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and let f –1 be the inverse function of f. Then (f –1)' (2) is equal to
Then F(e) equals- a)1
- b)2
- c)1/2
- d)0
Correct answer is option 'C'. Can you explain this answer?
Then F(e) equalsa)
1
b)
2
c)
1/2
d)
0
|
Veda Institute answered |
and limit for t = 1 ⇒ z = 1 and for t = 1/e ⇒ z = e
[∴ log1 = 0]
Equation (A) becomes
Let log t = x
[for limit t = 1, x = 0 and t = e, x = log e = 1]
Let log t = x
[for limit t = 1, x = 0 and t = e, x = log e = 1]
The area enclosed between the curves y2 = x and y = | x | is- a)1/6
- b)1/3
- c)2/3
- d)1
Correct answer is option 'A'. Can you explain this answer?
The area enclosed between the curves y2 = x and y = | x | is
a)
1/6
b)
1/3
c)
2/3
d)
1
|
Lavanya Menon answered |
The area enclosed between the curves y2 = x and y = | x |
From the figure, area lies between y2 = x and y = x
From the figure, area lies between y2 = x and y = x
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?- a)– 4
- b)– 2
- c)2
- d)4
Correct answer is option 'B,D'. Can you explain this answer?
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?
a)
– 4
b)
– 2
c)
2
d)
4
|
|
Lavanya Menon answered |
The two curves meet at
or (1 - m)3 = 27 ,
∴ m = -2
But if m >1 then 1– m is – ive, then 
Area of the region bounded by the curve y = ex and lines x = 0 and y = e is- a)e –1
- b)
- c)
- d)
Correct answer is option 'B,C,D'. Can you explain this answer?
Area of the region bounded by the curve y = ex and lines x = 0 and y = e is
a)
e –1
b)
c)
d)
|
Geetika Shah answered |
The area bounded by the curve y = ex and lines x = 0 and y = e is as shown in the graph.
Also required area
The option(s) with the values of a and L that satisfy the following equation is(are) 
- a)
- b)
- c)
- d)
Correct answer is option 'A,C'. Can you explain this answer?
The option(s) with the values of a and L that satisfy the following equation is(are)
a)
b)
c)
d)
|
|
Geetika Shah answered |
where ‘a’ can take any evenvalue.
Then which one of the following is true?- a)
- b)
- c)
- d)
Correct answer is option 'B'. Can you explain this answer?
Then which one of the following is true?a)
b)
c)
d)
|
|
Anand Kumar answered |
1 1
I = { [ (sinx)/√x ] dx J = { [ (cosx)/√x ] dx
0 0
replace √x by "y"
1 1
I = { 2(siny)dy J = { 2(cosy)dy
0 0
1 1
I = -2 cosy| J = 2sinx|
0 0
I = -2cos1 + 2 J = 2sin1
we know,
sin1 < sin60="" but="" cos1="" /> cos60
J < 2="" and="" i="" /><2 note:-="" 1="" radian="56.67" degree="" note:-="" 1="" radian="56.67">
I = { [ (sinx)/√x ] dx J = { [ (cosx)/√x ] dx
0 0
replace √x by "y"
1 1
I = { 2(siny)dy J = { 2(cosy)dy
0 0
1 1
I = -2 cosy| J = 2sinx|
0 0
I = -2cos1 + 2 J = 2sin1
we know,
sin1 < sin60="" but="" cos1="" /> cos60
J < 2="" and="" i="" /><2 note:-="" 1="" radian="56.67" degree="" note:-="" 1="" radian="56.67">
then the expression for l(m, n) in terms of l(m + 1, n – 1) is
The area enclosed between the curve y = loge (x +e) and the coordinate axes is- a)1
- b)2
- c)3
- d)4
Correct answer is option 'A'. Can you explain this answer?
The area enclosed between the curve y = loge (x +e) and the coordinate axes is
a)
1
b)
2
c)
3
d)
4
|
Dhruv Kulkarni answered |
The graph of the curve y = loge (x+e) is as shown in the fig.
e -e - 0 + 1=1
Hence the required area is 1 square unit.
Let f(x) = x – [x], for every real number x, where [x] is the integral part of x. Then 
- a)1
- b)2
- c)0
- d)1/2
Correct answer is option 'A'. Can you explain this answer?
