A spherical iron ball 10 cm in radius is coated with a layer of ice of uniform thickness that melts at a rate of 50 cm³/min. When the thickness of ice is 15 cm, then the rate at which the thickness of ice decreases, is
Area of the greatest rectangle that can be inscribed in the ellipse is
Let a point on the ellipse be P (a cosθ, b sinθ)
So, In the rectangle formed as this point as one of the vertices,
(Length)/2 = a cosθ
(Width)/2 = b sinθ
Area of the rectangle = (2a cosθ)(2b sinθ)
= 2ab sin(2θ)
Max (Area of the rectangle) = 2ab
Let f(x) be differentiable for all x. If f (1) =-2 and f '(x)³ ≥2 x ∈ (1/6) then
The normal to the curve x = a ( cos θ + θ sin θ) , y = a (sin θ - θ cos θ) at any point ' θ' is such that
Equation of normal is
(y -θ sinq- cosθ )
The function f (x) = x/2+2/x has a local minimum at
Signs of f ' (x )
f ' (x ) changes sign as x crosses 2.
f (x ) has minima at x = 2.
A value of c for which the conclusion of Mean value Theorem holds for the function f ( x ) = loge x on the interval [1, 3] is
The function f ( x ) = tan-1 (sin x cos x) is an increasing function in
Suppose the cubic x3 - px + q = 0 has three distinct real roots where p > 0 and q > 0. Then, which one of the following holds?
Graph of y = x2 - px + q cuts x-axis at three distinct pints
Maximum at x = -√p/3 , minima at x = √p/3
The shortest distance between the line y - x= 1 and the curve x =y2 is
Condition for shortest distance is slope of tangent to x = y² must be same as slope of line y = x + 1
shortest distance =
Given p ( x ) = x 4 + ax3 + bx 2 + cx + d such that x =0 is the only real root of p ' ( x ) = 0 .
If p ( -1) < p (1) , then in the interval [-1,1]
p ' (x ) has only one change of sign. x = 0 is a point of minima.
P(-1) =1- a + b + d P(0) = d
P(1) =1+ a + b + d
P ( -1) < P (1) , P ( 0) < P (1) , P ( -1) > P ( 0)
P (-1) is not minimum but P(1) is maximum
The equation of the tangent to the curve y= x+4/x2 that is parallel to the x-axis is
(2, 3) is point of contact thus y = 3 is tangent
Let be defined by If f has a local minimum at x = - 1, then a possible value of k is
∴f has a local minimum at x = - 1
Possible value of k is – 1
As f(x) is continuous and 1/3 belongs to range
The normal to the curve x2 + 2xy - 3y2 = 0 at 1,1
slope of normal at (1, 1) = - 1
∴ Equation of normal y -1 = -1(x -1)
y +x= 2
Solving we get point as (1, 1) and (1/-1/3)
So in forth quadrant
Consider the function, f ( x ) =| x - 2 | + | x - 5 | .x∈ R
Statement -1 : f ' ( 4) = 0
Statement -2 : f is continous in [2, 5], differentiable in (2, 5) and f ( 2) = f ( 5)
f (x ) is constant is [2, 5]
A spherical balloon is filled with 4500 π cubic meters of helium gas. If a leak in the balloon causes the gas to escape at the rate of 75 π cubic meters per minute, then the rate (in metres per minute) at which the radius of the balloon decreases 49 minutes after the leakage began is
Let a, b ∈ R be such that the function f given by f ( x ) = In | x | +bx 2 + ax, x ≠ 0 has extreme values at x =-1 and x = 2.
Statement – 1: f has local maximum at x = - 1 and at x = 2.
Statement-2 : a=1/2 and =-1/4
The real number k for which the equation, 2x3 + 3x + k = 0 has two distinct real roots in [0, 1]
f ' ( x ) = 6 x2+ 3
⇒ so f(x) is increasing equation cannot have 2 distinct roots.
The function f ( x ) = sin4 x+ cos4 x is increasing if
Let dt then f decreasing in the interval
The two curves x3 - 3xy2 + 2 = 0 and 3x 2 y - y3 - 2 = 0
mm = -1 ⇒ Curves are orthogonal
The greatest value of f (x) = (x + 1)1/3 - (x - 1)1/3 on [0, 1] is
is decreasing on [0, 1].
Greatest value of f (x ) is f ( 0) = 1 - ( -1) = 2
The function f ( x ) = cot -1 x+ x increases in the interval
The real number x (x > 0) when added to its reciprocal gives the minimum sum at x equals
Signs of f ' (x )
Minimum occurs at x = 1
If the function f ( x ) = 2x3 - 9ax2 + 12a 2x + 1, where a >0 , attains its maximum and minimum at p and q respectively such that p² = q, then 'a' equals
If 2a + 3b + 6c = 0, then at least one root of the equation ax 2 + bx + c = 0 lies in the interval
f ( 0) = df (1) (∵ 2a + 3b + 6x = 0)
By Rolls theorem, at least one root of f ' (x ) =0 lies in (0, 1)
A point on the parabola y2 = 18x at which the ordinate increases at twice the rate of the abscissa, is
The normal to the curve x = a (1 + cos θ) , y = a sin θ at 'θ ' always passes through the fixed point
x = a (1 + cosθ ) , ya = sinθ
Equation of normal at point ( a (1 + cosθ ) ,a sinθ ) is
y - a sinθ = sinθ/cosθ (x-a(1+cosθ))
y cosθ = x- a ) sinθ
It is clear that normal passes through fixed point (a, 0)
Angle between the tangents to the curve y = x 2 - 5x + 6 at the points (2, 0) and (3, 0) is
Product of slopes = -1 ⇒ angle between tangents is π/2
A function is matched below against an interval where it is supposed to be increasing, Which of the following pairs is incorrectly matched?
Let f (x) = x2 + 6x +6 ⇒ f (x) = 3x (x +4)
Increasing in (-∞,-4) Let ( ) 2 2
f2 (x) = 3x2 - 2x + 1