87036
A region in the \(x-y\) plane is bounded by the curve \(y=\sqrt{25-x^{2}}\) and the line \(y=0\). If the point \((a, a+1)\) lies in the interior of the region, then
1 \(\mathrm{a} \in(-4,3)\)
2 \(\mathrm{a} \in(-\infty,-1) \cup(3, \infty)\)
3 \(\mathrm{a} \in(-1,3)\)
4 none of these
Explanation:
(C) : Given, \(y=\sqrt{25-x^{2}}\) Given that point \((a, a+1)\) lies in the interior of the region \(\therefore \quad \mathrm{a}+1>0 \text {, i.e } \mathrm{a}>-1 \tag{i}\) Also, \((a+1)-\sqrt{25-a^{2}}\lt 0\) Squares on both sides - \(a^{2}+(a+1)^{2}-25\lt 0\) \(2 a^{2}+2 a-24\lt 0\) \(a^{2}+a-12\lt 0\) \((a-3)(a+4)\lt 0\) \(-4\lt a\lt 3 \tag{ii}\) From (i) and (ii) we get- \(\mathrm{a} \in(-1,3)\)
AMU-2018
Application of the Integrals
87037
The area of the region \(\left\{(x, y): x^{2}+y^{2} \leq 1 \leq x+y\right\} \text { is }\)
1 \(\frac{\pi^{2}}{2}\)
2 \(\frac{\pi}{4}\)
3 \(\frac{\pi}{4}-\frac{1}{2}\)
4 \(\frac{\pi^{2}}{3}\)
Explanation:
(C) :Given, Area of region, \(\left\{(\mathrm{x}, \mathrm{y}): \mathrm{x}^{2}+\mathrm{y}^{2} \leq 1 \leq \mathrm{x}+\mathrm{y}\right\}\) \(\therefore \mathrm{x}^{2}+\mathrm{y}^{2}=1\) and \(\mathrm{x}+\mathrm{y}=1\) Area of shaded region \(=\) Area of quadrant - Area of triangle \(=\frac{\pi}{4}(1)^{2}-\frac{1}{2} \times 1 \times 1=\frac{\pi}{4}-\frac{1}{2}\)
WB JEE-2020
Application of the Integrals
87038
The value of \(\int_{0}^{5} \max \left\{x^{2}, 6 x-8\right\} d x\) is
1 72
2 125
3 43
4 69
Explanation:
(C) : Given, for \(\mathrm{x}\lt 2\), \(\max \rightarrow \mathrm{x}^{2}\) function for \(2\lt x\lt 4, \quad \max \rightarrow 6 x-8\) function for \(4\lt x\lt 5 \quad \max \rightarrow x^{2}\) function Let \(I=\int_{0}^{5}\left\{x^{2}, 6 x-8\right\} d x\). \(=\int_{0}^{2} x^{2} d x+\int_{2}^{4}(6 x-8) d x+\int_{4}^{5} x^{2} d x\) \(=\left[\frac{\mathrm{x}^{3}}{3}\right]_{0}^{2}+\left[\frac{6 \mathrm{x}^{2}}{2}-8 \mathrm{x}\right]_{2}^{4}+\left[\frac{\mathrm{x}^{3}}{3}\right]_{4}^{5}\) \(\frac{8}{3}+\left[3 x^{2}-8 x\right]_{2}^{4}+\frac{125}{3}-\frac{64}{3}\) \(\frac{8}{3}+[48-32-12+16]+\frac{61}{3}\) \([48-32-12+16]+\frac{69}{3}=20+23=43\)
WB JEE-2021
Application of the Integrals
87040
If \(f(x)=x^{2 / 3}, x \geq 0\). Then, the area of the region enclosed by the curve \(y=f(x)\) and the three lines \(y=x, x=1\) and \(x=8\) is
1 \(\frac{63}{2}\)
2 \(\frac{93}{5}\)
3 \(\frac{105}{7}\)
4 \(\frac{129}{10}\)
Explanation:
(D) : Given, \(f(x)=x^{2 / 3}, x \geq 0\) and line \(y=x, x=1\) and \(x=8\) \(\therefore\) Required area \(A=\int_{1}^{8}\left(x-x^{2 / 3}\right) d x\) \(=\left[\frac{x^{2}}{2}-\frac{3}{5} x^{5 / 3}\right]_{1}^{8}=\left(32-\frac{3}{5} \times 32\right)-\left(\frac{1}{2}-\frac{3}{5}\right)\) \(=32 \times \frac{2}{5}-\frac{(5-6)}{10}=\frac{64}{5}+\frac{1}{10}=\frac{129}{10}\)