Let f(x) = x – [x], for every real number x, where [x] is the integral part of x. Then
a)
1
b)
2
c)
0
d)
1/2
|
Chirag Verma answered |
[∵ x is an odd function]
Thus, putting value in equation (1) we get 
Let f : [– 1, 2] → [0, ∞) be a continuous function such that f (x) = f (1 – x) for all x ∈ [–1, 2]
and R2 be the area of the region bounded by y = f (x), x = –1, x = 2, and the x-axis.
Then- a)R1 = 2R2
- b)R1 = 3R2
- c)2R1 = R2
- d)3R1 = R2
Correct answer is option 'C'. Can you explain this answer?
Let f : [– 1, 2] → [0, ∞) be a continuous function such that f (x) = f (1 – x) for all x ∈ [–1, 2]
and R2 be the area of the region bounded by y = f (x), x = –1, x = 2, and the x-axis.Then
a)
R1 = 2R2
b)
R1 = 3R2
c)
2R1 = R2
d)
3R1 = R2
|
|
Ankit Singh answered |
increases in
If f (x) is differentiable and 
equals- a)2/5
- b)–5/2
- c)1
- d)5/2
Correct answer is option 'A'. Can you explain this answer?
If f (x) is differentiable and equals
equalsa)
2/5
b)
–5/2
c)
1
d)
5/2
|
|
Harsh Singhal answered |
Differentiate & put t=2/5
Then the real roots of the equation x2 – f '(x) = 0 are 
The area of the region bounded by the parabola (y – 2)2 = x –1, the tangent of the parabola at the point (2, 3) and the x-axis is:- a)6
- b)9
- c)12
- d)3
Correct answer is option 'B'. Can you explain this answer?
The area of the region bounded by the parabola (y – 2)2 = x –1, the tangent of the parabola at the point (2, 3) and the x-axis is:
a)
6
b)
9
c)
12
d)
3
|
Prisha Das answered |
The given parabola is (y – 2)2 = x – 1 Vertex (1, 2) and it meets x–axis at (5, 0) Also it gives y2 – 4y – x + 5 = 0
So, that equation of tangent to the parabola at (2, 3) is
So, that equation of tangent to the parabola at (2, 3) is
which meets x-axis at (– 4, 0).
In the figure shaded area is the required area.
Let us draw PD perpendicular to y – axis.
In the figure shaded area is the required area.
Let us draw PD perpendicular to y – axis.
Then required area = Ar ΔBOA + Ar (OCPD) – Ar (ΔAPD)
The area bounded by the curves y = |x| –1 and y = –|x| + 1 is- a)1
- b)2
- c)2√2
- d)4
Correct answer is option 'B'. Can you explain this answer?
The area bounded by the curves y = |x| –1 and y = –|x| + 1 is
a)
1
b)
2
c)
2√2
d)
4
|
Anaya Patel answered |
The given lines are
y = x – 1; y = – x – 1;
y = x + 1 and y = – x + 1
which are two pairs of parallel lines and distance between the lines of each pair is √2 Also non parallel lines are perpendicular. Thus lines represents a square of side √2 Hence, area = (√2)2 = 2 sq. units.
y = x – 1; y = – x – 1;
y = x + 1 and y = – x + 1
which are two pairs of parallel lines and distance between the lines of each pair is √2 Also non parallel lines are perpendicular. Thus lines represents a square of side √2 Hence, area = (√2)2 = 2 sq. units.
then g (x + π) equals- a)
- b)g (x) + g (π)
- c)g (x) – g (π)
- d)g (x) . g (π)
Correct answer is option 'B,C'. Can you explain this answer?
