Mathematics-17.Unit-13 Definite Integration and Appl.

1. DEFINITE INTEGRATION 3 2.1 GEOMETRICAL INTERPRETATION OF DEFINITE INTEGRAL 3 2.1.1 FUNDAMENTAL THEOREM OF CALCULUS (NEWTON-LEIBNITZ FORMULA) 3 2.1.2 PROPERTIES OF DEFINITE INTEGRATION 5 2.2 DIFFERENTIATION UNDER THE INTEGRAL SIGN 9 2.2.1 DEFINITE INTEGRAL AS LIMIT OF A SUM 10 2. SECTION C – AREA AS DEFINITE INTEGRAL 11 3.1 WORKING RULE FOR FINDING THE AREA 12 3. OBJECTIVE PROBLEMS 15   SYLLABUS Integration as the inverse process of differentiation, indefinite integrals, of standard functions, definite integrals and their properties, application of the fundamental theorem of integral calculus. 1. DEFINITE INTEGRATION 2.1 GEOMETRICAL INTERPRETATION OF DEFINITE INTEGRAL If f(x) > 0 for all x ∈ [a, b]; then is numerically equal to the area bounded by the curve y = f(x), then x-axis and the straight lines x = a and x = b i.e. In general represents to algebraic sum of the figures bounded by the curve y = f(x), the x-axis and the straight line x = a and x = b. The areas above x-axis are taken place plus sign and the areas below x-axis are taken with minus sign i.e, i.e. area OLA − area AQM – area MRB + area BSCD Note: , represents algebraic sum of areas means, that if area of function y = f(x) is asked between a to b. ⇒Area bounded = dx e.g., If some one asks the area of y = x3 between −1 to 1. Then y = x3 could be plotted as; ∴ Area = or, using above definition Area = But if, we integrate x3 between −1 to 1. ⇒ which does not represent area. Thus, students are adviced to make difference between area and definite Integral. 2.1.1 FUNDAMENTAL THEOREM OF CALCULUS (NEWTON-LEIBNITZ FORMULA) This theorem state that If f(x) is a continuous function on [a, b] and F(x) is any anti derivative of f(x) on [a, b] i.e. (x) = f (x) ∀ x ∈ (a, b), then The function F(x) is the integral of f(x) and a and b are the lower and the upper limits of integration. Illustration 1: Evaluate directly as well as by the substitution x = 1 /t. Examine as to why the answer do not tally? Solution: On the other hand; if x = 1/t then, ∴ I = when x = In above two results l = − π/4 is wrong. Since the integrand and therefore the definite integral of this function cannot be negative. Since x = 1/t is discontinuous at t = 0, the substitution is not valid (∴ I = π/4). Note: It is important the substitution must be continuous in the interval of integration. Illustration 2: Let then show that Solution: put x = 1/t ⇒ dx = –1/t2 then 2.1.2 PROPERTIES OF DEFINITE INTEGRATION 1. Change of variable of integration is immaterial so long as limits of integration remain the same i.e. 2. 3. . Generally we break the limit first at the points where f(x) is discontinuous and second at the points where definition of f(x) changes. Illustration 3: Evaluate , where [.] is the greatest integer function. Solution: Let I = Value of tan x at x = is 2 + Value of tan x at x = 0 is 0 Integers between 0 and 2 + are 1, 2, 3 ∴ tan x = 1, tan x = 2, tan x = 3 ⇒ x = tan-1 1, x = tan-1 2, x = tan-1 3 ∴ I = + = + = 0 + = = π = -tan-1 (-1) = . 