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https://docs.google.com/document/d/1r4v6_8X4EZu8JjTLFMFltTjpjLyTLz8E/edit?usp=share_link&ouid=109474854956598892099&rtpof=true&sd=true 1. INTRODUCTION 3 2. CONCEPT OF LIMIT 3 2.1 RIGHT AND LEFT HAND LIMIT 4 2.2 REASON FOR NON-EXISTENCE OF THE LIMIT 4 2.3 BASIC THEOREMS ON LIMITS 5 3. EVALUATION OF LIMIT 5 3.1 METHOD TO EVALUATE THE LIMIT OF AN ALGEBRAIC FUNCTION 6 4. APPLICATION OF SOME BASIC LIMITS IN SOLVING THE LIMITS 8 5. METHOD OF SUBSTITUTION 9 6. METHOD FOR CALCULATING THE LIMITS OF THE FORM 10 7. METHOD TO SOLVE THE LIMIT OF THE FORM 11 8. L’HOSPITAL’S RULE 12 9. CONTINUITY OF A FUNCTION 13 10.1 REASONS FOR DISCONTINUITY OF A FUNCITON 13 10.2 CONTINUITY OF A FUNCTION IN AN INTERVAL 14 10.3 ALGEBRA OF CONTINUOUS FUNCTIONS 14 10.5 CONTINUITY OF COMPOSITE FUNCTIONS 14 10. TYPES OF DISCONTINUITY 15 OBJECTIVE ASSIGNMENT 16 SYLLABUS Limit and continuity of a function, limit and continuity of the sum, difference, product and quotient of two functions, I’Hospital rule of evaluation of limits of functions, continuity of composite functions, intermediate value property of continuous functions. INTRODUCTION Consider the function. Clearly f(x) is not defined at x = 1. At x = 1, , which is meaningless. X .9 .99 .999 .9999 .99999 f(x) 1.9 1.99 1.999 1.999 1.99999 From the above table it is clear that as x approaches to 1 i.e. x → 1 from the left hand side (means x approaches 1 from the values less than 1)f(x) approaches to 2 i.e. f(x)→2. The number 2 is called the left limit of f(x) and in symbol we shall write Again let us study the behaviour of f(x) where x approaches towards 1 from the right-hand side. X 1.1 1.01 1.001 1.0001 1.00001 f (x) 2.1 2.01 2.001 2.0001 2.00001 It is clear from the table that as x approaches to 2 i.e. x → 2, from the right-hand side (means x approaches 1 from the values greater than 1) f(x) approaches to 2 i.e. f(x) → 2. Here 2 is called the right-hand limit of f(x) and in symbol we will write Thus we see that f(x) is not defined at x = 1 but its left-hand limit and right-hand limit as x → 1 exist and are equal. When are equal we say exist and is equal to 2. Meaning of (x → a) Let x be a variable and ‘a’ be a constant. x assumes value nearer and nearer to ‘a’, then we say ‘x tends to a’ and write ‘x → a’ and it doesn’t mean x = a. CONCEPT OF LIMIT Let y = f(x) be a given function defined in the neighbourhood of x = a, but not necessarily at the point x = a. The limiting behaviour of the function in the neighbour of x = a is called the limit of the function when x approaches a. Mathematically we write this as . would simply mean that when we approach the point x = a from the value which are just greater than or just smaller than x = a, f (x) would have a tendency to move closer to the value . 2.1 RIGHT AND LEFT HAND LIMIT Right-hand limit means tendency of function when we approach x = a from the value which just greater than ‘a’ and we write . Working rule to evaluate Put x = a + h in f(x) to get Take the limit as h → 0 Left-hand limit means tendency of function when we approach x = a from the values which are just less than ‘a’ and we write . Working rule to evaluate Put x = a – h in f(x) to get Take the limit as h → 0 For example: =2 Thus for the existence of the limit of f(x) at x=a, it is necessary and sufficient that = , if these are finite or and both should be either + or -. 