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Mathematics Times February Mathematics Times February 19 By DHANANJAYA REDDY THANAKANTI (Bangalore) Introduction This article explains us to find a indefinite and definite integral of an inverse function when we are known the parent function , like finding integral Neither curve has any y- coordinates labelled so of ln x without knowing how to integrate ln x directly It can be used to review knowledge about and ln So the same numbers appears in both graphs except the x and y-coordinates have swapped over The swapping of coordinates might remind us that the inverse functions e x and ln x and to discuss how to find the area between a curve and the yaxis This method can be extended to other functions such as arc sinx, Once student can integrate sinx If f and f 1 are elementary on some closed interval, then integral f x dx is elementary i integral f 1 x dx is elementary Take a look at these graphs we can add these to the diagram For e x , the ycoordinates are and For y = ln x they are ln there is a relationship between e x and ln x They are inverse functions of each other, which we can think about graphically as a reflection in y x This symmetry around y x is not immediately obvious from the diagrams, as the scaling in the x and y directions is different on each sketch It is important not make assumptions about shape or symmetry based on sketch graphs There are two rectangles on each graph with areas of ln and ln respectively These areas are the same on both graphs as the x and y coordinates are reflected in y x The symmetry of the graphs implies that the two shaded areas are actually identical as they are reflected across y x They could be represented 3 by the integral ln x dx or the integral ln y dy Fig Fig Here are some things we have noticed There are two dierent graphs : each has a function and two x-coordinates given There is an area shaded on each graph but they are in different places One is between the curve and the y-axis and the other between the curve and the x-axis 2 We can also find the area of other regions bounded by the curves For example , the area A can be represented by the integral x e dx, Mathematics Times February 1 Putting all these together means the area represented by ln x dx can be found by ln 3 y ln x dx ln ln e dy ln 2 ln ln The answer can be written in several different forms If we combined the logarithms we will ended up As inverse functions have symmetry around y x, we know that area A is the same as area B , shaded in the diagram below 27 27 with ln ln or ln 4e Using a similar method we can find arcsinx dx The integral represents the area of the shaded region under the curve y ln x However , we also know that because these graphs are inverse functions, the shaded region rectangle between e x and the y- axis has the same area If we don‘t know how to integrate ln x directly , then we need to use other areas that we know how to find we have already calculated the areas of the large rectangle, ln and the smaller rectangle, ln In this case we find arcsinx dx 3 siny dy 6 3 12 We have seen that we can find the definite integral of any function if it has an inverse function that is easy to integrate Formula Suppose the function f is one-to-one and increasing Then, a geometric equivalence may be established: Therefore the L- Shaped region has area ln ln The area between the curve , the y-axis, ln and ln is given by the integral ln y e dy e y ln ln ln 32 f b b a f x dx f 1 x dx bf b af a f a Suppose the function f is one-to-one and decreasing Then, another geometric equivalence may be established: Mathematics Times February 19 f a b f x dx a f 1 x dx b a f b a f a f b f b Integral of inverse functions Inverse function integration is an indefinite integration technique While simple, it is an interesting application of integration by parts If f Let f x be a one-to-one continuous