85283
If \(f(x)=3 x^{4}+4 x^{3}-12 x^{2}+12\), then \(f(x)\) is
1 increasing in \((-\infty,-2)\) and in \((0,1)\)
2 increasing in \((-2,0)\) and in \((1, \infty)\)
3 decreasing in \((-2,0)\) and in \((0,1)\)
4 decreasing in \((-\infty,-2)\) and in \((1, \infty)\)
Explanation:
(B) : Given : \(\mathrm{f}(\mathrm{x})=3 \mathrm{x}^{4}+4 \mathrm{x}^{3}-12 \mathrm{x}^{2}+12\) On differentiating both side with respect to \(x\), we get \(f^{\prime}(x)=12 x^{3}+12 x^{2}-24 x\) For \(\mathrm{f}(\mathrm{x})\) to be increasing \(\mathrm{f}^{\prime}(\mathrm{x})>0\) \(\Rightarrow 12 \mathrm{x}^{3}+12 \mathrm{x}^{2}-24 \mathrm{x}>0\) \(\Rightarrow 12 \mathrm{x}\left(\mathrm{x}^{2}+\mathrm{x}-2\right)>0\) \(\Rightarrow 12 \mathrm{x}(\mathrm{x}-1)(\mathrm{x}+2)>0\) \(\Rightarrow \mathrm{x}(\mathrm{x}-1)(\mathrm{x}+2)>0\) \(\Rightarrow-2\lt \mathrm{x}\lt 0\) or \(\mathrm{x}>1\) It means \(x \in(-2,0) \cup(1, \infty)\). Hence, \(\mathrm{f}(\mathrm{x})\) is increasing in \((-2,0)\) and \((1, \infty)\).
BITSAT-2019
Application of Derivatives
85284
The function \(f(x)=\tan x-4 x\) is strictly decreasing on
85285
If \(f(x)=\left(\frac{a^{2}-1}{a^{2}+1}\right) x^{3}-3 x+5\) is a decreasing function of \(x\) in \(R\) then the set of possible values of a (independent of \(x\) ) is
1 \((1, \infty)\)
2 \((-\infty,-1)\)
3 \([-1,1]\)
4 None of these
Explanation:
(C) : Given function, \(f(x)=\left(\frac{a^{2}-1}{a^{2}+1}\right) x^{3}-3 x+5\) \(f^{\prime}(x)=3\left(\frac{a^{2}-1}{a^{2}+1}\right) x^{2}-3\) \(\mathrm{f}^{\prime}(\mathrm{x})\lt 0\) for all \(\mathrm{x}\) if \(\mathrm{a}^{2}-1 \leq 0\) \(-1 \leq \mathrm{a} \leq 1\) Hence, \([-1,1]\)
BITSAT-2012
Application of Derivatives
85286
The function \(f(x)=(x(x-2))^{2}\) is increasing in the set
1 \((-\infty, 0) \cup(2, \infty)\)
2 \((-\infty, 1)\)
3 \((0,1) \cup(2, \infty)\)
4 \((1,2)\)
Explanation:
(C) : Given, \(\mathrm{f}(\mathrm{x})=(\mathrm{x}(\mathrm{x}-2))^{2}\) \(\mathrm{f}^{\prime}(\mathrm{x})=2[\mathrm{x}(\mathrm{x}-2)][1 .(\mathrm{x}-2)+\mathrm{x}(1-0)]\) \(\mathrm{f}^{\prime}(\mathrm{x})=2 \mathrm{x}(\mathrm{x}-2)(2 \mathrm{x}-2)\) \(\mathrm{f}^{\prime}(\mathrm{x})=4 \mathrm{x}(\mathrm{x}-2)(\mathrm{x}-1)\) For \(\mathrm{f}(\mathrm{x})\) as increasing, \(\mathrm{f}^{\prime}(\mathrm{x})>0\) So, \(4 x(x-1)(x-2)>0\) From the above figure required interval is, \((0,1) \cup(2, \infty)\)
