Area Under the Curve Exercise # 5

EXERCISE # 5 Q.1 The slope of the tangent to the curve y = f(x) at a point (x, y) is 2x + 1 and the curve passes through (1, 2). The area of the region bounded by the curve, the x-axis and the line x = 1 is(A) 5/3 units (C) 6/5 units Sol. y =4 4 (B) 5/6 units (D) 6 units y2 = x [B] 4 dy = 2x +1 dx Integrating both side Here required area =  x dy 0  dy =  (2x  1) dx 4 4  y = x2 + x + c which passes through (1, 2) so 2=1+1+cc=0  y = x2 + x  y3  64 = y dy =   = 3  3  0 0  Q.3 2 The area of the triangle formed by joining the origin to the points of intersection of the line x 5 + 2y = 3 5 and circle x2 + y2 = 10, is (A) 5 Sol. –1 0 (B) 4 (C) 3 (D) 6 [A] x=1 A x2 + y2 = 10 M required area bounded by curve, x-axis and x = 1 is O 1 1 B  x3 x2  ( x 2  x ) dx =    3 2    0 0  = 1 1 5 + = sq. units. 3 2 6 x 5 + 2y = 3 5 We want to find the area of OAB for this, we draw a perpendicular from (0, 0) the given line Q.2 Sol. The area bounded by the parabola y2 = x, the line y = 4 and the y-axis is (A) 64/3 (B) 32/3 (C) 16/3 (D) 128/3  OM = [A] OM = 5 3 5 5 4 = 5 given that OB = 10  (MB)2 = (OB)2 – (OM)2 = 10 – 5 = 5  MB =  AB = 2MB = 2 5 Area of OAB 5 = 1 1 . AB. OM = . 2 5 . 2 2 1 5=5 required area = 2  (x  x 2 ) dx 0 Q.4 1  x2 x3    =2  3   2 0 For which of the following values of m, is the area of the region bounded by the curve y = x – x2 and the line y = mx equals 9/2. (A) –4 Sol. (B) –2 1 1 =2     2 3 (D) 4, – 2 (C) 2,–4 [C] y = x – x2 and y = mx  mx = x – x2  x2 + x (m –1) = 0  x = 0, x = 1 – m = 2. Q.6 1 m required area =  (x  x 2  mx) dx 1 1 = 6 3 The area of the triangle formed by the positive x-axis and the normal and the tangent to the circle x2 + y2 = 4 at (1, 3 ) is............ 0 1 m =  (x (1  m)  x 2 Sol. ) dx 0 A(1, 3 ) 1 m  x2 x3  =  (1  m)   3   2 0 (1  m) (1  m) (1  m) – = 2 3 6 3 = O 3 But given that area = 9 2 Equation of tangent at (1, 3 ) is x+ 9 (1  m) 3  =± 6 2 Solving, we get (1 – m)3 = ± 27 1 – m = ±3 m = 4, –2 Q.5 Sol. B(4, 0) 3 The area of the region bounded by the curves y = x2 and y = |x| is (A) 5/3 (B) 1/3 (C) 5/6 (D) 1/6 3y=4 OAB is a right triangle  Area of OAB =  AB = 1 (OA) (AB) 2 9  3 = 12 OA = 2 Area = Q.7 1 . 2. 12 = 2 3 2 Sketch the region bounded by the curves y = 5  x 2 and y = |x –1| and find its area. [B] y = x2 and y = |x| Sol. y= 5  x 2 and y = |x –1| point of intersection are 5 – x2 = (x –1)2  x = 2, –1 sketch is as follows  y = x2 and y = x  x(x – 1) = 0 ; x = 0, 1 5 3 Sol. – 5 4 –1 1 2 5 required area 2 =  1 2   5  x dx – (  x  1) dx – ( x  1) dx 2 1 1 y= 1 2 x 5 x  =  5  x 2  sin 1  – 2 2 5  1  y2 + 4y – 21 = 0 y2 + 7y – 3y – 21 = 0 (y – 3) (y + 7) = 0 1  x2   x   2  1 2 5 –1 2 5 sin + 1 – sin–1 2 2 5 =1+ 4  x2  –   x  1  2 A=  1  1   – 2 – 2  5 A= 2 – Find the area bounded by the curves: x2 + y2 = 4, 1 2 2 .y and x = y. 2  4  x 2 dx  0  0 = Sol. 2  0  x2 dx 2 2 (  2 , 2 ) | 4  x2 | dx 4 0  1 x x 25  x 2  25 sin 1   2 50 4 4  x2 x2  4 dx  dx 4 4 2  1 4 1 1 12  25 sin 1  – 2 + . 