85583
The maximum area of a rectangle inscribed in the circle \((x+1)^{2}+(y-3)^{2}=64\) is
1 64 sq. units
2 72 sq. units
3 128 sq. units
4 8 sq. units
Explanation:
(C) : Given, \((x+1)^{2}+(y-3)^{2}=64\) \((x+1)^{2}+(y-3)^{2}=(8)^{2} \tag{i}\) We known that, \((x-a)^{2}+(y-b)^{2}=r^{2} \tag{ii}\) On comparing equation (i) and (ii), we get- \(r=8\) The rectangle having diagonal as the diameter of circle Diagonal \((2 \mathrm{r})=2 \times 8=16\) The area of a rectangle inscribed in a circle is maximum, when it is square \((2 r)^{2}=(\text { side })^{2}+(\text { side })^{2}\) \((16)^{2}=2(\text { side })^{2}\) \((\text { side })^{2}=\frac{16 \times 16}{2}=\frac{256}{2}\) \(\therefore\) Area \(=\frac{256}{2}\) Area \(=128\) square units
Karnataka CET-2018
Application of Derivatives
85584
If \(x\) is real, then the minimum value of \(x^{2}-8 x+17\) is
1 2
2 4
3 1
4 3
Explanation:
(C) : Given, \(f(x)=x^{2}-8 x+17\) On differentiating both sides, we get- \(\mathrm{f}^{\prime}(\mathrm{x})=2 \mathrm{x}-8\) For minimum \(\mathrm{f}^{\prime}(\mathrm{x})=0\) \(2 x-8=0\) \(x=4\) \(\therefore\) Minimum value \(\mathrm{f}(4)=4^{2}-4 \times 8+17\) \(=16-32+17=1\) Hence minimum value is 1 .
Karnataka CET-2015
Application of Derivatives
85585
The maximum area of a rectangle that can be inscribed in a circle of radius 2 units is
1 \(8 \pi\) sq. units
2 4 sq. units
3 5 sq. units
4 8 sq. units
Explanation:
(D) : Given, Radius \((\mathrm{r})=2\) units \(\therefore \mathrm{AC}=2 \mathrm{r}=2 \times 2\) \(\mathrm{AC}=4\) units The rectangle having diagonal as the diameter of circle The area of rectangle inscribe in a circle is maximum when it is a square \((4)^{2}=(\text { side })^{2}+(\text { side })^{2}\) \(16=2(\text { side })^{2}\) \((\text { side })^{2}=\frac{16}{2}\) Maximum area \(=8\) square units.
Karnataka CET-2013
Application of Derivatives
85586
If a ball is thrown vertically upward and the height ' \(s\) ' reached in time ' \(t\) ' is given by \(s=22 t\) - \(11 t^{2}\), then the total distance travelled by the ball is
1 22 units
2 44 units
3 33 units
4 11 units
Explanation:
(A) : Given, Height ' \(s\) ' reached by the ball in time \(t\) is given by \(\mathrm{s}=22 \mathrm{t}-11 \mathrm{t}^{2}\) On differentiating both sides w.r.t. \(t\) on both side \(\frac{\mathrm{ds}}{\mathrm{dt}}=22-22 \mathrm{t}\) At maximum point \(\frac{\mathrm{ds}}{\mathrm{dt}}=0\) \(22-22 \mathrm{t}=0\) \(\mathrm{t}=1\) Distance travelled by upward \(\mathrm{s}=22-11 \times(1)^{2}\) \(\mathrm{~s}=11 \text { unit }\) The same distance travelled by the ball in return journey. Hence, the total distance \(11+11=22 \text { units. }\)
Karnataka CET-2012
Application of Derivatives
85587
The maximum value of \(\mathrm{xe}^{-\mathrm{x}} \mathrm{i}\)
1 \(-1 / \mathrm{e}\)
2 e
3 \(1 / \mathrm{e}\)
4 \(-\mathrm{e}\)
Explanation:
(C) : Given, \(y=x^{-x}\) On differencing both sides with respect to \(\mathrm{x}\) we get- \(\frac{d y}{d x}=e^{-x}+x\left(-e^{-x}\right)=e^{-x}(1-x)\) For maximum or minimum \(\frac{d y}{d x}=0\) \(e^{-x}(1-x)=0\) \(x=1\) Now, maximum value \(y=1 \cdot e^{-1}\) \(y=\frac{1}{e}\)
