85289
The set of all values of a for which the function \(f(x)=\left(a^{2}-3 a+2\right)\left(\cos ^{2} x / 4-\sin ^{2} x / 4\right)+(a-1) x+\sin 1\) does not possess critical points is
1 \([1, \infty)\)
2 \((0,1) \cup(1,4)\)
3 \((-2,4)\)
4 \((1,3) \cup(3,5)\)
Explanation:
(B) : Given, \(f(x)=\left(a^{2}-3 a+2\right)\left(\cos ^{2} \frac{x}{4}-\sin ^{2} \frac{x}{4}\right)+(a-1) x+\sin 1\) \(f(x)=(a-1)(a-2) \cos \frac{x}{2}+(a-1) x+\sin 1\) \(f^{\prime}(x)=\frac{-1}{2}(a-1)(a-2) \sin \frac{x}{2}+(a-1)\) \(f^{\prime}(x)=(a-1)\left[1-\frac{(a-2)}{2} \sin \frac{x}{2}\right]\) If \(f(x)\) does not possess critical points, then \(\mathrm{f}^{\prime}(\mathrm{x}) \neq 0\) for any \(\mathrm{x} \in \mathrm{R}\) \(a \neq 1\) and \(1-\left(\frac{a-2}{2}\right) \sin \frac{x}{2}=0\) must not have any solution in \(\mathrm{R}\) \(a \neq 1\) and \(\sin \frac{x}{2}=\frac{2}{a-2}\) is not solvable in \(R\). \(\mathrm{a} \neq 1\) and \(\left|\frac{2}{\mathrm{a}-2}\right|>1\) \([\) for \(\mathrm{a}=2, \mathrm{f}(\mathrm{x}) \cdot \mathrm{x}+\sin 1)\) \(\therefore\left[\mathrm{f}^{\prime}(\mathrm{x})=1 \neq 0\right]\) \(a \neq 1\) and \(|a-2|\lt 2\) \(a \neq 1\) and \(-2\lt a-2\lt 2\) \(\mathrm{a} \neq 1\) and \(0\lt \mathrm{a}\lt 4\) \(\mathrm{a} \in(0,1) \cup(1,4)\)
BITSAT-2016
Application of Derivatives
85290
If \(f(x)=x^{x}\), then \(f(x)\) is increasing in interval :
1 \([0, \mathrm{e}]\)
2 \(\left[0, \frac{1}{\mathrm{e}}\right]\)
3 \([0,1]\)
4 None of these
Explanation:
(B) : A function \(f(x)\) is said to be increasing function in \([a, b]\) if \(f^{\prime}(x)>0\) in \([a, b]\) Given \(f(x)=x^{x}\) Differentiate equation (i) \(\mathrm{f}^{\prime}(\mathrm{x})=\mathrm{x}^{\mathrm{x}}(1+\log \mathrm{x})\) Put, \(\mathrm{f}^{\prime}(\mathrm{x})=0\); \(0=\mathrm{x}^{\mathrm{x}}(1+\log \mathrm{x})\) \(\mathrm{x}=0, \log \mathrm{x}=-1\) \(\mathrm{x}=\mathrm{e}^{-1} \Rightarrow \mathrm{x}=\frac{1}{\mathrm{e}}, 0\) Now, in \(\left[0, \frac{1}{\mathrm{e}}\right], \mathrm{f}^{\prime}(\mathrm{x})>0\) \(\therefore \mathrm{f}(\mathrm{x})\) is increasing in interval \(\left[0, \frac{1}{\mathrm{e}}\right]\)
BITSAT-2017
Application of Derivatives
85291
If \(f(x)=x^{3}+b x^{2}+c x+d\) and \(0\lt b^{2}\lt c\), then in \((-\infty, \infty)\),
1 \(f(x)\) is a strictly increasing function
2 \(f(x)\) has local maxima
3 \(f(x)\) is a strictly decreasing function
