Practice Start Test

Area Under Curves

JEE Mathematics · 107 questions · Page 6 of 11 · Click an option or "Show Solution" to reveal answer

Q51
The area, enclosed by the curves y=sin⁡x+cos⁡xy = \sin x + \cos x and y=∣cos⁡x−sin⁡x∣y = \left| {\cos x - \sin x} \right| and the lines x=0,x=π2x = 0,x = {\pi \over 2}, is :
A 22(2−1)2\sqrt 2 (\sqrt 2 - 1)
B 2(2+1)2(\sqrt 2 + 1)
C 4(2−1)4(\sqrt 2 - 1)
D 22(2+1)2\sqrt 2 (\sqrt 2 + 1)
Correct Answer
Option A
Solution
A=∫0π2((sin⁡x+cos⁡x)−∣cos⁡x−sin⁡x∣) dxA = \int_0^{{\pi \over 2}} {\left( {(\sin x + \cos x) - \left| {\cos x - \sin x} \right|} \right)\,dx}
A=∫0π2((sin⁡x+cos⁡x)−(cos⁡x−sin⁡x)) dx+∫π4π2((sin⁡x+cos⁡x)−(sin⁡x−cos⁡x)) dxA = \int_0^{{\pi \over 2}} {\left( {(\sin x + \cos x) - (\cos x - \sin x)} \right)\,dx} + \int_{{\pi \over 4}}^{{\pi \over 2}} {\left( {(\sin x + \cos x) - (\sin x - \cos x)} \right)\,dx}
A=2∫0π2sin⁡x dx+2∫π4π2cos⁡x dxA = 2\int_0^{{\pi \over 2}} {\sin x\,dx + 2\int_{{\pi \over 4}}^{{\pi \over 2}} {\cos x\,dx} }
A=−2(12−1)+(1−12)A = - 2\left( {{1 \over {\sqrt 2 }} - 1} \right) + \left( {1 - {1 \over {\sqrt 2 }}} \right)
A=4−22=22(2−1)A = 4 - 2\sqrt 2 = 2\sqrt 2 (\sqrt 2 - 1)

Option (a)

Q52
The area enclosed by y2 = 8x and y = 2\sqrt2 x that lies outside the triangle formed by y = 2\sqrt2 x, x = 1, y = 22\sqrt2, is equal to:
A 1626{{16\sqrt 2 } \over 6}
B 1126{{11\sqrt 2 } \over 6}
C 1326{{13\sqrt 2 } \over 6}
D 526{{5\sqrt 2 } \over 6}
Correct Answer
Option C
Solution
A(2,22),B(1,22),C(1,2)A\left( {2,2\sqrt 2 } \right),B\left( {1,2\sqrt 2 } \right),C\left( {1,\sqrt 2 } \right)

Area

=∫042(y2−y28)dy−= \int\limits_0^{4\sqrt 2 } {\left( {{y \over {\sqrt 2 }} - {{{y^2}} \over 8}} \right)dy - }

area (

Δ\Delta

BAC)

=[y222−y324]042−12×AB×BC= \left[ {{{{y^2}} \over {2\sqrt 2 }} - {{{y^3}} \over {24}}} \right]_0^{4\sqrt 2 } - {1 \over 2} \times AB \times BC
=82−32×4224−12×1×2= 8\sqrt 2 - {{32 \times 4\sqrt 2 } \over {24}} - {1 \over 2} \times 1 \times \sqrt 2
=82−1623−22= 8\sqrt 2 - {{16\sqrt 2 } \over 3} - {{\sqrt 2 } \over 2}
=26(48−32−3)=1326= {{\sqrt 2 } \over 6}(48 - 32 - 3) = {{13\sqrt 2 } \over 6}
Q53
The area of the bounded region enclosed by the curve y=3−∣x−12∣−∣x+1∣y = 3 - \left| {x - {1 \over 2}} \right| - |x + 1| and the x-axis is :
A 94{9 \over 4}
B 4516{45 \over 16}
C 278{27 \over 8}
D 6316{63 \over 16}
Correct Answer
Option C
Solution
y={2x−72x<−132−1≤x≤1252−2xx>12y=\left\{\begin{array}{cc} 2 x-\frac{7}{2} & x<-1 \\\\ \frac{3}{2} & -1 \leq x \leq \frac{1}{2} \\\\ \frac{5}{2}-2 x & x>\frac{1}{2} \end{array}\right.
y=3−∣x−12∣−∣x+1∣y=3-\left|x-\frac{1}{2}\right|-|x+1|

Area of shaded region (required area)

=12(3+32)⋅32=278=\frac{1}{2}\left(3+\frac{3}{2}\right) \cdot \frac{3}{2}=\frac{27}{8}
Q54
The area of the region S = {(x, y) : y2 ≤\le 8x, y ≥\ge 2\sqrt2x, x ≥\ge 1} is
A 1326{{13\sqrt 2 } \over 6}
B 1126{{11\sqrt 2 } \over 6}
C 526{{5\sqrt 2 } \over 6}
D 1926{{19\sqrt 2 } \over 6}
Correct Answer
Option B
Solution

