Chapter 5.1 Overview 5.1.1 Continuity of a function at a point Let fbe a real function on a subset of the real numbers and let cbe a point in the domain of f. Then fis continuous at cif lim f() xf= () c xc→ More elaborately, if the left hand limit, right hand limit and the value of the function at x = c exist and are equal to each other, i.e., lim xc ( ) fx f ( ) c lim xc f ( ) x then f is said to be continuous at x = c. 5.1.2 Continuity in an interval (i) fis said to be continuous in an open interval (a, b) if it is continuous at every point in this interval. (ii) fis said to be continuous in the closed interval [a, b] if • fis continuous in (a, b) • lim x→a+ f(x) = f(a) • lim f(x) = f(b)x→b– 5.1.3 Geometrical meaning of continuity (i) Function f will be continuous at x = c if there is no break in the graph of the function at the point (cfc .,()) (ii) In an interval, function is said to be continuous if there is no break in the graph of the function in the entire interval. 5.1.4 Discontinuity The function f will be discontinuous at x = a in any of the following cases : (i) lim f (x) and lim f (x) exist but are not equal.x→a− x→a+ (ii) lim f (x) and lim f (x) exist and are equal but not equal to f (a).x→a− x→a+ (iii) f (a) is not defined. 5.1.5 Continuity of some of the common functions Function f (x) Interval in which f is continuous 1. The constant function, i.e. f (x) = c 2. The identity function, i.e. f (x) = x R 3. The polynomial function, i.e. f (x)= a0 xn + a1 xn–1 + ... + ax + a n–1 n 4. | x – a | 5. x–n, n is a positive integer 6. p (x) / q (x), where p (x) and q (x) are polynomials in x 7. sin x, cos x (– ∞ , ∞ ) (– ∞ , ∞ ) – {0} R – { x : q (x) = 0} R 8. tan x, sec x R– { (2 n + 1) π 2 : n ∈ Z} 9. cot x, cosec x R– { (nπ : n ∈ Z} 10. ex R 11. log x (0, ∞) 12. The inverse trigonometric functions, In their respective i.e., sin–1 x, cos–1 x etc. domains 5.1.6 Continuity of composite functions Let f and g be real valued functions such that (fog) is defined at a. If g is continuous at a and f is continuous at g (a), then (fog) is continuous at a. 5.1.7 Differentiability f (xh) x+−f () The function defined by f ′ (x) = lim , wherever the limit exists, ish→0 h defined to be the derivative of f at x. In other words, we say that a function f is f (ch) c+− f () differentiable at a point c in its domain if both lim − , called left handh→0 h f (ch) c+− f () derivative, denoted by Lf ′ (c), and lim + , called right hand derivative,h→0 h denoted by R f ′ (c), are finite and equal. (i) The function y = f (x) is said to be differentiable in an open interval (a, b) if it is differentiable at every point of (a, b) (ii) The function y = f (x) is said to be differentiable in the closed interval [a, b] if R f ′ (a) and L f ′ (b) exist and f ′ (x) exists for every point of (a, b). (iii) Every differentiable function is continuous, but the converse is not