MEDIUM
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Let the tangent to the graph of y =fx at the point x =a be parallel to x -axis and let f'a- h>0 and f'a + h < 0 where h is a very small +ve quantity. Then, the ordinate at x =a is -

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Important Questions on Application of Derivatives

MEDIUM
If x=-1 and x=2 are extreme points of fx=αlogx+βx2+x, then 
HARD
If a right circular cone, having maximum volume, is inscribed in a sphere of radius 3 cm, then the curved surface area (in cm2) of this cone is :
MEDIUM
From the top of a 64 metres high tower, a stone is thrown upwards vertically with the velocity of 48 m/s. The greatest height (in metres) attained by the stone, assuming the value of the gravitational acceleration g=32 m/s2, is:
MEDIUM
The maximum volume in cubic m of the right circular cone having slant height 3 m is:
EASY
If fx=xx2+1 is an increasing function then the value of x lies in
HARD
The maximum area of a rectangle that can be inscribed in a circle of radius 2 units is
HARD
Let k and K be the minimum and the maximum values of the function fx=1+x0.61+x0.6 in 0, 1, respectively, then the ordered pair (k, K) is equal to:
EASY
The maximum value of the function fx=2x3-15x2+36x+4 is attained at
HARD
The maximum area (in sq. units) of a rectangle having its base on the x- axis and its other two vertices on the parabola, y=12-x2 such that the rectangle lies inside the parabola, is :
HARD
If the function f given by fx=x3-3a-2x2+3ax+7, for some aR is increasing in 0, 1 and decreasing in 1, 5, then a root of the equation, fx-14x-12=0, x1 is :
MEDIUM
Twenty meters of wire is available for fencing off a flower-bed in the form of a circular sector. Then the maximum area (in sq. m) of the flower-bed, is:
MEDIUM
The difference between the greatest and the least value of fx=2sinx+sin2x, x0,3π2 is
HARD
The maximum value of fx=logxx (x0,x1) is
HARD
A solid hemisphere is mounted on a solid cylinder, both having equal radii. If the whole solid is to have a fixed surface area and the maximum possible volume, then the ratio of the height of the cylinder to the common radius is
EASY
Let M and m be respectively the absolute maximum and the absolute minimum values of the function, fx=2x3-9x2+12x+5 in the interval [0,3] . Then M-m is equal to
HARD
Among all sectors of a fixed perimeter, choose the one with maximum area. Then the angle at the center of this sector (i.e., the angle between the bounding radii) is-
HARD
Let fx=α x2-2+1x where α is a real constant. The smallest α for which fx0 for all x>0 is-
HARD
The maximum value of fx=x4+x+x2 on [-1, 1] is
MEDIUM
The least value of αR for which, 4αx2+1x 1, for all x>0, is 
EASY
If at x=1, the function x4-62x2+ax+9 attains its local maximum value, on the interval 0,2, then the value of a is