85269
The interval in which the function \(f(x)=x^{3}-6 x^{2}+9 x+10\) is increasing in
1 \((-\infty, 1) \cup(3, \infty)\)
2 \([1,3]\)
3 \((-\infty, 1] \cup[3, \infty)\)
4 \((-\infty,-1] \cup[3, \infty)\)
Explanation:
(C) : We know that, \(\mathrm{f}^{\prime}(\mathrm{x})\lt 0\) at each point in an interval 1 , then the function is said to be decreasing on 1 . Given, \(f^{\prime}(x)=x^{3}-6 x^{2}+9 x+10\) Differentiating, we get \(f^{\prime}(x)=3 x^{2}-12 x+9\) \(\mathrm{f}(\mathrm{x})\) is increasing function, \(\mathrm{f}^{\prime}(\mathrm{x}) \geq 0\) \(3 \mathrm{x}^{2}-12 \mathrm{x}+9 \geq 0\) \(\mathrm{x}^{2}-4 \mathrm{x}+3 \geq 0\) \((\mathrm{x}-3)(\mathrm{x}-1) \geq 0\) Hence, \(\mathrm{x} \in(-\infty, 1] \cup[3, \infty]\) The interval in which the function \(f(x)=x^{3}-6 x^{2}+9 x+\) 10 is increasing in \((-\infty, 1] \cup[3, \infty)\).
Karnataka CET-2019
Application of Derivatives
85270
The function \(f(x)=x^{2}+2 x-5\) is strictly increasing in the interval
1 \([-1, \infty)\)
2 \((-\infty,-1)\)
3 \((-\infty,-1)\)
4 \((-1, \infty)\)
Explanation:
(D) : Given, \(f(x)=x^{2}+2 x-5\) \(\therefore \mathrm{f}^{\prime}(\mathrm{x})=2 \mathrm{x}+2\) \(\mathrm{f}(\mathrm{x})\) is an increasing function, if \(\mathrm{f}^{\prime}(\mathrm{x})>0\) \(2 x+2>0\) \(2 x>-2\) \(\therefore \quad \mathrm{x}>-1\) Thus, \(f(x)\) is an increasing function for \(x>-1\) i.e, \((-1, \infty)\)
Karnataka CET-2017
Application of Derivatives
85271
The function \(f(x)=\frac{x}{3}+\frac{3}{x}\) decreases in the interva
1 \((-3,3)\)
2 \((-\infty, 3)\)
3 \((3, \infty)\)
4 \((-9,9)\)
Explanation:
(A) : Given that, \(f(x)=\frac{x}{3}+\frac{3}{x}\) On differentiating w.r.t to \(\mathrm{x}\) we get - \(\mathrm{f}^{\prime}(\mathrm{x})=\frac{1}{3}-\frac{3}{\mathrm{x}^{2}}\) Function is decreasing \(\mathrm{f}^{\prime}(\mathrm{x})\lt 0\) \(\frac{1}{3}-\frac{3}{\mathrm{x}^{2}}\lt 0 \Rightarrow \frac{1}{3}\lt \frac{3}{\mathrm{x}^{2}}\) \(\mathrm{x}^{2}\lt 9\) \(\mathrm{x} \in(-3,3)\) Which is the required interval in which, function \(f(x)\) decreases.
Karnataka CET-2013
Application of Derivatives
85272
The set of real values of \(x\) for which \(f(x)=\frac{x}{\log x}\) is increasing is
85269
The interval in which the function \(f(x)=x^{3}-6 x^{2}+9 x+10\) is increasing in
1 \((-\infty, 1) \cup(3, \infty)\)
2 \([1,3]\)
3 \((-\infty, 1] \cup[3, \infty)\)
4 \((-\infty,-1] \cup[3, \infty)\)
Explanation:
(C) : We know that, \(\mathrm{f}^{\prime}(\mathrm{x})\lt 0\) at each point in an interval 1 , then the function is said to be decreasing on 1 . Given, \(f^{\prime}(x)=x^{3}-6 x^{2}+9 x+10\) Differentiating, we get \(f^{\prime}(x)=3 x^{2}-12 x+9\) \(\mathrm{f}(\mathrm{x})\) is increasing function, \(\mathrm{f}^{\prime}(\mathrm{x}) \geq 0\) \(3 \mathrm{x}^{2}-12 \mathrm{x}+9 \geq 0\) \(\mathrm{x}^{2}-4 \mathrm{x}+3 \geq 0\) \((\mathrm{x}-3)(\mathrm{x}-1) \geq 0\) Hence, \(\mathrm{x} \in(-\infty, 1] \cup[3, \infty]\) The interval in which the function \(f(x)=x^{3}-6 x^{2}+9 x+\) 10 is increasing in \((-\infty, 1] \cup[3, \infty)\).
