85698
Let \(f(\mathrm{x})=\mathbf{a x}^{3}+\mathbf{b x}^{2}+\mathbf{c x}+1\) have extrema at \(\mathrm{x}\) \(=\alpha, \beta\) such that \(\alpha \beta\lt 0\) and \(f(\alpha) f(\beta)\lt 0\). Then the equation \(f(x)=0\) has
1 three equal roots
2 one negative root if \(f(\alpha)\lt 0\) and \(f(\beta)>0\)
3 one positive root if \(f(\alpha)>0\) and \(f(\beta)\lt 0\)
4 None of these
Explanation:
(D) : Given, \(f(x)=a x^{3}+b x^{2}+c x+1\) Then, \(f(x)=3 a x^{2}+2 b x+c\) It have extreme at \(x=\alpha, \beta\) such that \(\alpha \beta\lt 0\) Then, \(f^{\prime}(\alpha)=3 a \alpha^{2}+2 b \alpha+c\) and, \(f^{\prime}(\beta)=3 \alpha \beta^{2}+2 b \beta+c\) And also \(\mathrm{f}(\alpha) \mathrm{f}(\beta)\lt 0\) \(\left(3 \alpha^{3}+b \alpha^{2}+c \alpha+1\right)\left(a \beta^{3}+b \beta^{2}+c \beta+1\right)\lt 0\) As \(\alpha \beta\lt 0\), there will be a root \(\gamma\) between \(\alpha\) and \(\beta\). So, \(\gamma\) is the third root and may be positive or negative. Hence, none of these is the correct answer.
AMU-2019
Application of Derivatives
85699
The minimum value of \(\sec \theta+\operatorname{cosec} \theta\) is
(A) : Find maximum value of \(f(x)=\sin x+\cos x\) We know that, \(-\sqrt{a^{2}+b^{2}} \leq a \sin x+b \cos x \leq \sqrt{a^{2}+b^{2}}\) \(\sin x+\cos x \leq \sqrt{1^{2}+1^{2}}\) \(\sin x+\cos x \leq \sqrt{2}\) Maximum value of \(\sin x+\cos x\) is \(\sqrt{2}\)
AMU-2002
Application of Derivatives
85701
If the minimum value of \(f(x)=\frac{5 x^{2}}{2}+\frac{\alpha}{x^{5}}, x>0\), is 14 then the value of \(\alpha\) is equal to :
85698
Let \(f(\mathrm{x})=\mathbf{a x}^{3}+\mathbf{b x}^{2}+\mathbf{c x}+1\) have extrema at \(\mathrm{x}\) \(=\alpha, \beta\) such that \(\alpha \beta\lt 0\) and \(f(\alpha) f(\beta)\lt 0\). Then the equation \(f(x)=0\) has
1 three equal roots
2 one negative root if \(f(\alpha)\lt 0\) and \(f(\beta)>0\)
3 one positive root if \(f(\alpha)>0\) and \(f(\beta)\lt 0\)
4 None of these
Explanation:
(D) : Given, \(f(x)=a x^{3}+b x^{2}+c x+1\) Then, \(f(x)=3 a x^{2}+2 b x+c\) It have extreme at \(x=\alpha, \beta\) such that \(\alpha \beta\lt 0\) Then, \(f^{\prime}(\alpha)=3 a \alpha^{2}+2 b \alpha+c\) and, \(f^{\prime}(\beta)=3 \alpha \beta^{2}+2 b \beta+c\) And also \(\mathrm{f}(\alpha) \mathrm{f}(\beta)\lt 0\) \(\left(3 \alpha^{3}+b \alpha^{2}+c \alpha+1\right)\left(a \beta^{3}+b \beta^{2}+c \beta+1\right)\lt 0\) As \(\alpha \beta\lt 0\), there will be a root \(\gamma\) between \(\alpha\) and \(\beta\). So, \(\gamma\) is the third root and may be positive or negative. Hence, none of these is the correct answer.