Kerala CEE-2011
Application of Derivatives
85287
The interval in which the function \(2 x^{3}+15\) increases less rapidly than the function \(9 x^{2}-\) \(12 x\), is
1 \((-\infty, 1)\)
2 \((1,2)\)
3 \((2, \infty)\)
4 None of these
Explanation:
(B) : Let \(\mathrm{f}(\mathrm{x})=2 \mathrm{x}^{3}+15\) and \(\mathrm{g}(\mathrm{x})=9 \mathrm{x}^{2}-12 \mathrm{x}\) then \(\mathrm{f}^{\prime}(\mathrm{x})=6 \mathrm{x}^{2} \forall \mathrm{x} \in \mathrm{R}\) \(\therefore \mathrm{f}(\mathrm{x})\) is increasing function \(\forall \mathrm{x} \in \mathrm{R}\) \(\mathrm{g}^{\prime}(\mathrm{x})>0\) Also, \(18 \mathrm{x}-12>0\) \(x>\frac{2}{3}\) Thus, \(f(x)\) and \(g(x)\) both increases for \(x>\frac{2}{3}\) Let \(\mathrm{F}(\mathrm{x})=\mathrm{f}(\mathrm{x})-\mathrm{g}(\mathrm{x}), \mathrm{F}^{\prime}(\mathrm{x})\lt 0\) \((\because \mathrm{f}(\mathrm{x})\) increases less rapidly than the function \(\mathrm{g}(\mathrm{x}))\) \(6 \mathrm{x}^{2}-18 \mathrm{x}+12\lt 0\) \(6(\mathrm{x}-2)(\mathrm{x}-1)\lt 0\) \(1\lt \mathrm{x}\lt 2\)
85283
If \(f(x)=3 x^{4}+4 x^{3}-12 x^{2}+12\), then \(f(x)\) is
1 increasing in \((-\infty,-2)\) and in \((0,1)\)
2 increasing in \((-2,0)\) and in \((1, \infty)\)
3 decreasing in \((-2,0)\) and in \((0,1)\)
4 decreasing in \((-\infty,-2)\) and in \((1, \infty)\)
Explanation:
(B) : Given : \(\mathrm{f}(\mathrm{x})=3 \mathrm{x}^{4}+4 \mathrm{x}^{3}-12 \mathrm{x}^{2}+12\) On differentiating both side with respect to \(x\), we get \(f^{\prime}(x)=12 x^{3}+12 x^{2}-24 x\) For \(\mathrm{f}(\mathrm{x})\) to be increasing \(\mathrm{f}^{\prime}(\mathrm{x})>0\) \(\Rightarrow 12 \mathrm{x}^{3}+12 \mathrm{x}^{2}-24 \mathrm{x}>0\) \(\Rightarrow 12 \mathrm{x}\left(\mathrm{x}^{2}+\mathrm{x}-2\right)>0\) \(\Rightarrow 12 \mathrm{x}(\mathrm{x}-1)(\mathrm{x}+2)>0\) \(\Rightarrow \mathrm{x}(\mathrm{x}-1)(\mathrm{x}+2)>0\) \(\Rightarrow-2\lt \mathrm{x}\lt 0\) or \(\mathrm{x}>1\) It means \(x \in(-2,0) \cup(1, \infty)\). Hence, \(\mathrm{f}(\mathrm{x})\) is increasing in \((-2,0)\) and \((1, \infty)\).