2.2 REASON FOR NON-EXISTENCE OF THE LIMIT Following are the reasons when will not exist. Reason 1: when left and right tendencies of f(x) are not same in the neighbourhood of x = a ⇒ For example (when [.] denotes the greatest integer function) It is clear from the figure that left tendency of [x] at x = 2 is 1 while the right tendency of [x] at x = 2 is 2. This will not exist. Reason 2: If f(x) is not defined in the neighbourhood of x = a. For example Here f(x) is well defined at x = 0 But in the left neighbourhood of 0, means , when , similarly for right tendency , in this case also in the right neighbourhood of x=0, cosx<1. Hence sec-1(cosx) is not defined. Reason 3: When f(x) doesn’t have a unique tendency. For example we have that –1 ≤ sinx ≤ 1, that means for all non-zero values of x, sin 1/x would assume finite values. But when x becomes very near to zero sin 1/x would erratically oscillate between –1 and + 1. It implies that sin 1/x wouldn’t have unique tendency for very small value of x. Thus will not exist. 2.3 BASIC THEOREMS ON LIMITS Let f(x) = 1 and g(x) = 2, where 1 and 2 are finite, then the following theorems on limits can be used to evaluate the limits (i) (c1 f(x) ± c2 g(x)) = c1 1 ± c2 2, where c1and c2 are given constants. (ii) f(x). g(x) = f (x). g (x) = 1. 2 (iii) (iv) f (g(x)) = f (g(x)) = f(2), if and only if f(x) is continuous at x = 2. For example (where [.] denotes the greatest integer function) Here [x] is not continuous at x = 1. Also =0 and =1. (v) If f(x) ≤ g(x) ∀ x ∈ R, then f(x) ≤ g(x). Note: We have to be very careful while using these theorems. For example if we try to apply the theorems on =1 we get = sin x. 1/x, which does not exist. Which is an absurd result, because in this case the given limit can not be written as the product of two limits as 1/x does not exist. EVALUATION OF LIMIT When the given limit is not in the indeterminate form, limit can be evaluated by directly using the definition of limit or putting a in place of x if x→a. For exampleHere form is not indeterminate. When the form of the given function is indeterminate, then our motive will be to get rid of the indeterminate form. After getting rid of the indeterminate form we use the definition of limit or directly put the value. For example = 3.1 METHOD TO EVALUATE THE LIMIT OF AN ALGEBRAIC FUNCTION The following methods are useful for evaluating limits of algebraic function. Method of factorisation: If f(x) and g(x) are polynomials such that f(a) = g(a) = 0, then (x – a) is a factor of both f(x) and g(x). Now to solve , we cancel the common factor (x – a) from both the numerator and denominator, and again put x = a in the given expression. If we get a meaningful number, then the number is the limit of the given expression otherwise we repeat the process till we get rid of indeterminate form. Method of rationalisation: If in any limit denominator or numerator involve the radical sign, this method is useful. Example 1: Evaluate Solution: Example 2: Evaluate Common mistakes: How student used to solve this problem. Step1.→ Step2.→ But this is wrong, can you find out where is the mistake. Solution: Step1.→ Step2.