4. . In particular . Illustration 4: If f, g, h be continuous function on [0, a] such that f(a - x) = f(x), g(a - x) = - g(x) and 3h(x) - 4h(a - x) = 5, then prove that . Solution: I = = = – 7I = 3I + 4I = = 5 = 0, since f (a – x) g (a – x) = –f (x) g (x) ⇒ I = 0 Illustration 5: Solution: Let I = ……….(1) = = = ……….(2) Adding (1) & (2), we get 2I = ⇒ I = Put ⇒ = = = . 5. Special cases: If f (x) = f (a – x), then . If f (x) = - f (a – x), then . 6. Special case: Illustration 6: Evaluate Solution: Let I = 2I = dx + dx = dx + dx = I = dx = 4 – tan-11 Illustration 7: Find the value of is Solution: Let I = I1 + I2 Consider I1 = …(1) Now I1 = …(2) Adding (1) and (2) , we get 2I1 = I1 = π/2 Consider I2 = Let g(x) = Now g(-x) = ∴ g(x) is an odd function ∴ ⇒ I2 = 0 I = I1 + I2 = π/2 + 0 = π/2 7. Illustration 8: Evaluate Solution: …(i) Again let where I = I1 + 3I2 I = 0 =0 8. If f (x) is a periodic function with period T, then , In particular, (i) if a = 0, where n ∈ I (ii) If n = 1, Illustration 9: Evaluate . Solution: Let I = We know that |sinx| is a periodic function with period π Hence I = [ applying prop. 8 ] Illustration 10: If f(x) is a function satisfying f(x + a) + f(x) = 0 for all x ∈ R and constant a such that dx is independent of b, then find the least positive value of c. Solution: We have f(x + a) + f(x) for all x ∈ R ……(i) ⇒ f(x + a + a) + f(x + a) = 0 [Replacing x by x + a] ….(ii) ⇒ f(x + 2a) + f(x+ a) = 0 ….(iii) Subtracting (i) from (ii), we get f(x + 2a) − f(x) = 0 for all x ∈ R. ⇒ f(x + 2a) = f(x) for all x ∈ R So, f(x) is periodic with period 2a It is given that dx is independent of b. ∴ The minimum value of ‘c’ is equal to the period of f(x) i.e., 2a. 2.2 DIFFERENTIATION UNDER THE INTEGRAL SIGN Leibnitz’s Rule If g is continuous on [a, b] and f1 (x) and f2 (x) are differentiable functions whose values lie in [a , b], then Illustration 11: If f(x) = cos x - then show that f′′ (x) + f (x) = – cos x Solution: f′ (x) = - sin x – = - sin x - f′′ (x) = - cos x – f(x) Illustration 12 If a function f(x) is defined ∀x ∈ R such that , a ∈ R+ exist. If g(x) = . Prove that Solution: g(x) = Diffrentiate w.r.t. x g’(x) = F(x) = -x g’(x) Illustration 13 Determine a positive integer n ≤ 5, such that Solution: Let Integrating by parts, In = …(i) Also, From (i), I2 = ( −1)3 − 2I1 = −1 − 2(2 − e) = −5 + 2e and I3 = (−1)4 − 3I2 = 1 − 3(−5 + 2e) = 16 − 6e Which is given . ∴ n = 3 2.2.1 DEFINITE INTEGRAL AS LIMIT OF A SUM An alternative way of describing is that the definite integral is a limiting case of the summation of an infinite series, provided f(x) is continuous on [a, b] i.e., . The converse is also true i.e., if we have an infinite series of the above form, it can be expressed as a definite integral. Method to express the infinite series as definite integral: (i) Express the given series in the form (ii) Then the limit is its sum when , i.e. (iii) Replace by x and by dx and by the sign of . (iv) The lower and the upper limit of integration are the limiting values of for the first and the last term of r respectively. Some particular cases of the above are (a) (b) where (as r = 1) and (as r = pn) Illustration 14 Show that (A) = ln2. (B) = (p > 0) Solution: (A) Let I = = = Now α = ( as r = 1 ) and β = (as r = n) ⇒ ⇒ I = ln2. (B) Take