4 \(f(x)\) is bounded
Explanation:
(A) : Given, \(f(x)=x^{3}+b x^{2}+c x+d \text { and } 0\lt b^{2}\lt c\) \(f^{\prime}(x)=3 x^{2}+2 b x+c\) Discriminate \(=b^{2}-4 a c=(2 b)^{2}-4 \times 3 \times c\) \(\quad=4 b^{2}-12 c=4\left(b^{2}-3 c\right)\lt 0\) \(\therefore f^{\prime}(x)>0 \forall x \in R\) Thus, \(\mathrm{f}(\mathrm{x})\) is strictly increasing function
VITEEE-2019
Application of Derivatives
85292
The interval in which the function \(f(x)=\frac{4 x^{2}+1}{x}\) is decreasing is :
1 \(\left(-\frac{1}{2}, \frac{1}{2}\right)\)
2 \(\left[-\frac{1}{2}, \frac{1}{2}\right]\)
3 \((-1,1)\)
4 \([-1,1]\)
Explanation:
(A) : Given, \(f(x)=\frac{4 x^{2}+1}{x}=4 x+\frac{1}{x}\) \(f^{\prime}(x)=4-\frac{1}{x^{2}}\) \(\mathrm{f}(\mathrm{x})\) will be decreasing \(\mathrm{f}^{\prime}(\mathrm{x})\lt 0\) Thus, \(\quad 4-\frac{1}{\mathrm{x}^{2}}\lt 0 \Rightarrow \frac{-1}{\mathrm{x}^{2}}\lt -4\) \(\frac{1}{x^{2}}>4\) When multiple by (-) then inequality sign can be changed, \(\frac{-1}{2}\lt x\lt \frac{1}{2}\) Thus, interval in which \(\mathrm{f}(\mathrm{x})\) is decreasing, \(\text { is }\left(-\frac{1}{2}, \frac{1}{2}\right) \text {. }\)
85289
The set of all values of a for which the function \(f(x)=\left(a^{2}-3 a+2\right)\left(\cos ^{2} x / 4-\sin ^{2} x / 4\right)+(a-1) x+\sin 1\) does not possess critical points is
1 \([1, \infty)\)
2 \((0,1) \cup(1,4)\)
3 \((-2,4)\)
4 \((1,3) \cup(3,5)\)
Explanation:
(B) : Given, \(f(x)=\left(a^{2}-3 a+2\right)\left(\cos ^{2} \frac{x}{4}-\sin ^{2} \frac{x}{4}\right)+(a-1) x+\sin 1\) \(f(x)=(a-1)(a-2) \cos \frac{x}{2}+(a-1) x+\sin 1\) \(f^{\prime}(x)=\frac{-1}{2}(a-1)(a-2) \sin \frac{x}{2}+(a-1)\) \(f^{\prime}(x)=(a-1)\left[1-\frac{(a-2)}{2} \sin \frac{x}{2}\right]\) If \(f(x)\) does not possess critical points, then \(\mathrm{f}^{\prime}(\mathrm{x}) \neq 0\) for any \(\mathrm{x} \in \mathrm{R}\) \(a \neq 1\) and \(1-\left(\frac{a-2}{2}\right) \sin \frac{x}{2}=0\) must not have any solution in \(\mathrm{R}\) \(a \neq 1\) and \(\sin \frac{x}{2}=\frac{2}{a-2}\) is not solvable in \(R\). \(\mathrm{a} \neq 1\) and \(\left|\frac{2}{\mathrm{a}-2}\right|>1\) \([\) for \(\mathrm{a}=2, \mathrm{f}(\mathrm{x}) \cdot \mathrm{x}+\sin 1)\) \(\therefore\left[\mathrm{f}^{\prime}(\mathrm{x})=1 \neq 0\right]\) \(a \neq 1\) and \(|a-2|\lt 2\) \(a \neq 1\) and \(-2\lt a-2\lt 2\) \(\mathrm{a} \neq 1\) and \(0\lt \mathrm{a}\lt 4\) \(\mathrm{a} \in(0,1) \cup(1,4)\)