→ This is the correct answer Key concept: Using the concept of method of substitution put Method when : If given limit is in this form , where f(x) and g(x) are algebraic function in x we divide numerator and denominator by highest power of x in f(x) and g(x) Example 3: Evaluate Solution: Here highest power in f(x) and g(x) is x1/2. Hence divide numerator and denominator by x1/2 and then apply the limit. In this method basically we use the series expansion of sinx, cosx, tanx, log(1+x), ax, ex etc to evaluate the limit. Following are some of the frequently used series expansions: sin x = cos x = tan x = ex = ax = 1 + x.lna + (lna)2 +……. a ∈ R+ (1+x)n = 1 + nx + n ∈ R. |x|<1, n is any real number ln (1+x) = -1 < x ≤ 1. sin-1 x = tan-1 x = (1+x)1/x = Example 4: Evaluate Solution: = 1/6 APPLICATION OF SOME BASIC LIMITS IN SOLVING THE LIMITS Following are some basic limits which are used very frequently in solving the limits. (i) If p(x) is a polynomial, p(x) = p(a). (ii) cos x = 1 (where ‘x’ is in radians) (iii) = e (iv) = e (v) = 1 (vi) = , a ∈ R+. (vii) = n (viii) (ix) = 1 (x) = loga e, a > 0, ≠1 (xi) Now if then we can redefine the limits in the following manner (i) cos f(x) = 1 (ii) = 1 (iii) = ln b ( b> 0) (iv) = e Example 5: If and exist, then the value of k. Key concept: Since exist. That means Solution: Hence METHOD OF SUBSTITUTION To evaluate by the method of substitution we use different substitutions. For example if we substitute x by a + h, then the given limit becomes (as x→a h→0). Similarly we can use different substitution like substitute x by 2h, 1/h etc. For example, (putting x = a + h) Example 6: Evaluate Key concept: First of all substitute cos–1x = θ, use the method of substitution. Solution: Let cos–1x = θ, then x = cosθ Now as x → 1 θ → cos–1(1), i.e. θ → 0 METHOD FOR CALCULATING THE LIMITS OF THE FORM If the given limit is in this form then we express the given expression as power of e. For this use the formula . Here two cases arises Case I: When Let Hence Case II: When but f(x) is positive in the neighbourhood of x = a Case III: If f(x) is not throughout positive in the neighbour hood of x = a, then will not exist. Because in this case function will not be defined in the neighbour hood of x = a Example 7: Evaluate Solution: Clearly given limit is in the form of where Hence METHOD TO SOLVE THE LIMIT OF THE FORM To solve the limit of the form we use the concept ’ Case I: If , then ax will keep on decreasing with on increasing in x. Hence Case II: If a = 1, in this case given function becomes a constant function. Hence . Case III: If a > 1, then ax will keep on increasing with on increase in the value of x. Hence Case IV: If a < 0, then given function will not be defined. Hence will not exist. Misconception: Normally students have a confusion between andwhen a > 1 and they argue that should be infinite, because But we have already seen that Actually both the results can not be compared as in the case of , ‘a’ is a fixed number on the other hand in case of is clearly a variable number decreasing with increase in x. Example 8: Evaluate Solution: Given (∞° form) Example 9: Evaluate Solution: L’HOSPITAL’S RULE Let f(x) and g(x) be two functions differentiable in the neighbourhood of the point ‘a’, except at the point ‘a’ itself. If f(x)=g(x) = 0. Or, f(x) = g(x) = ∞. Thenprovided that the limit on the right exist as a finite number or is ± ∞ . Example 10: If = – 1 then find the value of a. Solution: Since the given limit is in the form of 0/0, we will use L’ Hospital’s Rule = It is satisfied only when a = 1. Remark: L’Hospitals rule is not always useful. Consider the example,. Here, if we apply Hospital’s rule, then Now, both the Nr and the Dr are undefined because cos x doesn’t exist. We can find this limit as follows: (since = 0) CONTINUITY OF A FUNCTION A function f(x) is said to be continuous at x = a if = f(a) i.e. L.H.L.