AMU-2019
Application of Derivatives
85699
The minimum value of \(\sec \theta+\operatorname{cosec} \theta\) is
(A) : Find maximum value of \(f(x)=\sin x+\cos x\) We know that, \(-\sqrt{a^{2}+b^{2}} \leq a \sin x+b \cos x \leq \sqrt{a^{2}+b^{2}}\) \(\sin x+\cos x \leq \sqrt{1^{2}+1^{2}}\) \(\sin x+\cos x \leq \sqrt{2}\) Maximum value of \(\sin x+\cos x\) is \(\sqrt{2}\)
AMU-2002
Application of Derivatives
85701
If the minimum value of \(f(x)=\frac{5 x^{2}}{2}+\frac{\alpha}{x^{5}}, x>0\), is 14 then the value of \(\alpha\) is equal to :
85698
Let \(f(\mathrm{x})=\mathbf{a x}^{3}+\mathbf{b x}^{2}+\mathbf{c x}+1\) have extrema at \(\mathrm{x}\) \(=\alpha, \beta\) such that \(\alpha \beta\lt 0\) and \(f(\alpha) f(\beta)\lt 0\). Then the equation \(f(x)=0\) has
1 three equal roots
2 one negative root if \(f(\alpha)\lt 0\) and \(f(\beta)>0\)
3 one positive root if \(f(\alpha)>0\) and \(f(\beta)\lt 0\)
4 None of these
Explanation:
(D) : Given, \(f(x)=a x^{3}+b x^{2}+c x+1\) Then, \(f(x)=3 a x^{2}+2 b x+c\) It have extreme at \(x=\alpha, \beta\) such that \(\alpha \beta\lt 0\) Then, \(f^{\prime}(\alpha)=3 a \alpha^{2}+2 b \alpha+c\) and, \(f^{\prime}(\beta)=3 \alpha \beta^{2}+2 b \beta+c\) And also \(\mathrm{f}(\alpha) \mathrm{f}(\beta)\lt 0\) \(\left(3 \alpha^{3}+b \alpha^{2}+c \alpha+1\right)\left(a \beta^{3}+b \beta^{2}+c \beta+1\right)\lt 0\) As \(\alpha \beta\lt 0\), there will be a root \(\gamma\) between \(\alpha\) and \(\beta\). So, \(\gamma\) is the third root and may be positive or negative. Hence, none of these is the correct answer.
AMU-2019
Application of Derivatives
85699
The minimum value of \(\sec \theta+\operatorname{cosec} \theta\) is
(A) : Find maximum value of \(f(x)=\sin x+\cos x\) We know that, \(-\sqrt{a^{2}+b^{2}} \leq a \sin x+b \cos x \leq \sqrt{a^{2}+b^{2}}\) \(\sin x+\cos x \leq \sqrt{1^{2}+1^{2}}\) \(\sin x+\cos x \leq \sqrt{2}\) Maximum value of \(\sin x+\cos x\) is \(\sqrt{2}\)
AMU-2002
Application of Derivatives
85701
If the minimum value of \(f(x)=\frac{5 x^{2}}{2}+\frac{\alpha}{x^{5}}, x>0\), is 14 then the value of \(\alpha\) is equal to :
85698
Let \(f(\mathrm{x})=\mathbf{a x}^{3}+\mathbf{b x}^{2}+\mathbf{c x}+1\) have extrema at \(\mathrm{x}\) \(=\alpha, \beta\) such that \(\alpha \beta\lt 0\) and \(f(\alpha) f(\beta)\lt 0\). Then the equation \(f(x)=0\) has
1 three equal roots
2 one negative root if \(f(\alpha)\lt 0\) and \(f(\beta)>0\)
3 one positive root if \(f(\alpha)>0\) and \(f(\beta)\lt 0\)
4 None of these
Explanation:
(D) : Given, \(f(x)=a x^{3}+b x^{2}+c x+1\) Then, \(f(x)=3 a x^{2}+2 b x+c\) It have extreme at \(x=\alpha, \beta\) such that \(\alpha \beta\lt 0\) Then, \(f^{\prime}(\alpha)=3 a \alpha^{2}+2 b \alpha+c\) and, \(f^{\prime}(\beta)=3 \alpha \beta^{2}+2 b \beta+c\) And also \(\mathrm{f}(\alpha) \mathrm{f}(\beta)\lt 0\) \(\left(3 \alpha^{3}+b \alpha^{2}+c \alpha+1\right)\left(a \beta^{3}+b \beta^{2}+c \beta+1\right)\lt 0\) As \(\alpha \beta\lt 0\), there will be a root \(\gamma\) between \(\alpha\) and \(\beta\). So, \(\gamma\) is the third root and may be positive or negative. Hence, none of these is the correct answer.
AMU-2019
Application of Derivatives
85699
The minimum value of \(\sec \theta+\operatorname{cosec} \theta\) is
(A) : Find maximum value of \(f(x)=\sin x+\cos x\) We know that, \(-\sqrt{a^{2}+b^{2}} \leq a \sin x+b \cos x \leq \sqrt{a^{2}+b^{2}}\) \(\sin x+\cos x \leq \sqrt{1^{2}+1^{2}}\) \(\sin x+\cos x \leq \sqrt{2}\) Maximum value of \(\sin x+\cos x\) is \(\sqrt{2}\)
AMU-2002
Application of Derivatives
85701
If the minimum value of \(f(x)=\frac{5 x^{2}}{2}+\frac{\alpha}{x^{5}}, x>0\), is 14 then the value of \(\alpha\) is equal to :
