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Chapter, , 6, , APPLICATION OF DERIVATIVES, 6.1 Overview, 6.1.1 Rate of change of quantities, For the function y = f (x),, , d, (f (x)) represents the rate of change of y with respect to x., dx, , Thus if ‘s’ represents the distance and ‘t’ the time, then, , ds, represents the rate of, dt, , change of distance with respect to time., 6.1.2 Tangents and normals, A line touching a curve y = f (x) at a point (x1, y1) is called the tangent to the curve at, ,  dy , that point and its equation is given y − y1 = ,  ( x , y ) ( x – x1 ) .,  dx  1 1, The normal to the curve is the line perpendicular to the tangent at the point of contact,, and its equation is given as:, y – y1 =, , –1, ( x − x1 ),  dy , ,  (x , y ),  dx  1 1, , The angle of intersection between two curves is the angle between the tangents to the, curves at the point of intersection., 6.1.3 Approximation, Since f ′(x) = lim, , ∆x →0, , f ( x + ∆x ) – f ( x), , we can say that f ′(x) is approximately equal, ∆x, , f ( x + ∆x) – f ( x ), ∆x, ⇒ approximate value of f (x + ∆ x) = f (x) + ∆x .f ′ (x)., , to, , 20/04/2018

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118, , MATHEMATICS, , 6.1.4 Increasing/decreasing functions, A continuous function in an interval (a, b) is :, (i), strictly increasing if for all x1, x2 ∈ (a, b), x1< x2 ⇒ f (x1) < f (x2) or for all, x ∈ (a, b), f ′ (x) > 0, (ii), strictly decreasing if for all x1, x2 ∈ (a, b), x1 < x2 ⇒ f (x1) > f (x2) or for all, x ∈ (a, b), f ′(x) < 0, 6.1.5 Theorem : Let f be a continuous function on [a, b] and differentiable in (a, b) then, (i) f is increasing in [a, b] if f ′ (x) > 0 for each x ∈ (a, b), (ii) f is decreasing in [a, b] if f ′ (x) < 0 for each x ∈ (a, b), (iii) f is a constant function in [a, b] if f ′ (x) = 0 for each x ∈ (a, b)., 6.1.6 Maxima and minima, Local Maximum/Local Minimum for a real valued function f, A point c in the interior of the domain of f, is called, (i), , local maxima, if there exists an h > 0 , such that f (c) > f (x), for all x in, (c – h, c + h)., , The value f (c) is called the local maximum value of f ., (ii), , local minima if there exists an h > 0 such that f (c) < f (x), for all x in, (c – h, c + h)., , The value f (c) is called the local minimum value of f., A function f defined over [a, b] is said to have maximum (or absolute maximum) at, x = c, c ∈ [a, b], if f (x) ≤ f (c) for all x ∈ [a, b]., Similarly, a function f (x) defined over [a, b] is said to have a minimum [or absolute, minimum] at x = d, if f (x) ≥ f (d) for all x ∈ [a, b]., 6.1.7 Critical point of f : A point c in the domain of a function f at which either, f ′ (c) = 0 or f is not differentiable is called a critical point of f., Working rule for finding points of local maxima or local minima:, (a), , First derivative test:, (i), , If f ′ (x) changes sign from positive to negative as x increases through, c, then c is a point of local maxima, and f (c) is local maximum value., , 20/04/2018

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APPLICATION OF DERIVATIVES, , (b), , 6.1.8, , 119, , (ii), , If f ′ (x) changes sign from negative to positive as x increases through, c, then c is a point of local minima, and f (c) is local minimum value., , (iii), , If f ′ (x) does not change sign as x increases through c, then c is, neither a point of local minima nor a point of local maxima. Such a, point is called a point of inflection., , Second Derivative test: Let f be a function defined on an interval I and, c ∈ I. Let f be twice differentiable at c. Then, (i), , x = c is a point of local maxima if f ′(c) = 0 and f ″(c) < 0. In this case, f (c) is then the local maximum value., , (ii), , x = c is a point of local minima if f ′ (c) = 0 and f ″(c) > 0. In this case, f (c) is the local minimum value., , (iii), , The test fails if f ′(c) = 0 and f ″ (c) = 0. In this case, we go back to, first derivative test., , Working rule for finding absolute maxima and or absolute minima :, Step 1 : Find all the critical points of f in the given interval., Step 2 : At all these points and at the end points of the interval, calculate the, values of f., Step 3 : Identify the maximum and minimum values of f out of the values, calculated in step 2. The maximum value will be the absolute maximum, value of f and the minimum value will be the absolute minimum, value of f., , 6.2 Solved Examples, Short Answer Type (S.A.), Example 1 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 =, , ⇒, , dy, = 5 – 6x2, dx, , d  dy , dx, ,  = –12x., dt  dx , dt, , 20/04/2018