=R.H.L. = value of the function at a i.e. . If f(x) is not continuous at x = a, we say that f(x) is discontinuous at x = a. Geometrical meaning: The function ‘f’ will be continuous at x = a if there is no break in the graph of the function y = f(x) at the point (a, f(a)). 10.1 REASONS FOR DISCONTINUITY OF A FUNCITON One of the following may be the reasons for the discontinuity of f(x) (i) exist but are not equal. For example f(x)=[x] is discontinuous at all integer points. (ii) exist and are equal but not equal to f(a). for example f(x)=[sinx] where at (iii) when f(x) is not defined at x=a. For example f(x)=1/x-1 (iv) At least one of the limits does not exist or atleast one of these limits is . 10.2 CONTINUITY OF A FUNCTION IN AN INTERVAL Continuity in an open interval: A function f(x) is said to be continuous in an open interval (a, b) if it is continuous at each point of the interval (a, b). Continuity in a closed interval: A function f(x) is said to be continuous in a closed interval [a, b] if (i) f(x) is continuous at each point of the interval (a, b). (ii) f(x) is continuous from right at x = a i.e. . (iii) f(x) is continuous from left at x = b i.e. and; Geometrical meaning : The function f(x) will be continuous in the closed interval [a, b] if the graph of y = f(x) is an unbroken line (curved or straight) from the point (a, f(a)) to (b, f(b)). 10.3 ALGEBRA OF CONTINUOUS FUNCTIONS Let f(x) and g(x) be two functions, then the following results holds true. Case I: If f and g both are continuous at x=a, then c1f(x) ± c2(x) and f(x) . g(x) will be continuous at x = a. And f(x)/g(x) will also be continuous at x = a, provided g(a) ≠ 0. Case II: When one of the function f or g is discontinuous at x=a, then c1f(x) ± c2(x) is definitely discontinuous, but nothing can be said about the continuity of f(x) . g(x) and f(x)/g(x). They may or may not be continuous at x=a. Case III: When f and g both are discontinuous at x=a, then nothing can be said about the continuity of c1f(x) ± c2(x), f(x) . g(x) and f(x)/g(x). 10.5 CONTINUITY OF COMPOSITE FUNCTIONS Let f(x) and g(x) are two functions, now we are interested in the continuity of f(g(x)). Case I: If f and g both are continuous, then f(g(x)) will also be continuous. Case II: If f is continuous and g is discontinuous. Here again two cases arises. (a) If points of discontinuity of g(x) are not lying in the domain, the f(g(x)) will be definitely discontinuous at those points. (b) If points of discontinuity lies in the domain, then nothing can be said about the continuity of f(g(x)) in general. Case III: If f and g both are discontinuous, the also nothing can be about the continuity of f(g(x)). Example 11: Find the points of discontinuity of g(f(x)) if g(x) = and f(x) = . Solution: The function f(x) = is discontinuous at the point x = 1. The function g(f(x))== is discontinuous at f(x) = -2 and f(x)=1. When f(x) = -2, =-2 ⇒ x = When f(x) = 1, = 1 ⇒ x = 2. Hence, the composite function y = g(f(x)) is discontinuous at three points x = 1/2, 1, 2. TYPES OF DISCONTINUITY Basically there are two types of discontinuity. Removable discontinuity: If f(x) exists but is not equal to f(a), then f(x) has a removable discontinuity at x = a and it can be removed by redefining f(x) for x = a. Example 12: Redefine the function f(x) =[sinx] where x in such a way that it could become continuous