BITSAT-2016
Application of Derivatives
85290
If \(f(x)=x^{x}\), then \(f(x)\) is increasing in interval :
1 \([0, \mathrm{e}]\)
2 \(\left[0, \frac{1}{\mathrm{e}}\right]\)
3 \([0,1]\)
4 None of these
Explanation:
(B) : A function \(f(x)\) is said to be increasing function in \([a, b]\) if \(f^{\prime}(x)>0\) in \([a, b]\) Given \(f(x)=x^{x}\) Differentiate equation (i) \(\mathrm{f}^{\prime}(\mathrm{x})=\mathrm{x}^{\mathrm{x}}(1+\log \mathrm{x})\) Put, \(\mathrm{f}^{\prime}(\mathrm{x})=0\); \(0=\mathrm{x}^{\mathrm{x}}(1+\log \mathrm{x})\) \(\mathrm{x}=0, \log \mathrm{x}=-1\) \(\mathrm{x}=\mathrm{e}^{-1} \Rightarrow \mathrm{x}=\frac{1}{\mathrm{e}}, 0\) Now, in \(\left[0, \frac{1}{\mathrm{e}}\right], \mathrm{f}^{\prime}(\mathrm{x})>0\) \(\therefore \mathrm{f}(\mathrm{x})\) is increasing in interval \(\left[0, \frac{1}{\mathrm{e}}\right]\)
BITSAT-2017
Application of Derivatives
85291
If \(f(x)=x^{3}+b x^{2}+c x+d\) and \(0\lt b^{2}\lt c\), then in \((-\infty, \infty)\),
1 \(f(x)\) is a strictly increasing function
2 \(f(x)\) has local maxima
3 \(f(x)\) is a strictly decreasing function
4 \(f(x)\) is bounded
Explanation:
(A) : Given, \(f(x)=x^{3}+b x^{2}+c x+d \text { and } 0\lt b^{2}\lt c\) \(f^{\prime}(x)=3 x^{2}+2 b x+c\) Discriminate \(=b^{2}-4 a c=(2 b)^{2}-4 \times 3 \times c\) \(\quad=4 b^{2}-12 c=4\left(b^{2}-3 c\right)\lt 0\) \(\therefore f^{\prime}(x)>0 \forall x \in R\) Thus, \(\mathrm{f}(\mathrm{x})\) is strictly increasing function
VITEEE-2019
Application of Derivatives
85292
The interval in which the function \(f(x)=\frac{4 x^{2}+1}{x}\) is decreasing is :
1 \(\left(-\frac{1}{2}, \frac{1}{2}\right)\)
2 \(\left[-\frac{1}{2}, \frac{1}{2}\right]\)
3 \((-1,1)\)
4 \([-1,1]\)
Explanation:
(A) : Given, \(f(x)=\frac{4 x^{2}+1}{x}=4 x+\frac{1}{x}\) \(f^{\prime}(x)=4-\frac{1}{x^{2}}\) \(\mathrm{f}(\mathrm{x})\) will be decreasing \(\mathrm{f}^{\prime}(\mathrm{x})\lt 0\) Thus, \(\quad 4-\frac{1}{\mathrm{x}^{2}}\lt 0 \Rightarrow \frac{-1}{\mathrm{x}^{2}}\lt -4\) \(\frac{1}{x^{2}}>4\) When multiple by (-) then inequality sign can be changed, \(\frac{-1}{2}\lt x\lt \frac{1}{2}\) Thus, interval in which \(\mathrm{f}(\mathrm{x})\) is decreasing, \(\text { is }\left(-\frac{1}{2}, \frac{1}{2}\right) \text {. }\)
85289