8– . 56 + 2  2 5 12 12 =6+ 1 . 2. 2 2 25  x dx  4 5 –1 1 5 1 sin (1) – = – 2 2 4 2 x2 = –  4 2 0 Solving we get Q.8 x2  4  x2 = 4y + 4 4 25 1 4 sin 4 2 5 Q.10 Find the area of the region bounded by the curve C : y = tan x, tangent drawn to C at x = /4 and the x-axis. Sol. y = tan x at x = /4  y = 1 dy  dy  = sec2 x    =2 dx  dx  (  / 41)     Hence tangent at  ,1 is (y – 1) = 2  x   4 4    2x – y = /2 – 1 Q.9 Find the area bounded by the curves x2 + y2 = 25, 4y = |4 – x2 |and x = 0 above the x-axis. /4 required area /4 =  /4  tan x dx – 0 0   x 4 2x 3 x 2     = ( x  2 x  x ).dx =   4  3 2  0 0   / 4 = log | sec x |0 / 4 – x 2  x   / 2 0 = log 2 2 (2 x  1   / 2) dx 3 2 16 16 2   =  4   2 = 6 – = sq. units. 3 3 3     2   2–    4 16    Q.12 2  1 = – – log 2 16 4 2 Compute the area of the region bounded by the curves y = ex n x and y = nx , where n e = ex 1. Q.11 Sol. Find all maxima and minima of the function y = x(x –1)2, 0  x  2. Also determine the area bounded by the curve y = x (x –1)2, the x-axis and the line x = 2. Sol. y = x(x–1)2 max. 1 O 1/2 1 min. n x ex n x 2 2 (e x – 1) = 0 ex 1 x = 1, x = e y = x (x – 1)2 4/27 ex n x = A= X    ex nx  1/ e dy = x. 2 (x– 1) + (x –1)2 dx = (x –1). {2x + x –1} = (x –1) (3x –1) + – + 1/3 1 nx   dx ex  Q.13 Sketch the curves and identify the region bounded by x = 1/2, x = 2, y = nx and y = 2x. Find the area of this region. Sol. The required area is the shaded portion in following figure.  y y = 2x 2  maximum at x = 1/3 ymax= 1  2 4   = 3  3 27 minimum at x = 1  ymin = 0 Now, to find the area bounded by the curve y = x (x –1)2, The y-axis and line x = 2. 2 4/27 O 1 X=2 y = logex O 1/2 x 1 2 1  x  2 the curve y = 2x lies above 2 as compared to y = loge x Hence, the required area In the region 2 =  (2 x  log x ) dx 1/ 2 2 2   Area = x ( x  1) 2 .dx 0  2x   ( x log x – x )  =   log 2   1 / 2 = –(x2 – 4x + 4 – 4) = – (x2 – 4x + 4) + 4 y = – (x –2)2 + 4 y – 4 = – (x – 2)2 y y = x2–x (2, 4) • 3 4 2 5 = – log 2 + 2 2 log 2 Q.14 Sketch the region bounded by the curves y = x2 and y = 2/(1 + x2), Find its area. Sol. The curve y = x2 is a parabola. It is symmetric about x-axis and has its vertex at (0, 0) and the 2 curve y = is a bell shaped curve, x-axis is 1 x2 its asymptote and it is symmetric about y-axis and its vertex is (0, 