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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, at the, 4, 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, r, ds, 2, d t = 2cm /sec, , Example 2 Water is dripping out from a conical funnel of semi-vertical angle, , π, π 2, ., l, l=, 4, 2, , s = π r.l = πl . sin, , h, l, /4, , Therefore,, , ds 2π dl, l. =, =, 2 dt, dt, , 2πl ., , dl, dt, , 1, 1, 2, dl, when l = 4 cm, dt = 2π.4 .2 = 2 2π = 4π cm/s ., , Fig. 6.1, , Example 3 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), Further y2 = x ⇒, , and, , x2 = y ⇒, , 2y, , dy, =1, dx, , ⇒, , 1, dy, =, 2y, dx, , dy, = 2x., dx, , 20/04/2018

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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., ⇒ angle of intersection =, , π, 2, , At (1, 1), slope of the tangent to the curve y2 = x is equal to, , 1, 2, 3, 1+ 1 = 4 ., , 2–, tan θ =, , 1, and that of x2 = y is 2., 2, ,  3, ⇒ θ = tan–1  4 , ,  –π π , , ., Example 4 Prove that the function f (x) = tanx – 4x is strictly decreasing on ,  3 3, Solution f (x) = tan x – 4x ⇒ f ′(x) = sec2x – 4, When, , –π, π, < x < , 1 < secx < 2, 3, 3, , Therefore, 1 < sec2x < 4 ⇒ –3 < (sec2x – 4) < 0, Thus for, , –π, π, < x < , f ′(x) < 0, 3, 3, ,  –π π , , ., Hence f is strictly decreasing on ,  3 3, , Example 5 Determine for which values of x, the function y = x4 –, , 4x3, is increasing, 3, , and for which values, it is decreasing., Solution y = x4 –, , 4 x3, 3, , ⇒, , dy, = 4x3 – 4x2 = 4x2 (x – 1), dx, , 20/04/2018

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122, , MATHEMATICS, , Now,, , dy, = 0 ⇒ x = 0, x = 1., dx, , 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, ∞)., Example 6 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, , 3, f ′ (x) = 0 ⇒ x = 2 (critical point), Since f ′ (x) > 0 for all x <, , Hence x =, , x=, , 3, 3, and for all x >, 2, 2, , 3, is a point of inflexion i.e., neither a point of maxima nor a point of minima., 2, , 3, is the only critical point, and f has neither maxima nor minima., 2, , Example 7 Using differentials, find the approximate value of, Solution Let f (x) =, Using f (x + ∆x), , 0.082, , 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), ⇒, , 0.082 =, ,  1 , 0.008, 0.09 – 0.008 .  2 0.09  = 0.3 – 0.6, , , , = 0.3 – 0.0133 = 0.2867., , 20/04/2018

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124, , MATHEMATICS, , Example 10 Show that the local maximum value of x +, , 1, is less than local minimum, x, , value., Solution Let y = x +, , dy, 1, 1, ⇒, =1– 2,, dx, x, x, , dy, 2, dx = 0 ⇒ x = 1 ⇒ x = ± 1., , d2y, d2y, 2, d2y, = + 3 , therefore, (at x = 1) > 0 and, (at x = –1) < 0., dx 2, dx 2, x, 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., Long Answer Type (L.A.), Example 11 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, angle of the conical vessel is, , Solution Given that, , π, ., 6, , dv, = 1 cm 3/s, where v is the volume of water in the, dt, , conical vessel., From the Fig.6.2, l = 4cm, h = l cos, , Therefore, v =, , π, π l, 3, l and r = l sin = ., =, 6, 6 2, 2, , 1 2, π l2 3, 3π 3, l=, l ., πr h =, 3, 3 4 2, 24, , 20/04/2018

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APPLICATION OF DERIVATIVES, , r, , dv, 3 π 2 dl, l, =, dt, 8, dt, , Therefore, 1 =, ⇒, , 125, , 3π, dl, 16., 8, dt, , h, l, /6, , dl, 1, =, cm/s., dt 2 3π, , Therefore, the rate of decrease of slant height =, , 1, cm/s., 2 3π, , Fig. 6.2, , Example 12 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., Solution Given that y = cos (x + y) ⇒, , dy, = – sin (x + y), dx, ,  dy , 1+ dx , , sin ( x + y ), dy, =–, dx, 1+ sin ( x + y ), , or, , 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, Since, , ...(i), , 1, 2, , .... (ii), , 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., Therefore, x = (2n + 1), , π, , n = 0, ± 1, ± 2..., 2, , 20/04/2018