BITSAT-2019
Application of Derivatives
85284
The function \(f(x)=\tan x-4 x\) is strictly decreasing on
85285
If \(f(x)=\left(\frac{a^{2}-1}{a^{2}+1}\right) x^{3}-3 x+5\) is a decreasing function of \(x\) in \(R\) then the set of possible values of a (independent of \(x\) ) is
1 \((1, \infty)\)
2 \((-\infty,-1)\)
3 \([-1,1]\)
4 None of these
Explanation:
(C) : Given function, \(f(x)=\left(\frac{a^{2}-1}{a^{2}+1}\right) x^{3}-3 x+5\) \(f^{\prime}(x)=3\left(\frac{a^{2}-1}{a^{2}+1}\right) x^{2}-3\) \(\mathrm{f}^{\prime}(\mathrm{x})\lt 0\) for all \(\mathrm{x}\) if \(\mathrm{a}^{2}-1 \leq 0\) \(-1 \leq \mathrm{a} \leq 1\) Hence, \([-1,1]\)
BITSAT-2012
Application of Derivatives
85286
The function \(f(x)=(x(x-2))^{2}\) is increasing in the set
1 \((-\infty, 0) \cup(2, \infty)\)
2 \((-\infty, 1)\)
3 \((0,1) \cup(2, \infty)\)
4 \((1,2)\)
Explanation:
(C) : Given, \(\mathrm{f}(\mathrm{x})=(\mathrm{x}(\mathrm{x}-2))^{2}\) \(\mathrm{f}^{\prime}(\mathrm{x})=2[\mathrm{x}(\mathrm{x}-2)][1 .(\mathrm{x}-2)+\mathrm{x}(1-0)]\) \(\mathrm{f}^{\prime}(\mathrm{x})=2 \mathrm{x}(\mathrm{x}-2)(2 \mathrm{x}-2)\) \(\mathrm{f}^{\prime}(\mathrm{x})=4 \mathrm{x}(\mathrm{x}-2)(\mathrm{x}-1)\) For \(\mathrm{f}(\mathrm{x})\) as increasing, \(\mathrm{f}^{\prime}(\mathrm{x})>0\) So, \(4 x(x-1)(x-2)>0\) From the above figure required interval is, \((0,1) \cup(2, \infty)\)
Kerala CEE-2011
Application of Derivatives
85287
The interval in which the function \(2 x^{3}+15\) increases less rapidly than the function \(9 x^{2}-\) \(12 x\), is
1 \((-\infty, 1)\)
2 \((1,2)\)
3 \((2, \infty)\)
4 None of these
Explanation:
(B) : Let \(\mathrm{f}(\mathrm{x})=2 \mathrm{x}^{3}+15\) and \(\mathrm{g}(\mathrm{x})=9 \mathrm{x}^{2}-12 \mathrm{x}\) then \(\mathrm{f}^{\prime}(\mathrm{x})=6 \mathrm{x}^{2} \forall \mathrm{x} \in \mathrm{R}\) \(\therefore \mathrm{f}(\mathrm{x})\) is increasing function \(\forall \mathrm{x} \in \mathrm{R}\) \(\mathrm{g}^{\prime}(\mathrm{x})>0\) Also, \(18 \mathrm{x}-12>0\) \(x>\frac{2}{3}\) Thus, \(f(x)\) and \(g(x)\) both increases for \(x>\frac{2}{3}\) Let \(\mathrm{F}(\mathrm{x})=\mathrm{f}(\mathrm{x})-\mathrm{g}(\mathrm{x}), \mathrm{F}^{\prime}(\mathrm{x})\lt 0\) \((\because \mathrm{f}(\mathrm{x})\) increases less rapidly than the function \(\mathrm{g}(\mathrm{x}))\) \(6 \mathrm{x}^{2}-18 \mathrm{x}+12\lt 0\) \(6(\mathrm{x}-2)(\mathrm{x}-1)\lt 0\) \(1\lt \mathrm{x}\lt 2\)
85283
If \(f(x)=3 x^{4}+4 x^{3}-12 x^{2}+12\), then \(f(x)\) is
1 increasing in \((-\infty,-2)\) and in \((0,1)\)
2 increasing in \((-2,0)\) and in \((1, \infty)\)
3 decreasing in \((-2,0)\) and in \((0,1)\)
4 decreasing in \((-\infty,-2)\) and in \((1, \infty)\)
Explanation:
(B) : Given : \(\mathrm{f}(\mathrm{x})=3 \mathrm{x}^{4}+4 \mathrm{x}^{3}-12 \mathrm{x}^{2}+12\) On differentiating both side with respect to \(x\), we get \(f^{\prime}(x)=12 x^{3}+12 x^{2}-24 x\) For \(\mathrm{f}(\mathrm{x})\) to be increasing \(\mathrm{f}^{\prime}(\mathrm{x})>0\) \(\Rightarrow 12 \mathrm{x}^{3}+12 \mathrm{x}^{2}-24 \mathrm{x}>0\) \(\Rightarrow 12 \mathrm{x}\left(\mathrm{x}^{2}+\mathrm{x}-2\right)>0\) \(\Rightarrow 12 \mathrm{x}(\mathrm{x}-1)(\mathrm{x}+2)>0\) \(\Rightarrow \mathrm{x}(\mathrm{x}-1)(\mathrm{x}+2)>0\) \(\Rightarrow-2\lt \mathrm{x}\lt 0\) or \(\mathrm{x}>1\) It means \(x \in(-2,0) \cup(1, \infty)\). Hence, \(\mathrm{f}(\mathrm{x})\) is increasing in \((-2,0)\) and \((1, \infty)\).