true 5.1.8 Algebra of derivatives If u, v are functions of x, then du( ±v) dudv d dv du (i) =± (ii) () =u +vdx dxdx dxuv dx dx du dv v −u⎛⎞(iii) du =dx dx ⎜⎟2dx ⎝⎠vv 5.1.9 Chain rule is a rule to differentiate composition of functions. Let f = vou. If dt dv df dvdt t = u (x) and both and exist then = . dxdtdx dtdx 5.1.10 Following are some of the standard derivatives (in appropriate domains) 1. –1(sin d dx )x = 1 1− 2x 2. –1(cos )d xdx = 2 1 1 x − − 3. –1 2 1(tan ) 1 = + d x dx x 4. –1 2 1(cot ) 1 d x dx x − = + 5. –1(sec )d xdx = 2 1 xx , 1− x 1> 6. –1(cosec )d x dx = 2 1 xx − , 1− x 1> 5.1.11 Exponential and logarithmic functions (i) The exponential function with positive base b > 1 is the function y = f (x) = bx. Its domain is R, the set of all real numbers and range is the set of all positive real numbers. Exponential function with base 10 is called the common exponential function and with base e is called the natural exponential function. (ii) Let b > 1 be a real number. Then we say logarithm of a to base b is x if bx=a, Logarithm of a to the base b is denoted by logba. If the base b = 10, we say it is common logarithm and if b = e, then we say it is natural logarithms. logx denotes the logarithm function to base e. The domain of logarithm function is R+, the set of all positive real numbers and the range is the set of all real numbers. (iii) The properties of logarithmic function to any base b > 1 are listed below: 1. logb (xy) = logbx + logby ⎛⎞x 2. logb ⎜⎟= logbx – logby3.logb xn = n logb x y⎝⎠ logcx 4. log b =x , where c > 1log cb 1 =5. logbx log xb 6. logbb = 1 and logb 1 = 0 d x x(iv) The derivative of ex w.r.t., x is exdx ()e . The derivative of logx, i.e. e 1 d 1 xw.r.t., x is ; i.e. (log ) . xdx x 5.1.12 Logarithmic differentiation is a powerful technique to differentiate functions of the form f (x) = (u (x))v(x), where both f and u need to be positive functions for this technique to make sense. 5.1.13 Differentiation of a function with respect to another function Let u = f (x) and v = g (x) be two functions of x, then to find derivative of f (x) w.r.t. du to g (x), i.e., to find , we use the formuladv du du dx= dv dv . dx 5.1.14 Second order derivative 2ddy dy 2 is called the second order derivative of y w.r.t. x. It is denoted by y′′ ordx dx dx y2 , if y = f (x). 5.1.15 Rolle’s Theorem Let f : [a, b] R be continuous on [a, b] and differentiable on (a, b), such that f (a) = f (b), where a and b are some real numbers. Then there exists at least one point c in (a, b) such that f ′ (c) = 0. Geometrically Rolle’s theorem ensures that there is at least one point on the curve y= f(x) at which tangent is parallel to x-axis (abscissa of the point lying in (a, b)). 