Class 12 Mathematics · 67 questions · 0 with answers
For the curve y = 5x – 2x3, if x increases at the rate of 2 units/sec, then how fast is the slope of curve changing when x = 3? Solution Slope of curve = = 5 – 6x2 d dy dx ⇒ = –12x. dt dx dt 120 MATHEMATICS = –12 . (3) . (2) = –72 units/sec. Thus, slope of curve is decreasing at the rate of 72 units/sec when x is increasing at the rate of 2 units/sec. π
Water is dripping out from a conical funnel of semi-vertical angle at the uniform rate of 2 cm2 /sec in the surface area, through a tiny hole at the vertex of the bottom. When the slant height of cone is 4 cm, find the rate of decrease of the slant height of water. Solution If s represents the surface area, then d t = 2cm /sec π π 2 s = π r.l = πl . sin . l= l
Find the angle of intersection of the curves y2 = x and x2 = y. Solution Solving the given equations, we have y2 = x and x2 = y ⇒ x4 = x or x4 – x = 0 ⇒ x (x3 – 1) = 0 ⇒ x = 0, x = 1 Therefore, y = 0, y = 1 i.e. points of intersection are (0, 0) and (1, 1) dy dy 1 Further y2 = x ⇒ 2y =1 ⇒ = dx dx 2y and x2 = y ⇒ = 2x. APPLICATION OF DERIVATIVES 121 At (0, 0), the slope of the tangent to the curve y2 = x is parallel to y-axis and the tangent to the curve x2 = y is parallel to x-axis. π ⇒ angle of intersection = At (1, 1), slope of the tangent to the curve y2 = x is equal to and that of x2 = y is 2. 2– 2 3 3 tan θ = 1+1 = 4 . ⇒ θ = tan–1 4 –π π
Prove that the function f (x) = tanx – 4x is strictly decreasing on , . 3 3 Solution f (x) = tan x – 4x ⇒ f ′(x) = sec2x – 4 –π π When < x < , 1 < secx < 2 3 3 Therefore, 1 < sec2x < 4 ⇒ –3 < (sec2x – 4) < 0 –π π Thus for < x < , f ′(x) < 0 3 3 –π π Hence f is strictly decreasing on , . 3 3 4 x3
Determine for which values of x, the function y = x4 – is increasing and for which values, it is decreasing. 4 x3 dy Solution y = x4 – ⇒ = 4x3 – 4x2 = 4x2 (x – 1) 3 dx 122 MATHEMATICS Now, = 0 ⇒ x = 0, x = 1. Since f ′ (x) < 0 ∀x ∈ (– ∞, 0) ∪ (0, 1) and f is continuous in (– ∞, 0] and [0, 1]. Therefore f is decreasing in (– ∞, 1] and f is increasing in [1, ∞). Note: Here f is strictly decreasing in (– ∞, 0) ∪ (0, 1) and is strictly increasing in (1, ∞).
Show that the function f (x) = 4x3 – 18x2 + 27x – 7 has neither maxima nor minima. Solution f (x) = 4x3 – 18x2 + 27x – 7 f ′ (x) = 12x2 – 36x + 27 = 3 (4x2 – 12x + 9) = 3 (2x – 3)2 f ′ (x) = 0 ⇒ x = 2 (critical point) 3 3 Since f ′ (x) > 0 for all x < and for all x > 2 2 Hence x = is a point of inflexion i.e., neither a point of maxima nor a point of minima. x= is the only critical point, and f has neither maxima nor minima.
Using differentials, find the approximate value of 0.082 Solution Let f (x) = x Using f (x + ∆x) f (x) + ∆x . f ′(x), taking x = .09 and ∆x = – 0.008, we get f (0.09 – 0.008) = f (0.09) + (– 0.008) f ′ (0.09) 1 0.008 ⇒ 0.082 = 0.09 – 0.008 . 2 0.09 = 0.3 – 0.6 = 0.3 – 0.0133 = 0.2867. APPLICATION OF DERIVATIVES 123 x2 y 2
Find the condition for the curves – = 1; xy = c2 to intersect a 2 b2 orthogonally. Solution Let the curves intersect at (x1, y1). Therefore, x2 y 2 2 x 2 y dy dy b 2 x – = 1 ⇒ – = 0 ⇒ = a 2 b2 a 2 b 2 dx dx a 2 y b 2 x1 ⇒ slope of tangent at the point of intersection (m1) = 2 a y1 dy dy – y − y1 Again xy = c2 ⇒ x + y=0 ⇒ = ⇒ m2 = x . dx dx x 1 b2 For orthoganality, m1 × m2 = – 1 ⇒ 2 = 1 or a2 – b2 = 0.