The set of all values of a for which the function \(f(x)=\left(a^{2}-3 a+2\right)\left(\cos ^{2} x / 4-\sin ^{2} x / 4\right)+(a-1) x+\sin 1\) does not possess critical points is
1 \([1, \infty)\)
2 \((0,1) \cup(1,4)\)
3 \((-2,4)\)
4 \((1,3) \cup(3,5)\)
Explanation:
(B) : Given, \(f(x)=\left(a^{2}-3 a+2\right)\left(\cos ^{2} \frac{x}{4}-\sin ^{2} \frac{x}{4}\right)+(a-1) x+\sin 1\) \(f(x)=(a-1)(a-2) \cos \frac{x}{2}+(a-1) x+\sin 1\) \(f^{\prime}(x)=\frac{-1}{2}(a-1)(a-2) \sin \frac{x}{2}+(a-1)\) \(f^{\prime}(x)=(a-1)\left[1-\frac{(a-2)}{2} \sin \frac{x}{2}\right]\) If \(f(x)\) does not possess critical points, then \(\mathrm{f}^{\prime}(\mathrm{x}) \neq 0\) for any \(\mathrm{x} \in \mathrm{R}\) \(a \neq 1\) and \(1-\left(\frac{a-2}{2}\right) \sin \frac{x}{2}=0\) must not have any solution in \(\mathrm{R}\) \(a \neq 1\) and \(\sin \frac{x}{2}=\frac{2}{a-2}\) is not solvable in \(R\). \(\mathrm{a} \neq 1\) and \(\left|\frac{2}{\mathrm{a}-2}\right|>1\) \([\) for \(\mathrm{a}=2, \mathrm{f}(\mathrm{x}) \cdot \mathrm{x}+\sin 1)\) \(\therefore\left[\mathrm{f}^{\prime}(\mathrm{x})=1 \neq 0\right]\) \(a \neq 1\) and \(|a-2|\lt 2\) \(a \neq 1\) and \(-2\lt a-2\lt 2\) \(\mathrm{a} \neq 1\) and \(0\lt \mathrm{a}\lt 4\) \(\mathrm{a} \in(0,1) \cup(1,4)\)
BITSAT-2016
Application of Derivatives
85290
If \(f(x)=x^{x}\), then \(f(x)\) is increasing in interval :
1 \([0, \mathrm{e}]\)
2 \(\left[0, \frac{1}{\mathrm{e}}\right]\)
3 \([0,1]\)
4 None of these
Explanation:
(B) : A function \(f(x)\) is said to be increasing function in \([a, b]\) if \(f^{\prime}(x)>0\) in \([a, b]\) Given \(f(x)=x^{x}\) Differentiate equation (i) \(\mathrm{f}^{\prime}(\mathrm{x})=\mathrm{x}^{\mathrm{x}}(1+\log \mathrm{x})\) Put, \(\mathrm{f}^{\prime}(\mathrm{x})=0\); \(0=\mathrm{x}^{\mathrm{x}}(1+\log \mathrm{x})\) \(\mathrm{x}=0, \log \mathrm{x}=-1\) \(\mathrm{x}=\mathrm{e}^{-1} \Rightarrow \mathrm{x}=\frac{1}{\mathrm{e}}, 0\) Now, in \(\left[0, \frac{1}{\mathrm{e}}\right], \mathrm{f}^{\prime}(\mathrm{x})>0\) \(\therefore \mathrm{f}(\mathrm{x})\) is increasing in interval \(\left[0, \frac{1}{\mathrm{e}}\right]\)
BITSAT-2017
Application of Derivatives
85291
If \(f(x)=x^{3}+b x^{2}+c x+d\) and \(0\lt b^{2}\lt c\), then in \((-\infty, \infty)\),
1 \(f(x)\) is a strictly increasing function
2 \(f(x)\) has local maxima
3 \(f(x)\) is a strictly decreasing function
4 \(f(x)\) is bounded
Explanation:
