Eamcet Mathematics - Areas Chapter Sample Question Paper With Answer Key
Exam Duration: 60 Mins Total Questions : 15
If An is the area bounded by y = (1- x2)n and coordinate axes, n\(\in\)N, then
- (a)
An = An-1
- (b)
An < An-1
- (c)
An > An-1
- (d)
An = 2An-1
Cut off x-axis, y = 0
0 = (1 - x2)n
\(\therefore\) x = \(\pm\)1
Then, \({ A }_{ n }=\int _{ -1 }^{ 1 }{ { (1-{ x }^{ 2 }) }^{ n }dx } \)
\(=0\int _{ 0 }^{ 1 }{ { (1-{ x }^{ 2 }) }^{ n }dx } \) (be property)
\(={ 2[\{ }{ (1-{ x }^{ 2 }) }^{ n }.{ x\} }_{ 0 }^{ 1 }\)
\(-\int _{ 0 }^{ 1 }{ { n(1-{ x }^{ 2 }) }^{ n-1 }(-2x).xdx] } \)
\(=2\{ 0-2n\int _{ 0 }^{ 1 }{ { n(1-{ x }^{ 2 }) }^{ n-1 } } (1-{ x }^{ 2 }-1)dx\} \)
\(=2\{ -n{ A }_{ n }+n{ A }_{ n-1 }\} \)
\(\Rightarrow \quad { A }_{ n }=\left( \frac { 2n }{ 2n+1 } \right) { A }_{ n-1 }\)
\(\because\) 2n+1>2n
\(\therefore \quad \left( \frac { 2n }{ 2n+1 } \right) <1\)
or An<An-1
Area bounded by the curve y = \(\sqrt { (sin[x]+[sinx]) } \) Where [.]denotes the greatest integer function, lines x = 1 and x = \(\frac { \pi }{ 2 } \): and the x-axis is
- (a)
\(\left( \frac { \pi }{ 2 } -1 \right) \) sq unit
- (b)
\(\sqrt { sin1 } \left( \frac { \pi }{ 2 } -1 \right) \) sq unit
- (c)
\(\sqrt { cos1 } \left( \frac { \pi }{ 2 } -1 \right) \) sq unit
- (d)
\(\sqrt { \frac { \pi }{ 2 } } \left( \frac { \pi }{ 2 } -1 \right) \) sq unit
\(\because\) 1 \(\le\) x < \(\pi\over2\)
\(\therefore\) [x] = 1 and sin 1 \(\le\) sin x < 1
and [sin x] = 0
Then y = \(\sqrt{sin 1+ 0}\)=\(\sqrt{sin1}\)
\(\therefore\) Requi.red area = \(\int _{ 2 }^{ \pi /2 }{ ydx } =\int _{ 2 }^{ \pi /2 }{ \sqrt { sin\quad 1 } dx } \)
\(=\sqrt { sin\quad 1 } \left( \frac { \pi }{ 2 } -1 \right) \) sq unit.
The area bounded by the x-axis, the curve y = f(x) and the lines x = 1and x = b is equal to \((\sqrt { ({ b }^{ 2 }+1) } -\sqrt { 2 } )\) b>1, then f(x) is
- (a)
\(\sqrt { (x-1) } \)
- (b)
\(\sqrt { (x+1) } \)
- (c)
\(\sqrt { ({ x }^{ 2 }+1) } \)
- (d)
\(\frac { x }{ \sqrt { (1+{ x }^{ 2 }) } } \)
\(\int _{ 1 }^{ b }{ f(x)dx } =\sqrt { ({ b }^{ 2 }+1) } -\sqrt { 2 } \)
Differentiating both sides W.r.t. b, then
\(f(b)=\frac { b }{ \sqrt { ({ b }^{ 2 }+1) } } \)
Hence \(f(x)=\frac { x }{ \sqrt { ({ x }^{ 2 }+1) } } \)
Let f(x) = min {x + \(\sqrt{(1-x)}\))(1- x)}, then area bounded by f(x) and x-axis is
- (a)
1/6 sq unit
- (b)
5/6 sq unit
- (c)
7/6 sq unit
- (d)
11/6 sq unit
f(x) = min {x + 1, \(\sqrt{(1-x)}\))
Required area
= \(\int _{ -2 }^{ -1 }{ |(x+1)dx|+|\int _{ 1 }^{ 0 }{ \sqrt { (1-x) } dx } | } \)
= 7/6 sq unit.
