85619
The maximum area of rectangle inscribed in a circle of diameter \(R\) i
1 \(R^{2}\)
2 \(\frac{R^{2}}{2}\)
3 \(\frac{R^{2}}{4}\)
4 \(\frac{R^{2}}{8}\)
Explanation:
(B) : The diagonal = \(\mathrm{R}\) Thus the area of rectangle \(=\frac{1}{2} \times \mathrm{R} \times \mathrm{R}=\frac{\mathrm{R}^{2}}{2}\)
BITSAT-2019
Application of Derivatives
85620
Suppose that \(f(0)=-3\) and \(f^{\prime}(x) \leq 5\) for all values of \(x\). Then, the largest value which \(f(2)\) can attain \(i\)
1 7
2 10
3 2
4 9
Explanation:
(A) : Using mean value theorem in \([0,2]\) \(\frac{\mathrm{f}(2)-\mathrm{f}(0)}{2-0}=\mathrm{f}^{\prime}(\mathrm{c})\), where \(\mathrm{c} \in(0,2)\) \(\frac{\mathrm{f}(2)+3}{2} \leq 5 \Rightarrow \mathrm{f}(2) \leq 7\).
BITSAT-2020
Application of Derivatives
85682
The maximum value of \(3 \cos \theta+4 \sin \theta\) is
1 3
2 4
3 5
4 None of these
Explanation:
(C) : Given, \(\therefore\) Maximum value of a \(\cos \theta+b \sin \theta=\sqrt{\mathrm{a}^{2}+\mathrm{b}^{2}}\) \(\therefore\) Maximum value of \(3 \cos \theta+4 \sin \theta=\sqrt{3^{2}+4^{2}}\) \(=\sqrt{25}=5\)
CG PET-2016
Application of Derivatives
85707
\(A B\) is a diameter of a circle and \(C\) is any point on the circumference of the circle. Then
1 The perimeter of \(\triangle \mathrm{ABC}\) is maximum when it is isosceles
2 The area of \(\triangle \mathrm{ABC}\) is minimum when it is isosceles
3 The area of \(\triangle \mathrm{ABC}\) is maximum when it is isosceles
4 None of these
Explanation:
(C) : \(\mathrm{AB}\) is the diameter of the circle and \(\mathrm{C}\) is any point on the circumference. So, \(\angle \mathrm{ACB}=90^{\circ}\) \(\therefore \triangle \mathrm{ACB}\) is a right angle triangle and area of right angled triangle is maximum when it is isosceles.
85619
The maximum area of rectangle inscribed in a circle of diameter \(R\) i
1 \(R^{2}\)
2 \(\frac{R^{2}}{2}\)
3 \(\frac{R^{2}}{4}\)
4 \(\frac{R^{2}}{8}\)
Explanation:
(B) : The diagonal = \(\mathrm{R}\) Thus the area of rectangle \(=\frac{1}{2} \times \mathrm{R} \times \mathrm{R}=\frac{\mathrm{R}^{2}}{2}\)
BITSAT-2019
Application of Derivatives
85620
Suppose that \(f(0)=-3\) and \(f^{\prime}(x) \leq 5\) for all values of \(x\). Then, the largest value which \(f(2)\) can attain \(i\)
1 7
2 10
3 2
4 9
Explanation:
(A) : Using mean value theorem in \([0,2]\) \(\frac{\mathrm{f}(2)-\mathrm{f}(0)}{2-0}=\mathrm{f}^{\prime}(\mathrm{c})\), where \(\mathrm{c} \in(0,2)\) \(\frac{\mathrm{f}(2)+3}{2} \leq 5 \Rightarrow \mathrm{f}(2) \leq 7\).
BITSAT-2020
Application of Derivatives
85682
The maximum value of \(3 \cos \theta+4 \sin \theta\) is
1 3
2 4
3 5
4 None of these
Explanation:
(C) : Given, \(\therefore\) Maximum value of a \(\cos \theta+b \sin \theta=\sqrt{\mathrm{a}^{2}+\mathrm{b}^{2}}\) \(\therefore\) Maximum value of \(3 \cos \theta+4 \sin \theta=\sqrt{3^{2}+4^{2}}\) \(=\sqrt{25}=5\)
CG PET-2016
Application of Derivatives
85707
\(A B\) is a diameter of a circle and \(C\) is any point on the circumference of the circle. Then
1 The perimeter of \(\triangle \mathrm{ABC}\) is maximum when it is isosceles
2 The area of \(\triangle \mathrm{ABC}\) is minimum when it is isosceles
3 The area of \(\triangle \mathrm{ABC}\) is maximum when it is isosceles
4 None of these
Explanation:
(C) : \(\mathrm{AB}\) is the diameter of the circle and \(\mathrm{C}\) is any point on the circumference. So, \(\angle \mathrm{ACB}=90^{\circ}\) \(\therefore \triangle \mathrm{ACB}\) is a right angle triangle and area of right angled triangle is maximum when it is isosceles.