5.1.16 Mean Value Theorem (Lagrange) Let f: [a, b] R be a continuous function on [a, b] and differentiable on (a, b). Then f()b () fa there exists at least one point cin (a, b) such that f′ (c) = .ba Geometrically, Mean Value Theorem states that there exists at least one point cin (a, b) such that the tangent at the point (c, f(c)) is parallel to the secant joining the points (a, f(a) and (b, f(b)). 5.2 Solved Examples Short Answer (S.A.) Example 1 Find the value of the constant kso that the function fdefined below is ⎧ ⎪⎪1– cos4 xfxcontinuous at x= 0, where () =⎨ 2, x≠0.8x⎪ ⎪k, x=0⎩ Solution It is given that the function fis continuous at x= 0. Therefore, lim f(x) = f(0)x→0 1– cos4 x⇒ lim 2 = k x→0 8x 2sin 2 2x⇒ lim 2 = k x→08x ⎛ sin 2 x⎞2 ⇒ lim = k⎜⎟ x→0 ⎝ 2x ⎠ ⇒ k= 1 Thus, fis continuous at x= 0 if k= 1. Example 2 Discuss the continuity of the function f(x) = sin x. cos x. Solution Since sin xand cos xare continuous functions and product of two continuous function is a continuous function, therefore f(x) = sin x. cos xis a continuous function. ⎧ x3 + x2 –16 x+ 20 ⎪ 2, x≠ 2 Example 3 If fx() =⎨ x(– 2) is continuous at x= 2, find ⎪⎩k , x=2 the value of k. Solution Given f(2) = k. x3 + x2 –16 x+ 20 Now, lim fx= lim fx = lim () () x 2x→2– x→2+→ (– 2) 2x (x 5)( x– 2) 2 = lim 2 lim( x 5) 7 2 x 2(–2) xx As fis continuous at x= 2, we have lim f() x= f(2) x→2 ⇒ k= 7. Example 4 Show that the function fdefined by ⎧ 1⎪xsin , x≠ 0fx() =⎨ x ⎪⎩ 0, x=0 is continuous at x= 0. Solution Left hand limit at x= 0 is given by fx lim lim () =xsin 1 = 0 [since, –1 < sin 1 < 1]–– xxx→0 x→0 fxSimilarly lim ( ) lim xsin 1 0 . Moreover f(0) = 0.x 0 x 0 x fx (0) Thus lim 0– f() x lim () f . Hence fis continuous at x= 0 xx 0 1Example 5 Given f(x) = . Find the points of discontinuity of the compositex–1 function y = f[f(x)]. 1Solution We know that f(x) = is discontinuous at x= 1 x–1 Now, for x 1, 1 x –1 f (f (x)) = f 1 = 1 2– x ,–1x–1 x –1 which is discontinuous at x = 2. Hence, the points of discontinuity are x = 1 and x = 2. Example 6 Let f(x) = x x , for all x ∈ R. Discuss the derivability of f(x) at x = 0 ⎧ x2,if x ≥ 0fxSolution We may rewrite f as () = ⎨⎪ ⎪−x2,if x <0⎩ f (0 + h)– f (0) –h2 –0 Now Lf ′ (0) = lim = lim = lim − h = 0 –– −h→0 hh→0 hh→0 f (0 + h)– f (0) h2 –0 Now Rf ′ (0) = lim = lim = lim h = 0 + +−h→0 hh→0 hh→0 Since the left hand derivative and right hand derivative both are equal, hence f is differentiable at x = 0. Example 7 Differentiate tan xw.r.t. x Solution Let y = tan x . Using chain rule, we have dy 1 d . (tan x)dx dx2tan x d1 .sec 2 x ( x)= dx2tan x 12 1(sec x)= 2 x2tan x dyExample 8 If y = tan(x + y), find dx . Solution Given y = tan (x + y). differentiating both sides w.r.t. x, we have dy 2dsec ( xy)(xy)dx dx dy = sec2 (x + y) 1 dx dyor [1 – sec2 (x + y] dx = sec2 (x + y) dy sec 2 (xy)Therefore, 2 = – cosec2 (x + y).dx 1sec (xy) Example 9 If ex + ey = ex+y, prove that dy =−eyx . − dx Solution Given that ex + ey = ex+y. Differentiating both sides w.r.t. x, we have dy dy = ex+y 1ex + ey dxdx dyor (ey – ex+y) = ex+y – ex,dx xyx xyxdy e – e eee – ewhich implies that yxy yxy yx .dxee e ee dy ⎛3xx− 3 ⎞ 11 sin x, so that cos x – sin x > 0, i.e.,4 f (x) = cos x – sin x f ′ (x) = – sin x – cos x 1Hence f ′ = – sin – cos = − (1+ 3) .6 662 Example 17 Verify Rolle’s theorem for the function, f (x) = sin 2x in 0,2. Solution Consider f (x) = sin 2x in 0, 2 . Note that: (i) The function f is continuous in 0, , as f is a sine function, which is2 always continuous. ⎛ π⎞(ii) f ′ (x) = 2cos 2x, exists in 0, , hence f is derivable in ⎜0, ⎟2 ⎝ 2 ⎠ . (iii) f (0) = sin0 = 0 and f = sinπ = 0 ⇒ f (0) = f .22 Conditions of Rolle’s theorem are satisfied. Hence there exists at least one c ∈ 0, such that f ′(c) = 0. Thus 2 cos 2c = 0 ⇒ 2c = ⇒ c = 4.2 Example 18 Verify mean value theorem for the function f (x) = (x – 3) (x – 6) (x – 9) in [3, 5]. Solution (i) Function f is continuous in [3, 5] as product of polynomial functions is a polynomial, which is continuous. (ii) f ′(x) = 3x2 – 36x + 99 exists in (3, 5) and hence derivable in (3, 5). Thus conditions of mean value theorem are satisfied. Hence, there exists at least one c ∈ (3, 5) such that f (5) f (3) ()fc 53 80⇒ 3c2 – 36c + 99 = = 42⇒ c = 6 13Hence c6 (since other value is not permissible).3 Long Answer (L.A.) 2cos x 1 Example 19 If f (x) = , x cot x 14 find the value of f so that f (x) becomes continuous at x =44. 2cos x 1 Solution Given, f (x) = , x cot x 14 2cos x 1lim fx( ) lim Therefore, xx cot x 1 44 x −1 sin x lim ) = x→π cos x −sin x 4 2cos x 1 2 cos x 1 cos x sin xlim . . .sin x= x 2cos x 1 cos x sin x cos x sin x 4 2cos2 x−1 cos x+sin xlim . . sin x()= x→πcos2 x−sin2 x 2cos x+1 4 cos2x cos x+sin xlim . . sin x ⎛ ⎞()= π⎜ x→ cos2x 2cos x+1 ⎟⎠4 ⎝ cos x sin xlim sin x = x 2cos x1 4 111 = 1 22. 1 2 1lim ()fxThus, x 2 4 π1 π⎛⎞If we define f⎜⎟=, then f(x) will become continuous at x=. Hence for fto be42⎝⎠ 4 continuous at x , f 1.4 42 1 xe 1 , ifx0() 1fxExample 20 Show that the function fgiven by xe 1 0, ifx0 is discontinuous at x= 0. Solution The left hand limit of fat x= 0 is given by 1 xe 101lim () 1 1 . fx lim x 0 x 0 01 xe 1 1 xe 1lim fx( ) lim Similarly, 1x 0 x 0 xe 1 111 1 ex x1 e 10lim x 01lim 1= 1 = x 0 1101 x1 e ex lim f () ≠lim f ( ), therefore, lim fxThus − xx ( ) does not exist. Hence f is discontinuous x→0 +x→0 x→0 at x = 0. 1cos 4 x 2, if x 0 x a , if x 0fxExample 21 Let () x , if x 0 16 x 4 For what value of a, f is continuous at x = 0? Solution Here f (0) = a Left hand limit of f at 0 is 1cos4 x 2sin 2 2xlim fx() lim 2 lim 2x 0 x 0 xx 0 x sin 2 x 2 lim 8 = 8 (1)2 = 8. 