2). y y y = x2 A 2 ____ B y= 2 2 1+x _ C O x M To obtain the area, we need point of intersection of y = x2 ... (1) 2 and y = .... (2) 1 x2 2  y= 1 y  y + y2 = 2  y2 + y – 2 = 0 2  y + 2y – y – 2 = 0  y(y + 2) – 1(y + 2) = 0  (y – 1) (y + 2) = 0 y = –2, 1 but y  0 so y = 1  x = ±1 Therefore coordinates of C are (–1, 1) and coordinates of B are (1, 1) The area is OBACO = 2 area curve OB/AO 1 1  2 dx  x 2 dx  =2  1 x2  0 0    1  x3  = 2[2 tan–1x]01 –    3  0 • y = 4x –x2 x O 5/2 (1/2, –1/4) • • Therefore, it is a vertically downward parabola with vertex at (2, 4) and its axis is x = 2. and y = x2 – x (given) y = x2 – x +  1 1  y = x   – 2 4   y+ 2 1 = (x – 1/2)2 4 1 1 This is a parabola having its vertex at  ,   2 4 1 and opening upwards. 2 To obtain the x-coordinate of the points of intersection we solve y = 4x – x2 and y = x2 – x  4x – x2 = x2 – x 2  2x = 5x  2x2 –5x = 0  x (2 – 5x) = 0 its axis at x = 5 2 Also y = x2 – x, meets x- axis at (0, 0) and (1, 0) Now area,  x = 0, 5/ 2 A1 =  [(4x – x 2 )] – [(x2 –x)] dx 0 5/ 2 = 2  2 1  =2   =– 4 3 3   1 1 – 4 4   (5x – 2x 2 ) dx 0 5/ 2 Q.15 Sol. In what ratio does the x-axis divide the area of the region bounded by the parabolas y = 4x – x2 and y = x2 – x? y = 4x – x2 (given)  5 2  =  x 2  x 3  2 3  0  2 = 5 5 2 5   – .  2 2 3 2 3   = 5 25 2 125 . – . 8 2 4 3 = 125  2  125 1   = 8  3  24 2 = 2 2 . x3/ 2 3 3 1 1  x 2 x3    A2 = – ( x  x ) dx =   2 3   0 0 2 Q.17 Consider a square with vertices at (1, 1), (–1, 1), (–1, –1) and (1, –1). Let S be the region consisting of all points inside the square which are nearer to the origin than to any edge. Sketch the region S and find its area. Sol. The equations of the sides of the square are as follows: AB : y = 1, BC : x = –1, CD : y = –1, DA : x = 1 Let the region be S and (x, y) is any point inside it. Then according to given conditions, 1 1 1 – = 2 3 6 Therefore net area above the x- axis = 125  4 121 = 24 24 Hence, ratio of area above the x- axis and area below x- axis = A1 – A2 = = 121 1 : = 121 : 4 24 6 Q.16 Find the area given by x + y  6, x2 + y2 6y and y2 8x. Sol. x + y  6, x2 + (y –3)2  9, y2  8x We take x + y = 6, x2 + (y –3)2 = 9, y2 = 8x (0, 6) (0, 3)• A y2=8x    2 0 2 3 • 6 2 x  (3  9  x 2  dx +  3  9  x 2 ) dx   (6  x)  (3  2 2 =2 2  0 3 x dx + A(1, 1) (0, 1/2) (–1/2, 0) O x (1/2, 0) (0, –1/2) C (–1, –1) D(1, –1) B Solving y2 = 8x and x + y = 6, we get A(2, 4) and solving x2 + (y –3)2 = 9 and x + y = 6, we get B (3, 3) required area = y B (–1, 1) x 2  y 2 < |1– x|, |1 + x|, |1 – y|, |1+ y| O 2 3  x2  + 6 x   2   2 x 9 x  – 3x  9  x 2  sin 1  2 2 3 0  Solving, we get 1 Area = (27 –2) 12 This area is