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APPLICATION OF DERIVATIVES, , Maximum Value of y at x = 0 is, , 1+0=1, , Maximum Value of y at x = π is, , –1 + 0 = –1, , Minimum Value of y at x =, , π, is, 3, , 2 + 2 log, , 1, = 2 (1 – log2), 2, , Minimum Value of y at x =, , 5π, is, 3, , 2 + 2 log, , 1, = 2 (1 – log2), 2, , 129, , Example 16 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, BC = 2y, where C (x, y) is a point on the ellipse, , x2 y 2, +, = 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 2 2, (a x – x4), Therefore, s = 16x  1– 2  . b =, a2,  a , 2, , (0, b), , D, , ⇒, , ds 16b 2, = 2 . [2a2x – 4x3]., dx, a, , C, y, (0, 0), , (–a, 0), A, , Again,, , a, b, ds, and y =, =0⇒ x=, 2, 2, dx, , Now,, , d 2 s 16b2, = 2 [2a2 – 12x2], dx2, a, , At, , x=, , a, 2, , ,, , (0, –b), , (a, 0), , x, B, , Fig. 6.3, , d 2 s 16b2, 16b2, 2, 2, a, a, =, [2, −, 6, ], =, ( − 4a 2 ) < 0, 2, 2, 2, dx, a, a, , 20/04/2018

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APPLICATION OF DERIVATIVES, , Example 21 The tangent to the curve given by x = et . cost, y = et . sint at t =, , 133, , π, makes, 4, , with x-axis an angle:, (A) 0, , Solution, , (B), , π, 4, , (C), , π, 3, , (D), , π, 2, , dx, dy, = – et . sint + etcost,, = etcost + etsint, dt, dt, ,  dy , cos t + sin t, 2, =, and hence the correct answer is (D)., Therefore,  dx  t = π =, cos, t, –, sin, t,   4, 0, , Example 22 The equation of the normal to the curve y = sinx at (0, 0) is:, (A) x = 0, , (B) y = 0, , (C) x + y = 0, , (D) x – y = 0, ,  –1 , dy, = cosx. Therefore, slope of normal =  cos x  = –1. Hence the equation, , x =0, dx, of normal is y – 0 = –1(x – 0) or x + y = 0, , Solution, , Therefore, correct answer is (C)., Example 23 The point on the curve y2 = x, where the tangent makes an angle of, , π, with x-axis is, 4, 1 1, (A)  , ,  2 4, , Solution, , 1 1, (B)  , ,  4 2, , (C) (4, 2), , (D) (1, 1), , dy 1, π, 1, 1, =, = tan = 1 ⇒ y =, ⇒x=, dx 2 y, 4, 2, 4, , Therefore, correct answer is B., , 20/04/2018

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134, , MATHEMATICS, , Fill in the blanks in each of the following Examples 24 to 29., Example 24 The values of a for which y = x2 + ax + 25 touches the axis of x, are______., , dy, = 0 ⇒ 2x + a = 0, dx, , Solution, , a2,  a, + a  −  + 25 = 0, 4,  2, , Therefore,, , i.e., , ⇒, , a, x= − ,, 2, , a = ± 10, , Hence, the values of a are ± 10., , Example 25 If f (x) =, , 1, , then its maximum value is _______., 4 x + 2 x +1, 2, , Solution For f to be maximum, 4x2 + 2x + 1 should be minimum i.e., , 4x2 + 2x + 1 = 4 (x +, , 1 2, ) +, 4, ,  1, 3, 1 −  giving the minimum value of 4x2 + 2x + 1 = ., 4, 4, , , , Hence maximum value of f =, , 4, ., 3, , Example 26 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 , Example 27 Minimum value of f if f (x) = sinx in  ,  is _____.,  2 2, Solution –1, Example 28 The maximum value of sinx + cosx is _____., Solution, , 2., , 20/04/2018

