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1 APPLICATIONS OF DERIVATIVES OBJECTIVES. The approimate increase in the area of a square plane when each side epands from c m to.0 cm is () 0.00 sq. cm () sq. cm () 0.06 sq. cm () None. If y log then 8y when,8 0.0 is () 0.0l () () () The approimate percentage error in the volume of a sphere is equal to () Percentage error in r () Double the percentage error in r () Treble the percentage error in r () None. If y ", then ratio of relative errors in y and is () : () : () : n () n : 5. Let P be the pressure and V the volume of a gas such that PV constant. If percentage error in P is k then percentage error in V is () k () /k () -k () None 6. If log.868 then log.0 7. ().968 ().898 ().89 () None correct to decimal places is 998 () 0. () () () None

2 8. The circumference of a circle is measured as 8 cm with an error of 0.0 cm. The percentage error in the area is () / () 0.0 () /7 () None 9. While measuring the side of an equilateral triangle an error of 0.5% is made. Percentage error in its area is () 0.5 () () 0 ().5 0. If there is an error of 0.0 cm in the measurement of the diameter of a sphere, then the percentage error in its volume when the radius 0 cm, is () 0. () 0. () 0. (). If the percentage error in the surface area of a sphere is α, then percentage error i n the volume is () (/) α () (/) α () α () None. If there is an error of % in measuring the length of a simple pendulum then percentage error in its period is () () () (). The radius of a closed cylinder is half of its height. If an error of 0.5% is made in measuring the radius, the percentage error in the surface area is () 0.5 () ().5 () none l. If T π then the ratio of the relative error in T to relative error in / is g () ½ () () / re () None

3 5. The voltage E of a thermocouple as a function of temperature is given by E 6.T T. When T changes from 00 to 0 the approimate change in E is () (). (). (). 6. In A ABC the sides b, c are given. If there is an error 8A i n measuring angle A then δ a. A a δ. A a δ. bc sin A. δ A. None 7. If the ratio of the base radius and height of a cone is : and percentage error in the radius is k, then percentage error in its volume is () k () k () k () None 8. A circular hole of mm in diameter and mm deep in a metal block is rebored to increase the diameter to. mm, and then the amount of metal removed is approimately ().88 π mm ().99 π mm ().79 π mm ().75 π mm 9. The semi-vertical angle of a cone is 5. If the height of the cone is 0.05, the approimate lateral surface area is.0 π.00 π.0 π.none 0. ABC is not right angled and is inscribed in a fied circle. If a, A, b, B be slightly varied keeping, c, C fied, then )R ) π )0 )None

4 . The focal length of a mirror is given by. If equal errors α are made v u f in measuring u and v then relative error is ) α ) α + u v ) α u v )None. Approimate value of cos6 given that sin and () 0.89 () 0.98 () () A stone moving vertically upwards has its equation of motion s 90t.9t. The maimum height reached by the stone is (a) 50 (b) 5 (c) 6750 (d) None of these. A particle moves in a straight line so that its velocity at any point is given by +, where a, b 0 are constants. The acceleration is v a b (a) Zero (c) Non-uniform (b) Uniform (d) Indeterminate 5. The maimum height is reached in 5 seconds by a stone thrown vertically upwards and moving under the equation 0s 0ut 9 t, where s is in metre and t is in second. The value of u is (a) (c).9m / sec (b) 9m / sec 98m / sec (d) None of these 6. A stone is falling freely and describes a distance s in t seconds given by equation s g t. The acceleration of the stone is (a) Uniform (b) Zero (c) Non-uniform (d) Indeterminate

5 7. A 0cm long rod AB moves with its ends on two mutually perpendicular straight lines OX and OY. If the end A be moving at the rate of cm / sec, then when the distance of A from O is 8 cm, the rate at which the end B is moving, is (a) (c) 8 cm / sec (b) cm / sec cm / sec (d) None of these 9 8. If the radius of a circle increases from cm to. cm, then the increase in the area of the circle is (a).π cm (b) π cm (c) 6π cm (d) None of these 9. The equation of motion of a particle is given by s t 9t + t +,where s and t are measured in cm and sec. The time when the particle stops momentarily is (a) sec (c), sec (b) sec (d) None of these 0. A particle is moving in a straight line according as s 5 t + t t then the time when it will come to rest, is (a) 9 seconds (c) 9 seconds (b) 5 seconds (d) 5 seconds. A ladder 5 m in length is resting against vertical wall. The bottom of the ladder is pulled along the ground away from the wall at the rate of of the highest point of the ladder when the foot of the ladder the wall decreases at the rate of (a) m/sec (c).5 m/sec (b) m/sec (d).5 m/sec.5 m / sec. The length.0 m away from