(A) : Given, \(f(x)=x^{3}+b x^{2}+c x+d \text { and } 0\lt b^{2}\lt c\) \(f^{\prime}(x)=3 x^{2}+2 b x+c\) Discriminate \(=b^{2}-4 a c=(2 b)^{2}-4 \times 3 \times c\) \(\quad=4 b^{2}-12 c=4\left(b^{2}-3 c\right)\lt 0\) \(\therefore f^{\prime}(x)>0 \forall x \in R\) Thus, \(\mathrm{f}(\mathrm{x})\) is strictly increasing function
VITEEE-2019
Application of Derivatives
85292
The interval in which the function \(f(x)=\frac{4 x^{2}+1}{x}\) is decreasing is :
1 \(\left(-\frac{1}{2}, \frac{1}{2}\right)\)
2 \(\left[-\frac{1}{2}, \frac{1}{2}\right]\)
3 \((-1,1)\)
4 \([-1,1]\)
Explanation:
(A) : Given, \(f(x)=\frac{4 x^{2}+1}{x}=4 x+\frac{1}{x}\) \(f^{\prime}(x)=4-\frac{1}{x^{2}}\) \(\mathrm{f}(\mathrm{x})\) will be decreasing \(\mathrm{f}^{\prime}(\mathrm{x})\lt 0\) Thus, \(\quad 4-\frac{1}{\mathrm{x}^{2}}\lt 0 \Rightarrow \frac{-1}{\mathrm{x}^{2}}\lt -4\) \(\frac{1}{x^{2}}>4\) When multiple by (-) then inequality sign can be changed, \(\frac{-1}{2}\lt x\lt \frac{1}{2}\) Thus, interval in which \(\mathrm{f}(\mathrm{x})\) is decreasing, \(\text { is }\left(-\frac{1}{2}, \frac{1}{2}\right) \text {. }\)
NEET Test Series from KOTA - 10 Papers In MS WORD
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Application of Derivatives
85289
The set of all values of a for which the function \(f(x)=\left(a^{2}-3 a+2\right)\left(\cos ^{2} x / 4-\sin ^{2} x / 4\right)+(a-1) x+\sin 1\) does not possess critical points is
1 \([1, \infty)\)
2 \((0,1) \cup(1,4)\)
3 \((-2,4)\)
4 \((1,3) \cup(3,5)\)
Explanation:
(B) : Given, \(f(x)=\left(a^{2}-3 a+2\right)\left(\cos ^{2} \frac{x}{4}-\sin ^{2} \frac{x}{4}\right)+(a-1) x+\sin 1\) \(f(x)=(a-1)(a-2) \cos \frac{x}{2}+(a-1) x+\sin 1\) \(f^{\prime}(x)=\frac{-1}{2}(a-1)(a-2) \sin \frac{x}{2}+(a-1)\) \(f^{\prime}(x)=(a-1)\left[1-\frac{(a-2)}{2} \sin \frac{x}{2}\right]\) If \(f(x)\) does not possess critical points, then \(\mathrm{f}^{\prime}(\mathrm{x}) \neq 0\) for any \(\mathrm{x} \in \mathrm{R}\) \(a \neq 1\) and \(1-\left(\frac{a-2}{2}\right) \sin \frac{x}{2}=0\) must not have any solution in \(\mathrm{R}\) \(a \neq 1\) and \(\sin \frac{x}{2}=\frac{2}{a-2}\) is not solvable in \(R\). \(\mathrm{a} \neq 1\) and \(\left|\frac{2}{\mathrm{a}-2}\right|>1\) \([\) for \(\mathrm{a}=2, \mathrm{f}(\mathrm{x}) \cdot \mathrm{x}+\sin 1)\) \(\therefore\left[\mathrm{f}^{\prime}(\mathrm{x})=1 \neq 0\right]\) \(a \neq 1\) and \(|a-2|\lt 2\) \(a \neq 1\) and \(-2\lt a-2\lt 2\) \(\mathrm{a} \neq 1\) and \(0\lt \mathrm{a}\lt 4\) \(\mathrm{a} \in(0,1) \cup(1,4)\)