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Application of the Integrals
87036
A region in the \(x-y\) plane is bounded by the curve \(y=\sqrt{25-x^{2}}\) and the line \(y=0\). If the point \((a, a+1)\) lies in the interior of the region, then
1 \(\mathrm{a} \in(-4,3)\)
2 \(\mathrm{a} \in(-\infty,-1) \cup(3, \infty)\)
3 \(\mathrm{a} \in(-1,3)\)
4 none of these
Explanation:
(C) : Given, \(y=\sqrt{25-x^{2}}\) Given that point \((a, a+1)\) lies in the interior of the region \(\therefore \quad \mathrm{a}+1>0 \text {, i.e } \mathrm{a}>-1 \tag{i}\) Also, \((a+1)-\sqrt{25-a^{2}}\lt 0\) Squares on both sides - \(a^{2}+(a+1)^{2}-25\lt 0\) \(2 a^{2}+2 a-24\lt 0\) \(a^{2}+a-12\lt 0\) \((a-3)(a+4)\lt 0\) \(-4\lt a\lt 3 \tag{ii}\) From (i) and (ii) we get- \(\mathrm{a} \in(-1,3)\)
AMU-2018
Application of the Integrals
87037
The area of the region \(\left\{(x, y): x^{2}+y^{2} \leq 1 \leq x+y\right\} \text { is }\)
1 \(\frac{\pi^{2}}{2}\)
2 \(\frac{\pi}{4}\)
3 \(\frac{\pi}{4}-\frac{1}{2}\)
4 \(\frac{\pi^{2}}{3}\)
Explanation:
(C) :Given, Area of region, \(\left\{(\mathrm{x}, \mathrm{y}): \mathrm{x}^{2}+\mathrm{y}^{2} \leq 1 \leq \mathrm{x}+\mathrm{y}\right\}\) \(\therefore \mathrm{x}^{2}+\mathrm{y}^{2}=1\) and \(\mathrm{x}+\mathrm{y}=1\) Area of shaded region \(=\) Area of quadrant - Area of triangle \(=\frac{\pi}{4}(1)^{2}-\frac{1}{2} \times 1 \times 1=\frac{\pi}{4}-\frac{1}{2}\)
WB JEE-2020
Application of the Integrals
87038
The value of \(\int_{0}^{5} \max \left\{x^{2}, 6 x-8\right\} d x\) is
1 72
2 125
3 43
4 69
Explanation:
(C) : Given, for \(\mathrm{x}\lt 2\), \(\max \rightarrow \mathrm{x}^{2}\) function for \(2\lt x\lt 4, \quad \max \rightarrow 6 x-8\) function for \(4\lt x\lt 5 \quad \max \rightarrow x^{2}\) function Let \(I=\int_{0}^{5}\left\{x^{2}, 6 x-8\right\} d x\). \(=\int_{0}^{2} x^{2} d x+\int_{2}^{4}(6 x-8) d x+\int_{4}^{5} x^{2} d x\) \(=\left[\frac{\mathrm{x}^{3}}{3}\right]_{0}^{2}+\left[\frac{6 \mathrm{x}^{2}}{2}-8 \mathrm{x}\right]_{2}^{4}+\left[\frac{\mathrm{x}^{3}}{3}\right]_{4}^{5}\) \(\frac{8}{3}+\left[3 x^{2}-8 x\right]_{2}^{4}+\frac{125}{3}-\frac{64}{3}\) \(\frac{8}{3}+[48-32-12+16]+\frac{61}{3}\) \([48-32-12+16]+\frac{69}{3}=20+23=43\)
WB JEE-2021
Application of the Integrals
87040
If \(f(x)=x^{2 / 3}, x \geq 0\). Then, the area of the region enclosed by the curve \(y=f(x)\) and the three lines \(y=x, x=1\) and \(x=8\) is
1 \(\frac{63}{2}\)
2 \(\frac{93}{5}\)
3 \(\frac{105}{7}\)
4 \(\frac{129}{10}\)
Explanation:
(D) : Given, \(f(x)=x^{2 / 3}, x \geq 0\) and line \(y=x, x=1\) and \(x=8\) \(\therefore\) Required area \(A=\int_{1}^{8}\left(x-x^{2 / 3}\right) d x\) \(=\left[\frac{x^{2}}{2}-\frac{3}{5} x^{5 / 3}\right]_{1}^{8}=\left(32-\frac{3}{5} \times 32\right)-\left(\frac{1}{2}-\frac{3}{5}\right)\) \(=32 \times \frac{2}{5}-\frac{(5-6)}{10}=\frac{64}{5}+\frac{1}{10}=\frac{129}{10}\)