85583
The maximum area of a rectangle inscribed in the circle \((x+1)^{2}+(y-3)^{2}=64\) is
1 64 sq. units
2 72 sq. units
3 128 sq. units
4 8 sq. units
Explanation:
(C) : Given, \((x+1)^{2}+(y-3)^{2}=64\) \((x+1)^{2}+(y-3)^{2}=(8)^{2} \tag{i}\) We known that, \((x-a)^{2}+(y-b)^{2}=r^{2} \tag{ii}\) On comparing equation (i) and (ii), we get- \(r=8\) The rectangle having diagonal as the diameter of circle Diagonal \((2 \mathrm{r})=2 \times 8=16\) The area of a rectangle inscribed in a circle is maximum, when it is square \((2 r)^{2}=(\text { side })^{2}+(\text { side })^{2}\) \((16)^{2}=2(\text { side })^{2}\) \((\text { side })^{2}=\frac{16 \times 16}{2}=\frac{256}{2}\) \(\therefore\) Area \(=\frac{256}{2}\) Area \(=128\) square units
Karnataka CET-2018
Application of Derivatives
85584
If \(x\) is real, then the minimum value of \(x^{2}-8 x+17\) is
1 2
2 4
3 1
4 3
Explanation:
(C) : Given, \(f(x)=x^{2}-8 x+17\) On differentiating both sides, we get- \(\mathrm{f}^{\prime}(\mathrm{x})=2 \mathrm{x}-8\) For minimum \(\mathrm{f}^{\prime}(\mathrm{x})=0\) \(2 x-8=0\) \(x=4\) \(\therefore\) Minimum value \(\mathrm{f}(4)=4^{2}-4 \times 8+17\) \(=16-32+17=1\) Hence minimum value is 1 .
Karnataka CET-2015
Application of Derivatives
85585
The maximum area of a rectangle that can be inscribed in a circle of radius 2 units is
1 \(8 \pi\) sq. units
2 4 sq. units
3 5 sq. units
4 8 sq. units
Explanation:
(D) : Given, Radius \((\mathrm{r})=2\) units \(\therefore \mathrm{AC}=2 \mathrm{r}=2 \times 2\) \(\mathrm{AC}=4\) units The rectangle having diagonal as the diameter of circle The area of rectangle inscribe in a circle is maximum when it is a square \((4)^{2}=(\text { side })^{2}+(\text { side })^{2}\) \(16=2(\text { side })^{2}\) \((\text { side })^{2}=\frac{16}{2}\) Maximum area \(=8\) square units.
Karnataka CET-2013
Application of Derivatives
85586
If a ball is thrown vertically upward and the height ' \(s\) ' reached in time ' \(t\) ' is given by \(s=22 t\) - \(11 t^{2}\), then the total distance travelled by the ball is
1 22 units
2 44 units
3 33 units
4 11 units
Explanation:
(A) : Given, Height ' \(s\) ' reached by the ball in time \(t\) is given by \(\mathrm{s}=22 \mathrm{t}-11 \mathrm{t}^{2}\) On differentiating both sides w.r.t. \(t\) on both side \(\frac{\mathrm{ds}}{\mathrm{dt}}=22-22 \mathrm{t}\) At maximum point \(\frac{\mathrm{ds}}{\mathrm{dt}}=0\) \(22-22 \mathrm{t}=0\) \(\mathrm{t}=1\) Distance travelled by upward \(\mathrm{s}=22-11 \times(1)^{2}\) \(\mathrm{~s}=11 \text { unit }\) The same distance travelled by the ball in return journey. Hence, the total distance \(11+11=22 \text { units. }\)
Karnataka CET-2012
Application of Derivatives
85587
The maximum value of \(\mathrm{xe}^{-\mathrm{x}} \mathrm{i}\)
1 \(-1 / \mathrm{e}\)
2 e
3 \(1 / \mathrm{e}\)
4 \(-\mathrm{e}\)
Explanation:
(C) : Given, \(y=x^{-x}\) On differencing both sides with respect to \(\mathrm{x}\) we get- \(\frac{d y}{d x}=e^{-x}+x\left(-e^{-x}\right)=e^{-x}(1-x)\) For maximum or minimum \(\frac{d y}{d x}=0\) \(e^{-x}(1-x)=0\) \(x=1\) Now, maximum value \(y=1 \cdot e^{-1}\) \(y=\frac{1}{e}\)
