Question

JEE MAINS MATHS QUESTION SEPTEMBER 2020...

Answer 1

The area encoded by [x](x - 1) ≤ y ≤ 2√x from x = 0 to 2 where [x] is the greatest integer less than or equal to x, is equal to ....

solution : [x](x - 1) ≤ y ≤ 2√x from x = 0 to 2

for x ≥ 0 ⇒y = 2√2x ⇒y² = 4x [ a parabolic graph ]

draw y² = 4x in interval x ≥ 0, diagram is shown in figure.

now let's solve the piece wise function I mean great integer function.

y = [x](x - 1)

case 1 : 0 ≤ x < 1 , y = 0

case 2 : 1 ≤ x < 2 , y = 1(x - 1) = x - 1

case 3 : x = 2 , y = 2(2 - 1) = 2

now draw y = [x](x - 1) with y² = 4x, you will get the enclosed area (shaded region) as shown in figure.

now area enclosed by the curves , A = $\int\limits^2_0{2\sqrt{x}}\,dx-\frac{1}{2}1\times1$

= 4/3[x³/²]²₀ - 1/2

= 4/3 × (2√2) - 1/2

= 8√2/3 - 1/2

Therefore the area enclosed by the curves is 8√2/3 - 1/2 i.e., option (3) is correct choice

Answer 2

Step-by-step explanation:

The area encoded by [x](x - 1) ≤ y ≤ 2√x from x = 0 to 2 where [x] is the greatest integer less than or equal to x, is equal to ....

solution : [x](x - 1) ≤ y ≤ 2√x from x = 0 to 2

for x ≥ 0 ⇒y = 2√2x ⇒y² = 4x [ a parabolic graph ]

draw y² = 4x in interval x ≥ 0, diagram is shown in figure.

now let's solve the piece wise function I mean great integer function.

y = [x](x - 1)

case 1 : 0 ≤ x < 1 , y = 0

case 2 : 1 ≤ x < 2 , y = 1(x - 1) = x - 1

case 3 : x = 2 , y = 2(2 - 1) = 2

now draw y = [x](x - 1) with y² = 4x, you will get the enclosed area (shaded region) as shown in figure.

now area enclosed by the curves , A = \int\limits^2_0{2\sqrt{x}}\,dx-\frac{1}{2}1\times1

0

∫

2

2

x

dx−

2

1

1×1

= 4/3[x³/²]²₀ - 1/2

= 4/3 × (2√2) - 1/2

= 8√2/3 - 1/2

Therefore the area enclosed by the curves is 8√2/3 - 1/2 i.e., option (3) is correct choice....

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