Exams Joint Entrance Examination Mathematics Area Under Curves Question 63 Question 63 of 230 Medium NUMERICAL
If the area of the larger portion bounded between the curves x 2 + y 2 = 25 {x}^{2}+{y}^{2}=25 x 2 + y 2 = 25 and y = ∣ x − 1 ∣ y=|x-1| y = ∣ x − 1∣ is 1 4 ( b π + c ) , b , c ∈ N \frac{1}{4}(b\pi +c),b,c\in N 4 1 ( bπ + c ) , b , c ∈ N , then b + c b+c b + c is equal to \_\_\_\_\_\_\_\_\_
Answer: 77
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x 2 + y 2 = 5 x 2 + ( x − 1 ) 2 = 25 ⇒ x = 4 x 2 + ( − x + 1 ) 2 = 5 ⇒ x = − 3 A = 25 π − ∫ 4 − 3 25 − x 2 d x + 1 2 × 4 × 4 + 1 2 × 3 × 3 A = 25 π + 25 2 − [ x 2 25 − x 2 + 25 2 sin − 1 x 5 ] 4 − 3 A = 25 π + 25 2 − [ 6 + 25 2 sin − 1 4 5 + 6 + 25 2 sin − 1 3 5 ] A = 25 π + 1 2 − 25 2 ⋅ π 2 A = 75 π 4 + 1 2 A = 1 4 ( 75 π + 2 ) b = 75 , c = 2 b + c = 75 + 2 = 77 \begin{gathered} {x}^{2}+{y}^{2}=5 \\ {x}^{2}+(x-1{)}^{2}=25\Rightarrow x=4 \\ {x}^{2}+(-x+1{)}^{2}=5\Rightarrow x=-3 \\ A=25\pi -{\int }_{4}^{-3}\sqrt{25-{x}^{2}}dx+\frac{1}{2}\times 4\times 4+\frac{1}{2}\times 3\times 3 \\ A=25\pi +\frac{25}{2}-{[\frac{x}{2}\sqrt{25-{x}^{2}}+\frac{25}{2}{\sin }^{-1}\frac{x}{5}]}_{4}^{-3} \\ A=25\pi +\frac{25}{2}-[6+\frac{25}{2}{\sin }^{-1}\frac{4}{5}+6+\frac{25}{2}{\sin }^{-1}\frac{3}{5}] \\ A=25\pi +\frac{1}{2}-\frac{25}{2}\cdot \frac{\pi }{2} \\ A=\frac{75\pi }{4}+\frac{1}{2} \\ A=\frac{1}{4}(75\pi +2) \\ b=75,c=2 \\ b+c=75+2=77 \end{gathered} x 2 + y 2 = 5 x 2 + ( x − 1 ) 2 = 25 ⇒ x = 4 x 2 + ( − x + 1 ) 2 = 5 ⇒ x = − 3 A = 25 π − ∫ 4 − 3 25 − x 2 d x + 2 1 × 4 × 4 + 2 1 × 3 × 3 A = 25 π + 2 25 − [ 2 x 25 − x 2 + 2 25 sin − 1 5 x ] 4 − 3 A = 25 π + 2 25 − [ 6 + 2 25 sin − 1 5 4 + 6 + 2 25 sin − 1 5 3 ] A = 25 π + 2 1 − 2 25 ⋅ 2 π A = 4 75 π + 2 1 A = 4 1 ( 75 π + 2 ) b = 75 , c = 2 b + c = 75 + 2 = 77 Answer the question to unlock