considering above and below x-axis both. Now for area below x-axis separately. We consider  2 0  2 3  (6  x ) dx – (3  9  x 2 ) dx 0  x2 + y2 < (1 – x)2, (1 + x)2, (1– y)2, (1 + y)2  x2 + y2 < x2 – 2x + 1, x2 + 2x + 1, y2 – 2y + 1, y2 + 2y + 1  y2 < 1 – 2x, y2 < 1 + 2x, x2 < 1 – 2y and x2 < 2y + 1 Now in y2 = 1 – 2x and y2 = 1 + 2x, the first equation represents a parabola with vertex at (1/2, 0) and second equation represents a parabola with vertex (–1/2, 0). And in x2 = 1 –2y and x2 = 1 + 2y, the first equation represents a parabola with vertex at (0, –1/2). Therefore, the region S is the region lying inside the four parabolas y2 = 1 – 2x, y2 = 1 + 2x, x2 = 1 + 2y, x2 = 1 – 2y y Q.18 Find the area of the region formed by 5 x2 + y2 – 6x – 4y + 12  0, y  x and x  . 2 Sol. x2 + y2 – 6x – 4y + 12  0, y  x, and x  A(1, 1) (0, 1) y2 = 2x –1 G I E O H F x2 = 2y –1 (1, 0) x  (x – 3)2 + (y –2)2 = 1, y = x, x = where S is the shaded region. Now, S is symmetrical in all four quadrants, therefore, S = 4 × area lying in the first quadrant. Now, y2 = 1 –2x and x2 = 1 –2y intersect on the line y = x. The point of intersection is P A E( 2 –1, 2 – 1) Area of the region OEFO = area of OEH + area of HEFH 1 = ( 2 –1)2 + 2 = 2 1  2 x dx 5/ 2 = 2 1 2 1 1   ( 2 –1)2 + (1  2x ) 3 / 2 . (1) 3 2 2   2 1    x   2  2 1/ 2 1 1 (2 + 1 –2 2 ) + (1+ 2 – 2 2 )3/2 3 2 1 1 = (3 – 2 2 ) + (3 –2 2 )3/2 3 2 1 1 = (3 – 2 2 ) + [( 2 –1)2]3/2 3 2 1 1 = (3 – 2 2 ) + ( 2 –1)3 3 2 1 1 = (3 – 2 ) + [2 2 –1 – 3 2 ( 2 –1)] 3 2 1 1 = (3 – 2 2 ) + [ 5 2 – 7] 3 2 1 1 = [9 – 6 2 + 10 2 –14] = [4 2 – 5] 6 6 1 Similarly, area OEGO = (4 2 –5) 6 Therefore, area of S lying in first quadrant 2 1 = (4 2 –5) = (4 2 –5) 6 3 4 1 Hence, S = (4 2 –5) = (16 2 – 20) 3 3 = 5 2 5/25/2 • 5/2 required area 1/ 2  5 2  1  ( x  3) 2  dx  5/ 2  x2  x 3 1 =   2x  1  ( x  3) 2  sin 1 ( x  3) 2 2  2  2 Q.19 =   25 3 –5– – –2+4– 8 8 12 4 = 1 3  – + 6 8 8 Let An be the area bounded by the curve y = (tan x)n and the lines x = 0, y = 0 and x = /4. Prove that for n  2, An + An–2 = deduce 1 1 < An < . 2n  2 2n  2 /4 Sol. We have, An =  (tan x ) n dx 0 Since, 0 < tan x < 1, when 0 < x < /4 We have 0 < (tan x)n + 1 < (tan x)n for each n  N /4   (tan x ) n 1 dx < 0  An+1 < An /4  (tan x ) 0 n dx 1 and n 1 y y = (tan x)n Sol. /4, 1) B /4 O x A= Now, for n > 2 /4  [(tan x ) An + An+2 = corresponding line. Sketch the region R and find its area. Point P lies inside OAB & closest to OA 1 1 1 3 1 3 A =  2 × tan 15° =  1 2 3 1 1 3 n  (tan x ) n  2 ] dx 42 3 2 A=2– 3 0 /4 =  0 /4 = Q.21 Indicate the region bounded by the curves x2 = y and y = x + 2 and obtain the area enclosed by them. Sol. x2 = y, y = x + 2, x- axis (tan x ) n  (1  tan 2 x ) dx  (tan x) n 2 sec x dx 0 C / 4  1  =  (tan x ) n 1   (n  1) 0  f ( x ) n 1  n  f ( x ) f ( x ) dx   n  1   1 1 = (1 – 0) = (n  1) n 1 Since An +2 < An+1 2 1 An + An < An + An–2 = n 1 1  2An < n 1 1  An < ... (2) 2n  2 1 1 From (1) and (2) < An < . 