2 x 02x and right hand limit of f at 0 is xlim fx( ) lim x 0 x 0 x (16 x 4) lim = x 0( 16 x 4)( 16 x 4) x(16 x 4) = lim lim 16 x 48 x 0 16 x 16 x 0 Thus, lim fx( ) lim fx 8( ) . Hence fis continuous at x= 0 only if a= 8. x 0 x 0 Example 22 Examine the differentiability of the function fdefined by 2x 3,if 3 x 2 () x1 ,if 2 x 0fx x 2,if 0 x 1 Solution The only doubtful points for differentiability of f(x) are x= – 2 and x= 0. Differentiability at x= – 2. f(–2 h) f(–2) Now L f′ (–2) = lim h 0 h 2(–2 h)3 (–2 1) 2h = lim lim lim 2 2. h 0 hh 0 hh 0 f(–2 h) f(–2) and R f′ (–2) = lim h 0 h –2 h 1 (2 1) = lim h 0 h h1(–1) h = lim lim 1 h 0 hh 0 h Thus R f′ (–2) ≠ L f′ (–2). Therefore fis not differentiable at x= – 2. Similarly, for differentiability at x= 0, we have f(0 h) f(0) L (f′(0)= lim h 0 h 0 h 1 (0 2) = lim h 0 h h 11lim lim 1= h 0 hh 0 h which does not exist. Hence fis not differentiable at x= 0. Example 23 Differentiate tan-1 21 x x with respect to cos-1 21x 2x , where x 1 ,12 . Solution Let u = tan-1 21 x x and v = cos-1 21x 2x . We want to find du dv du dx dv dx Now u = tan-1 21 x x . Put x = sinθ. 4 2 π π⎛ ⎞<θ< ⎜ ⎟⎝ ⎠. 1sin2 Then u = tan-1 = tan-1 (cot θ)sin⎧ ⎛π ⎫π = tan-1 ⎨tan ⎜ −θ⎟⎞⎬= −θ sin–1 x ⎩⎝ 2 ⎠⎭ 22 du 1 Hence 2 .dx 1 x Now v = cos–1 (2x = – sin–1 (2x2 2 π –1 = – sin–1 (2sinθ 1−sin θ)=−sin (sin2 ) θ2 2 π= – sin–1 {sin (π – 2θ)} [since < 2 θ < π]22 = (2) 222 ⇒ v = + 2sin–1x2 dv 2 ⇒ dx −1du du dx 1− x2 −1 .== = Hence dv dv 22 dx 21− x Objective Type Questions Choose the correct answer from the given four options in each of the Examples 24 to 35. sin x cos x,if x 0 xExample 24 The function f (x) = k ,if x 0 is continuous at x = 0, then the value of k is (A)3 (B) 2 (C) 1 (D) 1.5 Solution (B) is the Correct answer. Example 25 The function f (x) = [x], where [x] denotes the greatest integer function, is continuous at (A)4 (B) – 2 (C) 1 (D) 1.5 Solution (D) is the correct answer. The greatest integer function[x] is discontinuous at all integral values of x. Thus D is the correct answer. 1 Example 26 The number of points at which the function f (x) = is notx –[ ]xcontinuous is (A)1 (B) 2 (C) 3 (D) none of these Solution (D) is the correct answer. As x – [x] = 0, when x is an integer so f (x) is discontinuous for all x ∈ Z. Example 27 The function given by f (x) = tanx is discontinuous on the set (A) :Z (B) 2:Znn nn n (C) (2n 1) : n Z (D) : n Z22 Solution C is the correct answer. Example 28 Let f (x)= |cosx|. Then, (A) f is everywhere differentiable. (B) f is everywhere continuous but not differentiable at n = nπ, n Z . π(C) f is everywhere continuous but not differentiable at x = (2n + 1) ,2 n∈ Z . (D) none of these. Solution C is the correct answer. Example 29 The function f (x) = |x| + |x – 1| is (A) continuous at x = 0 as well as at x = 1. (B) continuous at x = 1 but not at x = 0. (C) discontinuous at x = 0 as well as at x = 1. (D) continuous at x = 0 but not at x = 1. Solution Correct answer is A. Example 30 The value of k which makes the function defined by 1sin , if x 0fx() x , continuous at x = 0 is k , if x0 (A)8 (B)1 (C) –1 (D) none of these Solution (D) is the correct answer. Indeed lim sin 1 does not exist. x→0 x Example 31 The set of points where the functions f given