f(x) = xp; Let h = so that as n → ∞; h → 0 ∴ = = 2. SECTION C – AREA AS DEFINITE INTEGRAL Let f (x) be a continuous function in (a, b). Then the area bounded by the curve y = f (x), x axis and lines x = a and x = b is given by the formula , provided f (x) ≥ 0 (or f(x) ≤ 0) ∀ x ∈ (a , b) It is sometimes convenient to use formula for area with respect to y i.e. regarding x as a function of y. The area between x = f(y), y axis and the lines y = c and y = d is given by If we have two functions f(x) and g(x) such that f(x) ≤ g(x) ∀ x ∈ [a, b], then the area bounded by the curves y = f(x), y = g(x) and lines x = a, x = b (a < b) is given by 3.1 WORKING RULE FOR FINDING THE AREA (i) If curve lies completely above the x axis, then the area is positive but when it lies completely below x axis, then the area is negative, however we have the convention to consider the magnitude only. (ii) If curve lies on both the sides of x axis i.e. above the x axis as well as below the x axis, then calculate both areas separately and add their modulus to get the total area. In general if curve y = f(x) crosses the x axis n times when x varies from a to b, then the area between y = f(x), x axis and lines x = a and x = b is given by (iii) If the curve is symmetrical about x axis, or y axis, or both, then calculate the area of one symmetrical part and multiply it by the number of symmetrical parts to get the whole area. Illustration 15 Find the area between the curves y = x2 + x –2 and y = 2x, for which |x2 + x –2| + | 2x | = |x2 + 3x –2| is satisfied. Solution: y = x2 + x − 2 ⇒ y = 2x |x2 + x − 2| + |2x| = |x2 + 3x − 2| (x2 + x − 2) and 2x have same sign Thus required area ar (PQR) + ar (ECD) = [2x − (x2 + x − 2)] dx + [2x − (x2 + x − 2)] dx = = Illustration 16 Find out the area enclosed by circle |z| = 2, parabola y = x2 + x + 1, the curve y = and x-axis (when [ . ] is the greatest integer function). Solution: For x ∈ [-2, 2] ⇒ 1 < sin2 < 2 ∴ [sin2 ] = 1 Now we have to find out the area enclosed by the circle |z| = 2, parabola (y - ) = , line y = 1and x-axis. ∴ Required area is shaded area in the figure. Hence required area = dx = sq. units Illustration 17: Let f(x) = Max. {sinx, cos x, } then determine the area of the region bounded by the curves y = f(x), x-axis, y-axis and x = 2π. Solution: f(x) = Max { sin x, cos x, } interval value of f(x) for 0 ≤ x < , cos x for ≤ x < , sin x for ≤ x < , for cos x Hence required area = = sq units. Illustration 18: Find area bounded by y = and y = 7 - |x| Solution: y = (I) y = 7 – |x| (II) Rewriting (I) and (II) y = x∈(- ∞, -2) ∪ (2, ∞) y = Required area (PQRSTUP) = 2 area (PQRSP) = 2 [area of (OVRSO) – area of (POQP) – area of (QRVQ) ] = 2 [ (7 + 3)4 - - = 2 18 + = 32 sq unit. Illustration 19: Let f be a real valued function satisfying f and = 3. Find the area bounded by the curve y = f(x), the y axis and the line y = 3 Solution: Given = f(x) – f(y) .... (1) Putting x = y = 1, we get f(1) = 0 Now, f’(x) = = (from(1)) = ⇒ f′(x) = ⇒ f(x) = 3 ln x + c Putting x = 1 ⇒ c = 0 ⇒ f(x) = 3 ln x = y (say) ∴ Required area = = = = 3 (e – 0) = 3e sq. units Illustration 20: Let An be the area bounded by y = tann x, x =0, y = 0 and x = /4. Prove that for n ≥ 2. (i) An+An-2 = (ii) Solution: (i) Obviously An = . An+An-2 = = = (ii) Obviously An+2< An< An-2 (from figure). Thus 2 An = An + An < An + An-2 = (by part (i) Thus An < . . . . (1) Also 2 An = An+ An > An + An+2 = , replacing n by n+2 in (i) An > . . . . (2) From (1) and (2), we get . 