Find all the points of local maxima and local minima of the function 3 4 3 45 2 f (x) = – x – 8 x – x +105 . 4 2 Solution f ′ (x) = –3x3 – 24x2 – 45x = – 3x (x2 + 8x + 15) = – 3x (x + 5) (x + 3) f ′ (x) = 0 ⇒ x = –5, x = –3, x = 0 f ″(x) = –9x2 – 48x – 45 = –3 (3x2 + 16x + 15) f ″(0) = – 45 < 0. Therefore, x = 0 is point of local maxima f ″(–3) = 18 > 0. Therefore, x = –3 is point of local minima f ″(–5) = –30 < 0. Therefore x = –5 is point of local maxima. 124 MATHEMATICS
Show that the local maximum value of x + is less than local minimum value. 1 dy 1 Solution Let y = x + ⇒ =1– 2, x dx x dx = 0 ⇒ x = 1 ⇒ x = ± 1. d2y 2 d2y d2y = + 3 , therefore (at x = 1) > 0 and (at x = –1) < 0. dx 2 x dx 2 dx 2 Hence local maximum value of y is at x = –1 and the local maximum value = – 2. Local minimum value of y is at x = 1 and local minimum value = 2. Therefore, local maximum value (–2) is less than local minimum value 2.
Water is dripping out at a steady rate of 1 cu cm/sec through a tiny hole at the vertex of the conical vessel, whose axis is vertical. When the slant height of water in the vessel is 4 cm, find the rate of decrease of slant height, where the vertical π angle of the conical vessel is . Solution Given that = 1 cm3/s, where v is the volume of water in the conical vessel. π 3 π l From the Fig.6.2, l = 4cm, h = l cos = l and r = l sin = . 6 2 6 2 1 2 π l2 3 3π 3 Therefore, v = πr h = l= l . 3 3 4 2 24 APPLICATION OF DERIVATIVES 125 dv 3 π 2 dl = l dt 8 dt 3π dl Therefore, 1 = 16.
Find the equation of all the tangents to the curve y = cos (x + y), –2π ≤ x ≤ 2π, that are parallel to the line x + 2y = 0. dy dy Solution Given that y = cos (x + y) ⇒ = – sin (x + y) 1+ ...(i) dx dx dy sin ( x + y ) or =– dx 1+ sin ( x + y ) Since tangent is parallel to x + 2y = 0, therefore slope of tangent = – sin ( x + y ) 1 Therefore, – 1+ sin x + y = – ⇒ sin (x + y) = 1 ( ) 2 .... (ii) Since cos (x + y) = y and sin (x + y) = 1 ⇒ cos2 (x + y) + sin2 (x + y) = y2 + 1 ⇒ 1 = y2 + 1 or y = 0. Therefore, cosx = 0. π Therefore, x = (2n + 1) , n = 0, ± 1, ± 2... 126 MATHEMATICS π 3π π –3π Thus, x = ± , ± , but x = , x = satisfy equation (ii) 2 2 2 2 π –3π Hence, the points are , 0 , ,0 . 2 2 π 1 π Therefore, equation of tangent at , 0 is y = – x – or 2x + 4y – π = 0, and 2 2 2 –3π 1 3π equation of tangent at ,0 is y = – x + or 2x + 4y + 3π = 0. 2 2 2