Karnataka CET-2019
Application of Derivatives
85270
The function \(f(x)=x^{2}+2 x-5\) is strictly increasing in the interval
1 \([-1, \infty)\)
2 \((-\infty,-1)\)
3 \((-\infty,-1)\)
4 \((-1, \infty)\)
Explanation:
(D) : Given, \(f(x)=x^{2}+2 x-5\) \(\therefore \mathrm{f}^{\prime}(\mathrm{x})=2 \mathrm{x}+2\) \(\mathrm{f}(\mathrm{x})\) is an increasing function, if \(\mathrm{f}^{\prime}(\mathrm{x})>0\) \(2 x+2>0\) \(2 x>-2\) \(\therefore \quad \mathrm{x}>-1\) Thus, \(f(x)\) is an increasing function for \(x>-1\) i.e, \((-1, \infty)\)
Karnataka CET-2017
Application of Derivatives
85271
The function \(f(x)=\frac{x}{3}+\frac{3}{x}\) decreases in the interva
1 \((-3,3)\)
2 \((-\infty, 3)\)
3 \((3, \infty)\)
4 \((-9,9)\)
Explanation:
(A) : Given that, \(f(x)=\frac{x}{3}+\frac{3}{x}\) On differentiating w.r.t to \(\mathrm{x}\) we get - \(\mathrm{f}^{\prime}(\mathrm{x})=\frac{1}{3}-\frac{3}{\mathrm{x}^{2}}\) Function is decreasing \(\mathrm{f}^{\prime}(\mathrm{x})\lt 0\) \(\frac{1}{3}-\frac{3}{\mathrm{x}^{2}}\lt 0 \Rightarrow \frac{1}{3}\lt \frac{3}{\mathrm{x}^{2}}\) \(\mathrm{x}^{2}\lt 9\) \(\mathrm{x} \in(-3,3)\) Which is the required interval in which, function \(f(x)\) decreases.
Karnataka CET-2013
Application of Derivatives
85272
The set of real values of \(x\) for which \(f(x)=\frac{x}{\log x}\) is increasing is
85269
The interval in which the function \(f(x)=x^{3}-6 x^{2}+9 x+10\) is increasing in
1 \((-\infty, 1) \cup(3, \infty)\)
2 \([1,3]\)
3 \((-\infty, 1] \cup[3, \infty)\)
4 \((-\infty,-1] \cup[3, \infty)\)
Explanation:
(C) : We know that, \(\mathrm{f}^{\prime}(\mathrm{x})\lt 0\) at each point in an interval 1 , then the function is said to be decreasing on 1 . Given, \(f^{\prime}(x)=x^{3}-6 x^{2}+9 x+10\) Differentiating, we get \(f^{\prime}(x)=3 x^{2}-12 x+9\) \(\mathrm{f}(\mathrm{x})\) is increasing function, \(\mathrm{f}^{\prime}(\mathrm{x}) \geq 0\) \(3 \mathrm{x}^{2}-12 \mathrm{x}+9 \geq 0\) \(\mathrm{x}^{2}-4 \mathrm{x}+3 \geq 0\) \((\mathrm{x}-3)(\mathrm{x}-1) \geq 0\) Hence, \(\mathrm{x} \in(-\infty, 1] \cup[3, \infty]\) The interval in which the function \(f(x)=x^{3}-6 x^{2}+9 x+\) 10 is increasing in \((-\infty, 1] \cup[3, \infty)\).