for x ∈(0, ). Solution: Here but . Hence, f(x) has a removable discontinuity at x = . To remove this we redefine f(x) as follows f(x) = [sinx], x ∈ (0,π/2) ∪ (π/2,) = 0 , x = . Now, f(x) is continuous for x∈(0, ). Non-removable discontinuity: If f(x) does not exist, then we can not remove this discontinuity. So this become a non- removable discontinuity or essential discontinuity. Example 13: Prove that f (x) = {x} has non removable discontinuity at any x∈I. Solution: Since does not exist for any a∈I. Hence, f(x)= {x} has non-removable discontinuity at any x ∈ I OBJECTIVE ASSIGNMENT 1. , where a > b >, is equal to: (A) –1 (B) 1 (C) 0 (D) none of these Solution: Hence (B) is the correct answer. 2. Let The value of is: (A) (B) (C) (D) none of these Solution: x2 + 2x + 3 = (x+1)2 + 2 ≥ 2. So a = 2 Now Hence (C) is the correct answer. 3. Let f: [1, 10] → Q be a continuous function and f(1) =10, then f(10) is equal to: (A) (B) 10 (C) 1 (D) Can’t be obtained. Solution: Using the intermediate value theorem of continuous function we get f(x) is a constant function , therefore ∀ x ∈ [1 ,10] , f(10)=10. Hence (B) is the correct answer. 4. is equal to: (A) (B) 1 (C) 0 (D) –1 Solution: = [Since sin2 n! always lies between 0 and 1. Also, since 1-k >0,hence n1-k Hence (C) is the correct answer. as n] 5. If f(x) At x=0, then (A) a ∈ (0, ∞ ) (B) a ∈ (1, ∞ ) (C) a ∈ (-1, ∞ ) (D) a ∈ (-∞, 1) Solution: f(x) is continuous at x=0 hence xa sin = f(0) =0 This is only possible when a>0, thus the required set of values of a is (0, ∞ ) Hence (A) is the correct answer. 6. Let f(x) =, then in [0, ]. (A) tan (f(x)) and both are continuous. (B) tan f(x) is continuous but f-1(x) is not continuous (C) tan (f–1(x)) and f-1(x) are discontinuous (D) none of these. Solution: x ∈ [0,] ⇒ Now which is discontinuations at x=2. tan (f(x)) is continuous for . f-1(x) = 2(x+1) which is clearly continuous but tan(f-1 (x)) is not continuous. Hence (B) is the correct answer. 7. The value of is: (A) 2/3 (B) 1/3 (C) 1 (D) 5/3 Solution: = = Hence (A) is the correct answer. 8. ( 1+cos2m n!x) is equal to (x is rational): (A) 2 (B) 1 (C) 0 (D) 3 Solution: We know that |cos | 1 for all . Also |cosn!x|<1, if x is irrational. Hence (1+ cos 2m n! x) =1 and if x is rational (i.e. x= p, q €I). n!x is an integral multiple of . Hence cos n!x =1 or -1 and cos2m n! x=1. Hence (1+ cos 2m n! x) =2 Hence (A) is the correct answer. 9. If (A) (B) λ = 2, μ = 1 (C) λ = 1, μ = is any real costant (D) λ = μ = 1 Solution: = = Hence (C) is the correct answer. 10. = (A) log 2 (B) log 2 (C) 0 (D) None of these Solution: = = dx let 1 + x3 = t 3x2 dx = dt ⇒ x2 dx = = log 2 Hence (A) is the correct answer. 11. If f(x) is a continuous function ∀ x ∈ R and the range of f(x) = ( 2, ) and g(x) = is continuous ∀ x ∈ R ( [.] denotes the greatest integer function), then the least positive integral value of a is: (A) 2 (B) 3 (C) 6 (D) 5 Solution: Since g(x) is continuous ∀ x ∈ R, g(x) should be constant. Since , a ≥, ( as = 0 ∀ x ∈ R) . So least integral value of a is 6. Hence (C) is the correct answer. 12. If the graph of the continuous function y = f(x) passes through (a, 0), then is equal to: (A) 1 (B) 0 (C) –1 (D) none of these Solution: Since f (a) = 0 ⇒ = = Hence (C) is the correct answer. 13. Among . . . . (1) and . . . . (2) (A) (1) exists, (2) does not exist (B) (1) does not exist , (2) exists (C) both (1) and (2) exist (D) neither (1) nor (2) exists Solution: is more than 1 in the neighbourhood of ‘0’. Hence exists while is less than 1 in the neighbourhood of ‘0’ . Hence does not exist. Hence (A) is the correct answer. 