6 . The equation of motion of a particle moving along a straight line is s 9 + t t t, where the units of s and t are cm and sec. The acceleration of the particle will be zero after (a) (c) sec sec (b) sec (d) Never. A particle is moving on a straight line, where its position s (in metre) is a function of time t (in seconds) given by s at + bt + 6, t 0. If it is known that the particle comes to rest after seconds at a distance of 6 metre from the starting position ( t 0), then the retardation in its motion is 5 m (a) m / sec (b) / sec m 5 m (c) / sec (d) / sec t. If the law of motion in a straight line is s v, then acceleration is (a) Constant (c) Proportional to v (b)proportional to t (d)proportional to s 5. If the distance travelled by a point in time t is s 80 t 6 t, then the rate of change in velocity is (a) 6 unit (c) unit (b) 8 unit (d) None of these 6. The edge of a cube is increasing at the rate of 5cm / sec. How fast is the volume of the cube increasing when the edge is cm long (a) (c) cm / sec (b) 60 cm / sec 80 cm / sec (d) None of these

7 7. A body moves according to the formula v + t, where v is the velocity at time t. The acceleration after sec will be (v in cm/sec) (a) cm / sec (b) cm / sec (c) 6 cm / sec (d) None of these 8. A point moves in a straight line during the time 0 s 5 t t. The average velocity is t to t according to the law (a) (b) 9 (c) 5 (d) 7 9. A man metre high walks at a uniform speed 5 metre/hour away from a lamp post 6 metre high. The rate at which the length of his shadow increases is (a) 5 m/h (c) 5 m/h (b) 5 m/h (d) 5 m/h 0. If the path of a moving point is the curve at any instant (a) Is constant (b) Varies as the distance from the ais of (c) Varies as the distance from the ais of y, y b sin at (d) Varies as the distance of the point from the origin, then its acceleration at. The rate of change of the surface area of a sphere of radius r when the radius is increasing at the rate of cm/sec is proportional (a) r (b) r (c) r (d) r

8 . A spherical iron ball 0 cm in radius is coated with a layer of ice of uniform thickness that melts at a rate of 50 cm /min. When the thickness of ice is 5 cm, then the rate at which the thickness of ice decreases, is (a) (c) 5π 6π cm/min cm/min (b) (d) 5 6π 8π cm/min cm/min. A ladder is resting with the wall at an angle of o 0. A man is ascending the ladder at the rate of ft/sec. His rate of approaching the wall is (a) ft/sec (c) ft/sec (b) ft/sec (d) ft/sec. Gas is being pumped into a spherical balloon at the rate of 0 ft /min. Then the rate at which the radius increases when it reaches the value 5 ft is 0π (a) ft/ min. 5π (b) ft/ min. ft/ (d) ft/ min 0 5. (c) min. 5. A ladder 0 m long rests against a vertical wall with the lower end on the horizontal ground. The lower end of the ladder is pulled along the ground away from the wall at the rate of cm/sec. The height of the upper end while it is descending at the rate of cm/sec is (a) m (b) 5 m (c) 5 m (d) 8 m 6. The speed v of a particle moving along a straight line is given by a + bv (where is its distance from the origin). The acceleration of the particle is (a) b (b) / a (c) / b (d) / ab

9 7. If the volume of a spherical balloon is increasing at the rate of 900cm persec, then the rate of change of radius of balloon at instant when radius is 5cm [in cm/sec] (a) 7 (b) (c) 7 (d) None of these 8. A spherical balloon is being inflated at the rate of 5 cc/min. The rate of increase of the surface area of the balloon when its diameter is cm is (a) 7 sq. cm/min (b) 0 sq. cm/min (c)7.5 sq. cm/min (d)8 sq. cm/min 9. The sides of an equilateral triangle are increasing at the rate of cm/sec. The rate at which the area increases, when the side is 0 cm is (a) sq. unit/sec (b) 0 sq. unit/sec (c) 0 sq. unit/sec (d) 0 sq. unit/sec 50. A point on the parabola y 8 at which the ordinate increases at twice the rate of the abscissa is (a) 9 9, (b) (, ) (c), (d) (, ) 5. The position of a point in time t is given by a + bt ct acceleration at time t is (a) b c (b) b + c (c) b c (d) b + c 5. A particle moves in a straight line so that s t proportional to, y at + bt. Its, then its acceleration is (a) Velocity (b) (Velocity) / (c)(velocity) (d) (Velocity)

