85654
If two sides of a triangle are given, then the area of the triangle will be maximum, if the angle between the given sides is
1 \(\frac{\pi}{3}\)
2 \(\frac{\pi}{4}\)
3 \(\frac{\pi}{6}\)
4 \(\frac{\pi}{2}\)
Explanation:
(D) : Let the sides be a, b, c and angles be A, B, C. Area \((\mathrm{A})=\frac{1}{2} \mathrm{absin} \mathrm{C}\) \(\frac{\mathrm{dA}}{\mathrm{dC}}=\frac{1}{2} \mathrm{ab} \cos \mathrm{C}\) A is maximum, when \(\frac{\mathrm{dA}}{\mathrm{dC}}=0\) \(\frac{1}{2} a b \cos C=0\) \(\cos C=0\) \(\cos C=\cos \frac{\pi}{2}\) \(C=\frac{\pi}{2}\)
UPSEE-2015
Application of Derivatives
85655
The maximum and minimum value of \(6 \sin x\) \(\cos x+4 \cos 2 x\) are respectively
1 5,5
2 \(-5,5\)
3 \(5,-5\)
4 None of these
Explanation:
(C) : Given, \(f(x)=6 \sin x \cos x+4 \cos 2 x\) \(=3 \cdot 2 \sin x \cos x+4 \cos 2 x\) \(=3 \sin 2 \mathrm{x}+4 \cos 2 \mathrm{x}\) \(=3 \sin 2 \mathrm{x}+4 \cos 2 \mathrm{x}\) We know that \(-\sqrt{a^{2}+b^{2}} \leq a \sin x+b \cos x \leq \sqrt{a^{2}+b^{2}}\) Comparing, \(\mathrm{a}=3\) and \(\mathrm{b}=4\) \(-\sqrt{3^{2}+4^{2}} \leq 3 \sin 2 x+4 \cos 2 x \leq \sqrt{3^{2}+4^{2}}\) \(-5 \leq 3 \sin 2 x+4 \cos 2 x \leq 5\)
UPSEE-2014
Application of Derivatives
85656
Let \(f(x)=x(x-1)^{2}\), the point at which \(f(x)\) assumes maximum and minimum are respectively
1 \(\frac{1}{3}, 1\)
2 \(1, \frac{1}{3}\)
3 3,1
4 None of these
Explanation:
(A) : \(\mathrm{f}(\mathrm{x})=\mathrm{x}(\mathrm{x}-1)^{2}\) \(=x\left(x^{2}+1-2 x\right)=x^{3}+x-2 x^{2}\) \(\mathrm{f}^{\prime}(\mathrm{x})=3 \mathrm{x}^{2}+1-4 \mathrm{x}\) For maxima or minima, \(\mathrm{f}^{\prime}(\mathrm{x})=0\) \(3 \mathrm{x}^{2}-4 \mathrm{x}+1=0\) \((\mathrm{x}-1)(3 \mathrm{x}-1)=0\) \(\mathrm{x}=1, \mathrm{x}=\frac{1}{3}\) Now, \(\quad \mathrm{f}^{\prime \prime}(\mathrm{x})=6 \mathrm{x}-4\) At \(\quad \mathrm{x}=1, \quad \mathrm{f}^{\prime \prime}(1)=6-4=2>0\) (minima) At \(\quad \mathrm{x}=\frac{1}{3} \quad \mathrm{f}^{\prime \prime}\left(\frac{1}{3}\right)=6 \times \frac{1}{3}-4=-2\lt 0\) (Maxima) \((\max , \min )=\frac{1}{3}, 1\)
UPSEE-2014
Application of Derivatives
85657
The value of \(\sin \theta+\cos \theta\) will be greatest, when
85654
If two sides of a triangle are given, then the area of the triangle will be maximum, if the angle between the given sides is
1 \(\frac{\pi}{3}\)
2 \(\frac{\pi}{4}\)
3 \(\frac{\pi}{6}\)
4 \(\frac{\pi}{2}\)
Explanation:
(D) : Let the sides be a, b, c and angles be A, B, C. Area \((\mathrm{A})=\frac{1}{2} \mathrm{absin} \mathrm{C}\) \(\frac{\mathrm{dA}}{\mathrm{dC}}=\frac{1}{2} \mathrm{ab} \cos \mathrm{C}\) A is maximum, when \(\frac{\mathrm{dA}}{\mathrm{dC}}=0\) \(\frac{1}{2} a b \cos C=0\) \(\cos C=0\) \(\cos C=\cos \frac{\pi}{2}\) \(C=\frac{\pi}{2}\)
UPSEE-2015
Application of Derivatives
85655
The maximum and minimum value of \(6 \sin x\) \(\cos x+4 \cos 2 x\) are respectively
1 5,5
2 \(-5,5\)
3 \(5,-5\)
4 None of these
Explanation:
(C) : Given, \(f(x)=6 \sin x \cos x+4 \cos 2 x\) \(=3 \cdot 2 \sin x \cos x+4 \cos 2 x\) \(=3 \sin 2 \mathrm{x}+4 \cos 2 \mathrm{x}\) \(=3 \sin 2 \mathrm{x}+4 \cos 2 \mathrm{x}\) We know that \(-\sqrt{a^{2}+b^{2}} \leq a \sin x+b \cos x \leq \sqrt{a^{2}+b^{2}}\) Comparing, \(\mathrm{a}=3\) and \(\mathrm{b}=4\) \(-\sqrt{3^{2}+4^{2}} \leq 3 \sin 2 x+4 \cos 2 x \leq \sqrt{3^{2}+4^{2}}\) \(-5 \leq 3 \sin 2 x+4 \cos 2 x \leq 5\)
UPSEE-2014
Application of Derivatives
85656
Let \(f(x)=x(x-1)^{2}\), the point at which \(f(x)\) assumes maximum and minimum are respectively
1 \(\frac{1}{3}, 1\)
2 \(1, \frac{1}{3}\)
3 3,1
4 None of these
Explanation:
(A) : \(\mathrm{f}(\mathrm{x})=\mathrm{x}(\mathrm{x}-1)^{2}\) \(=x\left(x^{2}+1-2 x\right)=x^{3}+x-2 x^{2}\) \(\mathrm{f}^{\prime}(\mathrm{x})=3 \mathrm{x}^{2}+1-4 \mathrm{x}\) For maxima or minima, \(\mathrm{f}^{\prime}(\mathrm{x})=0\) \(3 \mathrm{x}^{2}-4 \mathrm{x}+1=0\) \((\mathrm{x}-1)(3 \mathrm{x}-1)=0\) \(\mathrm{x}=1, \mathrm{x}=\frac{1}{3}\) Now, \(\quad \mathrm{f}^{\prime \prime}(\mathrm{x})=6 \mathrm{x}-4\) At \(\quad \mathrm{x}=1, \quad \mathrm{f}^{\prime \prime}(1)=6-4=2>0\) (minima) At \(\quad \mathrm{x}=\frac{1}{3} \quad \mathrm{f}^{\prime \prime}\left(\frac{1}{3}\right)=6 \times \frac{1}{3}-4=-2\lt 0\) (Maxima) \((\max , \min )=\frac{1}{3}, 1\)
UPSEE-2014
Application of Derivatives
85657
The value of \(\sin \theta+\cos \theta\) will be greatest, when
85654
If two sides of a triangle are given, then the area of the triangle will be maximum, if the angle between the given sides is
1 \(\frac{\pi}{3}\)
2 \(\frac{\pi}{4}\)
3 \(\frac{\pi}{6}\)
4 \(\frac{\pi}{2}\)
Explanation:
(D) : Let the sides be a, b, c and angles be A, B, C. Area \((\mathrm{A})=\frac{1}{2} \mathrm{absin} \mathrm{C}\) \(\frac{\mathrm{dA}}{\mathrm{dC}}=\frac{1}{2} \mathrm{ab} \cos \mathrm{C}\) A is maximum, when \(\frac{\mathrm{dA}}{\mathrm{dC}}=0\) \(\frac{1}{2} a b \cos C=0\) \(\cos C=0\) \(\cos C=\cos \frac{\pi}{2}\) \(C=\frac{\pi}{2}\)
UPSEE-2015
Application of Derivatives
85655
The maximum and minimum value of \(6 \sin x\) \(\cos x+4 \cos 2 x\) are respectively
1 5,5
2 \(-5,5\)
3 \(5,-5\)
4 None of these
Explanation:
(C) : Given, \(f(x)=6 \sin x \cos x+4 \cos 2 x\) \(=3 \cdot 2 \sin x \cos x+4 \cos 2 x\) \(=3 \sin 2 \mathrm{x}+4 \cos 2 \mathrm{x}\) \(=3 \sin 2 \mathrm{x}+4 \cos 2 \mathrm{x}\) We know that \(-\sqrt{a^{2}+b^{2}} \leq a \sin x+b \cos x \leq \sqrt{a^{2}+b^{2}}\) Comparing, \(\mathrm{a}=3\) and \(\mathrm{b}=4\) \(-\sqrt{3^{2}+4^{2}} \leq 3 \sin 2 x+4 \cos 2 x \leq \sqrt{3^{2}+4^{2}}\) \(-5 \leq 3 \sin 2 x+4 \cos 2 x \leq 5\)
UPSEE-2014
Application of Derivatives
85656
Let \(f(x)=x(x-1)^{2}\), the point at which \(f(x)\) assumes maximum and minimum are respectively
1 \(\frac{1}{3}, 1\)
2 \(1, \frac{1}{3}\)
3 3,1
4 None of these
Explanation:
(A) : \(\mathrm{f}(\mathrm{x})=\mathrm{x}(\mathrm{x}-1)^{2}\) \(=x\left(x^{2}+1-2 x\right)=x^{3}+x-2 x^{2}\) \(\mathrm{f}^{\prime}(\mathrm{x})=3 \mathrm{x}^{2}+1-4 \mathrm{x}\) For maxima or minima, \(\mathrm{f}^{\prime}(\mathrm{x})=0\) \(3 \mathrm{x}^{2}-4 \mathrm{x}+1=0\) \((\mathrm{x}-1)(3 \mathrm{x}-1)=0\) \(\mathrm{x}=1, \mathrm{x}=\frac{1}{3}\) Now, \(\quad \mathrm{f}^{\prime \prime}(\mathrm{x})=6 \mathrm{x}-4\) At \(\quad \mathrm{x}=1, \quad \mathrm{f}^{\prime \prime}(1)=6-4=2>0\) (minima) At \(\quad \mathrm{x}=\frac{1}{3} \quad \mathrm{f}^{\prime \prime}\left(\frac{1}{3}\right)=6 \times \frac{1}{3}-4=-2\lt 0\) (Maxima) \((\max , \min )=\frac{1}{3}, 1\)
UPSEE-2014
Application of Derivatives
85657
The value of \(\sin \theta+\cos \theta\) will be greatest, when
NEET Test Series from KOTA - 10 Papers In MS WORD
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Application of Derivatives
85654
If two sides of a triangle are given, then the area of the triangle will be maximum, if the angle between the given sides is
1 \(\frac{\pi}{3}\)
2 \(\frac{\pi}{4}\)
3 \(\frac{\pi}{6}\)
4 \(\frac{\pi}{2}\)
Explanation:
(D) : Let the sides be a, b, c and angles be A, B, C. Area \((\mathrm{A})=\frac{1}{2} \mathrm{absin} \mathrm{C}\) \(\frac{\mathrm{dA}}{\mathrm{dC}}=\frac{1}{2} \mathrm{ab} \cos \mathrm{C}\) A is maximum, when \(\frac{\mathrm{dA}}{\mathrm{dC}}=0\) \(\frac{1}{2} a b \cos C=0\) \(\cos C=0\) \(\cos C=\cos \frac{\pi}{2}\) \(C=\frac{\pi}{2}\)
UPSEE-2015
Application of Derivatives
85655
The maximum and minimum value of \(6 \sin x\) \(\cos x+4 \cos 2 x\) are respectively
1 5,5
2 \(-5,5\)
3 \(5,-5\)
4 None of these
Explanation:
(C) : Given, \(f(x)=6 \sin x \cos x+4 \cos 2 x\) \(=3 \cdot 2 \sin x \cos x+4 \cos 2 x\) \(=3 \sin 2 \mathrm{x}+4 \cos 2 \mathrm{x}\) \(=3 \sin 2 \mathrm{x}+4 \cos 2 \mathrm{x}\) We know that \(-\sqrt{a^{2}+b^{2}} \leq a \sin x+b \cos x \leq \sqrt{a^{2}+b^{2}}\) Comparing, \(\mathrm{a}=3\) and \(\mathrm{b}=4\) \(-\sqrt{3^{2}+4^{2}} \leq 3 \sin 2 x+4 \cos 2 x \leq \sqrt{3^{2}+4^{2}}\) \(-5 \leq 3 \sin 2 x+4 \cos 2 x \leq 5\)
UPSEE-2014
Application of Derivatives
85656
Let \(f(x)=x(x-1)^{2}\), the point at which \(f(x)\) assumes maximum and minimum are respectively
1 \(\frac{1}{3}, 1\)
2 \(1, \frac{1}{3}\)
3 3,1
4 None of these
Explanation:
(A) : \(\mathrm{f}(\mathrm{x})=\mathrm{x}(\mathrm{x}-1)^{2}\) \(=x\left(x^{2}+1-2 x\right)=x^{3}+x-2 x^{2}\) \(\mathrm{f}^{\prime}(\mathrm{x})=3 \mathrm{x}^{2}+1-4 \mathrm{x}\) For maxima or minima, \(\mathrm{f}^{\prime}(\mathrm{x})=0\) \(3 \mathrm{x}^{2}-4 \mathrm{x}+1=0\) \((\mathrm{x}-1)(3 \mathrm{x}-1)=0\) \(\mathrm{x}=1, \mathrm{x}=\frac{1}{3}\) Now, \(\quad \mathrm{f}^{\prime \prime}(\mathrm{x})=6 \mathrm{x}-4\) At \(\quad \mathrm{x}=1, \quad \mathrm{f}^{\prime \prime}(1)=6-4=2>0\) (minima) At \(\quad \mathrm{x}=\frac{1}{3} \quad \mathrm{f}^{\prime \prime}\left(\frac{1}{3}\right)=6 \times \frac{1}{3}-4=-2\lt 0\) (Maxima) \((\max , \min )=\frac{1}{3}, 1\)
UPSEE-2014
Application of Derivatives
85657
The value of \(\sin \theta+\cos \theta\) will be greatest, when