87036
A region in the \(x-y\) plane is bounded by the curve \(y=\sqrt{25-x^{2}}\) and the line \(y=0\). If the point \((a, a+1)\) lies in the interior of the region, then
1 \(\mathrm{a} \in(-4,3)\)
2 \(\mathrm{a} \in(-\infty,-1) \cup(3, \infty)\)
3 \(\mathrm{a} \in(-1,3)\)
4 none of these
Explanation:
(C) : Given, \(y=\sqrt{25-x^{2}}\) Given that point \((a, a+1)\) lies in the interior of the region \(\therefore \quad \mathrm{a}+1>0 \text {, i.e } \mathrm{a}>-1 \tag{i}\) Also, \((a+1)-\sqrt{25-a^{2}}\lt 0\) Squares on both sides - \(a^{2}+(a+1)^{2}-25\lt 0\) \(2 a^{2}+2 a-24\lt 0\) \(a^{2}+a-12\lt 0\) \((a-3)(a+4)\lt 0\) \(-4\lt a\lt 3 \tag{ii}\) From (i) and (ii) we get- \(\mathrm{a} \in(-1,3)\)
AMU-2018
Application of the Integrals
87037
The area of the region \(\left\{(x, y): x^{2}+y^{2} \leq 1 \leq x+y\right\} \text { is }\)
1 \(\frac{\pi^{2}}{2}\)
2 \(\frac{\pi}{4}\)
3 \(\frac{\pi}{4}-\frac{1}{2}\)
4 \(\frac{\pi^{2}}{3}\)
Explanation:
(C) :Given, Area of region, \(\left\{(\mathrm{x}, \mathrm{y}): \mathrm{x}^{2}+\mathrm{y}^{2} \leq 1 \leq \mathrm{x}+\mathrm{y}\right\}\) \(\therefore \mathrm{x}^{2}+\mathrm{y}^{2}=1\) and \(\mathrm{x}+\mathrm{y}=1\) Area of shaded region \(=\) Area of quadrant - Area of triangle \(=\frac{\pi}{4}(1)^{2}-\frac{1}{2} \times 1 \times 1=\frac{\pi}{4}-\frac{1}{2}\)
WB JEE-2020
Application of the Integrals
87038
The value of \(\int_{0}^{5} \max \left\{x^{2}, 6 x-8\right\} d x\) is
1 72
2 125
3 43
4 69
Explanation:
(C) : Given, for \(\mathrm{x}\lt 2\), \(\max \rightarrow \mathrm{x}^{2}\) function for \(2\lt x\lt 4, \quad \max \rightarrow 6 x-8\) function for \(4\lt x\lt 5 \quad \max \rightarrow x^{2}\) function Let \(I=\int_{0}^{5}\left\{x^{2}, 6 x-8\right\} d x\). \(=\int_{0}^{2} x^{2} d x+\int_{2}^{4}(6 x-8) d x+\int_{4}^{5} x^{2} d x\) \(=\left[\frac{\mathrm{x}^{3}}{3}\right]_{0}^{2}+\left[\frac{6 \mathrm{x}^{2}}{2}-8 \mathrm{x}\right]_{2}^{4}+\left[\frac{\mathrm{x}^{3}}{3}\right]_{4}^{5}\) \(\frac{8}{3}+\left[3 x^{2}-8 x\right]_{2}^{4}+\frac{125}{3}-\frac{64}{3}\) \(\frac{8}{3}+[48-32-12+16]+\frac{61}{3}\) \([48-32-12+16]+\frac{69}{3}=20+23=43\)
WB JEE-2021
Application of the Integrals
87040
If \(f(x)=x^{2 / 3}, x \geq 0\). Then, the area of the region enclosed by the curve \(y=f(x)\) and the three lines \(y=x, x=1\) and \(x=8\) is
1 \(\frac{63}{2}\)
2 \(\frac{93}{5}\)
3 \(\frac{105}{7}\)
4 \(\frac{129}{10}\)
Explanation:
(D) : Given, \(f(x)=x^{2 / 3}, x \geq 0\) and line \(y=x, x=1\) and \(x=8\) \(\therefore\) Required area \(A=\int_{1}^{8}\left(x-x^{2 / 3}\right) d x\) \(=\left[\frac{x^{2}}{2}-\frac{3}{5} x^{5 / 3}\right]_{1}^{8}=\left(32-\frac{3}{5} \times 32\right)-\left(\frac{1}{2}-\frac{3}{5}\right)\) \(=32 \times \frac{2}{5}-\frac{(5-6)}{10}=\frac{64}{5}+\frac{1}{10}=\frac{129}{10}\)