10 5. If + y 0, then the maimum value of y is (a) 5 (b) 0 (c) 5 (d) None of these 5. The necessary condition to be maimum or minimum for the function is (a) f' ( 0 and it is sufficient (b) f" ( 0 and it is sufficient (c) f' ( 0 but it is not sufficient (d) '( 0 f and f" ( ve 55. The value of a so that the sum of the squares of the roots of the equation ( a ) a + 0 assume the least value, is (a) (b) (c) (d) If f( + 5 and [, ], then the maimum value of function is at the following value of (a) (b) (c) (d) 57. If + y 6 and + y is minimum, then the values of and y are (a), (b), (c) 6, 0 (d) 8, A minimum value of (a) (b) (c) (d) 0 0 te t is

11 59. If two sides of a triangle be given, then the area of the triangle will be maimum if the angle between the given sides be (a) π (b) π (c) 6 π (d) π 60. The sufficient conditions for the function f : R R is to be maimum at a, will be (a) f' ( a) > 0 and f "( a) > 0 (b) f' ( a) 0 and f "( a) 0 (c) f' ( a) 0 and f "( a) < 0 (d) f' ( a) > 0 and f "( a) < 0 6. The maimum value of + 07 in the interval [, ] is (a) 75 (b) 89 (c) 5 (d) 9 6. If for a function (, f' ( a) 0, f"( a) 0 (a) Minimum (c) Not an etreme point f, ( a) > 0 f, then at a (b) Maimum (d) Etreme point, f ( 6. The area of a rectangle will be maimum for the given perimeter, when rectangle is a (a) Parallelogram (c) Square (b) Trapezium (d) None of these 6. and y be two variables such that > 0 and y + y is (a) (b) (c) (d) 0. Then the minimum value of

12 65. If from a wire of length 6 metre a rectangle of greatest area is made, then its two adjacent sides in metre are (a) 6, (b) 9, 9 (c) 0, 8 (d), The minimum value of (a) (c) e log log (b) Zero in the interval, ) (d) Does not eist 67. The minimum value of + y, when 6, (a) (b) 9 (c) 8 (d) 6 y is [ is 68. Area of the greatest rectangle that can be inscribed in the ellipse is (a) ab (c) ab (b) b a (d) ab 69. If y a log + b + has its etremum value at and, then (, b) (a) (c), (b),, (d), 6 a a 70. A cone of maimum volume is inscribed in a given sphere, then ratio of the height of the cone to diameter of the sphere is (a) / (b) / (c) / (d) ¼ + b y

13 7. If y c, then minimum value of a + by is (a) (c) c ab (b) c ab c ab (d) c ab π 7. If A + B, the maimum value of cos A cos B is (a) (b) (c) (d) 7. The minimum value of e + 9e is (a) (b) (c) 0 (d) 7. If PQ and PR are the two sides of a triangle, then the angle between them which gives maimum area of the triangle is (a) π (b) π / (c) π / (d) π / 75. If a + b y 6 c, then maimum value of y is (a) (c) c ab c ab c (b) ab (d) c ab ( + )sin 76. The minimum value of e is (a) e (b) /e (c) (d) 0

14 77. If is real, then greatest and least values of are (a), (b), (c), (d) None of these 78. The ratio of height of cone of maimum volume inscribed in a sphere to its radius is (a) (b) (c) 79. The maimum value of sin ( + cos will be at the (a) (c) π (b) π π 6 (d) π (d) 80. The perimeter of a sector is p. The area of the sector is maimum when its radius is (a) p (c) p (b) (d) p 8. The function y a( cos is maimum when (a) π (b) π / (c) π / (d) π / 6 8. If (, ) P, (, ) minimum at p Q and R is a point on -ais then the value of PR + RQ will be (a) 5, 0 (b), 0 (c) (, 0) (d) (, 0)

15 8. The function sin b + c will be increasing in the interval (, ), if (a) b (b) b 0 (c) b < (d) b 0 8. The function f defined by f( ( + ) e is (a) Decreasing for all (b) Decreasing in (, ) and increasing in (, ) (c) Increasing for all (d) Decreasing in (, ) and increasing in (, ) 85. If the function f( K sin + cos sin + cos (a) K < (b) K > (c) K < (d) K > is increasing for all values of, then 86. The interval for which the given function f ( is decreasing, is (a) (, ) (b) (, ) (c) (, ) (d) None of these 87. f( is an increasing function, when (a) < (b) > (c) (d) < 88. If f( k is monotonically increasing in each interval, (a) k < (b) k (c) k > (d) None of these