3. OBJECTIVE PROBLEMS 1: = (A) 0 (B) 100 (C) 200 (D) 400 Solution : I = = = 200 = 200 = –200 = – 200 (–1 – 1) = 400 2: , where [.] is the greatest integer function = (A) 0 (B) 1 (C) 2 (D) None of these. Solution : Let I = = 2 ( function is even) min (x –[x], -x –[-x]) = ∴ I = = 2 = = 1 Hence I = 1 3: (where [.] denotes greatest integer function) = (A) 2− (B) 2+ (C) 2 (D) Solution : dx = dx + dx + dx = ( − 1) + 2 = − 1 + 3 − 2 = 2 + − 2 = 2− 4: (A) 2 (B) 0 (C) 1 (D) None of these. Solution : I = sin 2x ln (tan x) dx … (1) I = sin 2 ln tan dx I = sin 2x ln (cot x) dx … (2) Adding (1) and (2) 2I = sin 2x [ln (tan x) + ln (cot x)] dx = sin 2x ln 1 dx = 0. 5: (A) 0 (B) 1 (C) 2 (D) Solution : Let I = Put z = x - Then I = as is odd function 6: The value of is (A) (B) (C) (D) Solution : Let I = Here the lower limit is zero hence , we can replace x by (a – x) i.e. by π/2 –x ∴ I = = Adding 2 I = ⇒ I = . 7: = (A) (B) (C) (D) Solution : = x, = dx dx = ln 2 ( t = πx/3) 8: The area bounded by the curve y = x (3 – x)2 , the x – axis and the ordinates of the maximum and minimum points of the curve is (A) 2 sq. units (B) 6 sq. units (C) 4 sq. units (D) 8 sq. units Solution : y = x (3 – x)2 After solving , we get x =1 and x = 3 are points of maximum and minimum respectively. Now the shaded region is the required region ∴ sq. units 9: What is the area of a plane figure bounded by the points of the lines max (x, y) = 1 and x2 + y2 = 1 ? (A) sq. units (B) sq. units (C) sq. units (D) sq. units Solution : By definition the lines max, (x, y) = 1 means. x = 1 and y ≤ 1 or y = 1 and x ≤ 1 Required area = = = 1 = 0 = 10: The area bounded by the curve y = (x –1) (x – 2) (x – 3) lying between the ordinates x = 0 and x = 3 is (A) sq. units (B) 4 sq. units (C) sq. units (D) 3 sq. units Solution : Reqd. area = =- + - = + sq units. 11: The area bounded by the curve y = x (3 – x)2 , the x – axis and the ordinates of the maximum and minimum points of the curve is (A) 2 sq. units (B) 4 sq. units (C) 3 sq. units (D) 8 sq. units Solution: y = x (3 – x)2 After solving , we get x =1 and x = 3 are points of maximum and minimum respectively. Now the shaded region is the required region ∴ sq. units 12: The area enclosed by the curve |x + 1| + |y + 1| = 2 is (A) 3 sq. units (B) 4 sq. units (C) 5 sq. units (D) 8 sq. units Solution: Shift the origin at the point (-1, -1) So |x + 1| + |y + 1| = 2 becomes |x| + | y | = 2 Hence required area is = 4 × × 2 × 2 = 8sq. units 13: The area common to the curves y = x3 and y = is (A) 2 (B) 4 (C) 8 (D) None of these Solution: A = dx = = 14: The area of the region consisting of points (x, y) satisfying |x ± y | ≤ 2 and x2 + y2 ≥ 2 is (A) 8 – 2π sq. units (B) 4 – 2π sq. units (C) 1 – 2π sq. units (D) 2π sq. units Solution: Shaded region is the required one. ∴Required Area = sq. unit 15: where x and [.] denotes the greatest