Find the angle of intersection of the curves y2 = 4ax and x2 = 4by. Solution Given that y2 = 4ax...(i) and x2 = 4by... (ii). Solving (i) and (ii), we get x2 = 4ax ⇒ x4 = 64 ab2 x 4b 1 2 or x (x – 64 ab ) = 0 ⇒ x = 0, 3 2 x = 4a 3 b 3 1 2 2 1 Therefore, the points of intersection are (0, 0) and 4a 3 b 3 , 4a 3 b 3 . dy 4a 2a dy 2 x x Again, y2 = 4ax ⇒ = = and x2 = 4by ⇒ = = dx 2 y y dx 4b 2b Therefore, at (0, 0) the tangent to the curve y2 = 4ax is parallel to y-axis and tangent to the curve x2 = 4by is parallel to x-axis. π ⇒ Angle between curves = 1 2 2 1 a 3 At 4 a 3 b 3 , 4a 3 b 3 , m1 (slope of the tangent to the curve (i)) = 2 b 1 1 2 1 2a 1 a 3 4a 3 b 3 a 3 = = , m (slope of the tangent to the curve (ii)) = =2 2 1 2 b 2 2b b 4a 3 b 3 APPLICATION OF DERIVATIVES 127 1 1 a 3 1 a 3 1 1 m2 – m1 2 – b 2 b 3a . b 3 Therefore, tan θ = 1+ m m = = 1 2 1 1 2 2 a 3 1 a 3 2 a 3 + b3 1+ 2 b 2 b 1 1 3a 3 . b 3 Hence, θ = tan–1 2 2 a 3 + b3
Show that the equation of normal at any point on the curve x = 3cos θ – cos3θ, y = 3sinθ – sin3θ is 4 (y cos3θ – x sin3θ) = 3 sin 4θ. Solution We have x = 3cos θ – cos3θ Therefore, = –3sin θ + 3cos2θ sinθ = – 3sinθ (1 – cos2θ) = –3sin3θ . dθ = 3cos θ – 3sin2θ cosθ = 3cosθ (1 – sin2θ) = 3cos3θ dθ dy cos3 θ sin 3 θ =– . Therefore, slope of normal = + dx sin 3 θ cos3 θ Hence the equation of normal is sin 3 θ y – (3sinθ – sin3θ) = [x – (3cosθ – cos3θ)] cos3 θ ⇒ y cos3θ – 3sinθ cos3θ + sin3θ cos3θ = xsin3θ – 3sin3θ cosθ + sin3θ cos3θ ⇒ y cos3θ – xsin3θ = 3sinθ cosθ (cos2θ – sin2θ) 128 MATHEMATICS = sin2θ . cos2θ = sin4θ or 4 (ycos3 θ – xsin3 θ) = 3 sin4θ.
Find the maximum and minimum values of f (x) = secx + log cos2x, 0 < x < 2π Solution f (x) = secx + 2 log cosx Therefore, f (x) = secx tanx – 2 tanx = tanx (secx –2) f (x) = 0 ⇒ tanx = 0 or secx = 2 or cosx = Therefore, possible values of x are x = 0, or x = π and π 5π x= or x= 3 3 Again, f ′ (x) = sec2x (secx –2) + tanx (secx tanx) = sec3x + secx tan2x – 2sec2x = secx (sec2x + tan2x – 2secx). We note that f ′ (0) = 1 (1 + 0 – 2) = –1 < 0. Therefore, x = 0 is a point of maxima. f ′ (π) = –1 (1 + 0 + 2) = –3 < 0. Therefore, x = π is a point of maxima. π π f ′ = 2 (4 + 3 – 4) = 6 > 0. Therefore, x = is a point of minima. 3 3 5π 5π f ′ = 2 (4 + 3 – 4) = 6 > 0. Therefore, x = is a point of minima. 3 3 APPLICATION OF DERIVATIVES 129 Maximum Value of y at x = 0 is 1+0=1 Maximum Value of y at x = π is –1 + 0 = –1 π 1 Minimum Value of y at x = is 2 + 2 log = 2 (1 – log2) 3 2 5π 1 Minimum Value of y at x = is 2 + 2 log = 2 (1 – log2) 3 2