BITSAT-2019
Application of Derivatives
85284
The function \(f(x)=\tan x-4 x\) is strictly decreasing on
85285
If \(f(x)=\left(\frac{a^{2}-1}{a^{2}+1}\right) x^{3}-3 x+5\) is a decreasing function of \(x\) in \(R\) then the set of possible values of a (independent of \(x\) ) is
1 \((1, \infty)\)
2 \((-\infty,-1)\)
3 \([-1,1]\)
4 None of these
Explanation:
(C) : Given function, \(f(x)=\left(\frac{a^{2}-1}{a^{2}+1}\right) x^{3}-3 x+5\) \(f^{\prime}(x)=3\left(\frac{a^{2}-1}{a^{2}+1}\right) x^{2}-3\) \(\mathrm{f}^{\prime}(\mathrm{x})\lt 0\) for all \(\mathrm{x}\) if \(\mathrm{a}^{2}-1 \leq 0\) \(-1 \leq \mathrm{a} \leq 1\) Hence, \([-1,1]\)
BITSAT-2012
Application of Derivatives
85286
The function \(f(x)=(x(x-2))^{2}\) is increasing in the set
1 \((-\infty, 0) \cup(2, \infty)\)
2 \((-\infty, 1)\)
3 \((0,1) \cup(2, \infty)\)
4 \((1,2)\)
Explanation:
(C) : Given, \(\mathrm{f}(\mathrm{x})=(\mathrm{x}(\mathrm{x}-2))^{2}\) \(\mathrm{f}^{\prime}(\mathrm{x})=2[\mathrm{x}(\mathrm{x}-2)][1 .(\mathrm{x}-2)+\mathrm{x}(1-0)]\) \(\mathrm{f}^{\prime}(\mathrm{x})=2 \mathrm{x}(\mathrm{x}-2)(2 \mathrm{x}-2)\) \(\mathrm{f}^{\prime}(\mathrm{x})=4 \mathrm{x}(\mathrm{x}-2)(\mathrm{x}-1)\) For \(\mathrm{f}(\mathrm{x})\) as increasing, \(\mathrm{f}^{\prime}(\mathrm{x})>0\) So, \(4 x(x-1)(x-2)>0\) From the above figure required interval is, \((0,1) \cup(2, \infty)\)
Kerala CEE-2011
Application of Derivatives
85287
The interval in which the function \(2 x^{3}+15\) increases less rapidly than the function \(9 x^{2}-\) \(12 x\), is
1 \((-\infty, 1)\)
2 \((1,2)\)
3 \((2, \infty)\)
4 None of these
Explanation:
(B) : Let \(\mathrm{f}(\mathrm{x})=2 \mathrm{x}^{3}+15\) and \(\mathrm{g}(\mathrm{x})=9 \mathrm{x}^{2}-12 \mathrm{x}\) then \(\mathrm{f}^{\prime}(\mathrm{x})=6 \mathrm{x}^{2} \forall \mathrm{x} \in \mathrm{R}\) \(\therefore \mathrm{f}(\mathrm{x})\) is increasing function \(\forall \mathrm{x} \in \mathrm{R}\) \(\mathrm{g}^{\prime}(\mathrm{x})>0\) Also, \(18 \mathrm{x}-12>0\) \(x>\frac{2}{3}\) Thus, \(f(x)\) and \(g(x)\) both increases for \(x>\frac{2}{3}\) Let \(\mathrm{F}(\mathrm{x})=\mathrm{f}(\mathrm{x})-\mathrm{g}(\mathrm{x}), \mathrm{F}^{\prime}(\mathrm{x})\lt 0\) \((\because \mathrm{f}(\mathrm{x})\) increases less rapidly than the function \(\mathrm{g}(\mathrm{x}))\) \(6 \mathrm{x}^{2}-18 \mathrm{x}+12\lt 0\) \(6(\mathrm{x}-2)(\mathrm{x}-1)\lt 0\) \(1\lt \mathrm{x}\lt 2\)