then g (x + π) equalsa)
b)
g (x) + g (π)
c)
g (x) – g (π)
d)
g (x) . g (π)
|
|
Rahul Kumar answered |
Integral of cos4t = (sin4t)/4 after limit process we got:-
(sin4x)/4 - (sin0)/4 = (sin4x)/4
hence:-
g(x) = (sin4x)/4
therefore:-
g(x + π) = (sin(4π + 4x))/4 = (sin4x)/4
from option (a):-
((sin4x)/4)/((sin4π)/4) = infinity
from option (b):-
((sin4x)/4) + ((sin4π)/4) = ((sin4x)/4) = g(x + π)
from option (c):-
((sin4x)/4) - ((sin4π)/4) = ( (sin4x)/4) = g( x +π)
from option (d):-
((sin4x)/4).((sin4π)/4) = 0
THUS WE CAN SAY THAT THE OPTION (b) AND (C) IS CORRECT
(sin4x)/4 - (sin0)/4 = (sin4x)/4
hence:-
g(x) = (sin4x)/4
therefore:-
g(x + π) = (sin(4π + 4x))/4 = (sin4x)/4
from option (a):-
((sin4x)/4)/((sin4π)/4) = infinity
from option (b):-
((sin4x)/4) + ((sin4π)/4) = ((sin4x)/4) = g(x + π)
from option (c):-
((sin4x)/4) - ((sin4π)/4) = ( (sin4x)/4) = g( x +π)
from option (d):-
((sin4x)/4).((sin4π)/4) = 0
THUS WE CAN SAY THAT THE OPTION (b) AND (C) IS CORRECT
The solution for x of the equation 
- a)
- b)
- c)2
- d)None
Correct answer is option 'D'. Can you explain this answer?
The solution for x of the equation
a)
b)
c)
2
d)
None
|
|
Anand Kumar answered |
Output of the integration will be sec inverse t i.e, sec^(-1)t
sec^(-1) x - sec^(-1)√2 = π/2
sec^(-1)x - π/4 = π/2
sec^(-1)x = π/2 + π/4 = 3π/4
x = sec(3π/4) = 1/cos(3π/4) = 1/cos(π - π/4)
x = - 1/cos(π/4) = -1/(1/√2) = - √2.
sec^(-1) x - sec^(-1)√2 = π/2
sec^(-1)x - π/4 = π/2
sec^(-1)x = π/2 + π/4 = 3π/4
x = sec(3π/4) = 1/cos(3π/4) = 1/cos(π - π/4)
x = - 1/cos(π/4) = -1/(1/√2) = - √2.
The area (in sq. units) of the region described by {(x, y) : y2 < 2x and y > 4x – 1} is - a)15/64
- b)9/32
- c)7/32
- d)5/64
Correct answer is option 'B'. Can you explain this answer?
The area (in sq. units) of the region described by {(x, y) : y2 < 2x and y > 4x – 1} is
a)
15/64
b)
9/32
c)
7/32
d)
5/64
|
Rohan Yadav answered |
Required area
= Area of ABCD – ar (ABOCD)
Then the correct expression(s) is(are)
Let S be the area of the region enclosed by 
, y = 0, x = 0 and x = 1; then- a)
- b)
- c)
- d)
Correct answer is option 'A,B,D'. Can you explain this answer?
Let S be the area of the region enclosed by , y = 0, x = 0 and x = 1; then
, y = 0, x = 0 and x = 1; thena)
b)
c)
d)
|
|
Gaurav Kumar answered |
First of all let us draw a rough sketch of y = e–x.
At x = 0, y = 1 and at x = 1, y = 1/e
At x = 0, y = 1 and at x = 1, y = 1/e
∴
is decreasing on (0, 1)Hence its graph is as shown in figure given below
Now, S = area exclosed by curve = ABRO
and area of rectangle ORBM = 1/e
Now S < area of rectangle APSO + area of rectangle CSRN
Now S < area of rectangle APSO + area of rectangle CSRN
For more accurate result for c ∈ (a, b), we can use 
that for 
then f(x) is of maximum degree
all x > 0. Then
The area enclosed between the curves y = ax2 and x = ay2 (a > 0) is 1 sq. unit, then the value of a is- a)1/√3
- b)1/2
- c)1
- d)1/3
Correct answer is option 'A'. Can you explain this answer?
The area enclosed between the curves y = ax2 and x = ay2 (a > 0) is 1 sq. unit, then the value of a is
a)
1/√3
b)
1/2
c)
1
d)
1/3
|
Nidhi Choudhary answered |
y = ax2 and x = ay2
Points of intersection are O (0, 0) and
The area bounded by the curves y = lnx, y = ln |x|,y=| ln x | and y = | ln |x| | is- a)4sq. units
- b)6 sq. units
- c)10 sq. units
- d)none of these
Correct answer is option 'A'. Can you explain this answer?
The area bounded by the curves y = lnx, y = ln |x|,y=| ln x | and y = | ln |x| | is
a)
4sq. units
b)
6 sq. units
c)
10 sq. units
d)
none of these
|
|
Rutuja Desai answered |
First we draw each curve as separate graph
NOTE : Graph of y = | f(x) | can be obtained from the graph of the curve y = f(x) by drawing the mirror image of the portion of the graph below x-axis, with respect to x-axis.