function such that f 1 and f 2, and assume 1 and f are inverses of each other on some closed interval, then integral f x dx 15 Calculate f x dx xf x f 1 f x f ' x dx, 1 f x dx 2 Evaluate so integral f x dx xf x G f x , where ln 1 x dx G x f 1 x dx Let a function f : R R be defined as Therefore, if it is possible to find an inverse f 1 of f, integrate f 1 , make the replacement x to f x , and subtract the result from x f x to obtain the 2 f x x sin x The value of f 1 x dx will be (a) 22 (b) 22 (c) 22 (d) 2 result for the original integral integral f x dx Examples (1) Assume that f x exp x , hence f 1 y ln y The formula above gives immediately ln y dy y ln y y C (2) Similarly, with f x cos x and f 1 y arccos y , 1.Sol: The region bounded by f , x 1, and y must have area 5, implying the integral in question corresponds to the area The above formula for decreasing functions provides the same answer 2.Sol: ln ln 22 ln 3.Sol: Using above formula, we get 2 arccos y dy y arc cos y sin arccos y C (3) with f x tan x and f 1 y arc tan y , arctan y dy y arctan y ln cos arctan y C 2 x sin x dx f 1 x dx 2 2 f 0 2 2 x2 cos x f 1 x dx 42 0 2 i.e., f 1 x dx 42 2 4 1 1 Mathematics Times February MANIPULATIONS OF TRIGONOMETRIC EXPRESSIONS Evaluate 256 sin10 sin 30 sin 50 sin 70 Find the value of Let a1 , a2 , , an be the sequence of all irreducible proper fractions with the denominator 24, arranged in ascending order Find the value of sin 1 sin 2 sin 3 sin 360 Find the smallest positive integer n such that n cos a 1 sin 45 sin 46 sin 47 sin 48 i i 1 Prove that 1 sin133 sin134 sin n cos( ) cos( 2 ) cos( n ) n n 1 sin cos( ) 2 sin Find the value of cot 25 1 cot 24 1 cot 23 1 cot 20 1 Prove tan 15 cot 15 must be an even positive integer for any positive integer n Prove that for any positive integer n , tan tan 2 tan 3 tan 3a n find the value of cos 11 Evaluate cos tan n n tan where tan 0and tan for 1, 2, , n Given , 2 If the equality 13 If 15 cos 2 4 7 cos cos 15 15 15 cos100 tan x, find x 4sin 25 cos 25 cos 50 1 sin1 sin 2 sin 2 sin 3 14 Prove that cos1 sin 89 sin 90 sin1 15 Prove that (i ) tan cos( x ) sin( x ) cos x holds for any x , find the value of and 12 Evaluate cos 36 cos 72 n tan(n 1) tan n , cos cos , 3 10 Given sin sin (ii ) tan tan tan 2 5; 2 10 5 Mathematics Times February 4.Sol: When 45, then tan 45 1.Sol: 256 sin10 sin 30 sin 50 sin 70 256 cos 20 cos 40 cos 60 cos80 128sin 20 cos 20 cos 40 cos80 sin 20 64 sin 40 cos 40 cos80 sin 20 32 sin 80 cos 80 sin 20 tan tan tan tan tan tan tan tan (cot 1)(cot 1) (1 tan )(1 tan ) tan tan tan tan tan tan Thus, (cot 25 1)(cot 24 1) (cot 20 1) [(cot 25 1)(cot 20 1)] [(cot 23 1) 16sin160 16 sin 20 2.Sol: All the irreducible proper fractions with denominator 24 are (cot 22 1)] 11 13 17 19 23 , , , , , , 24 24 24 24 24 24 24 24 23 19 24 24 24 24 since 17 11 13 and 24 24 24 24 23 5.Sol: tan15 tan 60 tan 45 1 tan 60 tan 45 3, and cot15 2 15 (2 3)n (2 3)n using the binomial expansion,it follows that cos cos( ) 0, it follows that (2 3) n 2n n 3, tan n 15 cot n 15 2 n n 1 3 3 n n2 cos(ai ) 1 2k sin cos( k ) sin 2 2k sin n implies that sin cos( k ) k 1 n 2k 2k [sin( ) sin( )] k 1 2 1 2n sin sin 2 2 sin n n 1 cos 2 n n (2 3) n 2n n 1 n ( 3) 2 ( 3) n Therefore in the sum (2 3) n (2 3) n Only the terms with even powers of appear,, and each of them appeared in pair, so the sum is an even positive integer 6.Sol: The formula tan(k 1) tan k tan gives tan tan k tan(k 1) tan k tan k tan(k 1) 1 tan tan n1 n1 tan(k 1) tan k (n 1) tan tan k 1 tank tan(k 1) k 1 ( 3)n i 1 3.Sol: For any k = 1, 2, , February Mathematics Times tan k n tan 7.Sol: Write the given equality in the form (cos sin 2) cos x (cos sin )sin x = 0, which holds for any real x ,so Multiplying both sides of the given equation by sin1 , we have sin1 (cot 45 cot 46) cot 47 cot 48) sin n (cot133 cot134) cot 45 (cot 46 cot134) cos sin 2) or cos sin 0, (cot 47 cot133) (cot 