16 89. Function f( sin + 6 cos sin + cos λ is monotonic increasing, if (a) λ > (b) λ < (c) λ < (d) λ > 90. The values of a for which the function ( a + ) a + 9a decreases monotonically throughout for all real, are (a) a < (b) a > (c) < a < 0 (d) < a 9. The function (a) bc > 0 a sin + b cos c sin + d cos ad (b) ad bc < 0 (c) cd > 0 ab (d) ab cd < 0 is decreasing, if 9. The function is decreasing in the interval (a) (, ] + (b) (, 0] (c) [, ) (d) ( 0, ) 9. The least value of k for which the function + k + is an increasing function in the interval < < is (a) (b) (c) (d) 9. If f ( be a decreasing function, then lies in (a) (, ) (, ) (b) (, ) (c) (, ) (d) None of these

17 95. The function f ( tan (sin + cos interval (a) 0, ), > 0 ( π (b) 0, / ) ( π (c) ( 0, π / ) (d) ( 0, π / ) 96. Let f + b + c + d,0 < b < c (. Then f (a) Is bounded (c) Has a local minima / 97. The function f ( is is always an increasing function on the (b) Has a local maima (d) Is strictly increasing (a) Increasing in (, ) (b) Decreasing in (, ) (c) Increasing in (, e ), decreasing in ( e, ) (d) Decreasing in (, e ), increasing in ( e, ) 98. For all (0,) (a) (c) < (b) log ( + e + e < sin > (d) log e e > 99. Given function f ( is e + (a) Increasing (c) Even (b) Decreasing (d) None of these 00. If f( + cot + log( +, then f ( ) (a) Increases in [0, ) (b) Decreases in [0, ) (c) Neither increases nor decreases in (0, ) (d) Increases in (, )

18 0. Rolle's theorem is not applicable to the function f ( defined on [, ] because (a) f is not continuous on [, ] (b) f is not differentiable on (,) (c) f( ) f() (d) f ( ) f() 0 0. From mean value theorem ( b) f( a) (a) ab (c) ab a + b (b) (d) a + b b a b + a f b a) f'( ); a < b ( < if f (, then ) 0. The function f( ( + ) e ( / satisfies all the conditions of Rolle's theorem in [, 0]. The value of c is (a) 0 (b) (c) (d) 0. The function f ( 6 + a + b satisfy the conditions of Rolle's theorem in [, ]. The values of a and b are (a), 6 (b) 6, (c), 6 (d) 6, f( b) f( a) b a 05. In the Mean-Value theorem f' ( c), value of c is 5 6 if 0, b (a) (b) + 5 (c) (d) If from mean value theorem, f ( 6 f( b) f( a) ) b a ', then a and f( ( )( ), the (a) a < b (b) a < b (c) a < < b (d) a b

19 07. If f ( satisfies the conditions of Rolle s theorem in [,] and f ( is continuous in [,] then '( d f is equal to (a) (b) 0 (c) (d) 08. Let f ( satisfy all the conditions of mean value theorem in [0, ]. If f (0) 0 and '( f for all, in [0, ] then (a) f ( (b) f ( (c) f ( (d) f( for at least one in [0, ] 09. If the function f ( 6 + a + b satisfies Rolle s theorem in the interval [, ] + and f ' 0, then (a) a (b) a 6 (c) 6 a (d) a 0. The radius of the cylinder of maimum volume, which can be inscribed in a sphere of radius R is (a) (c). Let R R α ln, > 0 f( 0, 0 (b) (d) R R (a) (b) (c) 0 (d), Rolle s theorem is applicable to f for [0,], if α