85583
The maximum area of a rectangle inscribed in the circle \((x+1)^{2}+(y-3)^{2}=64\) is
1 64 sq. units
2 72 sq. units
3 128 sq. units
4 8 sq. units
Explanation:
(C) : Given, \((x+1)^{2}+(y-3)^{2}=64\) \((x+1)^{2}+(y-3)^{2}=(8)^{2} \tag{i}\) We known that, \((x-a)^{2}+(y-b)^{2}=r^{2} \tag{ii}\) On comparing equation (i) and (ii), we get- \(r=8\) The rectangle having diagonal as the diameter of circle Diagonal \((2 \mathrm{r})=2 \times 8=16\) The area of a rectangle inscribed in a circle is maximum, when it is square \((2 r)^{2}=(\text { side })^{2}+(\text { side })^{2}\) \((16)^{2}=2(\text { side })^{2}\) \((\text { side })^{2}=\frac{16 \times 16}{2}=\frac{256}{2}\) \(\therefore\) Area \(=\frac{256}{2}\) Area \(=128\) square units
Karnataka CET-2018
Application of Derivatives
85584
If \(x\) is real, then the minimum value of \(x^{2}-8 x+17\) is
1 2
2 4
3 1
4 3
Explanation:
(C) : Given, \(f(x)=x^{2}-8 x+17\) On differentiating both sides, we get- \(\mathrm{f}^{\prime}(\mathrm{x})=2 \mathrm{x}-8\) For minimum \(\mathrm{f}^{\prime}(\mathrm{x})=0\) \(2 x-8=0\) \(x=4\) \(\therefore\) Minimum value \(\mathrm{f}(4)=4^{2}-4 \times 8+17\) \(=16-32+17=1\) Hence minimum value is 1 .
Karnataka CET-2015
Application of Derivatives
85585
The maximum area of a rectangle that can be inscribed in a circle of radius 2 units is
1 \(8 \pi\) sq. units
2 4 sq. units
3 5 sq. units
4 8 sq. units
Explanation:
(D) : Given, Radius \((\mathrm{r})=2\) units \(\therefore \mathrm{AC}=2 \mathrm{r}=2 \times 2\) \(\mathrm{AC}=4\) units The rectangle having diagonal as the diameter of circle The area of rectangle inscribe in a circle is maximum when it is a square \((4)^{2}=(\text { side })^{2}+(\text { side })^{2}\) \(16=2(\text { side })^{2}\) \((\text { side })^{2}=\frac{16}{2}\) Maximum area \(=8\) square units.
Karnataka CET-2013
Application of Derivatives
85586
If a ball is thrown vertically upward and the height ' \(s\) ' reached in time ' \(t\) ' is given by \(s=22 t\) - \(11 t^{2}\), then the total distance travelled by the ball is
1 22 units
2 44 units
3 33 units
4 11 units
Explanation:
(A) : Given, Height ' \(s\) ' reached by the ball in time \(t\) is given by \(\mathrm{s}=22 \mathrm{t}-11 \mathrm{t}^{2}\) On differentiating both sides w.r.t. \(t\) on both side \(\frac{\mathrm{ds}}{\mathrm{dt}}=22-22 \mathrm{t}\) At maximum point \(\frac{\mathrm{ds}}{\mathrm{dt}}=0\) \(22-22 \mathrm{t}=0\) \(\mathrm{t}=1\) Distance travelled by upward \(\mathrm{s}=22-11 \times(1)^{2}\) \(\mathrm{~s}=11 \text { unit }\) The same distance travelled by the ball in return journey. Hence, the total distance \(11+11=22 \text { units. }\)
Karnataka CET-2012
Application of Derivatives
85587
The maximum value of \(\mathrm{xe}^{-\mathrm{x}} \mathrm{i}\)
1 \(-1 / \mathrm{e}\)
2 e
3 \(1 / \mathrm{e}\)
4 \(-\mathrm{e}\)
Explanation:
(C) : Given, \(y=x^{-x}\) On differencing both sides with respect to \(\mathrm{x}\) we get- \(\frac{d y}{d x}=e^{-x}+x\left(-e^{-x}\right)=e^{-x}(1-x)\) For maximum or minimum \(\frac{d y}{d x}=0\) \(e^{-x}(1-x)=0\) \(x=1\) Now, maximum value \(y=1 \cdot e^{-1}\) \(y=\frac{1}{e}\)