Karnataka CET-2019
Application of Derivatives
85270
The function \(f(x)=x^{2}+2 x-5\) is strictly increasing in the interval
1 \([-1, \infty)\)
2 \((-\infty,-1)\)
3 \((-\infty,-1)\)
4 \((-1, \infty)\)
Explanation:
(D) : Given, \(f(x)=x^{2}+2 x-5\) \(\therefore \mathrm{f}^{\prime}(\mathrm{x})=2 \mathrm{x}+2\) \(\mathrm{f}(\mathrm{x})\) is an increasing function, if \(\mathrm{f}^{\prime}(\mathrm{x})>0\) \(2 x+2>0\) \(2 x>-2\) \(\therefore \quad \mathrm{x}>-1\) Thus, \(f(x)\) is an increasing function for \(x>-1\) i.e, \((-1, \infty)\)
Karnataka CET-2017
Application of Derivatives
85271
The function \(f(x)=\frac{x}{3}+\frac{3}{x}\) decreases in the interva
1 \((-3,3)\)
2 \((-\infty, 3)\)
3 \((3, \infty)\)
4 \((-9,9)\)
Explanation:
(A) : Given that, \(f(x)=\frac{x}{3}+\frac{3}{x}\) On differentiating w.r.t to \(\mathrm{x}\) we get - \(\mathrm{f}^{\prime}(\mathrm{x})=\frac{1}{3}-\frac{3}{\mathrm{x}^{2}}\) Function is decreasing \(\mathrm{f}^{\prime}(\mathrm{x})\lt 0\) \(\frac{1}{3}-\frac{3}{\mathrm{x}^{2}}\lt 0 \Rightarrow \frac{1}{3}\lt \frac{3}{\mathrm{x}^{2}}\) \(\mathrm{x}^{2}\lt 9\) \(\mathrm{x} \in(-3,3)\) Which is the required interval in which, function \(f(x)\) decreases.
Karnataka CET-2013
Application of Derivatives
85272
The set of real values of \(x\) for which \(f(x)=\frac{x}{\log x}\) is increasing is
NEET Test Series from KOTA - 10 Papers In MS WORD
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Application of Derivatives
85269
The interval in which the function \(f(x)=x^{3}-6 x^{2}+9 x+10\) is increasing in
1 \((-\infty, 1) \cup(3, \infty)\)
2 \([1,3]\)
3 \((-\infty, 1] \cup[3, \infty)\)
4 \((-\infty,-1] \cup[3, \infty)\)
Explanation:
(C) : We know that, \(\mathrm{f}^{\prime}(\mathrm{x})\lt 0\) at each point in an interval 1 , then the function is said to be decreasing on 1 . Given, \(f^{\prime}(x)=x^{3}-6 x^{2}+9 x+10\) Differentiating, we get \(f^{\prime}(x)=3 x^{2}-12 x+9\) \(\mathrm{f}(\mathrm{x})\) is increasing function, \(\mathrm{f}^{\prime}(\mathrm{x}) \geq 0\) \(3 \mathrm{x}^{2}-12 \mathrm{x}+9 \geq 0\) \(\mathrm{x}^{2}-4 \mathrm{x}+3 \geq 0\) \((\mathrm{x}-3)(\mathrm{x}-1) \geq 0\) Hence, \(\mathrm{x} \in(-\infty, 1] \cup[3, \infty]\) The interval in which the function \(f(x)=x^{3}-6 x^{2}+9 x+\) 10 is increasing in \((-\infty, 1] \cup[3, \infty)\).
Karnataka CET-2019
Application of Derivatives
85270
The function \(f(x)=x^{2}+2 x-5\) is strictly increasing in the interval
1 \([-1, \infty)\)
2 \((-\infty,-1)\)
3 \((-\infty,-1)\)
4 \((-1, \infty)\)
Explanation:
(D) : Given, \(f(x)=x^{2}+2 x-5\) \(\therefore \mathrm{f}^{\prime}(\mathrm{x})=2 \mathrm{x}+2\) \(\mathrm{f}(\mathrm{x})\) is an increasing function, if \(\mathrm{f}^{\prime}(\mathrm{x})>0\) \(2 x+2>0\) \(2 x>-2\) \(\therefore \quad \mathrm{x}>-1\) Thus, \(f(x)\) is an increasing function for \(x>-1\) i.e, \((-1, \infty)\)
Karnataka CET-2017
Application of Derivatives
85271
The function \(f(x)=\frac{x}{3}+\frac{3}{x}\) decreases in the interva
1 \((-3,3)\)
2 \((-\infty, 3)\)
3 \((3, \infty)\)
4 \((-9,9)\)
Explanation:
(A) : Given that, \(f(x)=\frac{x}{3}+\frac{3}{x}\) On differentiating w.r.t to \(\mathrm{x}\) we get - \(\mathrm{f}^{\prime}(\mathrm{x})=\frac{1}{3}-\frac{3}{\mathrm{x}^{2}}\) Function is decreasing \(\mathrm{f}^{\prime}(\mathrm{x})\lt 0\) \(\frac{1}{3}-\frac{3}{\mathrm{x}^{2}}\lt 0 \Rightarrow \frac{1}{3}\lt \frac{3}{\mathrm{x}^{2}}\) \(\mathrm{x}^{2}\lt 9\) \(\mathrm{x} \in(-3,3)\) Which is the required interval in which, function \(f(x)\) decreases.