85283
If \(f(x)=3 x^{4}+4 x^{3}-12 x^{2}+12\), then \(f(x)\) is
1 increasing in \((-\infty,-2)\) and in \((0,1)\)
2 increasing in \((-2,0)\) and in \((1, \infty)\)
3 decreasing in \((-2,0)\) and in \((0,1)\)
4 decreasing in \((-\infty,-2)\) and in \((1, \infty)\)
Explanation:
(B) : Given : \(\mathrm{f}(\mathrm{x})=3 \mathrm{x}^{4}+4 \mathrm{x}^{3}-12 \mathrm{x}^{2}+12\) On differentiating both side with respect to \(x\), we get \(f^{\prime}(x)=12 x^{3}+12 x^{2}-24 x\) For \(\mathrm{f}(\mathrm{x})\) to be increasing \(\mathrm{f}^{\prime}(\mathrm{x})>0\) \(\Rightarrow 12 \mathrm{x}^{3}+12 \mathrm{x}^{2}-24 \mathrm{x}>0\) \(\Rightarrow 12 \mathrm{x}\left(\mathrm{x}^{2}+\mathrm{x}-2\right)>0\) \(\Rightarrow 12 \mathrm{x}(\mathrm{x}-1)(\mathrm{x}+2)>0\) \(\Rightarrow \mathrm{x}(\mathrm{x}-1)(\mathrm{x}+2)>0\) \(\Rightarrow-2\lt \mathrm{x}\lt 0\) or \(\mathrm{x}>1\) It means \(x \in(-2,0) \cup(1, \infty)\). Hence, \(\mathrm{f}(\mathrm{x})\) is increasing in \((-2,0)\) and \((1, \infty)\).
BITSAT-2019
Application of Derivatives
85284
The function \(f(x)=\tan x-4 x\) is strictly decreasing on
85285
If \(f(x)=\left(\frac{a^{2}-1}{a^{2}+1}\right) x^{3}-3 x+5\) is a decreasing function of \(x\) in \(R\) then the set of possible values of a (independent of \(x\) ) is
1 \((1, \infty)\)
2 \((-\infty,-1)\)
3 \([-1,1]\)
4 None of these
Explanation:
(C) : Given function, \(f(x)=\left(\frac{a^{2}-1}{a^{2}+1}\right) x^{3}-3 x+5\) \(f^{\prime}(x)=3\left(\frac{a^{2}-1}{a^{2}+1}\right) x^{2}-3\) \(\mathrm{f}^{\prime}(\mathrm{x})\lt 0\) for all \(\mathrm{x}\) if \(\mathrm{a}^{2}-1 \leq 0\) \(-1 \leq \mathrm{a} \leq 1\) Hence, \([-1,1]\)
BITSAT-2012
Application of Derivatives
85286
The function \(f(x)=(x(x-2))^{2}\) is increasing in the set
1 \((-\infty, 0) \cup(2, \infty)\)
2 \((-\infty, 1)\)
3 \((0,1) \cup(2, \infty)\)
4 \((1,2)\)
Explanation:
(C) : Given, \(\mathrm{f}(\mathrm{x})=(\mathrm{x}(\mathrm{x}-2))^{2}\) \(\mathrm{f}^{\prime}(\mathrm{x})=2[\mathrm{x}(\mathrm{x}-2)][1 .(\mathrm{x}-2)+\mathrm{x}(1-0)]\) \(\mathrm{f}^{\prime}(\mathrm{x})=2 \mathrm{x}(\mathrm{x}-2)(2 \mathrm{x}-2)\) \(\mathrm{f}^{\prime}(\mathrm{x})=4 \mathrm{x}(\mathrm{x}-2)(\mathrm{x}-1)\) For \(\mathrm{f}(\mathrm{x})\) as increasing, \(\mathrm{f}^{\prime}(\mathrm{x})>0\) So, \(4 x(x-1)(x-2)>0\) From the above figure required interval is, \((0,1) \cup(2, \infty)\)
