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The maximum distance from origin of a point on the curve
Distance of origin from
∴ Maximum distance from origin = a + b
If 2a + 3b + 6c = 0, (a, b, c ∈ R) then the quadratic equation ax^{2} + bx + c = 0 has
Also f (x) is continuous and differentiable in [0, 1] and [0, 1]. So by Rolle’s theorem, f ' (x) = 0.
i.e ax^{2} + bx + c = 0 has at least one root in [0, 1].
If th e function f ( x) = 2x^{3}  9ax^{2} + 12a^{2}x+1 , whe re a > 0 , attains its maximum and minimum at p and q respectively such that p 2 = q , then a equals
f (x) = 2x^{3}  9ax^{2} + 12a^{2}x +1
f '(x) = 6x^{2}  18ax + 12a^{2}; f"(x) = 12x 18a
For max. or min.
6x^{2}  18ax + 12a^{2} = 0 ⇒ x^{2}  3ax + 2a^{2} = 0
⇒ x = aorx = 2a.At x = a max. and at x = 2a min
∴ p = a and q = 2a
As per question p^{2} = q
∴ a = 2a ⇒ a = 2ora = 0
but a > 0, therefore, a = 2.
A point on th e parabola y^{2} = 18x at which the ordinate increases at twice the rate of the abscissa is
A function y = f (x) h as a second or der der iva tive f "( x) = 6(x 1). If its graph passes through the point (2,1) and at that point the tangent to the graph is y = 3x – 5, then the function is
f ''(x) = 6(x  1). Inegrating, we get
f '( x) = 3x^{2} 6 x+c
Slope at (2, 1) = f '(2) = c = 3
[∴ slope of tangent at (2,1) is 3]
∴ f '( x) = 3x^{2}  6 x + 3 = 3(x 1)2
Inegrating again, we get f (x) = (x 1)^{3}+D
The curve passes through (2, 1)
⇒1= (21)^{3} +D ⇒ D = 0
∴ f (x) = (x – 1)^{3}
The normal to the curve x = a(1 + cosθ), y = a sinθ at ‘θ’ always passes through the fixed point
∴ The slope of the normal at θ = tan θ
∴ The equation of the normal at θ is
y a sin θ = tan θ( x a a cosθ)
which always passes through (a, 0)
If 2a + 3b + 6c = 0, then at least one root of the equation ax^{2} + bx +c= 0 lies in the interval
Let us defin e a function
Being polynomial, it is continuous and differentiable, also,
∴ f (x) satisfies all condition s of Rolle’s theor em therefore f ’(x) = 0 has a root in (0, 1)
i.e. ax^{2} + bx +c= 0 has at lease one root in (0, 1)
Area of the greatest rectangle that can be inscribed in the ellipse
Area of rectangle ABCD = 2a cosθ (2b sin θ) = 2ab sin2θ
⇒ Area of greatest rectangle is equal to 2ab When sin 2θ = 1 .
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
x =a (cos θ + θ sinθ
.....(1)
From equations (1) and (2), we get
Equation of normal at ' θ ' is y – a (sin θ – θ cos θ)
Clearly this is an equation of straight line which is at a constant distance ‘a’ from origin.
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 cm ^{3}/min.
When the thickness of ice is 5 cm,then the rate at which the thickness of ice decreases is
Given that
If the equation has a positive root x = a , then the equation has a positive root, which is
Again f (x) has root a, ⇒ f (a)=0
∴ f (0) = f (a)
∴ By Roll’s theorem f'(x) = 0 has root between (0, α)
Hence f '(x) has a positive root smaller than a.
The function has a local minimum at
has local min at x = 2.
A triangular park is enclosed on two sides by a fence and on the third side by a straight river bank. The two sides having fence are of same length x. The maximum area enclosed by the park is
Maximum value of sinq is 1 at
A value of c for which conclusion of Mean Value Theorem holds for the function f (x) = log_{e} x on the interval [1, 3] is
Using Lagrange's Mean Value Theorem
Let f (x) be a function defined on [a, b]
The function f (x) = tan^{–1}(sin x + cos x) is an increasing function in
Given f (x) = tan^{–1} (sin x + cos x)
if f ’ (x) > O then f (x) is increasing function.
Hence f (x) is increasing, if
Hence, f (x) is increasing when
If p and q are positive real numbers such that p^{2} + q^{2} = 1, then the maximum value of (p + q) is
Given that p^{2} + q^{2} = 1
∴ p = cosθ and q = sinθ
Then p +q = cos θ + sinθ
We know that
Suppose the cubic x^{3} – px + q has three distinct real roots where p > 0 and q > 0. Then which one of the following holds?
How many real solutions does the equation
x^{7} + 14x^{5} + 16x^{3} + 30x – 560 = 0 have?
Let f (x) = x^{7} + 14x^{5} + 16x^{3} + 30x –560
⇒The curve y = f (x) crosses xaxis only once.
∴ f (x) = 0 has exactly one real root.
Let f (x) = x  x  and g (x) = sin x.
Statement1 : gof is differentiable at x = 0 and its derivative is continuous at that point.
Statement2 : gof is twice differentiable at x = 0.