89 cot 91) cot 90 sin cos cos sin By taking squares to both sides of each equality and add up them, then so cos sin 2 2(sin sin cos cos ) , Further, implies that 2 , (cos cos ) By 3 adding up them, it is obtained that (sin sin )2 ( cos 2) sin which given the solution cos Therefore, sin n sin1 , and the least possible integer value for n is 10.Sol: 3 , 1 , hence cos 2 11.Sol: From the formulas for changing sum or difference to product, so cos( ) cos 7 Thus , since 2 8.Sol: By using the formula in Q3 of cos sin 1 sin 2 sin 3 sin 360 2(sin sin 2 sin 3 sin 180) sin 1 sin[( x 1) x] sin( x 1) cos x cos x sin( x 1) sin x cos( x 1) sin x sin( x 1) cot x cot( x 1) 4 cos cos 5 15 15 4 cos 180 sin180 cos181 / sin1 180 9.Sol: Note that 2 4 7 cos cos 15 15 15 4 cos cos cos 15 5 15 cos 180 (cos 2 cos 4 cos 6 cos 360) sin1 sin x sin( x 1) 15 cos 7 2 4 cos = cos cos cos 15 15 15 15 (for 2, n 180), cos( x 1) sin x cos sin 10 sin 10 sin since cos180 sin 72 sin 36 cos 36 sin18 cos18 cos 36 implies that sin18 cos 36 12.Sol: Note that cos 36 cos 72 2sin18 cos 36 Mathematics Times February 19 2(cos 36 cos 72)(cos 36 cos 72) 2(cos 36 cos 72) cos 36 cos 72 2(cos 36 cos 72) n , n 0,1, 2, 3, then each of the five roots satisfies the equation tan 3 tan 2 , therefore, by the muliple angle formulae,it satisfies the equation By the double -angle formulas, the above equality becomes cos 36 cos 72 cos 72 cos144 2(cos 36 cos 72) cos 72 cos 36 2(cos 36 cos 72) 13.Sol: By using the double angle formulas and the half angle formulas, cos100 sin 25 cos 25 cos 50 cos100 2sin 50 cos 50 cos 50 sin 50 (cos 50 sin 50)2 tan tan 2 tan tan tan Letting x tan , we have equivalently, (1) x( x 10 x 5) If consider non -zero roots , then it becomes x 10 x n , n 0,1, 2,3, are the four roots of (1).By the viete’s theorem, Thus, tan for cos 50 sin 50 cos 50 sin 50 tan tan 50 tan 50 tan 45 tan 50 tan 95 , x 95 tan 45 tan 50 14.Sol: The left-hand side of the desired equation equal to sin k sin(k 1) k 1 tan 2 3 4 tan tan 5 5 2 3 4 tan tan tan tan 5 5 k 1 cos1 cot1 , sin1 sin 1 15.Sol: We construct an equation with roots n , n 0,1, 2,3, as follows Since the tan Since tan 0, tan 2 and 3 2 4 , tan tan tan , 5 5 (2) gives tan 89 tan tan From (2), tan 2 5, 0, tan tan 2 , (i) is proven 2 2 2 tan tan tan tan 5 5 tan equation tan 5 for [0, ) has roots tan (2) 2 3 2 4 3 4 tan tan tan tan tan 5 5 5 [cot k cot(k 1)] tan tan 10 89 tan tan x x3 2 x ,or 3x x 2 2 tan tan tan 10 , 5 5 tan 2 10 , (ii) is proven Mathematics Times February 19 half as that of the same line, then the slope of L is: [2013] (a) 3 (b) 3 / (c) 3 / (d) 3 / 16 19 If the image of point P (2, 3) in a line L is Q (4, 5), then the image of point R (0, 0) in the same line is : [2013] (a) (2, 2) (b) (4, 5) (c) (3, 4) (d) (7, 7) 1.Sol: For A ; x cos 30 y 2 sin 30 20 Let 1 be the angle between two lines x y c1 and x y c2 and be the angle between two lines x y c1 and x y c3 0, where c1 , c2 , c3 are any real [2013] numbers : Statement-1: If c2 and c3 are proportional, then 1 Statement-2: 1 2 for all c2 and c3 (a) Statement-1 is true, Statement-2 is true ; Statement-2 is a correct explanation of Statement-1 (b) Statement-1 is true, Statement-2 is true ; Statement-2 is not a correct explanation of Statement-1 (c) Statement-1 is false ; Statement-2 is true (d) Statement-1 is true ; Statement-2 is false 21 If the three lines x y p, ax y q and ax y r form a right-angled triangle then : [2013] (a) a 9a 18 (b) a 6a 12 (c) a 6a 18 (d) a 9a 12 b a 11 d 16 c 21 a c c 12 b 17 b c c 13 c 18 d a d 14 b 19 d x and y 1 For vertex C, cos120 y sin120 2 x 1, y For vertex B, x x cos 75 y sin 75 2 x 1 and y 1 The sum of the x - coordinates is 1 1 2.Sol: Let slope of incident ray be m angle of incidence = angle of reflection d 10 b 15 d 20 a m7 2 7m 14 13 m7 m7 or m 13 7m 13 13m 91 63 or 13m 91 9 63m 50m 100 or 76m 82 52 Mathematics Times m 41 or m 38 5.Sol: Slope of x y is