20 t 5. The function f( t( e )( t )( t ) ( t ) has a local minimum at (a) 0 (b) (c) (d). If the function f( 6a + 5 satisfies the conditions of Lagrange's mean value theorem for the interval [, ] and the tangent to the curve y f( at 7 parallel to the chord that joins the points of intersection of the curve with the ordinates and. Then the value of a is 5 (a) 6 5 (b) 8 (c) 6 7 (d) 6 5. If f ( + b + c and g ( c + b such that min f( > ma g (, then the relation between b and c is (a) No real value of b and c (b) 0 < c < b (c) b c < (d) c > b 5. Let h ( f( ( f( ) + ( f( ) for every real number. Then (a) h is increasing whenever f is increasing (b) h is increasing whenever f is decreasing (c) h is decreasing whenever f is decreasing (d) Nothing can be said in general 6. The function ln( π + f( ln( e + (a) Increasing on [ 0, ) (b) Decreasing on [ 0, ) is is (c) Decreasing on (d) Increasing on π, e π e 0 and increasing on, π, e π, e 0 and decreasing on

21 7. In [0, ] Lagrange's mean value theorem is NOT applicable to (a), <, f (b) ( (c) ( sin, f (, f (d) f ( On the interval [0, ], the function ( takes its maimum value at the point (a) 0 (b) / (c) / (d) /

22 APPLICATIONS OF DERIVATIVES. (). (). (). () 5. () Let side and area A then A. da δ A d d dy δ y δ d HINTS AND SOLUTIONS For a sphere of radius r and volume V, we have V π log V log + log r δv δr 0 + V r δv δr V r n y log y n log δy δ n n : y PV Constant r π

23 log P + log V log C δp δv P V 6. () Take f( log so that f ( /. We have and δ 0.0 f ( + δ f ( + f ( δ (0.0) () 8. () Consider f ( / so that / / f ( ( / ) (/ )( ) Take 000, δ Then f( 0. and f ( 0 f ( + δ Ω 0. ( ) 0 0. (0.6666)0 If r is radius then C πr and A πr C A π By log differentiation δa δc (0.0) A C 8

24 9. () Let side and A area for the equilateral triangle. Then A log A log + log δa 00 (0.5) A 0. () Let d diameter and V volume for a sphere. Then,. () V π d 6 π log V log + log d 6 δv δd V d (0.0) Let radius r, surface area S and volume V for a sphere. Then S πr and V π S V π / r π log V log(constant) + log S δv δs V S

25 . () l T π g log T log π + logl log g δt δl (). T l. () Let r radius and S surface area. Given height r. () 5. () S π r(r) + π r 6πr δs δr (0.5). S r T π l g log T log π + logl log g δt δl T l de T dt + Take T 00, δt δ E [6. + (0.0006)(00) ].

26 6. () a b + c bccos A a δ a bc( sin A) δa bcsin A δ a δ A δa a a 7. () Let r radius, h height and V volume of a cone. Given that r : h : h r 8. () 9. () V π r (r) πr δv δr k V r Let r radius and h depth. Then r, δr 0.06 and h. Volume V πr h πr dv δv Ω δ r π r δ r π 0.06 dr Semi vertical angle 5 r h and l h Take h 0 and δh 0.05 Let S lateral surface area. Then, S π h(h ) π h S + δ S π (0) + π(0)(0.05) 0 π

27 0. (). (). () We have a R sin A, b R sin B, c R sin C. δa R cos AδA, δb R cos B δb, δc 0 δa δb + R( δ A + δb) cos A cos B R( δ A + δ B + δ C) R δ( π ) 0 v u f δ v + δ u δf v u f δf f α + α f v u v u Let f( cos so that f ( sin. Take 60 and δ Then cos , sin cos 6 Ω cos 60 + ( sin 60 ) (0.0075) (a) Here u 90, g 9. 8 (downward) u g Therefore, S 50.. (b) v a + b dv d dv dv b v b v bv Hence acceleration is constant or uniform. 5. (b) Given equation is 0 s 0ut 9 t or s ut.9t ds u 9. 8t v

28 When stone reached the maimum height, then v 0 u 9.8t 0 u 9. 8t But time t 5 sec So the value of u m/sec Hence initial velocity 9 m/sec. 6. (a) Given s gt gt ds d s ; Again g 7. (a) By figure, + y 00...(i) d dy + y (ii) Therefore by (i) and (ii), dy 6 8 cm / sec. 6 8 cm / sec. 8. (a) We know that area of a circle is A πr da dr π R.πcm. ds 9. (c) 6t 8 t + velocity 0 6t 8 t + 0 ( t )( t ) 0 Hence time, sec. ds 0. (c) velocity 5 + t t (when particle stopped) When particle will come to rest, then v 0 5 t t 5 0 t 9, since t.. (a) According to fig. + y 5...(i) Differentiate (i) w.r.t.t, we get y Y B O 0 cm A X

