Q&A - Ask Doubts and Get Answers

Sort by :
Clear All
Q

Q : 9   Find the area of the smaller region bounded by the ellipse  \small \frac{x^2}{a^2}+\frac{y^2}{b^2}=1  and the line  \small \frac{x}{a}+\frac{y}{b}=1.
 

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  9 y^2 = x ^3, where the normal to the curve makes equal intercepts with the axes are

A ) \left ( 4 , \pm \frac{8}{3} \right )\\\\ .\: \: \: \: \: B ) \left ( 4 , \frac{-8}{3} \right ) \\\\ . \: \: \: \: \: C) \left ( 4 , \pm \frac{3}{8} \right ) \\\\ . \: \: \: \: D ) \left ( \pm 4 , \frac{3}{8} \right )

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 x^2 = 4 y 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,...

22) The normal at the point (1,1) on the curve 2y + x ^2 = 3  is


(A) x + y = 0

(B) x – y = 0

(C) x + y +1 = 0

(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 (1,1) Now, the equation of normal with point (1,1) and slope = 1 Hence, the correct answer is (B)

21) The line y is equal to  mx+1 is a tangent to the curve y^2 = 4xif the value of m is
(A) 1

(B) 2

(C) 3

(D)1/2

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 x = t^2 + 3t - 8, y = 2t^2 - 2t - 5 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

19) A cylindrical tank of radius 10 m is being filled with wheat at the rate of 314
cubic metre per hour. Then the depth of the wheat is increasing at the rate of

(A) 1 m/h

(B) 0.1 m/h

(C) 1.1 m/h

(D) 0.5 m/h

It is given that Volume of cylinder (V) =  Hence, (A) is correct answer

18) Show that height of the cylinder of greatest volume which can be inscribed in a right circular cone of height h and semi vertical angle a is one-third that of the cone and the greatest volume of cylinder is

\frac{4}{27}\pi h ^3 \tan ^2 \alpha
 

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   

17) Show that the height of the cylinder of maximum volume that can be inscribed in a sphere of radius R is   \frac{2 R }{\sqrt 3 }

 . Also, find the maximum volume.

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

16) Let f be a function defined on [a, b] such that f (x) > 0, for allx \: \: \epsilon \: \: ( a,b). Then prove that f is an increasing function on (a, b).

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)

15) Show that the altitude of the right circular cone of maximum volume that can be inscribed in a sphere of radius r is  4r/3 

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    

14) Find the absolute maximum and minimum values of the function f given by
f (x) = \cos ^2 x + \sin x , x \epsilon [ 0 , \pi ]

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

13) Find the points at which the function f given byf(x) = (x-2)^4(x+1)^3has  

(i) local maxima (ii) local minima (iii) point of inflexion

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...

12) A point on the hypotenuse of a triangle is at distance a and b from the sides of the triangle.

Show that the minimum length of the hypotenuse is

( a ^{\frac{2}{3}}+ b ^\frac{2}{3}) ^ \frac{3}{2}

 

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

11) A window is in the form of a rectangle surmounted by a semicircular opening. The total perimeter of the window is 10 m. Find the dimensions of the window to admit maximum light through the whole opening.

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...

10) The sum of the perimeter of a circle and square is k, where k is some constant. Prove that the sum of their areas is least when the side of square is double the radius of the circle.

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

9) A tank with rectangular base and rectangular sides, open at the top is to be constructed so that its depth is 2 m and volume is 8 m3. If building of tank costs Rs 70 per sq metres for the base and Rs 45 per square metre for sides. What is the cost of least expensive tank?

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...

8) Find the maximum area of an isosceles triangle inscribed in the ellipse \frac{x ^2 }{a^2}+ \frac{y^2}{b^2 } = 1 with its vertex at one end of the major axis.

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...

7) Find the intervals in which the function f given by f ( x) = x ^3 + \frac{1}{x^3} , x \neq 0 

ii) decreasing 

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)

7) Find the intervals in which the function f given by f (x) = x ^3 + \frac{1}{x^3}, x \neq 0

i)increasing

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)
Exams
Articles
Questions