85583
The maximum area of a rectangle inscribed in the circle \((x+1)^{2}+(y-3)^{2}=64\) is
1 64 sq. units
2 72 sq. units
3 128 sq. units
4 8 sq. units
Explanation:
(C) : Given, \((x+1)^{2}+(y-3)^{2}=64\) \((x+1)^{2}+(y-3)^{2}=(8)^{2} \tag{i}\) We known that, \((x-a)^{2}+(y-b)^{2}=r^{2} \tag{ii}\) On comparing equation (i) and (ii), we get- \(r=8\) The rectangle having diagonal as the diameter of circle Diagonal \((2 \mathrm{r})=2 \times 8=16\) The area of a rectangle inscribed in a circle is maximum, when it is square \((2 r)^{2}=(\text { side })^{2}+(\text { side })^{2}\) \((16)^{2}=2(\text { side })^{2}\) \((\text { side })^{2}=\frac{16 \times 16}{2}=\frac{256}{2}\) \(\therefore\) Area \(=\frac{256}{2}\) Area \(=128\) square units
Karnataka CET-2018
Application of Derivatives
85584
If \(x\) is real, then the minimum value of \(x^{2}-8 x+17\) is
1 2
2 4
3 1
4 3
Explanation:
(C) : Given, \(f(x)=x^{2}-8 x+17\) On differentiating both sides, we get- \(\mathrm{f}^{\prime}(\mathrm{x})=2 \mathrm{x}-8\) For minimum \(\mathrm{f}^{\prime}(\mathrm{x})=0\) \(2 x-8=0\) \(x=4\) \(\therefore\) Minimum value \(\mathrm{f}(4)=4^{2}-4 \times 8+17\) \(=16-32+17=1\) Hence minimum value is 1 .
Karnataka CET-2015
Application of Derivatives
85585
The maximum area of a rectangle that can be inscribed in a circle of radius 2 units is
1 \(8 \pi\) sq. units
2 4 sq. units
3 5 sq. units
4 8 sq. units
Explanation:
(D) : Given, Radius \((\mathrm{r})=2\) units \(\therefore \mathrm{AC}=2 \mathrm{r}=2 \times 2\) \(\mathrm{AC}=4\) units The rectangle having diagonal as the diameter of circle The area of rectangle inscribe in a circle is maximum when it is a square \((4)^{2}=(\text { side })^{2}+(\text { side })^{2}\) \(16=2(\text { side })^{2}\) \((\text { side })^{2}=\frac{16}{2}\) Maximum area \(=8\) square units.
Karnataka CET-2013
Application of Derivatives
85586
If a ball is thrown vertically upward and the height ' \(s\) ' reached in time ' \(t\) ' is given by \(s=22 t\) - \(11 t^{2}\), then the total distance travelled by the ball is
1 22 units
2 44 units
3 33 units
4 11 units
Explanation:
(A) : Given, Height ' \(s\) ' reached by the ball in time \(t\) is given by \(\mathrm{s}=22 \mathrm{t}-11 \mathrm{t}^{2}\) On differentiating both sides w.r.t. \(t\) on both side \(\frac{\mathrm{ds}}{\mathrm{dt}}=22-22 \mathrm{t}\) At maximum point \(\frac{\mathrm{ds}}{\mathrm{dt}}=0\) \(22-22 \mathrm{t}=0\) \(\mathrm{t}=1\) Distance travelled by upward \(\mathrm{s}=22-11 \times(1)^{2}\) \(\mathrm{~s}=11 \text { unit }\) The same distance travelled by the ball in return journey. Hence, the total distance \(11+11=22 \text { units. }\)
Karnataka CET-2012
Application of Derivatives
85587
The maximum value of \(\mathrm{xe}^{-\mathrm{x}} \mathrm{i}\)
1 \(-1 / \mathrm{e}\)
2 e
3 \(1 / \mathrm{e}\)
4 \(-\mathrm{e}\)
Explanation:
(C) : Given, \(y=x^{-x}\) On differencing both sides with respect to \(\mathrm{x}\) we get- \(\frac{d y}{d x}=e^{-x}+x\left(-e^{-x}\right)=e^{-x}(1-x)\) For maximum or minimum \(\frac{d y}{d x}=0\) \(e^{-x}(1-x)=0\) \(x=1\) Now, maximum value \(y=1 \cdot e^{-1}\) \(y=\frac{1}{e}\)