The area bounded by the graph y = |[x - 3]|, the x-axis and the lines x = -2 and x = 3 is ([.] denotes the greatest integer function)
- (a)
7 sq unit
- (b)
15 sq unit
- (c)
21 sq unit
- (d)
28 sq unit
Required area = \(\int _{ -2 }^{ 3 }{ |[x-3]|dx } \)
\(=\int _{ -1 }^{ -2 }{ |[x-3]|dx+\int _{ -1 }^{ 0 }{ |[x-3]|dx } } +\)
\(\int _{ 0 }^{ 1 }{ |[x-3]|dx+\int _{ 1 }^{ 2 }{ |[x-3]|dx } } +\int _{ 2 }^{ 3 }{ |[x-3]|dx } \)
\(=\int _{ -2 }^{ -1 }{ 5.dx } +\int _{ -1 }^{ 0 }{ 4.dx } +\int _{ -1 }^{ 0 }{ 3.dx } +\int _{ -1 }^{ 0 }{ 2.dx } +\int _{ -1 }^{ 0 }{ 1.dx } \)
=5(1) + 4(1) + 3(1) + 2(1) + 1(1) = 15 sq unit
The value of c for which the area of the figure bounded by the curve y = 8x2- x5, the straight lines x = 1 and x = c an d the x-axis is equal to16/3 is
- (a)
2
- (b)
\(\sqrt{8-{\sqrt17}}\)
- (c)
3
- (d)
-1
For C<1, \(\int _{ c }^{ 1 }{ (8{ x }^{ 2 }-{ x }^{ 3 })dx=\frac { 16 }{ 3 } } \)
\(\Rightarrow \frac { 8 }{ 3 } -\frac { 1 }{ 6 } -\frac { 8{ c }^{ 3 } }{ 3 } +\frac { { c }^{ 6 } }{ 6 } =\frac { 16 }{ 3 } \)
\(\Rightarrow { c }^{ 3 }\left[ -\frac { 8 }{ 3 } +\frac { { c }^{ 3 } }{ 6 } \right] =\frac { 16 }{ 3 } -\frac { 8 }{ 3 } +\frac { 1 }{ 6 } =\frac { 17 }{ 6 } \)
\(\Rightarrow\) c = -1 satisfy the above eqation. For c \(\ge \) 1, none of the values of c satisfy the required condition that
\(\int _{ c }^{ 1 }{ (8{ x }^{ 2 }-{ x }^{ 3 })dx=\frac { 16 }{ 3 } } \)
The slope of the tangent to a curve y = f(x) at (x, f(x)) is 2x + 1. If the curve passes through the point (1, 2), then the area of the region bounded by the curve, the x-axis and the line x = 1 is
- (a)
5/6 sq unit
- (b)
6/5 sq unit
- (c)
1/6 sq unit
- (d)
6 sq unit
We have, dy/dx = 2x + 1
\(\Rightarrow\) Y = X2 + X + c, it passes through (1, 2)
C = 0
Then, y = X2 + x
Required area = \(\int _{ 0 }^{ 1 }{ ({ x }^{ 2 }+x)dx } \)
5/6 sq unit
The area of the region bounded by the curve a4y2 = (2a - x) X5 is to that of the circle whose radius is a, is given by the ratio
- (a)
4 : 5
- (b)
5 : 8
- (c)
2 : 3
- (d)
3 : 2
Given curve a4y2 = (2a - x) x5
cut off x-axis, when y = 0
0=(2a-x)x5
\(\therefore\) x = 0, 2a
Hence, the area bounded by the curve
a4y2 = (2a - x) x5 is
\({ A }_{ 1 }=\int _{ 0 }^{ 2a }{ \frac { \sqrt { (2a-x) } { x }^{ 5/2 } }{ { a }^{ 2 } } } dx\)
Put x = 2 a sin2 \(\theta\)
\(\therefore\) dx = 4 a sin \(\theta\) cos \(\theta\) d\(\theta\)
\(\therefore\) \({ A }_{ 1 }=\int _{ 0 }^{ \pi /2 }{ \frac { \sqrt { 2a } cos\theta { (2a) }^{ 5/2 }{ sin }^{ 5 }\theta 4asin\theta cos\theta }{ { a }^{ 2 } } } d\theta \)
=32a2 \({ A }_{ 1 }=\int _{ 0 }^{ \pi /2 }{ { sin }^{ 6 }\theta { cos }^{ 2 }\theta } d\theta \)
\(=32{ a }^{ 2 }.\frac { (5.3.1)(1) }{ 8.6.4.2 } .\frac { \pi }{ 2 } \)
\(=\frac { 5\pi { a }^{ 2 } }{ 8 } \)
Area of circle, A2 = \(\pi\)a2
\(\therefore\) \(\frac { { A }_{ 1 } }{ { A }_{ 2 } } =\frac { 5 }{ 8 } \)
A1 : A2 = 5 : 8
The area bounded by y = x e1xl and lines Iex|= y = 0 is
- (a)
4 sq unit
- (b)
6 sq unit
