The area of the shaded region ACB is to be found
The given ellipse and the line intersect at following points
Y will always be positive since the shaded region lies above x axis
The required area is

24) The points on the curve , where the normal to the curve makes equal intercepts with the axes are

Given the equation of the curve
We know that the slope of the tangent at a point on a given curve is given by
We know that
At point (a,b)
Now, the equation of normal with point (a,b) and
It is given that normal to the curve makes equal intercepts with the axes
Therefore,
point(a,b) also satisfy the given equation of the curve
Hence, The points on the curve , where the...

23) The normal to the curve passing (1,2) is

(A) x + y = 3

(B) x – y = 3

(C) x + y = 1

(D) x – y = 1

Given the equation of the curve
We know that the slope of the tangent at a point on the given curve is given by
We know that
At point (a,b)
Now, the equation of normal with point (a,b) and
?
It is given that it also passes through the point (1,2)
Therefore,
-(i)
It also satisfies equation -(ii)
By comparing equation (i) and (ii)
Now,...

Given the equation of the curve
We know that the slope of the tangent at a point on the given curve is given by
We know that
At point (1,1)
Now, the equation of normal with point (1,1) and slope = 1
Hence, the correct answer is (B)

Standard equation of the straight line
y = mx + c
Where m is lope and c is constant
By comparing it with equation , y = mx + 1
We find that m is the slope
Now,
we know that the slope of the tangent at a given point on the curve is given by
Given the equation of the curve is
Put this value of m in the given equation
Hence, value of m is 1
Hence, (A) is correct answer

20) The slope of the tangent to the curve at the point

(2,– 1) is

A ) 22/7

B ) 6/7

C ) 7/6

D ) -6 /7

Given curves are
At point (2,-1)
Similarly,
The common value between two is t = 2
Hence, we find the slope of the tangent at t = 2
We know that the slope of the tangent at a given point is given by
Hence, (B) is the correct answer

Let's take radius and height of cylinder = r and h ' respectively
Let's take radius and height of cone = R and h respectively
Volume of cylinder =
Volume of cone =
Now, we have
Now, since are similar
Now,
Now,
Now,
at
Hence, is the point of maxima
Hence proved
Now, Volume (V) at and is
hence proved

The volume of the cylinder (V) =
By Pythagoras theorem in
h = 2OA
Now,
Hence, the point is the point of maxima
Hence, the height of the cylinder of maximum volume that can be inscribed in a sphere of radius R is
and maximum volume is

Let's do this question by taking an example
suppose
Now, also
Hence by this, we can say that f is an increasing function on (a, b)

The volume of a cone (V) =
The volume of the sphere with radius r =
By Pythagoras theorem in we ca say that
V =
Now,
Hence, the point is the point of maxima
Hence, the altitude of the right circular cone of maximum volume that can be inscribed in a sphere of radius r is

Given function is
Now,
Hence, the point is the point of maxima and the maximum value is
And
Hence, the point is the point of minima and the minimum value is

Given function is
Now, for value x close to and to the left of , ,and for value close to and to the right of
Thus, point x = is the point of maxima
Now, for value x close to 2 and to the Right of 2 , ,and for value close to 2 and to the left of 2
Thus, point x = 2 is the point of minima
There is no change in the sign when the value of x is -1
Thus x = -1 is the point...

It is given that
A point on the hypotenuse of a triangle is at a distance a and b from the sides of the triangle
Let the angle between AC and BC is
So, the angle between AD and ED is also
Now,
CD =
And
AD =
AC = H = AD + CD
= +
Now,
When
Hence, is the point of minima
and
AC = =
Hence proved

Let l and bare the length and breadth of rectangle respectively and r will be the radius of circle
The total perimeter of window = perimeter of rectangle + perimeter of the semicircle
=
Area of window id given by (A) =
Now,
Hence, b...

It is given that
the sum of the perimeter of a circle and square is k =
Let the sum of the area of a circle and square(A) =
Now,
Hence, is the point of minima
Hence proved that the sum of their areas is least when the side of the square is double the radius of the circle

Let l , b and h are length , breath and height of tank
Then, volume of tank = l X b X h = 8
h = 2m (given)
lb = 4 =
Now,
area of base of tank = l X b = 4
area of 4 side walls of tank = hl + hl + hb + hb = 2h(l + b)
Total area of tank (A) = 4 + 2h(l + b)
Now,
Hence, b = 2 is the point of minima
So, l = 2 , b = 2 and h = 2 m
Area of base = l X B = 2 X 2 =
building of tank...

Given the equation of the ellipse
Now, we know that ellipse is symmetrical about x and y-axis. Therefore, let's assume coordinates of A = (-n,m) then,
Now,
Put(-n,m) in equation of ellipse
we will get
Therefore, Now
Coordinates of A =
Coordinates of B =
Now,
Lenghth AB(base) =
And height of triangle ABC = (a+n)
Now,
Area of triangle =
Now,
Now,
but n cannot be...

Given function is
Hence, three intervals are their
In interval
Hence, given function is increasing in interval
In interval (-1,1) ,
Hence, given function is decreasing in interval (-1,1)

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