29 d dy + y 0..(ii) d Here and. 5 From (i), + y 5 y dy From (ii), ()(.5) + () 0 So, dy m / sec Hence, length of the highest point decreases at the rate of m/sec. ds. (a) 6t 8 t + d Again t 8 acceleration s If acceleration becomes zero, then 0 t 8 t sec. Hence acceleration will be zero after sec.. (b) Given equation s at + bt + 6 (i) Differentiating w.r.t. time, we get Velocity (v) at + b (ii) After sec, v 0 and distance s 6 metres 0 + b 8a + b 0 a..(iii) and b a 8. (a) s vt a 6 6 a + ( 8a) But retardation in its motion is, a m / sec 5 Retardation m / s (Retardation itself means ve). ds dv s vt v + t. d s dv d v + t. + dv But dv acceleration (a) y O Y B 5 m A X

30 da da a a + t. + a 0 or t 0 But for whole notation t 0 is impossible so that 0 d 5. (c) unit. s 6. (b) Let velocity v 5 cm / sec (Increasing the rate/sec is called the velocity) da 5 Where a is distance and t is time....(i) But if a is edge of a cube, then V a Differentiating w.r.t. time t, so dv da a a.5 5a 5 () 60 cm / sec ( edge a cm). dv da i.e.,a is constant. 7. (c) Acceleration f t, then acceleration after second 6cm / sec 8. (b) Motion of a particle s 5 t t Therefore, velocity ds 5 t 0 ds ds 5 t and t 5 +. Therefore, average 9 dy d 9. (b) 5,? From figure, 6 y + y y y 6.

31 d dy 5. Hence metre / hour d 0. (c) v a 0 a d a is acceleration in -ais d y ba sin at ay a y Hence, a y changes as y changes. dr. (c) Surface area s πr and. (d) ( + 0). (b) ds dr ds π r 8 πr 6πr r. V π where is thickness of ice. dv d π (0 +. (a) Given that dv / 0 ft / min and r 5 ft V πr ; dv dr πr dr dv / 0 πr π 5 5 0π 5. (b) We have + y 0 Y B 0 y X O A ft/min

32 d dy + y 0. + y.( ) 0 y. Thus, + y 0 y 6 6. (c) a bv 7. (c) V π y m. dv d b v.. dv d dv d v. b., v. b r Differentiate with respect to t, dv dr dr dv π r.. πr dr 900 π 5 5 dv r 8. (b) Volume V π 5 cc/min π(7) Surface area, S π r ds dr 8π πr 0 8π dr 7 π dr πr. dr dr 5 8π., at r 7 cm cm /min. 9. (c) If is the length of each side of an equilateral triangle and A is its area, then da d A Here, 50. (a) y 8 d 0 cm and cm / sec Differentiate both sides w.r.t.t dy d y 8 d d dy d y 8,

33 y 8 9 or y and y Hence the required point is 9 9, (d) Acceleration in direction of -ais c and acceleration in direction of y-ais d y b Resultant acceleration is ( c ) + (b) b + c 5. (a) Given s t. Now Also a / ds v t dv a ( t) t or v t a. 5. (c) + y 0 ; y 0..(i) Now f( y (0 0 f' ( 0 For maimum value of f (, f' ( 0 and y 5 5 So maimum value of y. 5. (c) The necessary condition to be maimum or minimum for function f ( 0 and for maimum f ve ( and for minimum ve Hence f '( 0, but it is not sufficient. 55. (b) Let α, β be the roots of the equation 56. (d) f '( 6 6 f '( 0 ( )( + ) 0, Here f ( ) f ( ) d f ( ) +.

34 f ( ) f ( ) Therefore the maimum value of function is 7 at. 57. (d) + y y y + ( 6 Let z + (6 z' To be minimum of z, z' ' > 0, and it is. Therefore 0 8 y 8 t 58. (d) f( te f' ( e f ( e ( ); f (0) > 0 Minimum value ( 0) 0 f. 59. (d) Let a and b are given, then area A ab sin C da dc Hence Ais maimum, when 0 C 90. da ab cos C dc 60. (c) Given function f : R R is to be maimum, if f ( a) 0 and ( a) < 0 6. (d) Let f ( + 07 At, f( ) ( ) ( ) At, f() () () For maima or minima, f ( 6 0, So at, f() () () At, f( ) ( ) ( ) f. Thus the maimum value of the given function in [, ] is (c) It is a fundamental property. 6. (c) We know that perimeter of a rectangle S ( + y), where and y are adjacent sides