- (c)
1 sq unit
- (d)
2 sq unit
Since, / x] = 1
\(\therefore\) x=±1
y = xe|x| = \(\left| { { xe }^{ x } }-{ { e }^{ x } }\} ^{ 1 }_{ 0 } \right| \)
\(\therefore\) Required area = \(=\left| \int _{ -1 }^{ 0 }{ { x }e^{ -x } } dx \right| +\left| \int _{ 1 }^{ 0 }{ { xe }^{ x }\quad dx } \right| \)
= \(\left| { { -xe }^{ -x } }-{ { e }^{ -x } }\} ^{ 0 }_{ -1 } \right| +\left| { { xe }^{ x } }-{ { e }^{ x } }\} ^{ 1 }_{ 0 } \right| \)
= 2 sq unit
Let f(x) be a continuous function such that the area bounded by the curve y = f(x), the x-axis and the two ordinates x = 0 and x = a is \(\left( \frac { { a }^{ 2 } }{ 2 } +\frac { a }{ 2 } sina+\frac { \pi }{ 2 } cosa \right) \)sq unit, then f\(\left\{ \frac { \pi }{ 2 } \right\} \)is
- (a)
1/2
- (b)
\(f\left\{ \frac { { \pi }^{ 2 } }{ 8 } \right\} \)
- (c)
\({\pi+1}\over2\)
- (d)
none of these
Since \(\int _{ 0 }^{ a }{ f(x)dx=\frac { { a }^{ 2 } }{ 2 } +\frac { a }{ 2 } sin\quad a+\frac { \pi }{ 2 } cos\quad a } \)
Differentiating both sides w.r.t. a, we get
\(\Rightarrow f(a)=a+\frac { a }{ 2 } cos\quad a+\frac { 1 }{ 2 } sin\quad a-\frac { \pi }{ 2 } sin\quad a\)
\(\therefore \quad f\left( \frac { \pi }{ 2 } \right) =\frac { \pi }{ 2 } +0+\frac { 1 }{ 2 } .1-=\frac { \pi }{ 2 } .1=\frac { 1 }{ 2 } \)
The area bounded by the curve y = f(x), the x-axis and the ordinates x = 1 and x = b is (b - 1) cos (3b + 4) sq unit. Then f(x) is given by
- (a)
(x - 1) sin (3x + 4)
- (b)
3 (x - 1) sin (3x + 4) + cos (3x + 4)
- (c)
cos (3x + 4) - 3(x - 1) sin (3x + 4)
- (d)
none of the above
\(\int _{ 1 }^{ b }{ f(x) } dx=(b-1)cos(3b+4)\)(given)
Differentiating bath sides W.r.t. b, then
f(b) = (b -1): 3 (- sin (3b + 4)) + cos (3b + 4)·1
or f(b) = cos (3b + 4) - 3 (b - 1) sin (3b + 4)
f(x) = cos (3x + 4)- 3(x -1) sin (3x + 4)
polynomial P is positive for x > 0 and the area of the region bounded by P(x), the x-axis and the vertical lines x=0 and x = \(\lambda\) sq unit. en polynomial p(x) is
- (a)
X2 + 2x
- (b)
X2 + 2x + 1
- (c)
X2 + X + 1
- (d)
x3 + 2X2 + 2
\(\int _{ 0 }^{ \lambda }{ p(x)dx\frac { { \lambda }^{ 2 }(\lambda +3) }{ 3 } } \) (given)
Differentiating both sides W.r.t.\(\lambda\) , then
\(p(\lambda )=\frac { { 3\lambda }^{ 2 }+6\lambda }{ 3 } ={ \lambda }^{ 2 }+2\lambda \)
\(\therefore\) p(x) = x2 + 2x
Alternative meth
\(\int _{ 0 }^{ \lambda }{ p(x)dx\frac { { \lambda }^{ 2 }(\lambda +3) }{ 3 } } \)
Since, RHS is a polynomial of degree 3. So, P(x) is a polynomial of degree 2
Let P(x) = ax2 + bx + c
Then \(\int _{ 0 }^{ \lambda }{ (a{ x }^{ 2 }+bx+c) } dx=\frac { { a\lambda }^{ 3 } }{ 3 } +\frac { { b\lambda }^{ 2 } }{ 2 } +c\)
but given \(\frac { { a\lambda }^{ 3 } }{ 3 } +\frac { { b\lambda }^{ 2 } }{ 2 } +c\lambda =\frac { { \lambda }^{ 3 } }{ 3 } +{ \lambda }^{ 2 }\)
On compaflng, we get a =1, b =2, c =0
p(x) =x2+2x
For which of the following values of m, is the area of the region bounded by the curve y = x - x2 and the line y = mx equals 9/2 sq unit?