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APPLICATION OF DERIVATIVES, , 135, , Example 29 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, v=, , 4 3 dv, ds, dv r, πr ⇒ = 4πr 2 , s = 4πr 2 ⇒, = = 1 at r = 2., = 8πr ⇒, 3, dr, dr, ds 2, , 6.3 EXERCISE, Short Answer (S.A.), 1., , 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., , 2., , If the area of a circle increases at a uniform rate, then prove that perimeter, varies inversely as the radius., , 3., , 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., , 4., , 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.., , 5., , Find an angle θ, 0 < θ <, , 6., , Find the approximate value of (1.999)5., , 7., , Find the approximate volume of metal in a hollow spherical shell whose internal, and external radii are 3 cm and 3.0005 cm, respectively., , 8., , π, 2, , , which increases twice as fast as its sine., , 2, A man, 2m tall, walks at the rate of 1 m/s towards a street light which is, 3, 1, 5 m above the ground. At what rate is the tip of his shadow moving? At what, 3, , 20/04/2018

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136, , MATHEMATICS, , 1, rate is the length of the shadow changing when he is 3 m from the base of, 3, the light?, 9., , 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?, , 10., , 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., , 11., , 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., , 12., , Find the condition that the curves 2x = y2 and 2xy = k intersect orthogonally., , 13., , Prove that the curves xy = 4 and x2 + y2 = 8 touch each other., , 14., , Find the co-ordinates of the point on the curve, , x+, , y = 4 at which tangent, , is equally inclined to the axes., 15., , Find the angle of intersection of the curves y = 4 – x2 and y = x2., , 16., , Prove that the curves y2 = 4x and x2 + y2 – 6x + 1 = 0 touch each other at the, point (1, 2)., , 17., , Find the equation of the normal lines to the curve 3x2 – y2 = 8 which are, parallel to the line x + 3y = 4., , 18., , At what points on the curve x2 + y2 – 2x – 4y + 1 = 0, the tangents are parallel, to the y-axis?, , 19., , Show that the line, , 20., , Show that f (x) = 2x + cot–1x + log, , –x, x y, + = 1, touches the curve y = b . e a at the point where, a b, the curve intersects the axis of y., , ( 1+ x − x) is increasing in R., 2, , 20/04/2018

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APPLICATION OF DERIVATIVES, , 137, , 21., , Show that for a ≥ 1, f (x) =, , 22., , Show that f (x) = tan–1(sinx + cosx) is an increasing function in 0,, , 23., , At what point, the slope of the curve y = – x3 + 3x2 + 9x – 27 is maximum?, Also find the maximum slope., , 24., , Prove that f (x) = sinx +, , 3 sinx – cosx – 2ax + b is decreasing in R., , π, 4, , ., , π, 3 cosx has maximum value at x = 6 ., , Long Answer (L.A.), 25., , 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, , π, 3, , ., , 26., , 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., , 27., , 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?, , 28., , 29., , 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, of area c2. Show that the maximum volume of the box is, , 30., , c3, 6 3, , cubic units., , 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., , 20/04/2018

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138, , MATHEMATICS, , 31., , 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?, , 32., , 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., , 33., , 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., , 34., , The sum of the surface areas of a rectangular parallelopiped with sides x, 2x, , x, and a sphere is given to be constant. Prove that the sum of their volumes, 3, 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., and, , Objective Type Questions, Choose the correct answer from the given four options in each of the following questions, 35 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:, (A) 10 cm2/s, , 36., , (B), , 3 cm2/s, , (C) 10 3 cm2/s, , (D), , 10 2, cm /s, 3, , 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, radian/sec, 10, (D) 10 radian/sec, (A), , (B), , 1, radian/sec, 20, , (C) 20 radian/sec, , 1, , 37., , The curve y = x 5 has at (0, 0), , 20/04/2018

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APPLICATION OF DERIVATIVES, , 139, , (A) a vertical tangent (parallel to y-axis), (B) a horizontal tangent (parallel to x-axis), (C) an oblique tangent, (D) no tangent, 38., , 39., , The equation of normal to the curve 3x2 – y2 = 8 which is parallel to the line, x + 3y = 8 is, (A) 3x – y = 8, , (B) 3x + y + 8 = 0, , (C) x + 3y ± 8 = 0, , (D) x + 3y = 0, , If the curve ay + x2 = 7 and x3 = y, cut orthogonally at (1, 1), then the value of, a is:, (A) 1, , 40., , 42., , 43., , (D) .6, , (B) .032, , (C) 5.68, , (D) 5.968, , The equation of tangent to the curve y (1 + x2) = 2 – x, where it crosses x-axis, is:, (A) x + 5y = 2, , (B) x – 5y = 2, , (C) 5x – y = 2, , (D) 5x + y = 2, , The points at which the tangents to the curve y = x3 – 12x + 18 are parallel to, x-axis are:, (A) (2, –2), (–2, –34), , (B) (2, 34), (–2, 0), , (C) (0, 34), (–2, 0), , (D) (2, 2), (–2, 34), , The tangent to the curve y = e2x at the point (0, 1) meets x-axis at:, , (A) (0, 1), 44., , (C) – 6, , If y = x4 – 10 and if x changes from 2 to 1.99, what is the change in y, (A) .32, , 41., , (B) 0, , (B), , 1, – ,0, 2, , (C) (2, 0), , (D) (0, 2), , The slope of tangent to the curve x = t2 + 3t – 8, y = 2t2 – 2t – 5 at the point, (2, –1) is:, , 20/04/2018