Required area

=∫14(8x−2x)dx= \int\limits_1^4 {\left( {\sqrt {8x} - \sqrt 2 x} \right)dx}
=283x32−x22∣14= \left. {{{2\sqrt 8 } \over 3}{x^{{3 \over 2}}} - {{{x^2}} \over {\sqrt 2 }}} \right|_1^4
=1633−162−283+12= {{16\sqrt 3 } \over 3} - {{16} \over {\sqrt 2 }} - {{2\sqrt 8 } \over 3} + {1 \over {\sqrt 2 }}
=1126= {{11\sqrt 2 } \over 6}

sq. units

Q55
The area bounded by the curve y = |x2 −- 9| and the line y = 3 is :
A 4(23+6−4)4(2\sqrt 3 + \sqrt 6 - 4)
B 4(43+6−4)4(4\sqrt 3 + \sqrt 6 - 4)
C 8(43+36−9)8(4\sqrt 3 + 3\sqrt 6 - 9)
D 8(43+26−9)8(4\sqrt 3 + 2\sqrt 6 - 9)
Correct Answer
Option D
Solution
y=3y = 3

and

y=∣x2−9∣y = |{x^2} - 9|

Intersect in first quadrant at

x=6x = \sqrt 6

and

x=12x = \sqrt {12}

Required area

=2[23(6×6)+∫63(3−(9−x2))dx+∫312(3−(x2−9))dx]= 2\left[ {{2 \over 3}\left( {6 \times \sqrt 6 } \right) + \int\limits_{\sqrt 6 }^3 {\left( {3 - \left( {9 - {x^2}} \right)} \right)dx + \int\limits_3^{\sqrt {12} } {\left( {3 - \left( {{x^2} - 9} \right)} \right)dx} } } \right]
=2[46+(x33−6x)∣63+(12x−x33)∣312]= 2\left[ {4\sqrt 6 + \left. {\left( {{{{x^3}} \over 3} - 6x} \right)} \right|_{\sqrt 6 }^3 + \left. {\left( {12x - {{{x^3}} \over 3}} \right)} \right|_3^{\sqrt {12} }} \right]
=2[46+(46−9)+(812−27)]= 2\left[ {4\sqrt 6 + \left( {4\sqrt 6 - 9} \right) + \left( {8\sqrt {12} - 27} \right)} \right]
=2[86+163−36]=8[26+43−9]= 2\left[ {8\sqrt 6 + 16\sqrt 3 - 36} \right] = 8\left[ {2\sqrt 6 + 4\sqrt 3 - 9} \right]
Q56
The area of the region bounded by y2 = 8x and y2 = 16(3 −- x) is equal to:
A 323{{32} \over 3}
B 403{{40} \over 3}
C 16
D 19
Correct Answer
Option C
Solution
c1:y2=8x{c_1}:{y^2} = 8x
c2:y2=16(3−x){c_2}:{y^2} = 16(3 - x)

Solving c1 and c2

48−16x=8x48 - 16x = 8x
x=2x = 2

∴\therefore

y=± 4y = \pm \,4

∴\therefore Area of shaded region

=2∫04{(48−y216)−(y28)}dy= 2\int\limits_0^4 {\left\{ {\left( {{{48 - {y^2}} \over {16}}} \right) - \left( {{{{y^2}} \over 8}} \right)} \right\}dy}
=18[48y−y3]04=16= {1 \over 8}\left[ {48y - {y^3}} \right]_0^4 = 16
Q57
The area of the region given by A={(x,y):x2≤y≤min⁡{x+2,4−3x}}A=\left\{(x, y): x^{2} \leq y \leq \min \{x+2,4-3 x\}\right\} is :
A 318\dfrac{31}{8}
B 176\dfrac{17}{6}
C 196\dfrac{19}{6}
D 278\dfrac{27}{8}
Correct Answer
Option B
Solution

A={(x,y):x2≤y≤min⁡{x+2,4−3x}A=\left\{(x, y): x^{2} \leq y \leq \min \{x+2,4-3 x\}\right. So area of required region