- (a)
-4
- (b)
-2
- (c)
2
- (d)
4
The two curves meet at m.x = x - X2 or X2 = x (1 - m)
\(\therefore\) x = 0,1- m
\(\int _{ 0 }^{ 1-m }{ ({ y }_{ 1 }-{ y }_{ 2 })dx } =\int _{ 0 }^{ 1-m }{ (x-{ x }^{ 2 }-mx) } dx\)
\(={ \left[ (1-m)\frac { { x }^{ 2 } }{ 2 } -\frac { { x }^{ 3 } }{ 3 } \right] }_{ 0 }^{ 1-m }=\frac { 9 }{ 2 } \)
if m<1
or (1-m)3 \(\left[ \frac { 1 }{ 2 } -\frac { 1 }{ 3 } \right] =\frac { 9 }{ 2 } \)
or (1-m)3 = 27
m = -2
Now If m>1, then 1 - m is -ive, then
\(\left[ (1-m)\frac { { x }^{ 2 } }{ 2 } -\frac { { x }^{ 3 } }{ 3 } \right] =\frac { 9 }{ 2 } \)
-(1-m)3 \(\left( \frac { 1 }{ 2 } -\frac { 1 }{ 3 } \right) =\frac { 9 }{ 2 } \)
\(\therefore\) -(1-m)3 = -27
or 1-m = -3
\(\Rightarrow\) m = 4
The value (s) of 'a' for which the area of the triangle included between the axes and any tangent to the curve xay = \(\lambda\)a is constant, is/are
- (a)
-1/2
- (b)
-1
- (c)
1/2
- (d)
1
GIven curve, xay = \(\lambda\) ... (i)
(\(\lambda\), 1) is a point on the given curve.
Now, differentiating Eq. (i) W.r.t. x, we get
axa-1y + y + xadx/dy = 0
\(\Rightarrow\) dx/dy = \(\frac { { -ax }^{ a-1 }y }{ { x }^{ a } } =-\frac { ay }{ x } \)
At (\(\lambda\) , 1)dx/dy = -\(a\over\lambda\)
Equation of tangent at (\(\lambda\), 1)
y-1 = -\(a\over\lambda\) (x-\(\lambda\))
Now, x =0
\(\Rightarrow\) y = 1 + a
y = 0
\(\Rightarrow\) x=\(\lambda \over a\) + \(\lambda\) =\(\lambda(1+a)\over a\)
Area, A = 1/2 x (1+a) \((1+a)\lambda\over a\)
Now, dA/da = 1/2\(\lambda\) \(\left[ \frac { a.2(1+a)-{ (1+a) }^{ 2 } }{ { a }^{ 2 } } \right] =0\)
\(\Rightarrow\) (2a - 1 - a)(1 + a) = 0
\(\Rightarrow\) (a-1)(a + 1) = 0
\(\Rightarrow\) a = 1, a = -1
Let f(x) = x2 - 3x + 2 be a function, \(\forall \) x \(\in\) R .
if |f(|xl)l = a has 6 real roots, then
- (a)
\(a\in \left\{ 0,\frac { 1 }{ 4 } \right\} \)
- (b)
\(a\in \left( \frac { 1 }{ 4 } ,2 \right) \)
- (c)
\(a\in \left( -\frac { 1 }{ 4 } ,2 \right) \)
- (d)
None of these
For six real roots of
|f(|x|)| = a
a=1/4