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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  

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G Gautam harsolia
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...

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G Gautam harsolia
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,...

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G Gautam harsolia
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)

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G Gautam harsolia
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

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G Gautam harsolia
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

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G Gautam harsolia
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   

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G Gautam harsolia
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

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G Gautam harsolia
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)

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G Gautam harsolia
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    

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G Gautam harsolia
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

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G Gautam harsolia
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...

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G Gautam harsolia
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

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G Gautam harsolia
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...

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G Gautam harsolia
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

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G Gautam harsolia
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...

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G Gautam harsolia
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...

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G Gautam harsolia
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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G Gautam harsolia
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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