Find the area of greatest rectangle that can be inscribed in an ellipse x2 y 2 + = 1. a 2 b2 Solution Let ABCD be the rectangle of maximum area with sides AB = 2x and x2 y 2 BC = 2y, where C (x, y) is a point on the ellipse + = 1 as shown in the Fig.6.3. a 2 b2 The area A of the rectangle is 4xy i.e. A = 4xy which gives A2 = 16x2y2 = s (say) x2 2 16b 2 Therefore, s = 16x 1– 2 . b = (a2x2 – x4) a a2 ds 16b 2 ⇒ = 2 . [2a2x – 4x3]. dx a ds a b Again, =0⇒ x= and y = dx 2 2 d 2 s 16b 2 Now, = 2 [2a2 – 12x2] dx 2 a a d 2 s 16b 2 16b 2 At x= , 2 = 2 [2 a 2 − 6 a 2 ] = 2 ( − 4a 2 ) < 0 2 dx a a 130 MATHEMATICS a b Thus at x = ,y= , s is maximum and hence the area A is maximum. 2 2 Maximum area = 4.x.y = 4 . . = 2ab sq units. 2 2
Find the difference between the greatest and least values of the π π function f (x) = sin2x – x, on – , . 2 2 Solution f (x) = sin2x – x ⇒ f ′(x) = 2 cos2 x – 1 1 −π π π π Therefore, f ′(x) = 0 ⇒ cos2x = ⇒ 2x is 3 or 3 ⇒ x = – or 2 6 6 π π π f – = sin (– π) + = 2 2 2 π 2π π f – = sin – + = – 3 + π 6 6 6 2 6 π 2π π f = sin – = 3 – π 6 6 6 2 6 π π π f = sin ( π ) – = – 2 2 2 π π Clearly, is the greatest value and – is the least. 2 2 π π Therefore, difference = + =π 2 2 APPLICATION OF DERIVATIVES 131
An isosceles triangle of vertical angle 2θ is inscribed in a circle of radius π a. Show that the area of triangle is maximum when θ = . Solution Let ABC be an isosceles triangle inscribed in the circle with radius a such that AB = AC. AD = AO + OD = a + a cos2θ and BC = 2BD = 2a sin2θ (see fig. 16.4) Therefore, area of the triangle ABC i.e. ∆ = BC . AD = 2a sin2θ . (a + a cos2θ) = a2sin2θ (1 + cos2θ) 1 2 ⇒ ∆ = a2sin2θ + a sin4θ d∆ Therefore, = 2a2cos2θ + 2a2cos4θ dθ = 2a2(cos2θ + cos4θ) d∆ = 0 ⇒ cos2θ = –cos4θ = cos (π – 4θ) dθ π Therefore, 2θ = π – 4θ ⇒ θ = d 2∆ π 2 = 2a (–2sin2θ – 4sin4θ) < 0 (at θ = ). dθ 6 π Therefore, Area of triangle is maximum when θ = . 132 MATHEMATICS
The abscissa of the point on the curve 3y = 6x – 5x3, the normal at which passes through origin is: 1 1
The two curves x3 – 3xy2 + 2 = 0 and 3x2y – y3 = 2
The tangent to the curve given by x = et . cost, y = et . sint at t = makes with x-axis an angle: π π π
The equation of the normal to the curve y = sinx at (0, 0) is:
The point on the curve y2 = x, where the tangent makes an angle of π with x-axis is 1 1 1 1
The values of a for which y = x2 + ax + 25 touches the axis of x are______. dy a Solution = 0 ⇒ 2x + a = 0 i.e. x= − , dx 2 a2 a Therefore, + a − + 25 = 0 ⇒ a = ± 10 4 2 Hence, the values of a are ± 10.
If f (x) = , then its maximum value is _______. 4 x + 2 x +1 Solution For f to be maximum, 4x2 + 2x + 1 should be minimum i.e. 1 2 1 3 4x2 + 2x + 1 = 4 (x + ) + 1 − giving the minimum value of 4x2 + 2x + 1 = . 4 4 4 Hence maximum value of f = .