2n  2 2n  2 B A –2   Q.20  1   be the vertices Let O(0, 0), A(2, 0) and B 1, 3  of a triangle. Let R be the region consisting of all those points P inside OAB which satisfy d(P, OA)  min {d(P, OB), d(P, AB)}, where d denotes the distance from the point to the –1 O 0 D 2 required area = area of ABD + Area of region BOD 1 = × |x| + 2 0 x 2 dx 1 0  x3  1 1 1 5 = +   = + = 2 2 3 6  3  1 Q.22 Find all possible values of b > 0, so that the area of the bounded region enclosed between the parabola y = x – bx2 and y = x2/b is maximum. Sol. Eliminating y from y = x2 and y = x – bx2, we b get x2 = bx – b2x2  x = 0, b 1 b2 y y (0, 1) Q –1 O x b 1+b2 Thus, the area enclosed between the parabolas,  A= 0 2   x  bx 2  x  dx,  b   b / 1 b 2 b2 1 . 6 (1  b 2 ) 2 For maximum, value of A, dA =0 db 1 b (1  b 2 ) . 3 (1  b 2 ) 3 Hence, 2/3 = dA Sign scheme for around b = 1 is as below db – +  2/3  (1  x ) 2 dx + 0 1 Sol. 1/ 3 2 dx 2/3 2 /3  2x 3   1  =  (1  x ) 3  + x 2  +  3   3 0  1/ 3 1/ 3  1 1 1 3  3 x    2/3 Let f(x) = maximum {x2, (1 – x)2, 2x (1 – x)} where 0  x  1. Determine the area of the region bounded by the curves y = f(x), x-axis, x = 0 and x = 1. We can draw the graph of y = x2, y = (1 –x2) and y = 2x (1 –x) in following fig. Now, to get the point of intersection of y = x2 and y = 2x (1 –x). We solve both the equations. x 2 x (1  x ) dx + from scheme it is clear A is maximum at b = 1. Q.23 x 1  f (x ) dx 0 1/ 3  we consider only b = 1 0 1/3 1 A= dA = 0 gives b= –1, 0, 1 since b > 0 db – (1/2, 1/2) A B  (1  x ) 2 , 0  x  1 / 3  f(x) = 2x (1  x ) , 1 / 3  x  2 / 3  x2 , 2/3  x 1  The required area A is given by dA 1 (1  b 2 ) 2 .2b  2b 2 .(1  b 2 ).2b But = . db 6 (1  b 2 ) 4 = y = x2 We get x2 = 2x (1 – x)  x2 = 2x – 2x2  3x2 = 2x 2  3x – 2x = 0  x(3x –2) = 0  x = 0, 2/3 Similarly, we can find the coordinate of the points of intersection of y = (1 – x2) and y = 2x (1 – x) are x = 1/3 and x = 1. From the figure it is clear that  x 2 x 3 1  b 2  =   .  