Kerala CEE-2011
Application of Derivatives
85287
The interval in which the function \(2 x^{3}+15\) increases less rapidly than the function \(9 x^{2}-\) \(12 x\), is
1 \((-\infty, 1)\)
2 \((1,2)\)
3 \((2, \infty)\)
4 None of these
Explanation:
(B) : Let \(\mathrm{f}(\mathrm{x})=2 \mathrm{x}^{3}+15\) and \(\mathrm{g}(\mathrm{x})=9 \mathrm{x}^{2}-12 \mathrm{x}\) then \(\mathrm{f}^{\prime}(\mathrm{x})=6 \mathrm{x}^{2} \forall \mathrm{x} \in \mathrm{R}\) \(\therefore \mathrm{f}(\mathrm{x})\) is increasing function \(\forall \mathrm{x} \in \mathrm{R}\) \(\mathrm{g}^{\prime}(\mathrm{x})>0\) Also, \(18 \mathrm{x}-12>0\) \(x>\frac{2}{3}\) Thus, \(f(x)\) and \(g(x)\) both increases for \(x>\frac{2}{3}\) Let \(\mathrm{F}(\mathrm{x})=\mathrm{f}(\mathrm{x})-\mathrm{g}(\mathrm{x}), \mathrm{F}^{\prime}(\mathrm{x})\lt 0\) \((\because \mathrm{f}(\mathrm{x})\) increases less rapidly than the function \(\mathrm{g}(\mathrm{x}))\) \(6 \mathrm{x}^{2}-18 \mathrm{x}+12\lt 0\) \(6(\mathrm{x}-2)(\mathrm{x}-1)\lt 0\) \(1\lt \mathrm{x}\lt 2\)
85283
If \(f(x)=3 x^{4}+4 x^{3}-12 x^{2}+12\), then \(f(x)\) is
1 increasing in \((-\infty,-2)\) and in \((0,1)\)
2 increasing in \((-2,0)\) and in \((1, \infty)\)
3 decreasing in \((-2,0)\) and in \((0,1)\)
4 decreasing in \((-\infty,-2)\) and in \((1, \infty)\)
Explanation:
(B) : Given : \(\mathrm{f}(\mathrm{x})=3 \mathrm{x}^{4}+4 \mathrm{x}^{3}-12 \mathrm{x}^{2}+12\) On differentiating both side with respect to \(x\), we get \(f^{\prime}(x)=12 x^{3}+12 x^{2}-24 x\) For \(\mathrm{f}(\mathrm{x})\) to be increasing \(\mathrm{f}^{\prime}(\mathrm{x})>0\) \(\Rightarrow 12 \mathrm{x}^{3}+12 \mathrm{x}^{2}-24 \mathrm{x}>0\) \(\Rightarrow 12 \mathrm{x}\left(\mathrm{x}^{2}+\mathrm{x}-2\right)>0\) \(\Rightarrow 12 \mathrm{x}(\mathrm{x}-1)(\mathrm{x}+2)>0\) \(\Rightarrow \mathrm{x}(\mathrm{x}-1)(\mathrm{x}+2)>0\) \(\Rightarrow-2\lt \mathrm{x}\lt 0\) or \(\mathrm{x}>1\) It means \(x \in(-2,0) \cup(1, \infty)\). Hence, \(\mathrm{f}(\mathrm{x})\) is increasing in \((-2,0)\) and \((1, \infty)\).
