Here, logarithmic function is not defined for negative values of a.
$\Rightarrow\text{a}^\text{x}>0$
$\therefore\ \text{a}^{\text{x}}\log\text{a}<0$
It can be possible when $\log\text{a}<0,\forall\ \text{x}\in\text{R}.$
$\Rightarrow0<\text{a}>1$
View full question & answer→Question 354 Marks
Find the intervals in which f(x) is increasing or decreasing:
$\text{f}(\text{x})=\sin\text{x}(1+\cos\text{x}),0<\text{x}<\frac{\pi}{2}$
AnswerConsider the function,
$\text{f}(\text{x})=\sin\text{x}(1+\cos\text{x}),0<\text{x}<\frac{\pi}{2}$
$\Rightarrow\text{f}'(\text{x})=\cos\text{x}+\sin\text{x}(-\sin\text{x})+\cos\text{x}(\cos\text{x})$
$\Rightarrow\text{f}'(\text{x})=\cos\text{x}-\sin^2\text{x}+\cos\text{x}(\cos\text{x})$
$\Rightarrow\text{f}(\text{x})=\cos\text{x}+\big(\cos^2\text{x}-1\big)+\cos^2\text{x}$
$\Rightarrow\text{f}'(\text{x})=\cos\text{x}+2\cos^2\text{x}-1$
$\Rightarrow\text{f}'(\text{x})=2\cos^2\text{x}+\cos\text{x}-1$
$\Rightarrow\text{f}'(\text{x})=2\cos^2\text{x}+2\cos\text{x}-\cos\text{x}-1$
$\Rightarrow\text{f}'(\text{x})=2\cos\text{x}(\cos\text{x}+1)-1(\cos\text{x}+1)$
$\Rightarrow\text{f}'(\text{x})=(2\cos\text{x}-1)(\cos\text{x}+1)$
For f(x) to be increasing, we must have,
$\text{f}'(\text{x})>0$
$\Rightarrow\text{f}'(\text{x})=(2\cos\text{x}-1)(\cos\text{x}+1)>0$
$\Rightarrow0<\text{x}<\frac{\pi}{3}$
$\Rightarrow\text{x}\in\Big(0,\frac{\pi}{3}\Big)$
So, f(x) is increasing in $\Big(0,\frac{\pi}{3}\Big)$
For f(x) to be decreasing, we must have,
$\text{f}'(\text{x})<0$
$\Rightarrow\text{f}'(\text{x})=(2\cos\text{x}-1)(\cos\text{x}+1)<0$
$\Rightarrow\frac{\pi}{3}<\text{x}<\frac{\pi}{3}$
$\Rightarrow\text{x}\in\Big(\frac{\pi}{3},\frac{\pi}{2}\Big)$
So, f(x) is decreasing in $\Big(\frac{\pi}{3},\frac{\pi}{2}\Big)$
View full question & answer→Question 364 Marks
Find the intervals in which the following functions are increasing or decreasing.
f(x) = 2x3 + 9x2 + 12x + 20
Answerf(x) = 2x3 + 9x2 + 12x + 20
f'(x) = 6x2 + 18x + 12
= 6(x2 + 3x + 2)
= 6(x + 1)(x + 2)
For f(x) to be increasing, we must have
f'(x) > 0
⇒ 6(x + 1)(x + 2) > 0
⇒ (x + 1)(x + 2) > 0
[Since, 6 > 0, 6(x + 1)(x + 2) > 0 ⇒ (x + 1)(x + 2) > 0]
⇒ x < -2 or x > -1
$\Rightarrow\text{x}\in(-\infty,-2)\cup(-1,\infty)$
So, f(x) is increasing on $\text{x}\in(-\infty,-2)\cup(-1,\infty).$
For f(x) to be decreasing, we must have,
f'(x) < 0
⇒ 6(x + 1)(x + 2) < 0
⇒ (x + 1)(x + 2) < 0
[Since, 6 > 0, 6(x + 1)(x + 2) < 0 ⇒ (x + 1)(x + 2) < 0]
⇒ -2 < x < -1
$\Rightarrow\text{x}\in(-2,-1)$
So, f(x) is decreasing on $\text{x}\in(-2,-1).$
View full question & answer→Question 374 Marks
Find the intervals in which the following functions are increasing or decreasing.