35 6. (a) y S. Now area of rectangle, A y ( S ( S y y and let z + y dz z + d 65. (b) Given ( + b) (d) Let a, a + b 8 Area of rectangle ab a( 8 a) da A 8 a a, 8 a da. log log dy y d dy d log Put 0 0 log 0 e At ) log d y + log d and d y e, < 0 d e In [, ) the function eist. 67. (a) f( + y when y 6 8 f( + y + 8 f ( ± 0 6 f and f ( () > 0 log Putting +, we get the minimum value to be. will be maimum and minimum value does not 68. (c) Concept

36 a 69. (d) + b + dy dy a + b + d d 0 a b dy d a and + b + 0 b + b + 0 b + b + 0 b b (a) Standard Problem 7. (b) y c and a. c bc y f( a + by a + Differentiate with respect to f ( a Put ( 0 At f a bc 0 bc ± c b / a a + c b a, a + by The minimum value / will be minimum. bc a a bc f c a. c +. b b c c ab. π 7. (a) Let f( A) cos A cos B cos A cos A cos A sin A f ( A) cos A sin A cos A Now, f ( A) 0 cos A 0 π π π A A Now f ( A) sin A sin ( ve ) 7. (b) Let f( e + 9e f ( 8e 8e Put f ( 0 8e 8 e 0 e / log( / ) / b a

37 Again f ( 6e + 6e > 0 ) ) Now /.(log( / / ) ) (log( / / f (log( / ) ) e + 9e Hence minimum value. ab 7. (d) Let PQ a and PR b, then sinθ sinθ Since, area is maimum when sin θ (c) a b y c Hence 6 + c a y b 6 ( ) c f f ( y a b 6 c b / 8 a / / Differentiate f ( with respect to, then 6 c f ( Put ( 0 b a 8 / c b c 8 a f, 0 b b c ± / a 6 c At c / / a 6 / 7 a a b the f ( will be maimum ( + )sin 76. (c) Given y e e dy d For minima or maima, 0 ( + )sin [( )sin + ( + )sin cos ] 0 [( )sin + ( + )sin cos ] 0 π θ.

38 77. (b) Let sin [( )sin + ( + )cos ] 0 sin 0 y is minimum for sin 0 y dy ( + + )( ) ( d ( + + ) + )( + ) dy 0 0, + d ( + + ) d y ( d + + ) (b) Standard Problem sin + sin 79. (c) y sin ( + cos dy d y cos + cos and sin sin d d dy d On putting 0, cos + cos 0 cos cos cos( π π ; sin π sin π π d y d. π / 80. (d) Standard Problem 8. (a) Standard Problem 8. (a) Standard Problem 8. (c) Let f ( sin b + c '( cos b > 0 f or b cos > or b <., which is negative.

39 8. (d) f( ( + ) e f' ( e e ( + ) 85. (d) Since f( K sin + cos sin + cos is increasing for all, therefore f' ( > 0 for all K (sin + cos > 0 > 0 K > K. 86. (a) f ( for all f' ( but for decreasing f '( < 0 6 < 0 ( )( + ) < 0 < < Hence the required interval is (, ). 87. (b) To be increasing f '( 7 > 0 > 9 >. 88. (c) f'( k [ k 6 + ] > 0, R b ac < 0, k > 0. e., 6 k < 0 i or > k. 89. (d) The function is monotonic increasing, if f ( > 0 ( sin + cos ( λ cos 6 sin ( λ sin + 6 cos ( cos sin ( sin + cos ( sin + cos λ (sin + cos (sin + cos > 0 λ > 0 λ >. > 90. (d) If f ( ( a + ) a + 9a decreases monotonically for all R, then f' ( 0 for all R ( + ) 6a + 9a 0 a for all R ( + ) a + a 0 a for all R + < 0 a and Discriminant 0 a <, 8a a 0 < a and a ( a + ) 0 a <, a or a 0 a < a 0

40 9. (b) Let a sin + b cos y c sin + d cos dy d The function will be decreasing, when < 0. ( c sin + d cos ( acos b sin ( a sin + b cos ( c cos d sin < 0 ( c sin + d cos ac sin cos bc sin + ad cos bd sin cos ac sin cos + ad sin bc cos + bd sin cos < 0 ad (sin + cos bc(sin + cos < 0 ( bc) < 0 9. (d) y ad. dy + d ( + ) To be decreasing, 0 d d ( + ) < 9. (d) To be increasing, ( +k + ) > 0 > 0 (0, ). + k > 0 For (, ), the least value of k is. 9. (b) f ( , For decreasing f ( < < 0 + < 0 ( ) ( ) < 0, (, ). π 95. (c) f ( y tan sin + tan y π sin + sec dy y d cos + π dy π > 0 cos + > 0. 0, π. d 96. (d) Given f ( + b + c + d f '( + b + c Now its discriminant ( b c) ( c) 8c < 0, b as b < c and c 0 Therefore, f' ( > 0 for all R >