BITSAT-2016
Application of Derivatives
85290
If \(f(x)=x^{x}\), then \(f(x)\) is increasing in interval :
1 \([0, \mathrm{e}]\)
2 \(\left[0, \frac{1}{\mathrm{e}}\right]\)
3 \([0,1]\)
4 None of these
Explanation:
(B) : A function \(f(x)\) is said to be increasing function in \([a, b]\) if \(f^{\prime}(x)>0\) in \([a, b]\) Given \(f(x)=x^{x}\) Differentiate equation (i) \(\mathrm{f}^{\prime}(\mathrm{x})=\mathrm{x}^{\mathrm{x}}(1+\log \mathrm{x})\) Put, \(\mathrm{f}^{\prime}(\mathrm{x})=0\); \(0=\mathrm{x}^{\mathrm{x}}(1+\log \mathrm{x})\) \(\mathrm{x}=0, \log \mathrm{x}=-1\) \(\mathrm{x}=\mathrm{e}^{-1} \Rightarrow \mathrm{x}=\frac{1}{\mathrm{e}}, 0\) Now, in \(\left[0, \frac{1}{\mathrm{e}}\right], \mathrm{f}^{\prime}(\mathrm{x})>0\) \(\therefore \mathrm{f}(\mathrm{x})\) is increasing in interval \(\left[0, \frac{1}{\mathrm{e}}\right]\)
BITSAT-2017
Application of Derivatives
85291
If \(f(x)=x^{3}+b x^{2}+c x+d\) and \(0\lt b^{2}\lt c\), then in \((-\infty, \infty)\),
1 \(f(x)\) is a strictly increasing function
2 \(f(x)\) has local maxima
3 \(f(x)\) is a strictly decreasing function
4 \(f(x)\) is bounded
Explanation:
(A) : Given, \(f(x)=x^{3}+b x^{2}+c x+d \text { and } 0\lt b^{2}\lt c\) \(f^{\prime}(x)=3 x^{2}+2 b x+c\) Discriminate \(=b^{2}-4 a c=(2 b)^{2}-4 \times 3 \times c\) \(\quad=4 b^{2}-12 c=4\left(b^{2}-3 c\right)\lt 0\) \(\therefore f^{\prime}(x)>0 \forall x \in R\) Thus, \(\mathrm{f}(\mathrm{x})\) is strictly increasing function
VITEEE-2019
Application of Derivatives
85292
The interval in which the function \(f(x)=\frac{4 x^{2}+1}{x}\) is decreasing is :
1 \(\left(-\frac{1}{2}, \frac{1}{2}\right)\)
2 \(\left[-\frac{1}{2}, \frac{1}{2}\right]\)
3 \((-1,1)\)
4 \([-1,1]\)
Explanation:
(A) : Given, \(f(x)=\frac{4 x^{2}+1}{x}=4 x+\frac{1}{x}\) \(f^{\prime}(x)=4-\frac{1}{x^{2}}\) \(\mathrm{f}(\mathrm{x})\) will be decreasing \(\mathrm{f}^{\prime}(\mathrm{x})\lt 0\) Thus, \(\quad 4-\frac{1}{\mathrm{x}^{2}}\lt 0 \Rightarrow \frac{-1}{\mathrm{x}^{2}}\lt -4\) \(\frac{1}{x^{2}}>4\) When multiple by (-) then inequality sign can be changed, \(\frac{-1}{2}\lt x\lt \frac{1}{2}\) Thus, interval in which \(\mathrm{f}(\mathrm{x})\) is decreasing, \(\text { is }\left(-\frac{1}{2}, \frac{1}{2}\right) \text {. }\)