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140, , MATHEMATICS, , (A), 45., , π, 4, , 48., , 6, 7, , (B), , π, , (C), , 3, , (B) [–2, –1], , π, , (D), , 2, , (C), , (– ∞ , –2], , π, 6, , (D) [–1, 1], , (B) has a maximum, at x = 0, , (C) is a decreasing function, , (D) is an increasing function, , y = x (x – 3)2 decreases for the values of x given by :, (B) x < 0, , (C) x > 0, , (D) 0 < x <, , 3, 2, , The function f (x) = 4 sin3x – 6 sin2x + 12 sinx + 100 is strictly, , (C) decreasing in, , –π π, ,, 2 2, ,  , (B) decreasing in  ,  , 2 , (D) decreasing in 0,, , Which of the following functions is decreasing on 0,, (A) sin2x, , 51., , (D) – 6, , (A) has a minimum at x = π, ,  3 , (A) increasing in  ,, , 2 , , , 50., , –6, 7, , Let the f : R → R be defined by f (x) = 2x + cosx, then f :, , (A) 1 < x < 3, 49., , (C), , The interval on which the function f (x) = 2x3 + 9x2 + 12x – 1 is decreasing is:, (A) [–1, ∞ ), , 47., , (B), , The two curves x3 – 3xy2 + 2 = 0 and 3x2y – y3 – 2 = 0 intersect at an angle of, (A), , 46., , 22, 7, , (B) tanx, , π, 2, , π, 2, , (C) cosx, , (D) cos 3x, , The function f (x) = tanx – x, (A) always increases, , (B) always decreases, , (C) never increases, , (D) sometimes increases and sometimes, decreases., , 20/04/2018

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APPLICATION OF DERIVATIVES, , 52., , If x is real, the minimum value of x2 – 8x + 17 is, (A) –1, , 53., , (B) 0, , 55., , 57., , (C), , 135, , (D) 160, , (B) two points of local minimum, , (C) one maxima and one minima, , (D) no maxima or minima, , The maximum value of sin x . cos x is, , 1, 4, , At x =, , (B), , 1, 2, , (C), , (D) 2 2, , 2, , 5π, , f (x) = 2 sin3x + 3 cos3x is:, 6, , (A) maximum, , (B) minimum, , (C) zero, , (D) neither maximum nor minimum., , Maximum slope of the curve y = –x3 + 3x2 + 9x – 27 is:, (B) 12, , (C) 16, , (D) 32, , f (x) = xx has a stationary point at, (A) x = e, , 59., , (B) 0, , (A) two points of local maximum, , (A) 0, 58., , (D) 2, , The function f (x) = 2x3 – 3x2 – 12x + 4, has, , (A), , 56., , (C) 1, , The smallest value of the polynomial x3 – 18x2 + 96x in [0, 9] is, (A) 126, , 54., , 141, , (B) x =, , 1, e, , 1, The maximum value of, x, , (A) e, , e, , (B) e, , (C) x = 1, , (D) x =, , e, , x, , is:, , (C) e, , 1, e, , (D), , 1, e, , 1, e, , 20/04/2018

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142, , MATHEMATICS, , Fill in the blanks in each of the following Exercises 60 to 64:, 60., , The curves y = 4x2 + 2x – 8 and y = x3 – x + 13 touch each other at the, point_____., , 61., , The equation of normal to the curve y = tanx at (0, 0) is ________., , 62., , The values of a for which the function f (x) = sinx – ax + b increases on R are, ______., , 63., , The function f (x) =, , 64., , The least value of the function f (x) = ax +, , 2 x 2 –1, , x > 0, decreases in the interval _______., x4, , b, (a > 0, b > 0, x > 0) is ______., x, , 20/04/2018