integer function is equal to. (A) −nπ (B) −(n +1)π (C) −2nπ (D) −2(n +1)π Solution: 16: If f(π) = 2 and then f(0) is equal to, (it given that f(x) is continuous in [0, π]) (A) 7 (B) 3 (C) 5 (D) 1 Solution: + sinx . f′ 17: Let f(x) is a continuous function for all real values of x and satisfies + a then value of ‘a’ is equal to (A) (B) (C) (D) None of these Solution: For x = 1, Diff. both sides of (i) w. r. t. x we get; f(x) = 0 − x2 f(x) + 2x15 + 2x5 18: . Then complete set of values of x for which f(x) ≤ lnx is (A) (0, 1] (B) [1, ∞) (C) (0, ∞) (D) None of these Solution: Let g(x) = f(x) − ln(x). x ∈ R + ⇒ g ’(x) = f ’(x) − = ⇒ g (1) = f(1) − ln1 = 0 − 0 = 0 ⇒ g(x) > 0 ∀ x > 1 and g(x) ≥ 0 ∀ x ∈ (0, 1] ⇒ ln x ≥ f(x) ∀ x ∈ (0, 1] 19: , where [.] denotes the greatest integer function, and x ∈ R+, is equal to (A) (B) (C) (D) Solution: Let n ≤ x < n + 1 where n ∈ I, I ≥ 0 20: Let I1 = , I2 = , I3 = , I4 = then (A) I1 > I2 > I3 > I4 (B) I2 > I3 > I4 > I1 (C) I3 > I4 > I1 > I2 (D) I2 > I1 > I3 > I4 Solution: (b) x > x2 ∀ x ∈ ⇒ ex > ∀ x ∈ ⇒ Since cos x > sinx ∀∈ ⇒ .cosx > sinx ⇒ ex > > .cos x > sinx ∀ x ∈ ⇒ I2 > I1 > I3 > I4 21: Area bounded by y = g(x), x-axis and the lines x = -2, x = 3, where and f(x) = x2 - , is equal to (A) (B) (C) (D) Solution: (A) Clearly 22: Area of the region which consists of all the points satisfying the conditions , is equal to (A) 4(7 – ln8)sq. units (B) 4(9 – ln8)sq. units (C) 2(7 – ln8)sq. units (D) 2(9 – ln8)sq. units Solution: The expression , represents the interior region of the square formed by the lines x = 4, y = 4 and xy 2. represents the region lying inside the hyperbola xy =2 Required area 23: Area bounded by the parabola (y - 2)2 = x – 1, the tangent to it at the point P (2, 3) and the x-axis is equal to (A) 9 sq. units (B) 6 sq. units (C) 3 sq. units (D) None of these Solution: (y - 2)2 = (x - 1) ⇒ 2(y - 2). = 1 ⇒ Thus equation of tangent at P(2, 3) is, (y - 3) = Required area = 9 sq. units 24: Two lines draw through the point P(4, 0) divide the area bounded by the curves , between the linea x = 2, and x = 4, in to three equal parts. Sum of the slopes of the drawn lines is equal to (A) (B) (C) (D) Solution: Area bounded by y = and x-axis between the lines x = 2 and x = 4, = Let the drawn lines are L1: y – m1(x - 4) = 0 and L2: y – m2(x - 4) = 0, meeting the line x = 2 at the points A and B respectively Clearly A = (2, - 2m1); B= (2, -2m2) Now Also 25: If A = , then is equal to (A) (B) (C) (D) Solution: Given A = A = ⇒ Now let I = put x = 2t ⇒ dx = 2dt ⇒ I = ⇒ I = . 26. The value of the integral is (A) 1 (B) (C) (D) none of these Solution: Using the property f (x) d x = f (a + b – x) dx, the given integral I = = = Here 2 I = ⇒ I = = . Hence (B) is the correct answer. 27. If I = dx then (A) 0 (B) 2 (C) (D) 2 – Solution: I = I = d x = 2 = 2 – 2 tan–1 x = 2 - . Hence (D) is the correct answer. 28. If I = , then (A) 0 < I < 1 (B) I > (C) I < (D) I > 2 π Solution: Since x ∈ ⇒ 1 ≤ 1 + sin 3 x ≤ 2 ⇒ ≤ ≤ 1⇒ ⇒ . Hence (C) is the correct answer. 29. If f (a + b –x) = f (x) then f (x) dx is equal to (A) d x (B) d x (C) 0 (D) none of these Solution: I = f (x) dx = (a+b-x) f (a +b-x) dx = (a + b) f(a +b –x) - x f (a + b –x) dx = ( a + b) f (x) dx - x f (x) d x Hence I = f(x) dx. Hence (B) is the correct answer. 