tan sin 41 y x or y x 38 sin i.e., x y or 38 y 38 41x 41x 38 y 38 3.Sol: Length of perpendicular to x y 10 from origin (0, 0) is 10 P1 2 Length of perpendicular to 8x y from origin (0, 0) is February i.e., ,cos 1 2 ,cos or 1 Q sin ,1 2 cos ,1 The equation of required line is x y 6.Sol: Equation of line L is 10 Lines are parallel to each other ratio will be : or : 4.Sol: Given that P2 L1 : x y 12 L2 : 3x y 12 x y 1 i.e., L1 L2 we have equation of a line passing through the 2x y (1) intersection of given L1 For line x y 12 3x y 12 x y 4 i.e., (2) Solving equation (1) and (2) ; we get point of x 3 y 4 12 1 now, this line cuts x - axis at 12 1 A ,0 3 now then line also cuts y - axis at 12 1 B 0, 4 12 intersection , 5 8 7.Sol: Given that A 0, , B 1, C 89,30 3 Slope of Eliminating from h, and k, we get i.e., 3 So, the points A,B,C are collinear Slope of 1 Mid point (h, k) is 3 , 4 AB BC 8.Sol: Given equation of line is y 3x h k 7hk y 3x x y xy is the locus i.e., m 53 Mathematics Times February Let the slope other line be m1 tan 60 we have 3m m1 m1 0, m2 given that line L is passing through 3, 2 i.e., y 2 x y x 3 y 3x 3 9.Sol: Given that and Circumcenter is 0, 0 a 12 a 12 , Centroid 2 We know the circumcenter O , Centroid G and orthocenter H are related as HG ; GO :1 a and b The coordinates of B is (6, 0) and coordinates of A is (0, 9) now, slope of line is 3 Equation of required line is y 3 x 6 i.e., 3x y 18 11.Sol: Equation of RQ is x y (1) Since, R is on x - axis x 2, y as PQ is parallel to x-axis, y-coordinates of Q is also putting value of y in equation (1), we get Q (8, 3) 85 33 , Centroid of PQR 5, 3 Only x y satisfy the given co-ordinates 12.Sol: Let equation of line L, perpendicular to x y be x y c HG GO Coordinate of orthocenter is i.e., a 1 a 12 , 2 3 2 a 1 and k a 1 2 Let h a 1 h 2 a 1 k k a 1 a 1 h 2 Locus is y a 1 a 1 x 10.Sol: Given that A divider CB in 2:1 1 a 1 b i.e , and 1 54 2a b and 3 Given that area of AOB is c c i.e., 5 c 50 Equation of line L is x y 50 Distance between L and line x y is Mathematics Times d 50 12 52 50 26 Let be x,0 , then we have February 13 tan 13.Sol: Given that x 2ay a (1) x 3by b (2) x 4ay a are concurrent (3) 07 03 and tan 180 x6 x 1 we have tan 180 tan i.e , 3 7 x x 1 x 16.Sol: Let C x1 , y1 2a a 3a b i.e., 4a a R2 R2 R1 and R3 R3 R1 2a a 3b 2a b a i.e., 2a 0 2a or b a Locus of a,b k or y x 14.Sol: Since the point (1, 2) does not lies on the base line 3x y i.e., the perpendicular distance from the paid to the base is altitude h 1 4 And we have h a, where a is the side of an x1 y1 , Centroid, E Given that centroid lies on the line 3x y x1 y 3 4 i.e., 2 Hence vertex x1 , y1 lies on the line equilateral triangle a 2h 4 15 3x1 y1 3x y 17.Sol: 15.Sol: 55 Mathematics Times February Now AB 2a 2 a 5a 15a Required line is We know D is midpoint of AB 2a 0 a a , i.e., coordinates of D is a, 2 we have BC = AC i.e., 2a 0 y1 a 2 2a 2a y,0 4a y1 a y1 5a Now CD a 2a 5a a 2 a 4a =15 a Area of a triangle is AB CD 5a 5a 5a i.e., 2 18.Sol: Given line 3x y 12 can be written as 3x y x y 1 1 12 12 x-intercept and y-intercept Let the required line be L: 3x 16 y 24 y 3 24 x 16 16 Hence, required slope 3 16 19.Sol: Lies on L i.e., equation of a line is 5a 2ay1 y1 x 2y 1 x y where a b a x-intercept and b y-intercept According to the question a and b / 1 y x 3 42 y x image of R(0, 0) with respect to the line x y is (7, 7) 20.Sol: Two lines x y c2 and x y c3 are parallel to each other Hence statement-1 is true, statement-2 is true and statement-2 is the correct explanation of statement1 21.Sol: Since three lines x y p, ax y q and ax y r form a right angled triangle i.e., product of slopes of any two lines 1 Suppose ax y q and x y p perpendicular to each other a 1 a Now, consider option one by one a satisfies only option (a) Required answer is a 9a 18 56 Mathematics Times February Prof Iyer, head of department, - Math ibwithiyer@gmail.com | 9323040686 a IB WITH IYER Takes pride in announcing to tie up with MATHEMATICS TIMES to provided dedicated support to students who study international curriculums INTERNATIONALBACCALAUREATE DIPLOMA CAMBRIDGALEVEL IGCSE MYP PEARSON ED EXCEL Dedicated support includes a Mathematical concepts b Challenging problems c Exam Corner d Effective use of Graphical Display Calculator e Mathematics HL – Paper problem f Math Exploration - IA tips g Model Papers / Model questions h Test Prep page- SAT / ACT / AP About IB WITH IYER (IBI) – a division of IYER EXPERT GUIDANCE PVT LTD started by Prof Iyer is a teachers’ organization Prof Iyer started IB WITH IYER in 2005 moving from teaching CBSE / IIT international curriculums IBI is a teacher-led organization Prof Iyer-founder director is still a teacher, working as head of department for an international school He is one of the authors for the current Higher Level math text book recommended by IBO He is Examiner for IB Higher Level Math He is currently authoring a single authored Question bank for IB Math and Cambridge Al level Math to be published in Fall 2019.Parallely he is developing IB Math App for the New curriculum compatible to Android and iOS Platforms He is currently consulting to IB schools as an expert resource His support to schools include IB Math Exam cram program for Grade improvement b Cambridge A level / Add Math / Intl Math Exam cram programs c Conducting Assessments and Moderation d Teacher workshop e Advance Placement (AP) coaching – AP Calculus AB / BC for US departures f Test Prep – SAT/ ACT / SAT Subject coaching Understanding F’(x) curve Technology partner: Casio , fx - Graphical Display calculator Let y f ( x ) be a function Let us consider f ( x) sin x I have taken a simple and standard function The above graph represents f ( x) sin x drawn between x and x 2 we know f '( x ) means derivative of the function or it is a rate change function otherwise called as dy - instantaneous rate change of y with respect dx 57 Mathematics Times February 19 to x for the curve y f ( x ) The curve f ( x ) is increasing when f '( x) f ( x ) is decreasing when f '( x) Now let us connect the sign of f '( x ) to a graph when the derivative of f ( x ) is positive ( 0) then the graph is ascending or increasing.What does it mean for x1 x2 for f ( x1 ) f ( x2 ) or we call here as Now let us look at the sketch of f ( x ) for the x in blue colour if you notice,f (x) is decreasing or descending curve.The rate change or derivative is negative A graph is negative when it is below x-axis The second sub interval black graph represent f '( x ) For x , f '( x ) graph is below x-axis f '( x) When the derivative of (x) is negative ( 0) then the graph is descending or decreasing What does it mean? Now I shall draw the full curve for x 2 of both f (x) and f '( x ) on same graph You can notice the above concepts illustrated below for x1 x2 for f ( x1 ) f ( x2 ) or we call here as f '( x) The blue graph is increasing for x or Now let us look at the sketch of f ( x ) for the sub interval x in blue color If you notice, f ( x ) is increasing or ascending curve.The rate change or derivative is positive.A graph is positive when it is above x-axis The black graph represents f '( x ) For x above x - axis , f '( x ) graph is 3 x in this interval if you notice black graph i,e f '( x ) graph is above x-axis The blue graph is decreasing for x 3 in this interval if you notice black graph ie f '( x ) graph is below x-axis Interpreting stationary point using f '( x ) graph we know stationary points exist for a function f(x) when f '( x) looking at the f(x) in blue colour we know f ( x) sin x has a stationary maxima at x and stationary minima at x 3 in graph 4, we also have f '( x ) in black colour.Taking a closer look at it f '( x) at x x 58 and at 3 At these 2points, the graph f '( x ) cut Mathematics Times the x -axis So these are