$\text{f}(\text{x})=\frac{\text{x}^4}{4}+\frac{2}{3}\text{x}^3-\frac{5}{4}\text{x}^2-6\text{x}+7$
Answer$\text{f}(\text{x})=\frac{\text{x}^4}{4}+\frac{2}{3}\text{x}^3-\frac{5}{4}\text{x}^2-6\text{x}+7$
$\therefore$ f'(x) = x3 + 2x2 - 5x - 6
Critical points
f'(x) = 0
⇒ x3 + 2x2 - 5x - 6 = 0
⇒ (x + 1)(x + 3)(x - 2) = 0
⇒ x = -1, -3, 2
Clearly, f'(x) > 0 if -3 x < -1 and x > 2
f'(x) < 0 if x < -3 and -1 < x < 2
Thus, f(x) increases in $(-3,-1)\cup(2,\infty),$ decreases in $(-\infty,-3)\cup(-1,2).$
View full question & answer→Question 384 Marks
Find the intervals in which the following functions are increasing or decreasing.
f(x) = 10 - 6x - 2x2
Answerf(x) = 10 - 6x - 2x2
f'(x) = -6 - 4x
For f(x) to be increasing, we must have
f'(x) > 0
⇒ -6 - 4x > 0
⇒ -4x > 6
$\Rightarrow\text{x}<\frac{-3}{2}$
$\Rightarrow\text{x}\in\Big(-\infty,\frac{-3}{2}\Big)$
So, f(x) is increasing on $\Big(-\infty,\frac{-3}{2}\Big).$
For f(x) to be decreasing, we must have
f'(x) < 0
⇒ -6 - 4x < 0
⇒ -4x < 6
$\Rightarrow\text{x}>\frac{-6}{4}$
$\Rightarrow\text{x}>\frac{-3}{2}$
$\Rightarrow\text{x}\in\Big(\frac{-3}{2},\infty\Big)$
So, f(x) is decreasing on $\Big(\frac{-3}{2},\infty\Big).$
View full question & answer→Question 394 Marks
Find the intervals in which the following functions are increasing or decreasing.
f(x) = 5 + 36x + 3x2 - 2x3
Answerf(x) = 5 + 36x + 3x2 - 2x3
$\therefore$ f'(x) = 36 + 6x - 6x2
Critical point
f'(x) = 0
⇒ 36 + 6x - 6x2 = 0
⇒ -6(x2 - x - 6) = 0
⇒ (x - 3)(x + 2) = 0
$\therefore$ x = 3, -2
Clearly f'(x) > 0 if -2 < x < 3
Also f'(x) < 0 if x < -2 and x > 3
Thus increases if $\text{x}\in(-2,3),$ decreases if $\text{x}\in(-\infty,-2)\cup(3,\infty)$
View full question & answer→Question 404 Marks
Find the intervals in which the following functions are increasing or decreasing.