Let f have second deriative at c such that f ′(c) = 0 and f″(c) > 0, then c is a point of ______. Solution Local minima. –π π
Minimum value of f if f (x) = sinx in , is _____. 2 2 Solution –1
The maximum value of sinx + cosx is _____. Solution 2. APPLICATION OF DERIVATIVES 135
The rate of change of volume of a sphere with respect to its surface area, when the radius is 2 cm, is______. Solution 1 cm3/cm2 4 3 dv ds dv r v= πr ⇒ = 4πr 2 , s = 4πr 2 ⇒ = 8πr ⇒ = = 1 at r = 2. 3 dr dr ds 2
2 ds 2π dl dl Therefore, = l. = 2πl . dt 2 dt dt dl 1 1 2 when l = 4 cm, dt = 2π.4 .2 = 2 2π = 4π cm/s .
dt dl 1 ⇒ = cm/s. dt 2 3π Therefore, the rate of decrease of slant height = cm/s. 2 3π
to 23.
A spherical ball of salt is dissolving in water in such a manner that the rate of decrease of the volume at any instant is propotional to the surface. Prove that the radius is decreasing at a constant rate.
If the area of a circle increases at a uniform rate, then prove that perimeter varies inversely as the radius.
A kite is moving horizontally at a height of 151.5 meters. If the speed of kite is 10 m/s, how fast is the string being let out; when the kite is 250 m away from the boy who is flying the kite? The height of boy is 1.5 m.
Two men A and B start with velocities v at the same time from the junction of two roads inclined at 45° to each other. If they travel by different roads, find the rate at which they are being seperated..
Find an angle θ, 0 < θ < , which increases twice as fast as its sine.
Find the approximate value of (1.999)5.
Find the approximate volume of metal in a hollow spherical shell whose internal and external radii are 3 cm and 3.0005 cm, respectively.
A man, 2m tall, walks at the rate of 1 m/s towards a street light which is 5 m above the ground. At what rate is the tip of his shadow moving? At what 136 MATHEMATICS rate is the length of the shadow changing when he is 3 m from the base of the light?
A swimming pool is to be drained for cleaning. If L represents the number of litres of water in the pool t seconds after the pool has been plugged off to drain and L = 200 (10 – t)2. How fast is the water running out at the end of 5 seconds? What is the average rate at which the water flows out during the first 5 seconds?
The volume of a cube increases at a constant rate. Prove that the increase in its surface area varies inversely as the length of the side.
x and y are the sides of two squares such that y = x – x2 . Find the rate of change of the area of second square with respect to the area of first square.
Find the condition that the curves 2x = y2 and 2xy = k intersect orthogonally.
Prove that the curves xy = 4 and x2 + y2 = 8 touch each other.
Find the co-ordinates of the point on the curve x+ y = 4 at which tangent is equally inclined to the axes.
Find the angle of intersection of the curves y = 4 – x2 and y = x2.
Prove that the curves y2 = 4x and x2 + y2 – 6x + 1 = 0 touch each other at the point (1, 2).
Find the equation of the normal lines to the curve 3x2 – y2 = 8 which are parallel to the line x + 3y = 4.
At what points on the curve x2 + y2 – 2x – 4y + 1 = 0, the tangents are parallel to the y-axis? –x
Show that the line + = 1, touches the curve y = b . e a at the point where the curve intersects the axis of y.
Show that f (x) = 2x + cot–1x + log ( 1+ x − x ) is increasing in R. APPLICATION OF DERIVATIVES 137
Show that for a ≥ 1, f (x) = 3 sinx – cosx – 2ax + b is decreasing in R.
Show that f (x) = tan–1(sinx + cosx) is an increasing function in 0, .
At what point, the slope of the curve y = – x3 + 3x2 + 9x – 27 is maximum? Also find the maximum slope.
Prove that f (x) = sinx + 3 cosx has maximum value at x = 6 .
If the sum of the lengths of the hypotenuse and a side of a right angled triangle is given, show that the area of the triangle is maximum when the angle between them is .
Find the points of local maxima, local minima and the points of inflection of the function f (x) = x5 – 5x4 + 5x3 – 1. Also find the corresponding local maximum and local minimum values.