zeroes of the graph f '( x ) which is stationary of f(x) How we determine nature of stationary from f '( x ) graph? At x f '( x ) graph moves from positive to negative.By “first derivative test’’,f(x) has a mixima February b number line c interval inequality representative : x kUx k number line representation : interval representation: x [, k ]U (k , ) The curve bracket indicates open interval This means x is between k and k but not equal to either of the limits if the sign of f '( x ) changes from negative to positive This is evident from the fact that graph of (4) x k where k is a non-zero real constant f '( x ) moves from above x-axis to below x- axis Inequality representation: x k Ux k Number line representation : 3 , f '( x ) graph moves from negative to positive By“first derivative test’’,f(x) has a At x minima if the sign of f '( x ) changes from negative to positive.This is evident from the fact that graph of f '( x ) moves from below x-axis to above x-axis soon you will hear interval representation: x , k U k , The box bracket indicates closed interval The open bracket indicates open interval I mentioned that k has to be positive real constant and not merely real constant Why k has to be positive real constant ? Let us revisit x k Interpreting f ( x ) from f '( x ) graph let us assume k ie negative understanding f " ( x) graph if k is negative, how will x will be less than a Interpreting f ' '( x) graph from f '( x ) graph Absolute value - Inequalities (1) x k where k is a non-zero positive real constant we can x k in three ways a inequality b number line c interval Inequality representation : k x k Number line representation : negative real constant , as the result of x is always positive or zero(when x ) As this contradicts ,our assumption k fails So k has to be a positive real constant Graphs of inequalities A sample problem: x interval representation : x (k , k ) The curve bracket indicates open interval This means x is between -k and k but not equal to -k or k (2) x k where k is a non-zero real constant we sketch x in casio graphical display inequality representation : k x k number line representation : calculator using the method of interval representation: x [ k , k ] The box bracket indicates closed interval This means x is between k and k inclusive of k and k By inspecting the point of intersection we obtain the following (3) x k where k is a non -zero real constant we can x k in three ways First sketching y x and y x for 3 x x for x 3 or 3 x for x 3 or x a inequality 59 Mathematics Times February 19 APPLICATIONS OF TRIGONOMETRY [ONLINE QUESTIONS] In a ABC , metres) of the tower is : a and C 60 Then the b ordered pair A, B is equal to : [2014] sin sin (a) sin [2015] (a) 45 , 75 (c) 15 ,105 (b) 105 ,15 (d) 75 , 45 h cos sin sin (b) h sin a cos sin h cos a sin h sin a cos (d) cos cos The angle of elevation of the top of a vertical tower from a point P on the horizontal ground was observed to be After moving a distance meters from P towards the foot of the tower, the angle of elevation changes to Then the height (in (c) 60 2sin sin cos (d) cos sin sin Let 10 vertical poles standing at equal distances on a straight line, subtend the same angle of elevation at a point O on this line and all the poles are on the same side of O If the height of the longest pole is ‘h’ and the distance of the foot of the smallest pole from O is ‘a’ ; then the distance between two consecutive poles is : [2015] (a) (c) sin sin (b) cos [OFFLINE QUESTIONS] Let a vertical tower AB have its end A on the level ground Let C be the mid-point of AB and P be a point on the ground such that AP=2AB If BPC , then tan is equal to : [2017] (b) (c) (d) 9 A man is walking towards a vertical pillar in a straight at a uniform speed At a certain point A on the path, he observes that the angle of elevation of (a) the top of the pillar is 30 After walking for 10 minites from