30. The value of ( where {x} is the fractional part of x) is (A) 50 (B) 1 (C) 100 (D) none of these Solution: Given integral = ( by the def. of {x} ) = - = where t2 = x = = = . Hence (D) is the correct answer. 31. The value of |sin 2  x| dx is equal to (A) 0 (B) (C) (D) 2 Solution: Since |sin 2  x | is periodic with period , I = |sin 2 π x| dx= 2 sin 2 π x dx = 2 = Hence (B) is the correct answer. 32. Let f : R → R, f(x) = , where [.] denotes greatest integer function, then is equal to (A) 5/2 (B) 3/2 (C) 5 (D) 3 Solution: x – [x] = {x} x – [x +1] ={x} – 1 = = 3 Hence (D) is the correct answer. 33. is equal to (A) 0 (B) 2 (C) e (D) none of these Solution: I = property (f (–x) = –f (x), odd function) Hence I = 0 Hence (A) is the correct answer. 34. The value of is equal to (where [.] denotes greatest integer function) : (A) 20 (B) (C) (D) none of these Solution: I = dx = 20 dx = 20 dx = 20 = 20 (3 − 1) = . Hence (B) is the correct answer. 35. Values of is : (A) 1/2 (B) – 1/2 (C) 0 (D) none of these Solution: I = dx f (x) = cos x ln f (− x) = cox (− x) ln = − cos (x) ln = − f (x) f (x) is an odd function hence I = 0 Hence (C) is the correct answer. 36. f (x) = min (tan x, cot x), 0 ≤ x ≤ , then dx is equal to : (A) ln2 (B) ln (C) 2 ln (D) none of these Solution: f (x) = min (tan x, cot x), ∈ f (x) = tan x, 0 ≤ x ≤ = cot x, < x ≤ Hence = I = = (ln − 0) + 2 ln = ln 2. Hence (A) is the correct answer. 37. The value of is equal to : (A) (B) 2π (C) π (D) Solution: I = dx = dx = 2π . Hence (B) is the correct answer. 38. The value of is equal to : (A) 2 – (B) 2 + (C) (D) none of these Solution: I = = (1 − e−1) − (0 − 1) = 2 − e−1 Hence (A) is the correct answer. 39. has the value is : (A) 0 (B) 1/2 (C) 1 (D) 1/4 Solution: = = 0 Hence (A) is the correct answer. 40. is : (A) 0 (B) 1 (C) (D) Solution: I = dx I = dx ⇒ 2 I = dx ⇒ I = Hence (D) is the correct answer. 41. The value of depends on : (A) p (B) q (C) r (D) p and q Solution: I = dx = q (Since sin3 x and sin5 x are odd functions) Hence (B) is the correct answer. 42. The value of is equal to : (A) 2 (B) –2 (C) 1 (D) 0 Solution: I = = (1 − 0) − (0 − 1) = 2. Hence (A) is the correct answer. 43. Value of (where n ∈ N) is (A) (B) (C) n (n – 1) (D) None of these. Sol: Hence (B) is the correct answer. 44. is equal to (A) 0 (B) 2 (8295 + 1) (C) 8295 + 2 (D) None of these. Sol: sin193 x + x295 is an odd function of x.  The given integral is zero. Hence (A) is the correct answer. 45. If and , then (A) (B) (C) F' (1) = 1 (D) None of these. Sol: . Hence (B) is the correct answer. 46. The value of depends on (A) b (B) c (C) a (D) a and b. Sol: ( is an odd function of x). Hence (B) is the correct answer. 47. (m ≠ n and m, n are integers) = (A) 0 (B) π (C) π/2 (D) 2π Sol: . Hence (A) is the correct answer. 48. (A) (B) log a – log b (C) log a + log b (D) Sol: . Hence (A) is the correct answer. 49. (A) (B) (C) 1 (D) 0 Sol: sin11 x is odd function of x. So, integral is zero. Hence (D) is the correct answer. 50. If , then (A) (B) (C) (D) Sol: . Hence (D) is the correct answer. 