Given that f (x) = x  x  and g (x) = sin x
So that go f (x) = g (f (x)) = g (x  x ) = sin x  x 
Here we observe L (go f )' (0) = 0 = R (go f)' (0)
⇒ go f is differentiable at x = 0 and (go f)' is continuous at x = 0
∴ L(go f)'' (0) ≠ R (go f )'' (0)
⇒ go f (x) is not twice differentiable at x = 0.
∴ Statement  1 is true but statement 2 is false.
Given P(x) = x^{4} + ax^{3} + 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] :
We have P (x) = x^{4} + ax^{3} + bx^{2} + cx + d
⇒ P' (x) = 4x^{3} + 3ax^{2} + 2bx + c
But P' (0) = 0 ⇒ c = 0
∴ P(x) = x^{4} + ax^{3} + bx^{2} + d
As given that P (– 1) < P (a)
⇒ 1 – a + b + d < 1 + a + b + d ⇒ a > 0
Now P ' (x) = 4x^{3} + 3ax^{2} +2bx = x (4x^{2} + 3ax + 2b)
As P' (x) = 0, there is only one solution x = 0, therefore 4x^{2} + 3ax + 2b = 0 should not have any real roots i.e. D < 0
∴ P (x) is an increasing function on (0,1) ∴ P (0) < P (a)
Similarly we can prove P (x) is decreasing on (– 1, 0)
∴ P (– 1) > P (0)
So we can conclude that
Max P (x) = P (1) and Min P (x) = P (0) ⇒ P(–1) is not minimum but P (1) is the maximum of P.
The equation of the tangent to the curve that is parallel to the xaxis, is
Since tangent is parallel to xaxis,
Equation of tangent is y – 3 = 0 (x – 2) ⇒ y = 3
If f has a local minimum at x = – 1 , then a possible value of k is
This is true where k = – 1
Let f : R → R be a continuous function defined by
The shortest distance between line y – x = 1 and curve x = y^{2} is
Shortest distance between two curve occurred along the common normal Slope of normal to y^{2} = x at point P(t^{2}, t) is – 2t and slope of line y – x = 1 is 1.
As they are perpendicular to each other
and shortest distance
So shortest distance between them is
A spherical balloon is filled with 4500π cubic meters of helium gas. If a leak in the balloon causes the gas to escape at the rate of 72π cubic meters per minute, then the rate (in meters per minute) at which the radius of the balloon decreases 49 minutes after the leakage began is :
Volume of spherical balloon
Differentiating both the sides, w.r.t 't' we get,
Let a, b ∈ R be such that the function f given by f (x) = ln x+ bx^{2} + ax, x ≠ 0 has extreme values at x = –1 and x = 2
Statement1 : f has local maximum at x = –1 and at x = 2.
...(ii)
On solving (i) and (ii) we get
So maxima at x = –1,2
Hence both the statements are true and statement 2 is a correct explanation for 1.
A line is drawn through the point (1,2) to meet the coordinate axes at P and Q such that it forms a triangle OPQ, where O is the origin. If the area of the triangle OPQ is least, then the slope of the line PQ is :
Equation of a line passing through (x_{1},y_{1}) having slope m is given by y – y_{1} = m (x–x_{1})
Since the line PQ is passing through (1,2) therefore its equation is
(y – 2) = m (x – 1)
where m is the slope of the line PQ.
Now, point P (x,0) will also satisfy the equation of PQ
∴ y –2 = m (x –1)
⇒ 0 – 2 = m (x – 1)
Similarly, point Q (0,y) will satisfy equation of PQ
The intercepts on xaxis made by tangents to the curve, which are parallel to the line y = 2x, are equal to :
∴ equation of tangent is
y – 2 = 2(x – 2) or y + 2 = 2(x + 2)
⇒ xintercept = ± 1.
If f and g are differentiable functions in [0, 1] satisfying f (0) = 2 = g(1), g(0) = 0 and f (1) = 6, then for some c ∈[0,1]
Since, f and g both are continuous functions on [0, 1] and differentiable on (0, 1) then
Let f(x) be a polynomial of degree four having extreme values at x = 1 and x = 2. then f(2) is equal to :
Hence, f(x) = 2x^{2} + a_{2}x^{3} + a_{3}x^{4}
f '(x) = 4x + 3a_{2}x^{2} + 4a_{3}x^{3}
As given : f '(1) = 0 and f '(2) = 0
Hence, 4 + 3a_{2} + 4a_{3} = 0 ...(1)
and 8 + 12a_{2} + 32a_{3} = 0 ...(2)
By 4x (eq1) – eq (2), we get
16 + 12a_{2} + 16a_{3} – (8 + 12a_{2} + 32a_{3}) = 0
⇒ 8 – 16a_{3} = 0 ⇒ a_{3} = 1/2
and by eqn. (1), 4 + 3a_{2} + 4/2 = 0 ⇒ a_{2} = –2
Consider :
A normal to y = f(x) at also passes through the point :
This equation is satisfied only by the point
A wire of length 2 units is cut into two parts which are bent respectively to form a square of side = x units and a circle of radius = r units. If the sum of the areas of the square and the circle so formed is minimum, then:
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