f(x) = 2x3 - 24x + 107
Answerf(x) = 2x3 - 24x + 107
f'(x) = 6x2 - 24
= 6(x2 - 4)
= 6(x + 2)(x - 2)
For f(x) to be increasing, we must have
f'(x) > 0
⇒ 6(x + 2)(x - 2) > 0
⇒ (x + 2)(x - 2) > 0
[Since, 6 > 0, 6(x + 2)(x - 2) > 0 ⇒ (x + 2)(x - 2) > 0]
⇒ x < -2 or x > 2
$\Rightarrow\text{x}\in(-\infty,-2)\cup(2,\infty)$
So, f(x) is increasing on $\text{x}\in(-\infty,-2)\cup(2,\infty).$
For f(x) to be decreasing, we must have,
f'(x) < 0
⇒ 6(x + 2)(x - 2) < 0
⇒ (x + 2)(x - 2) < 0
[Since, 6 > 0, 6(x + 2)(x - 2) < 0 ⇒ (x + 2)(x - 2) < 0]
⇒ -2 < x < 2
$\Rightarrow\text{x}\in(-2,2)$
So, f(x) is decreasing on $\text{x}\in(-2,2).$
View full question & answer→Question 414 Marks
Find the intervals in which the following functions are increasing or decreasing.
$\text{f}(\text{x})=\frac{3}{10}\text{x}^4-\frac{4}{5}\text{x}^3-3\text{x}^2+\frac{36}{5}\text{x}+11$
Answer$\text{f}(\text{x})=\frac{3}{10}\text{x}^4-\frac{4}{5}\text{x}^3-3\text{x}^2+\frac{36}{5}\text{x}+11$
$=\frac{3\text{x}^4-8\text{x}^3-30\text{x}^2+72\text{x}+110}{10}$
$\text{f}'(\text{x})=\frac{12\text{x}^3-24\text{x}^2+60\text{x}+72}{10}$
$=\frac{12}{10}\big(\text{x}^3-2\text{x}^2-5\text{x}+6\big)$
$=\frac{(\text{x}-1)(\text{x}^2-\text{x}-6)}{10}$
$=\frac{12}{10}(\text{x}-1)(\text{x}+2)(\text{x}-3)$
Here, 1, 2, 3 are the Critical points.
The possible intervals are $(-\infty,-2),(-2,-1),(1,3)$ and $(3,\infty).$
For f(x) to be increasing, we must have
$\text{f}'(\text{x})>0$
$\Rightarrow\frac{12}{10}(\text{x}-1)(\text{x}+2)(\text{x}-3)>0$
$\Rightarrow(\text{x}-1)(\text{x}+2)(\text{x}-3)>0$
$\Rightarrow\text{x}\in(-2,1)\cup(3,\infty)$
So, f(x) is increasing on $\text{x}\in(-2,1)\cup(3,\infty).$
For f(x) to be decreasing, we must have
$\text{f}'(\text{x})<0$
$\Rightarrow\frac{12}{10}(\text{x}-1)(\text{x}+2)(\text{x}-3)<0$
$\Rightarrow(\text{x}-1)(\text{x}+2)(\text{x}-3)<0$
$\Rightarrow\text{x}\in(-\infty,-2)\cup(1,3)$
So, f(x) is decreasing on $\text{x}\in(-\infty,-2)\cup(1,3).$
View full question & answer→Question 424 Marks
Find the intervals in which f(x) = (x + 2)e-x is increasing or decreasing.
Answerf(x) = (x + 2)e-x
f'(x) = -e-x(x + 2) + e-x
= -xe-x - 2e-x + e-x
= -xe-x - e-x
= e-x(-x - 1)
For f(x) to be increasing, we must have
f'(x) > 0
⇒ e-x(-x - 1) > 0
⇒ -x - 1 > 0 $\big[\because\ \text{e}^{-\text{x}}>0,\forall\ \text{x}\in\text{R}\big]$
⇒ -x > 1
⇒ x < -1
$\Rightarrow\text{x}\in(-\infty,-1)$
So, f(x) is increasing on $(-\infty,-1).$
For f(x) to be decreasing, we must have
f'(x) < 0
⇒ e-x(-x - 1) < 0
⇒ -x - 1 < 0 $\big[\because\ \text{e}^{-\text{x}}>0,\forall\ \text{x}\in\text{R}\big]$
⇒ -x < 1
⇒ x < -1
$\Rightarrow\text{x}\in(-1,\infty)$
So, f(x) is decreasing on $(-1,\infty).$
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