51. (A) (B) (C) (D) Sol: . Hence (C) is the correct answer. 52. (A) (B) (C) πab (D) π2ab Sol: Divide Num. and Den. by cos2 x and put tan x = t etc. Hence (B) is the correct answer. 53. If g (x) = , then g (x + π) = (A) g(x) + g(π) (B) g(x) – g(π) (C) g(x) g(π) (D) Sol: (Put t = π + θ is second integral) . Hence (A) is the correct answer. 54. (A) 0 (B) (C) (D) Sol: [ first integrand is an odd function and second is an even function.] Hence (C) is the correct answer. 55. (where a, b are integers) = (A) – π (B) 0 (C) π (D) 2π Sol: . Hence (D) is the correct answer. 56. The smallest interval in which the value of lies is (A) (B) [0, 1] (C) (D) None of these. Sol: The function is increasing in [0, 1].  Min. value of f(x) is f(0) = 0 and Max. value of f(x) is f(1) = . Since , . Hence (A) is the correct answer. 57. If then (A) (B) (C) (D) Sol: For 0 < x < 1, x4 < x3 and for 1 < x < 2, x4 > x3. and . and and Hence (D) is the correct answer. 58. (A) (B) (C) 100 (e – 1) (D) Sol: Since the function x – [x] is periodic with period 1, therefore . Hence (C) is the correct answer. 59. (A) (B) (C) (D) Sol: (Integrating by parts taking (log x)n as first function) . Hence (C) is the correct answer. 60. (A) (B) (C) (D) Sol: (1) + (2) gives . Hence (D) is the correct answer. 61. The function satisfies (A) f(x + y) = f(x) + f(y) (B) (C) f(xy) = f(x) + f(y) (D) None of these. Sol: . Hence (C) is the correct answer. 62. If A is the area between the curves y = sin x and x-axis in the interval , then in the same interval, area between the curve y = cos x and x-axis is (A) A (B) (C) 1 – A (D) None of these. Sol: . . Hence (C) is the correct answer. 63. The ratio of the areas between the curves = cos x and y = cos 2x and x-axis from x = 0 to x = is (A) 1 : 3 (B) 2 : 1 (C) (D) None of these. Sol: . Hence (B) is the correct answer. 64. The area formed by triangular shaped region bounded by the curves y = sin x, y = cos x and x = 0 is (A) (B) 1 (C) (D) Sol: Area = . ( y = sin x and y = cos x meet when sin x = cos x ) Hence (A) is the correct answer. 65. The area of the figure bounded by the curves y = | x – 1| and y = 3 – | x | is (A) 2 (B) 3 (C) 4 (D) None of these. Sol: Required area = 4 sq. units. Hence (C) is the correct answer. 66. If the ordinate x = a divides the area bounded by the curve and the ordinates x = 2, x = 4 into two equal parts, then a = (A) (B) (C) 3 (D) None of these. Sol: Area = Also . Hence (B) is the correct answer. 67. The slope of the tangent to a curve y = f(x) at [x, f(x)] is 2x + 1. If the curve passes through the point (1, 2), then area bounded by the curve, x-axis and the line x = 1 is (A) (B) (C) (D) 6 Sol: . It passes through (1, 2)  c = 0  . It is a parabola with vertex  Required area . Hence (A) is the correct answer. 68. The triangle formed by the tangent to the curve f(x) = x2 + bx – b at the point (1, 1) and the coordinate axes, lies in the first quadrant. If its area is 2, then the value of b is (A) – 1 (B) 3 (C) – 3 (D) 1 Sol:  Equation of tangent at (1, 1) is y – 1 = (2 + b) (x – 1) (b + 2) x – y – (b + 1) = 0  Length of x-intercept = Length of y-intercept = – (b + 1).  Area of triangle (given) . Hence (C) is the correct answer.