by f (x) = |x – 3| cosx is differentiable is (A) R (B) R – {3} (C) (0, ∞) (D) none of these Solution B is the correct answer. Example 32 Differential coefficient of sec (tan–1x) w.r.t. x is x x (A) (B) 21+x21+x 1 (C) x 1+x2 (D) 21+x Solution (A) is the correct answer. ⎛2x ⎞⎛2x ⎞–1 –1 duExample 33 If u = sin ⎜ 2 ⎟and v = tan ⎜ 2 ⎟, then is1+x 1−x dv⎝⎠ ⎝⎠1 1– x2 (A) (B) x (C) 2 (D)121+x Solution (D) is the correct answer. Example 34 The value of c in Rolle’s Theorem for the function f (x) = ex sinx, [0, ]isx ∈ππππ 3π(A) (B) (C) (D)642 4 Solution (D) is the correct answer. Example 35 The value of c in Mean value theorem for the function f (x) = x (x – 2), x ∈ [1, 2] is 321 3(A) (B) (C) (D)232 2 Solution (A) is the correct answer. Example 36 Match the following COLUMN-I COLUMN-II sin 3 x ,if x 0 x(A) If a function fx() (a) |x|k ,if x 02 is continuous at x = 0, then k is equal to (B) Every continuous function is differentiable (b) True (C) An example of a function which is continuous (c) 6 everywhere but not differentiable at exactly one point (D) The identity function i.e. f (x) = x xR (d) False∀∈ is a continuous function Solution A → c, B → d,C → a, D → b Fill in the blanks in each of the Examples 37 to 41. 1 Example 37 The number of points at which the function f (x) = islog | x| discontinuous is ________. Solution The given function is discontinuous at x = 0, ± 1 and hence the number of points of discontinuity is 3. ⎧ax +1if x≥1 Example 38 If () =is continuous, then a should be equal to _______.fx ⎨ x +2if x <1⎩ Solution a = 2 Example 39 The derivative of log10x w.r.t. x is ________. Solution (log10 e )1 . x ⎛ x +1 ⎞⎛ x –1 ⎞ dysin–1Example 40 If y =sec–1 ⎜ , then is equal to ______.⎝ x −1 ⎟⎠ + ⎜⎝ x +1 ⎟⎠dx Solution 0. Example 41 The deriative of sin x w.r.t. cos x is ________. Solution – cot x State whether the statements are True or False in each of the Exercises 42 to 46. f ( ) lim f ()Example 42 For continuity, at x = a, each of x lim →+ and →ax xa – x is equal to f (a). Solution True. Example 43 y = |x – 1| is a continuous function. Solution True. Example 44 A continuous function can have some points where limit does not exist. Solution False. Example 45 |sinx| is a differentiable function for every value of x. Solution False. Example 46 cos |x| is differentiable everywhere. Solution True. 5.3 EXERCISE Short Answer (S.A.) 1. Examine the continuity of the function f (x) = x3 + 2x2 – 1 at x = 1 Find which of the functions in Exercises 2 to 10 is continuous or discontinuous at the indicated points: 1cos 2 x⎧−⎧3x+5, if x≥2 , if x ≠0⎪ ⎪ 2() = fx =x2. fx ⎨2 3. () ⎨⎪x ,if x<2⎩ ⎪⎩5, if x =0 at x=2 at x=0 x−4⎧2x2 −− 3x 2 ⎧ , if x ≠4⎪ , if x ≠2 ⎪ 4. fx() =⎨ x−2 5. fx ⎨2(x−4) ()=⎪⎪⎩5, if x =2 ⎩0, if x =4 at x=2 at x = 4 11 ⎧⎧ xa−sin , if x ≠0x cos , if x ≠0 ⎪⎪ xa−6. () =fx ⎨ () =x 7. fx ⎨ ⎪⎪⎩0, if x =0 ⎩0, if x =a at x = 0 at x = a ⎧ 12 x ⎧x⎪e ,0 x 1if ≤≤⎪ 1, if x ≠0 ⎪⎪2() = () =fx ⎨ fx8. x 9. ⎨1+e⎪ 232x 3x ,if 1 4 ⎪x −4⎩ is a continuous function at x = 4. 