3 b   2 0 = y = (1–x) y = 2x(1– x) O b / 1 b 2 (1, 1) 2 3 =– 2 2 2 1 2 1 2   +   –   3 3 3 3 3 3 2 2 3 3 2 1 1 1 2 17 1 –   +   + (1) –   = sq. units 3 3 3 3 3 27 3 Q.24 Indicate the region bounded by the curves y = x log x and y = 2x – 2x2 and obtain the area enclosed by them. Sol. y = x log x, y = 2x – 2x2 2 1 1 1    y = x log x,  x   = – y  2 2 2   x  t3  =   –  3  0 y= x lox x x3 = – 3 y = 2x –2x 2 0 x x2 0 0  f (t ) dt –  2 dt 2  x log x dx ) dx + 0 2 0  area (OPQO) = 1 = 1  x2  1x  2 2 3 + x  x log x dx     3  0   0 0 2  2     2 t dt – 0 2   2 dt t 0 2 2 3/ 2 x 1 2 x t t 0 0  4 3 1 2 = x3 – x4 3 4 according to the given condition, = 1  x2  2 = 1 – +   3  4  0 =1– t the lower curve : x = y/2 lower limit of y : 0 and upper limit of y : x2 1 1  (2x  2x  f (t ) dt For the area (OPQO) the upper curve: x = y required area = x x x4 2 1 3 x – f ( t ) dt = x3 – 4 3 3 7 2 1 + = 3 4 12  0 Q.25 Differentiating both sides w.r.t. x, get x2 – f(x). 1 = 2x2 – x3  f(x) = x3 – x2, 0  x  1 Let C1 and C2 be the graphs of the functions y = x2 and y = 2x, 0  x  1 respectively. Let C3 be the graph of a function y = f(x), 0  x  1, f(0) = 0. For a point P on C1, let the Q.26 lines through P, parallel to the axes meet C2 and C3 at Q and R respectively (see in figure). Find the area of the region lying inside x2 + (y – 1)2 = 1 and outside c2x2 + y2 = c2 where c = (2 – 1) If for every position of P(on C1), the areas of the shaded regions OPQ and ORP are equal, determine the function f(x). (½,1) (0,1) Sol. (1,1) C2 C1 Q P (y – 1)2 – O (0,0) Sol. C3 (1,0) R Refer to the fig. in the question. Let the coordinates of P be (x, x2), where 0  x  1. For the area (OPRO), upper boundary y = x2 lower boundary : y = f(x) lower limit of x : 0 upper limit of x : x x  area (OPRO) =  0 y–1= y  0 f ( t ) dt y c & y–1=– y c (c  1) 1 c 2 1 1  c  y=  = 1  = 2  c 1 2 x t 2 dt – y2 0 c2 x=± 1 2 1/ 2  A=c  1/ 2 1  2 1/ 2  1. dx   1  x dx  2 1 2  1  x 2 dx P(2,3) 1 2 1 A= (c  1)  x 1  x 2  sin 1 x  21 – 2   2  –4 2 (c  1)    1    2 2  2 c 1  =  (c  1)  2 2 4 = Q.27 Sol.  A= the region in the third quadrant bounded by the curves x = –2y2 and y = f (x) lying on the left of the line 8x + 1 = 0 2=1+a+b a+b=1 …..(1) –2 = 1 – a + b a–b=3 …..(2) a = 2 & b = –1 x2 + 2x – 1   1  (2 y  4) dy 2 0 Q.29  2 2   2 2  2 2 4 1   = 1 2   2 2 2 2 Let f(x) be a continuous function given by 2x, | x | 1  f(x) =  2 . Find the area of x  ax  b, | x |  1 =  (y  2) 3 (c  1)  1  1   =      2 2 2 4 2 4  Consider the collection of all curve of the form y = a – bx2 that pass through the point (2, 1), where a and b are positive constants. Determine the value of a and b that will minimise the area of the region bounded by y = a – bx2 and x-axis. Also find the minimum area. Sol. 1 = a – 4b 4 b 1 b a b A=  (a  bx 2 ) dx =  (1  4b  bx 2 ) dx 4 b 1  b a  b Diff. w.r. to b a b  A=  a 1 a  2  (4  x ) dx = 4.2 b 3 b a 3/ 2 2 b a 2  a  8     0 b 3  b  Q.28 Sol. Find the area enclosed by the parabola (y – 2)2 = x – 1, the tangent to the parabola at (2, 