A=∫−112(x+2−x2)dx+∫121(4−3x−x2)dx=[x22+2x−x33]−112+[4x−3x22−x33]121=(18+1−124)−(12−2+13)+(4−32−13)−(2−38−124)=176\begin{aligned} &A=\int_{-1}^{\frac{1}{2}}\left(x+2-x^{2}\right) d x+\int_{\frac{1}{2}}^{1}\left(4-3 x-x^{2}\right) d x \\\\ &=\left[\frac{x^{2}}{2}+2 x-\frac{x^{3}}{3}\right]_{-1}^{\frac{1}{2}}+\left[4 x-\frac{3 x^{2}}{2}-\frac{x^{3}}{3}\right]_{\frac{1}{2}}^{1} \\\\ &=\left(\frac{1}{8}+1-\frac{1}{24}\right)-\left(\frac{1}{2}-2+\frac{1}{3}\right)+\left(4-\frac{3}{2}-\frac{1}{3}\right)-\left(2-\frac{3}{8}-\frac{1}{24}\right) \\\\ &=\frac{17}{6} \end{aligned}
Q58
Let the locus of the centre (α,β),β>0(\alpha, \beta), \beta>0, of the circle which touches the circle x2+(y−1)2=1x^{2}+(y-1)^{2}=1 externally and also touches the xx-axis be L\mathrm{L}. Then the area bounded by L\mathrm{L} and the line y=4y=4 is:
A 3223 \dfrac{32 \sqrt{2}}{3}
B 4023 \dfrac{40 \sqrt{2}}{3}
C 643\dfrac{64}{3}
D 323 \dfrac{32}{3}
Correct Answer
Option C
Solution

Radius of circle SS touching xx-axis and centre (α,β)(\alpha, \beta) is ∣β∣|\beta|. According to given conditions

α2+(β−1)2=(∣β∣+1)2α2+β2−2β+1=β2+1+2∣β∣α2=4β as β>0\begin{aligned} &\alpha^{2}+(\beta-1)^{2}=(|\beta|+1)^{2} \\\\ &\alpha^{2}+\beta^{2}-2 \beta+1=\beta^{2}+1+2|\beta| \\\\ &\alpha^{2}=4 \beta \text{ as } \beta>0 \end{aligned}

∴\therefore \quad Required louse is L:x2=4yL: x^{2}=4 y The area of shaded region =2∫042ydy=2 \int_{0}^{4} 2 \sqrt{y} d y

=4.[y3232]04=643 square units. \begin{aligned} &=4 . {\left[\frac{y^{\frac{3}{2}}}{\frac{3}{2}}\right]_{0}^{4} } \\\\ &=\frac{64}{3} \text{ square units. } \end{aligned}
Q59
The odd natural number a, such that the area of the region bounded by y = 1, y = 3, x = 0, x = ya is 3643{{364} \over 3}, is equal to :
A 3
B 5
C 7
D 9
Correct Answer
Option B
Solution
a\mathrm{a}

is a odd natural number and

∣∫13yady∣=3643\left| {\int\limits_1^3 {{y^a}dy} } \right| = {{364} \over 3}
⇒∣1a+1(ya+1)13∣=3643\Rightarrow \left| {{1 \over {a + 1}}\left( {{y^{a + 1}}} \right)_1^3} \right| = {{364} \over 3}
⇒3a+1−1a+1= ± 3643\Rightarrow {{{3^{a + 1}} - 1} \over {a + 1}} = \, \pm \,{{364} \over 3}

Solving with

(−)( - )

sign,

3a+1−1a+1=3643⇒(a=5){{{3^{a + 1}} - 1} \over {a + 1}} = {{364} \over 3} \Rightarrow (a = 5)

Solving with

(+)( + )

sign,

3a+1−1a+1=−3643{{{3^{a + 1}} - 1} \over {a + 1}} = {{ - 364} \over 3}

, No a exist ∴\therefore

(a=5)(a = 5)
Q60
The area bounded by the curves y=∣x2−1∣y=\left|x^{2}-1\right| and y=1y=1 is
A 23(2+1)\dfrac{2}{3}(\sqrt{2}+1)
B 43(2−1)\dfrac{4}{3}(\sqrt{2}-1)
C 2(2−1)2(\sqrt{2}-1)
D 83(2−1)\dfrac{8}{3}(\sqrt{2}-1)
Correct Answer
Option D
Solution

Area

=2∫02(1−∣x2−1∣)dx= 2\int\limits_0^{\sqrt 2 } {(1 - |{x^2} - 1|)dx}
2[∫01(1−(1−x2))dx+∫12(2−x2)dx]2\left[ {\int\limits_0^1 {\left( {1 - (1 - {x^2})} \right)dx + \int\limits_1^{\sqrt 2 } {\left( {2 - {x^2}} \right)dx} } } \right]
=2[[x33]01+[2x−x33]12]]= 2\left[ {\left. {\left[ {{{{x^3}} \over 3}} \right]_0^1 + \left[ {2x - {{{x^3}} \over 3}} \right]_1^{\sqrt 2 }} \right]} \right]
=2(42−43)=83(2−1)= 2\left( {{{4\sqrt 2 - 4} \over 3}} \right) = {8 \over 3}\left( {\sqrt 2 - 1} \right)
Ready for a full JEE mock test? Timed · full syllabus · instant results
Take a Mock Test →