Clearly the bounded area is as shown in the following figure.
Clearly the bounded area is as shown in the following figure.
The parabolas y2 = 4x and x2 = 4y divide the square region bounded by the lines x = 4, y = 4 and the coordinate axes. If S1 , S2 , S3 are respectively the areas of these parts numbered from top to bottom; then S1 : S2 : S3 is - a)1 : 2 : 1
- b)1 : 2 : 3
- c)2 : 1 : 2
- d)1 : 1 : 1
Correct answer is option 'D'. Can you explain this answer?
The parabolas y2 = 4x and x2 = 4y divide the square region bounded by the lines x = 4, y = 4 and the coordinate axes. If S1 , S2 , S3 are respectively the areas of these parts numbered from top to bottom; then S1 : S2 : S3 is
a)
1 : 2 : 1
b)
1 : 2 : 3
c)
2 : 1 : 2
d)
1 : 1 : 1
|
Tanvi Bajaj answered |
Intersection points of x2 = 4y and y2 = 4x are (0, 0) and (4, 4). The graph is as shown in the figure.
∴ S1 : S2 :S3 = 1 : 1:1
PASSAGE - 4 f (x) = 1 + 2x + 3x2 + 4x3.
Let s be the sum of all distinct real roots of f (x) and let t = |s|.Q. The real numbers lies in the interval- a)
- b)
- c)
- d)
Correct answer is option 'C'. Can you explain this answer?
PASSAGE - 4
f (x) = 1 + 2x + 3x2 + 4x3.
Let s be the sum of all distinct real roots of f (x) and let t = |s|.
Let s be the sum of all distinct real roots of f (x) and let t = |s|.
Q. The real numbers lies in the interval
a)
b)
c)
d)
|
Pranav Ghosh answered |
f ( x) = 4x3 + 3x2 + 2x+1
∵ f (x) is a cubic polynomial
∴ It has at least one real root.
Also f '(x) =12x2 + 6x+ 2 = 2(6x2 +3x+1)
∵ f (x) is a cubic polynomial
∴ It has at least one real root.
Also f '(x) =12x2 + 6x+ 2 = 2(6x2 +3x+1)
∴ f (x) is strictly increasing function
⇒ There is only one real root of f (x) = 0
⇒ There is only one real root of f (x) = 0
∴ Real root lies between 
and hence 
and hence PASSAGE - 2 Consider the functions defined implicitly by the equation y3 – 3y + x = 0 on various intervals in the real line. If x ∈(-∞, - 2) ∪ (2,∞) , the equation implicitly defines a unique real valued differentiable function y = f (x). If x ∈(-2, 2) , the equation implicitly defines a unique real valued differentiable function y = g(x) satisfying g(0) = 0.Q. 
- a)2g (–1)
- b)0
- c)–2g (1)
- d)2g (1)
Correct answer is option 'D'. Can you explain this answer?
PASSAGE - 2
Consider the functions defined implicitly by the equation y3 – 3y + x = 0 on various intervals in the real line. If x ∈(-∞, - 2) ∪ (2,∞) , the equation implicitly defines a unique real valued differentiable function y = f (x). If x ∈(-2, 2) , the equation implicitly defines a unique real valued differentiable function y = g(x) satisfying g(0) = 0.
Q.
a)
2g (–1)
b)
0
c)
–2g (1)
d)
2g (1)
|
Mahi Sengupta answered |
For y = g(x), we have y3 – 3y + x = 0
⇒ [g (x) ]3 - 3[ g (x)] + x=0 ...(1)
Putting x = –x, we get
⇒ [g (-x)]3 - 3[ g (-x)] - x = 0 ...(2)
Adding equations (1) and (2) we get
⇒ [g (x) ]3 - 3[ g (x)] + x=0 ...(1)
Putting x = –x, we get
⇒ [g (-x)]3 - 3[ g (-x)] - x = 0 ...(2)
Adding equations (1) and (2) we get
For g(0) = 0, we should have g(x) + g(–x) = 0
[∵ From other factor we get g(0) = ±√3]
[∵ From other factor we get g(0) = ±√3]
⇒ g(x) is an odd function
equal to a.