3) and x-axis. Tangent at P(2, 3) 2y = x + 4 a  12  a = 12b b 1 3 b= , a= 2 8 a b  A  a b 3 1 2 3  x dx = .2 2 8 2 = 3 12  1 12 (12)3/2 a 2 a    b 24  b  3/ 2 Q.32 = 2 12  4 3 Q.30 For what value of ‘a’ is the area bounded by the curve y = a2x2 + ax + 1 and the straight line y = 0, x = 0 and x = 1 the least ? Sol. A= 1  y dx = 0 a2 a  1 3 2 Sol. dA 2a 1   0 da 3 2 3 a=– 4 Q.31 Sol. For what value of ‘a’ is the area of the figure 1 1 bounded by the lines, y  , y  ,x=2 x 2x  1 4 and x = a equal to n ? 5 1 1  x 2x  1 x=1 1 The tangent drawn from the origin to the curve, y = 2x2 + 5x + 2 meets the curve at a point whose y-coordinate is negative. Find the area of the figure bounded by the tangent between the point of contact and origin, the x-axis and the parabola. y = 2x2 + 5x + 2 = (2x + 1) (x + 2) 2 a Case 1 a > 2 a a 2a  1 3a P(x1, y1) tangent at P y  y1 5 = 2x x1 + (x + x1) + 2 2 2 2  n 2  n 3  4 1  4 5 15a2 = 64(2a – 1) 1   x  2x  1 dx  n a 2 2a  1 O lies on it y1 = 5x1 + 4 5x1 + 4 = 2x12 + 5x1 + 2 x1 = ± 1 x1 = –1, y1 = –1 (–1, –1) y=x  4 4 5 10a – 5 = 12a2 3a 5 2  12a – 10a + 5 = 0 D< 0 1 Case 3 < a< 1 2 1 2 1 1  4  1 1  dx     dx  n 5  2x  1 x   x 2x  1  a 1  1 / 2 2 5 2 2a  1 5 2  15a – 128a + 64 = 0 a = 8 Case 2. 1 < a < 2 –1/2  (2x 4 a –1 1 .1.1+ 2   x  4 n   n 2x  1  2 5  2 A= 1 2 n –2 1   x  2x  1 dx  n A=  5x  2) dx  1 Q.33 Let f be a real valued function satisfying x f (1  x ) f    f ( x )  f ( y) and lim  2. The x0 x y   area bounded by curve y = f(x), y-axis and the e line y = 2 is , then  = 3 Sol. 2 e y/2 dy  1 Put   y = t then 2 e t dt = 2 e t 2  1   e =2.e=  =6 3 Sol. Let f : R  R be a differentiable function such that f(0) = 1 and f(xy + 1) = f(x) f(y) – f(y) – x + 2 x,y R then area bounded by f(x) and g(x) = x2 + 1 can be expressed as p/q where p and q are relatively prime find (p + q). f ' (xy + 1) (y) = f ' (x) – 1 = f(y) f ' (x) – 1 put y = 0 f ' (x) = 1 f(x) = x + c c=1 f(x) = x + 1 g(x) = x2 + 1 2  1 dx 1 1 1 p    2 3 6 q p–q=7 Let f(x) = lim xn n n ( x  1) lim 2 x 0 x n=2 f(x) = n x2 = 2n x A=  x 1 x 0 = 2 Q.34 1 A= EXERCISE # 5 1. B 2. A 6. 2 3 sq. unit 7. 10. 2  1 – + log 2 16 4 2 14.  – 18. 2 3 5 1 – 4 2 11. 2 sq. units 3 15. 121 : 4 3. A 4. D 8. + (1/3) sq. units 9. 25. sin–1 (4/5) + 4 sq. units 12. (e2–5)/4e units2 13. 16. 1/12 (27 – 2) 17. 4/3 (4 2 –5) 3(1  3 )  4 unit2 20. (2 – 3 ) sq. unit 24 23. 17/27 sq. units 4 32. a = 8 or 27. 2  2 (6  21 ) 5 30. a   33. 6 22. b = 1 25. f (x) = –x2 + x3 24. 7/12 sq. units 29. b = 1/8, Aminimum = 4 3 sq. units 5 3 4 2 – log 2 + 2 2 log 2 21. 9/2 sq. unit  2 1  26.    units2    5. B 3 4 31. 257 sq. units 192 5 24 34. 7 28. 9 sq. units