41 Hence f is strictly increasing. / 97. (c) Let y log y log dy log log dy / log y d d log log / Now, > 0 for all and > 0 in (, e) and < 0 in ( e, ) ( f is increasing in (, e) and decreasing in ( e, ). 98. (b) Both e and + are increasing and e +, because e. 65 nearly. so the 99. (a) answer (a) is not correct. Since log < because log is negative. e f ( e (i) f( Again f( + f e e ( e + + e is an odd function. π π sin < 6 6 e f( f( e + e e f( f ( > 0 n R e + ( + e ) is an increasing function 00. (d) We have f( + cot + log( + f '( ( + + ) because <. So, (c) is not correct ( + ) 0 for all Hence f( is an increasing function on (, ) and in particularon 0, ) [.

42 0. (b), when < 0 f (, when 0 Clearly f ( ) f() But f(0 + h) f(0) h h Rf '(0) lim lim lim h 0 h h 0 h h 0 h f(0 h) f(0) h Lf '(0) lim lim h 0 h h 0 h Rf '(0) Lf '(0) h lim h 0 h Hence it is not differentiable on, ) 0. (a) f '( ) b a b a ab ab. (. 0. (c) To determine 'c' in Rolle's theorem, '( c) 0 Here f'( ( + e ( / ). + ( + ) e ( / ) e ( f '( c) 0 c ( / ) e { But c [,0]. 6} c 6 0 c,, ( / ) f. 0. (a) f ( ) f() a + b 5 a + b 7 a, which is given in option (a) only. 05. (c) From mean value theorem a 0, f( a) 0 b, f( b) 8 f ( c) f ( ( )( ) + ( ) + ( ) f ( c) ( c )( c ) + c( c ) + c( c ) c c + + c c + c c f( b) f( a) b a

43 f ( c) c 6c + According to mean value theorem, f ( c) f( b) f( a) b a ( / 8) 0 5 c 6c + c 6c + 0 ( / ) 0 6 ± ± c 6 ± (c) According to mean value theorem, In interval [a, b] for f ( f( b) f( a) f'( c) b a a < < b., where a < c < b 07. (b) f ( d [ f( ] f() f() (b) f ( ) f()] ) 09. (d) f ( 6 + a + b f ( + a f ( c) 0 f a a a 0 a. 0. (b) Standard Problem. (d) For Rolle s theorem to be applicable to f, for [0,], we should have (i) ( ) f(0) (ii) f is continuous for [0,] and f is differentiable for (0,) f,

44 From (i), f ( ) 0, which is true. From (ii), 0 f(0) f(0 + ) 0 + lim α ln. (b,d) f( f' ( ( e Which is true only for positive values of α, thus (d) is correct t 5 t( e )( t )( t ) ( t ) )( )( ) ( ). (b) f( b) f() 8 a a f( a) f() 6a a f '( a + 5 From Lagrange's mean value theorem, f( b) f( a) 8 a 6 + 6a f'( b a f' ( 8 a At 7 9 7, a + 5 8a a 7 a a (d) f( ( + b) + c b is minimum at b and g ( b + c ( + c) is maimum at c c b > b + c c > b. 5. (a,c) h ( f( [ f( ] + [ f( ] h '( f'( f( f'( + [ f( ] f'( f '( [ f( + [ f( ] ] f '( f( '( + 9 h and f '( have same sign.

45 6. (b) Let ln( π + f( ln( e + ln( e + ln( π + f' ( π + e + ln ( e + ( e + ln( e + ( π + ln( π + ln ( e + ( e + ( π + f for all 0, { > e} '( < 0 Hence f( is decreasing in 0, ) π [. 7. (a) The function defined in option (a) is not differentiable at (d) f( ( f' ( 5 (75)( 7 ( ) + 5 For maima and minima, 75 5 ( ( 7 5 ( [( ] 0 Either 0 or 75 or ( 0 At, f' h > 0 and f ' + h < 0 f( is maimum at. 75.