14. The values of A and B so that function f (x) defined by f (x) = become continuous, respectively are (A) , (B) , (C) , (D) , Solution: f (x) = L.H. limit at x < = x + Asin x = + A sin = + A× = A + R.H. limit = 2 x cot x + B = . cot + B = + B A + = B + ⇒ A − B = … (1) L.H. limit at x < = − (2x cot x + B) = 2× cot + B = B RH limit = x+ A cos 2x − B sin x = A cos π − B sin = − A − B − A − B = B ⇒ a = − 2B … (2) − 3B = ⇒ B = , A = + = Hence (B) is the correct answer. 15. The value of , n ∈I is: (A) 1 (B) 0 (C) n (D) n(n –1) Solution: = = . = = = = = . Hence (B) is the correct answer. 16. = (A) 1 (B) 0 (C) –1 (D) does not exist Solution: x-1 ≤[x] ≤x ⇒ ⇒ Hence (B) is the correct answer. 17. is (A) 1 (B) e2 (C) elm (D) log 6m Solution: Hence (B) is the correct answer. 18. The value of is usual to (A) 0 (B) 1 (C) -1 (D) none of these Solution: Put Hence (A) is the correct answer. 19. e1/x equals (A) does not exist (B) 1 (C) 0 (D) infinity Solution: , Hence (A) is the correct answer. 20. (A) 9/2 (B) 3/4 (C) 9/4 (D) 1/4 Solution: Hence (C) is the correct answer. 21. If f(x) = x - |x –x2| , x ∈ [-1 , 1]. Then the number of points at which f(x) is discontinuous (A) 0 (B) 2 (C) 1 (D) 3 Solution: Given function can be written as Hence (A) is the correct answer. 22. (A) √2 (B) 1/√2 (C) 1/2 (D) 2 Solution: Hence (B) is the correct answer. 23. where [⋅] denotes the integral part of x (A) Is equal to – 1 (B) Is equal to – 2 (C) Is equal to – 3 (D) Does not exist Solution: Given limit is Hence (B) is the correct answer. 24. Let f(x) = and g(x) = , the is, (where [.] denotes the greatest integer function) (A) 1 (B) 2 (C) 3 (D) 4 Solution: Hence (B) is the correct answer. 25. equals to (A) /2 (B) 2 / (C) 2 (D) 1 Solution: Hence (D) is the correct answer. 26. equals to (A) sin 2 (B) 2 sin 2 (C) cos 2 (D) 2 cos 2 Solution: Hence (B) is the correct answer. 27. The number of points at which the function f(x) = is discontinuous is (A) 1 (B) 2 (C) 3 (D) none of these Solution: The given function is discontinuous at all integral points Hence (D) is the correct answer. 28. (A) 2 (B) –2 (C) 1 (D) –1 Solution: Put x - 2 = t Hence (C) is the correct answer. 29. Evaluating gives (A) e (B) e2 (C) e–1 (D) e–2 Solution: Hence (A) is the correct answer. 30. f(x) = (Where [.] denotes greatest integer function) If f(x) is continuous at x = 0 then β is equal to (A) α + 1 (B) α –1 (C) α + 2 (D) α – 2 Solution: f(0) = 2 Hence (A) is the correct answer. 31. Ifα, β are the roots of the equation lx2 + mx + n = 0 then is (A) α - β (B) 2ln|α - β| (C) e2l(α - β) (D) el(α - β) Solution: α, β are roots of lx2 + mx + n = 0, then the given equation can be written as lx2 + mx + n = 0 = l (x - α) (x - β). Hence Hence (D) is the correct answer. 32. equals (A) - π (B) π (C) (D) 1 Solution: = = π Hence (B) is the correct answer. 33. (A) 2 (B) –2 (C) (D) – Solution: Hence (D) is the correct answer. 34. If f(x) = . Then f(x) is (A) 0 (B) 1 (C) 2 (D) does not exist Solution: Hence (A) is the correct answer. 35. = (A) -1 (B) 1 (C) 16 (D) 32 Solution: Directly apply the limit, we get 32 Hence (D) is the correct answer. 36. (A) 0 (B) 1 (C) e4 (D) e5 Solution: 1 ∞ form, given limit equal to Hence (B) is the correct answer. 37. = (A) 1 (B) –1 (C) 0 (D) none of these. Solution: Given limit is and Hence (D) is the correct answer. 