A telephone company in a town has 500 subscribers on its list and collects fixed charges of Rs 300/- per subscriber per year. The company proposes to increase the annual subscription and it is believed that for every increase of Re 1/- one subscriber will discontinue the service. Find what increase will bring maximum profit? x2 y2
If the straight line x cosα + y sinα = p touches the curve 2 + = 1, then a b2 prove that a2 cos2α + b2 sin2α = p2.
An open box with square base is to be made of a given quantity of card board c3 of area c2. Show that the maximum volume of the box is cubic units. 6 3
Find the dimensions of the rectangle of perimeter 36 cm which will sweep out a volume as large as possible, when revolved about one of its sides. Also find the maximum volume. 138 MATHEMATICS
If the sum of the surface areas of cube and a sphere is constant, what is the ratio of an edge of the cube to the diameter of the sphere, when the sum of their volumes is minimum?
AB is a diameter of a circle and C is any point on the circle. Show that the area of ∆ ABC is maximum, when it is isosceles.
A metal box with a square base and vertical sides is to contain 1024 cm3. The material for the top and bottom costs Rs 5/cm2 and the material for the sides costs Rs 2.50/cm2 . Find the least cost of the box.
The sum of the surface areas of a rectangular parallelopiped with sides x, 2x and and a sphere is given to be constant. Prove that the sum of their volumes is minimum, if x is equal to three times the radius of the sphere. Also find the minimum value of the sum of their volumes.
to 39: 35. The sides of an equilateral triangle are increasing at the rate of 2 cm/sec. The rate at which the area increases, when side is 10 cm is: 10 2
A ladder, 5 meter long, standing on a horizontal floor, leans against a vertical wall. If the top of the ladder slides downwards at the rate of 10 cm/sec, then the rate at which the angle between the floor and the ladder is decreasing when lower end of ladder is 2 metres from the wall is: 1 1
The curve y = x 5 has at (0, 0) APPLICATION OF DERIVATIVES 139
The equation of normal to the curve 3x2 – y2 = 8 which is parallel to the line x + 3y = 8 is
If the curve ay + x2 = 7 and x3 = y, cut orthogonally at (1, 1), then the value of a is:
If y = x4 – 10 and if x changes from 2 to 1.99, what is the change in y
The equation of tangent to the curve y (1 + x2) = 2 – x, where it crosses x-axis is:
The points at which the tangents to the curve y = x3 – 12x + 18 are parallel to x-axis are:
The tangent to the curve y = e2x at the point (0, 1) meets x-axis at: 1
The slope of tangent to the curve x = t2 + 3t – 8, y = 2t2 – 2t – 5 at the point (2, –1) is: 140 MATHEMATICS 22 6 –6
The two curves x3 – 3xy2 + 2 = 0 and 3x2y – y3 – 2 = 0 intersect at an angle of
The interval on which the function f (x) = 2x3 + 9x2 + 12x – 1 is decreasing is:
Let the f : R → R be defined by f (x) = 2x + cosx, then f :
y = x (x – 3)2 decreases for the values of x given by :
The function f (x) = 4 sin3x – 6 sin2x + 12 sinx + 100 is strictly 3π π
Which of the following functions is decreasing on 0,
The function f (x) = tanx – x
If x is real, the minimum value of x2 – 8x + 17 is
The smallest value of the polynomial x3 – 18x2 + 96x in [0, 9] is
The function f (x) = 2x3 – 3x2 – 12x + 4, has
The maximum value of sin x . cos x is 1 1
At x = , f (x) = 2 sin3x + 3 cos3x is:
Maximum slope of the curve y = –x3 + 3x2 + 9x – 27 is:
f (x) = xx has a stationary point at
The maximum value of is: x 1 1 e
The curves y = 4x2 + 2x – 8 and y = x3 – x + 13 touch each other at the point_____.
The equation of normal to the curve y = tanx at (0, 0) is ________.
The values of a for which the function f (x) = sinx – ax + b increases on R are ______. 2 x 2 –1
The function f (x) = , x > 0, decreases in the interval _______. x4
The least value of the function f (x) = ax + (a > 0, b > 0, x > 0) is ______.