85619
The maximum area of rectangle inscribed in a circle of diameter \(R\) i
1 \(R^{2}\)
2 \(\frac{R^{2}}{2}\)
3 \(\frac{R^{2}}{4}\)
4 \(\frac{R^{2}}{8}\)
Explanation:
(B) : The diagonal = \(\mathrm{R}\) Thus the area of rectangle \(=\frac{1}{2} \times \mathrm{R} \times \mathrm{R}=\frac{\mathrm{R}^{2}}{2}\)
BITSAT-2019
Application of Derivatives
85620
Suppose that \(f(0)=-3\) and \(f^{\prime}(x) \leq 5\) for all values of \(x\). Then, the largest value which \(f(2)\) can attain \(i\)
1 7
2 10
3 2
4 9
Explanation:
(A) : Using mean value theorem in \([0,2]\) \(\frac{\mathrm{f}(2)-\mathrm{f}(0)}{2-0}=\mathrm{f}^{\prime}(\mathrm{c})\), where \(\mathrm{c} \in(0,2)\) \(\frac{\mathrm{f}(2)+3}{2} \leq 5 \Rightarrow \mathrm{f}(2) \leq 7\).
BITSAT-2020
Application of Derivatives
85682
The maximum value of \(3 \cos \theta+4 \sin \theta\) is
1 3
2 4
3 5
4 None of these
Explanation:
(C) : Given, \(\therefore\) Maximum value of a \(\cos \theta+b \sin \theta=\sqrt{\mathrm{a}^{2}+\mathrm{b}^{2}}\) \(\therefore\) Maximum value of \(3 \cos \theta+4 \sin \theta=\sqrt{3^{2}+4^{2}}\) \(=\sqrt{25}=5\)
CG PET-2016
Application of Derivatives
85707
\(A B\) is a diameter of a circle and \(C\) is any point on the circumference of the circle. Then
1 The perimeter of \(\triangle \mathrm{ABC}\) is maximum when it is isosceles
2 The area of \(\triangle \mathrm{ABC}\) is minimum when it is isosceles
3 The area of \(\triangle \mathrm{ABC}\) is maximum when it is isosceles
4 None of these
Explanation:
(C) : \(\mathrm{AB}\) is the diameter of the circle and \(\mathrm{C}\) is any point on the circumference. So, \(\angle \mathrm{ACB}=90^{\circ}\) \(\therefore \triangle \mathrm{ACB}\) is a right angle triangle and area of right angled triangle is maximum when it is isosceles.
85619
The maximum area of rectangle inscribed in a circle of diameter \(R\) i
1 \(R^{2}\)
2 \(\frac{R^{2}}{2}\)
3 \(\frac{R^{2}}{4}\)
4 \(\frac{R^{2}}{8}\)
Explanation:
(B) : The diagonal = \(\mathrm{R}\) Thus the area of rectangle \(=\frac{1}{2} \times \mathrm{R} \times \mathrm{R}=\frac{\mathrm{R}^{2}}{2}\)
BITSAT-2019
Application of Derivatives
85620
Suppose that \(f(0)=-3\) and \(f^{\prime}(x) \leq 5\) for all values of \(x\). Then, the largest value which \(f(2)\) can attain \(i\)
1 7
2 10
3 2
4 9
Explanation:
(A) : Using mean value theorem in \([0,2]\) \(\frac{\mathrm{f}(2)-\mathrm{f}(0)}{2-0}=\mathrm{f}^{\prime}(\mathrm{c})\), where \(\mathrm{c} \in(0,2)\) \(\frac{\mathrm{f}(2)+3}{2} \leq 5 \Rightarrow \mathrm{f}(2) \leq 7\).
BITSAT-2020
Application of Derivatives
85682
The maximum value of \(3 \cos \theta+4 \sin \theta\) is
1 3
2 4
3 5
4 None of these
Explanation:
(C) : Given, \(\therefore\) Maximum value of a \(\cos \theta+b \sin \theta=\sqrt{\mathrm{a}^{2}+\mathrm{b}^{2}}\) \(\therefore\) Maximum value of \(3 \cos \theta+4 \sin \theta=\sqrt{3^{2}+4^{2}}\) \(=\sqrt{25}=5\)
CG PET-2016
Application of Derivatives
85707
\(A B\) is a diameter of a circle and \(C\) is any point on the circumference of the circle. Then
1 The perimeter of \(\triangle \mathrm{ABC}\) is maximum when it is isosceles
2 The area of \(\triangle \mathrm{ABC}\) is minimum when it is isosceles
3 The area of \(\triangle \mathrm{ABC}\) is maximum when it is isosceles
4 None of these
Explanation:
(C) : \(\mathrm{AB}\) is the diameter of the circle and \(\mathrm{C}\) is any point on the circumference. So, \(\angle \mathrm{ACB}=90^{\circ}\) \(\therefore \triangle \mathrm{ACB}\) is a right angle triangle and area of right angled triangle is maximum when it is isosceles.