BITSAT-2019
Application of Derivatives
85284
The function \(f(x)=\tan x-4 x\) is strictly decreasing on
85285
If \(f(x)=\left(\frac{a^{2}-1}{a^{2}+1}\right) x^{3}-3 x+5\) is a decreasing function of \(x\) in \(R\) then the set of possible values of a (independent of \(x\) ) is
1 \((1, \infty)\)
2 \((-\infty,-1)\)
3 \([-1,1]\)
4 None of these
Explanation:
(C) : Given function, \(f(x)=\left(\frac{a^{2}-1}{a^{2}+1}\right) x^{3}-3 x+5\) \(f^{\prime}(x)=3\left(\frac{a^{2}-1}{a^{2}+1}\right) x^{2}-3\) \(\mathrm{f}^{\prime}(\mathrm{x})\lt 0\) for all \(\mathrm{x}\) if \(\mathrm{a}^{2}-1 \leq 0\) \(-1 \leq \mathrm{a} \leq 1\) Hence, \([-1,1]\)
BITSAT-2012
Application of Derivatives
85286
The function \(f(x)=(x(x-2))^{2}\) is increasing in the set
1 \((-\infty, 0) \cup(2, \infty)\)
2 \((-\infty, 1)\)
3 \((0,1) \cup(2, \infty)\)
4 \((1,2)\)
Explanation:
(C) : Given, \(\mathrm{f}(\mathrm{x})=(\mathrm{x}(\mathrm{x}-2))^{2}\) \(\mathrm{f}^{\prime}(\mathrm{x})=2[\mathrm{x}(\mathrm{x}-2)][1 .(\mathrm{x}-2)+\mathrm{x}(1-0)]\) \(\mathrm{f}^{\prime}(\mathrm{x})=2 \mathrm{x}(\mathrm{x}-2)(2 \mathrm{x}-2)\) \(\mathrm{f}^{\prime}(\mathrm{x})=4 \mathrm{x}(\mathrm{x}-2)(\mathrm{x}-1)\) For \(\mathrm{f}(\mathrm{x})\) as increasing, \(\mathrm{f}^{\prime}(\mathrm{x})>0\) So, \(4 x(x-1)(x-2)>0\) From the above figure required interval is, \((0,1) \cup(2, \infty)\)
Kerala CEE-2011
Application of Derivatives
85287
The interval in which the function \(2 x^{3}+15\) increases less rapidly than the function \(9 x^{2}-\) \(12 x\), is
1 \((-\infty, 1)\)
2 \((1,2)\)
3 \((2, \infty)\)
4 None of these
Explanation:
(B) : Let \(\mathrm{f}(\mathrm{x})=2 \mathrm{x}^{3}+15\) and \(\mathrm{g}(\mathrm{x})=9 \mathrm{x}^{2}-12 \mathrm{x}\) then \(\mathrm{f}^{\prime}(\mathrm{x})=6 \mathrm{x}^{2} \forall \mathrm{x} \in \mathrm{R}\) \(\therefore \mathrm{f}(\mathrm{x})\) is increasing function \(\forall \mathrm{x} \in \mathrm{R}\) \(\mathrm{g}^{\prime}(\mathrm{x})>0\) Also, \(18 \mathrm{x}-12>0\) \(x>\frac{2}{3}\) Thus, \(f(x)\) and \(g(x)\) both increases for \(x>\frac{2}{3}\) Let \(\mathrm{F}(\mathrm{x})=\mathrm{f}(\mathrm{x})-\mathrm{g}(\mathrm{x}), \mathrm{F}^{\prime}(\mathrm{x})\lt 0\) \((\because \mathrm{f}(\mathrm{x})\) increases less rapidly than the function \(\mathrm{g}(\mathrm{x}))\) \(6 \mathrm{x}^{2}-18 \mathrm{x}+12\lt 0\) \(6(\mathrm{x}-2)(\mathrm{x}-1)\lt 0\) \(1\lt \mathrm{x}\lt 2\)