Karnataka CET-2013
Application of Derivatives
85272
The set of real values of \(x\) for which \(f(x)=\frac{x}{\log x}\) is increasing is
85269
The interval in which the function \(f(x)=x^{3}-6 x^{2}+9 x+10\) is increasing in
1 \((-\infty, 1) \cup(3, \infty)\)
2 \([1,3]\)
3 \((-\infty, 1] \cup[3, \infty)\)
4 \((-\infty,-1] \cup[3, \infty)\)
Explanation:
(C) : We know that, \(\mathrm{f}^{\prime}(\mathrm{x})\lt 0\) at each point in an interval 1 , then the function is said to be decreasing on 1 . Given, \(f^{\prime}(x)=x^{3}-6 x^{2}+9 x+10\) Differentiating, we get \(f^{\prime}(x)=3 x^{2}-12 x+9\) \(\mathrm{f}(\mathrm{x})\) is increasing function, \(\mathrm{f}^{\prime}(\mathrm{x}) \geq 0\) \(3 \mathrm{x}^{2}-12 \mathrm{x}+9 \geq 0\) \(\mathrm{x}^{2}-4 \mathrm{x}+3 \geq 0\) \((\mathrm{x}-3)(\mathrm{x}-1) \geq 0\) Hence, \(\mathrm{x} \in(-\infty, 1] \cup[3, \infty]\) The interval in which the function \(f(x)=x^{3}-6 x^{2}+9 x+\) 10 is increasing in \((-\infty, 1] \cup[3, \infty)\).
Karnataka CET-2019
Application of Derivatives
85270
The function \(f(x)=x^{2}+2 x-5\) is strictly increasing in the interval
1 \([-1, \infty)\)
2 \((-\infty,-1)\)
3 \((-\infty,-1)\)
4 \((-1, \infty)\)
Explanation:
(D) : Given, \(f(x)=x^{2}+2 x-5\) \(\therefore \mathrm{f}^{\prime}(\mathrm{x})=2 \mathrm{x}+2\) \(\mathrm{f}(\mathrm{x})\) is an increasing function, if \(\mathrm{f}^{\prime}(\mathrm{x})>0\) \(2 x+2>0\) \(2 x>-2\) \(\therefore \quad \mathrm{x}>-1\) Thus, \(f(x)\) is an increasing function for \(x>-1\) i.e, \((-1, \infty)\)
Karnataka CET-2017
Application of Derivatives
85271
The function \(f(x)=\frac{x}{3}+\frac{3}{x}\) decreases in the interva
1 \((-3,3)\)
2 \((-\infty, 3)\)
3 \((3, \infty)\)
4 \((-9,9)\)
Explanation:
(A) : Given that, \(f(x)=\frac{x}{3}+\frac{3}{x}\) On differentiating w.r.t to \(\mathrm{x}\) we get - \(\mathrm{f}^{\prime}(\mathrm{x})=\frac{1}{3}-\frac{3}{\mathrm{x}^{2}}\) Function is decreasing \(\mathrm{f}^{\prime}(\mathrm{x})\lt 0\) \(\frac{1}{3}-\frac{3}{\mathrm{x}^{2}}\lt 0 \Rightarrow \frac{1}{3}\lt \frac{3}{\mathrm{x}^{2}}\) \(\mathrm{x}^{2}\lt 9\) \(\mathrm{x} \in(-3,3)\) Which is the required interval in which, function \(f(x)\) decreases.
Karnataka CET-2013
Application of Derivatives
85272
The set of real values of \(x\) for which \(f(x)=\frac{x}{\log x}\) is increasing is