85583
The maximum area of a rectangle inscribed in the circle \((x+1)^{2}+(y-3)^{2}=64\) is
1 64 sq. units
2 72 sq. units
3 128 sq. units
4 8 sq. units
Explanation:
(C) : Given, \((x+1)^{2}+(y-3)^{2}=64\) \((x+1)^{2}+(y-3)^{2}=(8)^{2} \tag{i}\) We known that, \((x-a)^{2}+(y-b)^{2}=r^{2} \tag{ii}\) On comparing equation (i) and (ii), we get- \(r=8\) The rectangle having diagonal as the diameter of circle Diagonal \((2 \mathrm{r})=2 \times 8=16\) The area of a rectangle inscribed in a circle is maximum, when it is square \((2 r)^{2}=(\text { side })^{2}+(\text { side })^{2}\) \((16)^{2}=2(\text { side })^{2}\) \((\text { side })^{2}=\frac{16 \times 16}{2}=\frac{256}{2}\) \(\therefore\) Area \(=\frac{256}{2}\) Area \(=128\) square units
Karnataka CET-2018
Application of Derivatives
85584
If \(x\) is real, then the minimum value of \(x^{2}-8 x+17\) is
1 2
2 4
3 1
4 3
Explanation:
(C) : Given, \(f(x)=x^{2}-8 x+17\) On differentiating both sides, we get- \(\mathrm{f}^{\prime}(\mathrm{x})=2 \mathrm{x}-8\) For minimum \(\mathrm{f}^{\prime}(\mathrm{x})=0\) \(2 x-8=0\) \(x=4\) \(\therefore\) Minimum value \(\mathrm{f}(4)=4^{2}-4 \times 8+17\) \(=16-32+17=1\) Hence minimum value is 1 .
Karnataka CET-2015
Application of Derivatives
85585
The maximum area of a rectangle that can be inscribed in a circle of radius 2 units is
1 \(8 \pi\) sq. units
2 4 sq. units
3 5 sq. units
4 8 sq. units
Explanation:
(D) : Given, Radius \((\mathrm{r})=2\) units \(\therefore \mathrm{AC}=2 \mathrm{r}=2 \times 2\) \(\mathrm{AC}=4\) units The rectangle having diagonal as the diameter of circle The area of rectangle inscribe in a circle is maximum when it is a square \((4)^{2}=(\text { side })^{2}+(\text { side })^{2}\) \(16=2(\text { side })^{2}\) \((\text { side })^{2}=\frac{16}{2}\) Maximum area \(=8\) square units.
Karnataka CET-2013
Application of Derivatives
85586
If a ball is thrown vertically upward and the height ' \(s\) ' reached in time ' \(t\) ' is given by \(s=22 t\) - \(11 t^{2}\), then the total distance travelled by the ball is
1 22 units
2 44 units
3 33 units
4 11 units
Explanation:
(A) : Given, Height ' \(s\) ' reached by the ball in time \(t\) is given by \(\mathrm{s}=22 \mathrm{t}-11 \mathrm{t}^{2}\) On differentiating both sides w.r.t. \(t\) on both side \(\frac{\mathrm{ds}}{\mathrm{dt}}=22-22 \mathrm{t}\) At maximum point \(\frac{\mathrm{ds}}{\mathrm{dt}}=0\) \(22-22 \mathrm{t}=0\) \(\mathrm{t}=1\) Distance travelled by upward \(\mathrm{s}=22-11 \times(1)^{2}\) \(\mathrm{~s}=11 \text { unit }\) The same distance travelled by the ball in return journey. Hence, the total distance \(11+11=22 \text { units. }\)
Karnataka CET-2012
Application of Derivatives
85587
The maximum value of \(\mathrm{xe}^{-\mathrm{x}} \mathrm{i}\)
1 \(-1 / \mathrm{e}\)
2 e
3 \(1 / \mathrm{e}\)
4 \(-\mathrm{e}\)
Explanation:
(C) : Given, \(y=x^{-x}\) On differencing both sides with respect to \(\mathrm{x}\) we get- \(\frac{d y}{d x}=e^{-x}+x\left(-e^{-x}\right)=e^{-x}(1-x)\) For maximum or minimum \(\frac{d y}{d x}=0\) \(e^{-x}(1-x)=0\) \(x=1\) Now, maximum value \(y=1 \cdot e^{-1}\) \(y=\frac{1}{e}\)