38. equals (A) 2 (B) 0 (C) –2 (D) none of these Solution: = . = LHL ≠ RHL Limit does not exist. Hence (D) is the correct answer. 39. equals (A) –1 (B) 0 (C) 2 (D) 1 Solution: = = – 1 Hence (A) is the correct answer. 40. is equal to (A) 2 (B) 1 (C) –1 (D) 1/2 Solution: = ( L’ Hospital rule ) = = = . Hence (D) is the correct answer. 41. If = – 1 then (A) a = 1 (B) a = 0 (C) a = e (D) none of these Solution: L’ Hospital Rule = It is satisfied only when a = 1. Hence (A) is the correct answer. 42. In order that the function f(x) = (x +1)cotx is continuous at x = 0, f(0) must be defined as (A) f(0) = 0 (B) f(0) = e (C) f(0) = 1/ e (D) none of these Solution: For continuity actual value must be equal to limiting value A = logA = = = (By L’ Hospital Rule) log A = 1 ⇒ A = e1 = e . For f(0) must be defined as f(0) = e . Hence (B) is the correct answer. 43. Which of the following is not the point of discontinuity of the function f(x) = ? (A) x = 0 (B) x = (C) π (D) x = Solution: f(x) is continuous except at the points where 1 – cos4x = 0. Thus x = 0, x = and x = π are point of discontinuity. Hence (D) is the correct answers. 44. (A) 1 (B) 2 (C) 0 (D) 3 Solution: = Hence (B) is the correct answer. 45. is equal to (A) 2 (B) 3 (C) 4 (D) 5 Solution: = 2 Hence (A) is the correct answer. 46. If = e2, then (A) λ = 1, μ = 2 (B) λ = 2, μ = 1 (C) λ = μ = 1 (D) none of these Solution: e2λ = e2 ⇒ λ = 1. Hence (D) is the correct answer. 47. is equal to (A) log2 e (B) loge 2 (C) 1 (D) none of these Solution: = Hence (A) is the correct answer. 48. If f (x) = cos , 1 < x < 2 (where [.] denotes the greatest integer function) then (A) 0 (B) 3 (C) –3 (D) none of these Solution: x = ⇒ [x] = 1 So, f (x) = cos = sin x3 f′ (x) = 3x2 cos x3 f′ = 3 cos = 0 Hence (A) is the correct answer. 49. is equal to (A) 0 (B) 3 (C) –3 (D) does not exist Solution: RH Limit = = = –3 LH Limit = Hence (C) is the correct answer. 50. f(x) = , then is (A) 1 (B) 0 (C) – 1 (D) none of these Solution: f (x) = = 1 not defined Hence (D) is the correct answer. 51. The function f (x) = 1 + |sin x| is (A) continuous nowhere (B) continuous everywhere (C) discontinuous at x = 0 (D) None of these Solution: f (x) = 1 + |sin x| where sin x > 0 ⇒ f (x) = 1 + sin x sin x < 0 ⇒ f (x) = 1 − sin x so continuous every where Hence (B) is the correct answer. 52. f(x) = Value of ‘a’ for which f(x) is continuous, is (A) 1 (B) 2 (C) –1 (D) –2 Solution: f (x) = (x + 1) = 2 = (3 − ax2) = 3 − a 3 − a = 2 ⇒ a = 3 − 2 ⇒ a = 1 Hence (A) is the correct answer. 53. If f (x) = , then is (A) e4 (B) e3 (C) e2 (D) 24 Solution: f (x) = log f (x) = = = = = = = 4 ⇒ f (x) = e4 Hence (A) is the correct answer. 54. is (A) 2 (B) 1 (C) 0 (D) none of these Solution: , let x = cos θ ⇒ (sin 2θ) = 2θ = 2 tan−1 x = 2 = 2 Hence (A) is the correct answer. 55. If f (x) = , then is (A) e–4 (B) e–6 (C) e (D) Solution: f (x) = log f (x) = 2x log = 2x = − 1 × 2x = − 2x = = − 4 f (x) = e−4 Hence (A) is the correct answer. 56. = (A) (B) 2 (C) (D) 1 Solution: so Hence (C) is the correct answer. 57. = (A) 1 (B) 2 (C) 0 (D) none of these Solution: = = 1 Hence (A) is the correct answer. 58. is equal to (A) 0 (B) 1 (C) 2 (D) 3 Solution: tan x. loge sin x = = − = − sin x = 1 Hence (B) is the correct answer. 59. = (A) 1 (B) 2 (C) –1 (D) 0 Solution: y = log y = = = log y = 0 y = e0 = 1 Hence (A) is the correct answer. 60. = (A) 1 (B) 2 (C) 0 (D) none of these Solution: = = dx = = log 2 Hence (D) is the correct answer.