A in the same direction, at a point B, he observes that the angle of elevation of the top of the pillar is 60 Then the time taken (in minutes) by him, from B to reach the pillar, is : [2016] Mathematics Times (a) 20 (b) (c) (d) 10 If the angles of elevation of the top of a tower from three collinear points A,B and C, on a line leading sin 105 sin 15 February cos15 2 sin15 to the foot of the tower are 30 , 45 and 60 respectively Then the ratio, AB : BC, is : [2015] (a) 1: (b) : (c) :1 (d) : A bird is sitting on the top of a vertical pole 20 m high and its elevation from a point O on the ground 2.Sol: is 45 It flies off horizontally straight away from the point O After one second, the elevation of the bird from O is reduced to 30 Then the speed (in m/s) of the bird is [2014] (c) 40 (d) 40 (b) 20 (a) 20 1 3 CD q, then AB is equal to : (a) p q sin p cos q sin p2 q2 (c) p cos q sin (b) Since, h10 h a10 tan (1) and a1 a h1 a tan (2) h a 9d tan where d is distance between a sin h h a tan cos d d sin tan cos q sin p cos q sin a10 a 9d h a tan 9d tan p q cos p cos q sin p (d) h h1 h2 h3 10 tan a1 a2 a3 a10 poles [2013] 1 ABCD is a trapezium such that AB and CD are parallel and BC CD If ADB , BC p and OA1 B1 , OA2 B2 , A3 B3 , , OA10 B10 all are similar triangles h cos a sin 9sin 3.Sol: Let AB be the tower of height ‘h’ d [ONLINE QUESTIONS] b a a [OFFLINE QUESTIONS] d b c b [ONLINE QUESTIONS] 1.Sol: sin A 2 sin B a Given : In ABP tan or AB PB sin h cos x 61 Mathematics Times February 19 x sin h cos h x sin sin cos AB BC Now, In ABC , tan As tan h cos sin h x cos x sin Putting the value of x in eq.(2) to eq (1), we get h cos sin 2sin sin h cos h cos sin 2sin sin h sin cos h sin cos cos sin 2sin sin h sin sin AB tan tan tan 1 AP 1 tan tan tan From 1 tan 1 tan tan h h 3h x a xa xa (1) 2.Sol: tan 30 tan 60 2sin sin h sin tan tan tan tan h h 3 a a h 3a [OFFLINE QUESTIONS] 1.Sol: Since AP AB Let AB AP APC tan AC AB AP AP C is the midpoint AC AB tan From (1) and (2) 3a x a x 2a Here, the speed is uniform So, time taken to cover x (time taken to cover a) Time taken to cover a 10 minutes =5 minutes 3.Sol: PB bisects APC , therefore AB : BC PA : PC 62 Mathematics Times February 20 and from BOD, tan 30 x y (2) From (1) and (2), we have x 20 and Also in APQ,sin 30 and in CPQ,sin 60 AB : BC 2h : h PA 2h PA So, h 2h PC PC 20 x y 20 20 y 20 20 y y 20 i.e., speed 20 1 1 m / s 5.Sol: From sine rule 2h :1 4.Sol: Let the speed be m/sec Let AC be the vertical pole of height 20 m Let O be the point on the ground such that AOC 45 Let OC x Time p q sin p q sin AB sin cos cos sin q sin p cos t 1s From AOC , tan 45 p2 q2 AB sin sin 20 x (1) cos q p2 q2 and sin p q p 63 Mathematics Times February 19 64 NEET PHYSICS NEET PHYSICS NEET PHYSICS NEET PHYSICS VOL-4 VOL-3 VOL-2 VOL-1 NEET NEET NEET NEET PHYSICS PHYSICS PHYSICS PHYSICS MCQ’s with Full Explanations MCQ’s with Full Explanations VOL-4 VOL-3 VOL-2 VOL-1 MCQ’s with Full Explanations MCQ’s with Full Explanations Ray Optics| Wave Optics| Dual Nature Of Matter & Photo Electric Effect| Atoms X-rays| Nuclear Physics| Electronic Devices|Current Communication Systems Electrostatics| Capacitors| Electricity| Magnetic Effects Of Current| Magnetism & Gravitation| Fluid Mechanics| MechanicalWaves Properties Of Matter| Thermal Properties Of Matter| 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Mathematics Times February Mathematics Times February 19 By DHANANJAYA REDDY THANAKANTI (Bangalore) Introduction This article explains us to find a indefinite and definite integral... write it as aRb If (a, b) R, we write it as aRb (I)Total number of relations: Let A and B be two non empty finite sets consisting of m and n elements respectively 17 Mathematics Times February. .. 12, 14 Median is 10 15 Mathematics Times February SETS & RELATIONS (1) The properties of inclusion () A, A A (reflexivity); A B and B C A C ( transitivity); () A, A If