87036
A region in the \(x-y\) plane is bounded by the curve \(y=\sqrt{25-x^{2}}\) and the line \(y=0\). If the point \((a, a+1)\) lies in the interior of the region, then
1 \(\mathrm{a} \in(-4,3)\)
2 \(\mathrm{a} \in(-\infty,-1) \cup(3, \infty)\)
3 \(\mathrm{a} \in(-1,3)\)
4 none of these
Explanation:
(C) : Given, \(y=\sqrt{25-x^{2}}\) Given that point \((a, a+1)\) lies in the interior of the region \(\therefore \quad \mathrm{a}+1>0 \text {, i.e } \mathrm{a}>-1 \tag{i}\) Also, \((a+1)-\sqrt{25-a^{2}}\lt 0\) Squares on both sides - \(a^{2}+(a+1)^{2}-25\lt 0\) \(2 a^{2}+2 a-24\lt 0\) \(a^{2}+a-12\lt 0\) \((a-3)(a+4)\lt 0\) \(-4\lt a\lt 3 \tag{ii}\) From (i) and (ii) we get- \(\mathrm{a} \in(-1,3)\)
AMU-2018
Application of the Integrals
87037
The area of the region \(\left\{(x, y): x^{2}+y^{2} \leq 1 \leq x+y\right\} \text { is }\)
1 \(\frac{\pi^{2}}{2}\)
2 \(\frac{\pi}{4}\)
3 \(\frac{\pi}{4}-\frac{1}{2}\)
4 \(\frac{\pi^{2}}{3}\)
Explanation:
(C) :Given, Area of region, \(\left\{(\mathrm{x}, \mathrm{y}): \mathrm{x}^{2}+\mathrm{y}^{2} \leq 1 \leq \mathrm{x}+\mathrm{y}\right\}\) \(\therefore \mathrm{x}^{2}+\mathrm{y}^{2}=1\) and \(\mathrm{x}+\mathrm{y}=1\) Area of shaded region \(=\) Area of quadrant - Area of triangle \(=\frac{\pi}{4}(1)^{2}-\frac{1}{2} \times 1 \times 1=\frac{\pi}{4}-\frac{1}{2}\)
WB JEE-2020
Application of the Integrals
87038
The value of \(\int_{0}^{5} \max \left\{x^{2}, 6 x-8\right\} d x\) is
1 72
2 125
3 43
4 69
Explanation:
(C) : Given, for \(\mathrm{x}\lt 2\), \(\max \rightarrow \mathrm{x}^{2}\) function for \(2\lt x\lt 4, \quad \max \rightarrow 6 x-8\) function for \(4\lt x\lt 5 \quad \max \rightarrow x^{2}\) function Let \(I=\int_{0}^{5}\left\{x^{2}, 6 x-8\right\} d x\). \(=\int_{0}^{2} x^{2} d x+\int_{2}^{4}(6 x-8) d x+\int_{4}^{5} x^{2} d x\) \(=\left[\frac{\mathrm{x}^{3}}{3}\right]_{0}^{2}+\left[\frac{6 \mathrm{x}^{2}}{2}-8 \mathrm{x}\right]_{2}^{4}+\left[\frac{\mathrm{x}^{3}}{3}\right]_{4}^{5}\) \(\frac{8}{3}+\left[3 x^{2}-8 x\right]_{2}^{4}+\frac{125}{3}-\frac{64}{3}\) \(\frac{8}{3}+[48-32-12+16]+\frac{61}{3}\) \([48-32-12+16]+\frac{69}{3}=20+23=43\)
WB JEE-2021
Application of the Integrals
87040
If \(f(x)=x^{2 / 3}, x \geq 0\). Then, the area of the region enclosed by the curve \(y=f(x)\) and the three lines \(y=x, x=1\) and \(x=8\) is
1 \(\frac{63}{2}\)
2 \(\frac{93}{5}\)
3 \(\frac{105}{7}\)
4 \(\frac{129}{10}\)
Explanation:
(D) : Given, \(f(x)=x^{2 / 3}, x \geq 0\) and line \(y=x, x=1\) and \(x=8\) \(\therefore\) Required area \(A=\int_{1}^{8}\left(x-x^{2 / 3}\right) d x\) \(=\left[\frac{x^{2}}{2}-\frac{3}{5} x^{5 / 3}\right]_{1}^{8}=\left(32-\frac{3}{5} \times 32\right)-\left(\frac{1}{2}-\frac{3}{5}\right)\) \(=32 \times \frac{2}{5}-\frac{(5-6)}{10}=\frac{64}{5}+\frac{1}{10}=\frac{129}{10}\)