The area of the region described by
- a)
- b)
- c)
- d)
Correct answer is option 'C'. Can you explain this answer?
The area of the region described by
a)
b)
c)
d)
|
|
Rishika Khanna answered |
Given curves are x2 + y2 = 1 and y2 = 1 –x. Intersecting points are x = 0, 1
Area of shaded portion is the required area.
So, Required Area = Area of semi-circle + Area bounded by parabola
So, Required Area = Area of semi-circle + Area bounded by parabola
(∵ radius of circle = 1)
The value of the integral 
- a)
- b)
- c)-1
- d)1
Correct answer is option 'B'. Can you explain this answer?
The value of the integral
a)
b)
c)
-1
d)
1
|
Kolli Lokesh Reddy answered |
Option b is the correct answer .
If for a real number y, [y] is the greatest integer less than or equal to y, then the value of the integral 
- a)– π
- b)0
- c)– π / 2
- d) π /2
Correct answer is option 'C'. Can you explain this answer?
If for a real number y, [y] is the greatest integer less than or equal to y, then the value of the integral
a)
– π
b)
0
c)
– π / 2
d)
π /2
|
|
Harshad Unni answered |
In the range 
we have to find the value of
we have to find the value of
and 
then constants A and B are
The value of 
- a)π
- b)απ
- c)π/2
- d)2π
Correct answer is option 'C'. Can you explain this answer?
The value of
a)
π
b)
απ
c)
π/2
d)
2π
|
Vikash Yadav answered |
Substitute tan x = sin x / cos x in the given integral.
the take LCM
After this use property of definite integral i.e., integral 0 to a f(x) dx = intergal 0 to a f(a-x)dx
After this add 2 eqns and u will get
2I = integral (0 to pi/2) 1 dx = x | from 0 to pi/2
then 2I = pi/2 - 0
I=pi/4
try it by yourself !!!!
Let a, b, c be non-zero real numbers such that
Then the quadratic equation ax2 + bx +c= 0 has- a)no root in (0, 2)
- b)at least one root in (0, 2)
- c)a double root in (0, 2)
- d)two imaginary roots
Correct answer is option 'B'. Can you explain this answer?
Let a, b, c be non-zero real numbers such that
Then the quadratic equation ax2 + bx +c= 0 has
a)
no root in (0, 2)
b)
at least one root in (0, 2)
c)
a double root in (0, 2)
d)
two imaginary roots
|
Rajeev Choudhary answered |
Now we know that if
then it means thatf (x) is + ve on some part of (α, β) and – ve on other part of (α, β).
But here 1 + cos8 x is always + ve,
∴ ax2 + bx + c is + ve on some part of [1 , 2] and – ve on other part [1, 2]
∴ ax2 + bx + c= 0 has at least one root in (1, 2).
⇒ ax2 + bx + c = 0 has at least one root in (0,2).
then the possible values of m and M are
- a)1/2
- b)0
- c)1
- d)–1/2
Correct answer is option 'A'. Can you explain this answer?
a)
1/2
b)
0
c)
1
d)
–1/2
|
|
Anand Kumar answered |
Question seems to be incomplete.
PASSAGE - 4 f (x) = 1 + 2x + 3x2 + 4x3.
Let s be the sum of all distinct real roots of f (x) and let t = |s|.Q. The area bounded by the curve y = f (x) and the lines x = 0, y = 0 and x = t, lies in the interval- a)
- b)
- c)(9, 10)
- d)
Correct answer is option 'A'. Can you explain this answer?
PASSAGE - 4
f (x) = 1 + 2x + 3x2 + 4x3.
Let s be the sum of all distinct real roots of f (x) and let t = |s|.
Let s be the sum of all distinct real roots of f (x) and let t = |s|.
Q. The area bounded by the curve y = f (x) and the lines x = 0, y = 0 and x = t, lies in the interval
a)
b)
c)
(9, 10)
d)
|
|
Anand Roy answered |
y = f (x), x = 0,y= 0 and x = t bounds the area as shown in the figure
∴ Required area is given by
denotes the greatest integer function, is equal to :- a)1
- b)-1
- c)
- d)
Correct answer is option 'C'. Can you explain this answer?
denotes the greatest integer function, is equal to :a)
1
b)
-1
c)
d)
|
Devika Roy answered |
Adding two values of I in eqn s (1) & (2), We get
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