1 17. Given the function f (x) = . Find the points of discontinuity of the compositex +2 function y = f (f (x)). 1118. Find all points of discontinuity of the function ft()= , where t = 2 −. tt 2 x+− 1 19. Show that the function f (x) = sin x +cos x is continuous at x = π. Examine the differentiability of f, where f is defined by [], , if 0 x 2⎧xx ≤< 20. f (x) = ⎨ x 3⎩(x −1) x, if 2≤< at x = 2. ⎧21⎪x sin , if x ≠0 ⎨ x21. f (x) = ⎪⎩0, if x =0 at x = 0. ⎧1+x , if x ≤2 22. f (x) = ⎨5−x , if x >2⎩ at x = 2. 23. Show that f (x) = x −5 is continuous but not differentiable at x = 5. 24. A function f : R →R satisfies the equation f ( x + y) = f (x) f (y) for all x, y ∈R, f (x) ≠0. Suppose that the function is differentiable at x = 0 and f ′ (0) = 2. Prove that f ′(x) = 2 f (x). Differentiate each of the following w.r.t. x (Exercises 25 to 43) : 225. 2cos x 26. 8x 8 27. log x + x ( 228. log ⎡⎣log log x5 ⎤⎦ 29.() sin x +cos x 30. sin ( ax ++cn 2 bx ) ⎛⎞ ⎟31. cos (tan x +1) 32. sinx2 + sin2x + sin2(x2) 33. sin–1 ⎜⎜ 1 x +1 ⎟⎝ ⎠ cos x34. (sin x)35. sinmx . cosnx 36. (x + 1)2 (x + 2)3 (x + 3)4 − ⎞π–1 ⎛sin x +cos x ⎞−π π –1 ⎛1cos x , x π−<<, x tan<<37. cos ⎜ ⎟ 38. ⎜ ⎟ 1cos x 4⎝ 2 ⎠44 ⎝+ ⎠ ππ39. tan–1 (sec x +tan x),−< –1 41. –1 ⎛ 1 ⎞ 4x3 −3x xsec ⎜ ⎟, 0<< ⎝ ⎠ 1 2 42. –1tan 2 3 a3 3ax2 3ax x , 1 3 x a 1 3 43. ⎛ 2 2 –1 1 x 1 ⎜2 2 tan ⎜ , 1 x 1, x 0 1 x 1 ++ −x ⎞ ⎟−<< ≠ ⎝ +− −x ⎟⎠ dy dx11 ⎛ 1 ⎞ −θ⎛ 1 ⎞ =θ44. x = t + , y = t – 45. x = eθ ⎜θ+⎟, ye⎜−⎟tt ⎝θ⎠ ⎝θ⎠ 46. x = 3cosθ – 2cos3θ, y = 3sinθ – 2sin3θ. 2t 2tsin x = , tan y =47. 22− .1+t 1 t 1logt 3+2logt+ 48. x = 2, y = .tt dy −y log x =49. If x = ecos2t and y = esin2t, prove that .dx x log y dy b⎛⎞ =⎜⎟50. If x = asin2t (1 + cos2t) and y = b cos2t (1–cos2t), show that dx π a .⎝⎠at t= 4 dy π51. If x = 3sint – sin 3t, y = 3cost – cos 3t, find at t = .dx3 x52. Differentiate w.r.t. sinx.sin x⎛ 2 ⎞1+ x −1⎜ ⎟53. Differentiate tan–1 w.r.t. tan–1 x when x ≠ 0.⎜ x ⎟⎝⎠dyFind when x and y are connected by the relation given in each of the Exercises 54 to 57.dxx 54. sin (xy) + = x2 – yy55. sec (x + y) = xy 56. tan–1 (x2 + y2) = a 57. (x2 + y2)2 = xy dy dx . =158. If ax2 + 2hxy + by2 + 2gx + 2fy + c = 0, then show that .dx dy x dyx − y=59. If x =ey , prove that .dx x log x − dy 1log )2 x yx (+ y60. If y =e , prove that = .dx log y (cos ) (cos x )..... ∞ y2 tan x dyx =61. If y =(cos x) , show that .dx y log cos x −1 2dy sin ( a + y)62. If x sin (a + y) + sin a cos (a + y) = 0, prove that = .dx sin a dy 1− y2 63. If 1− x2+ 1− y2= a (x – y), prove that = .dx 1− x2 2dy64. If y = tan–1x, find 2 in terms of y alone.dx Verify the Rolle’s theorem for each of the functions in Exercises 65 to 69. 65. f(x) = x(x– 1)2 in [0, 1]. ⎡ π⎤66. f(x) = sin4x+ cos4xin ⎢0, ⎥ .⎣ 2 ⎦ 67. f(x) = log (x2 + 2) – log3 in [–1, 1]. –x68. f(x) = x(x+ 3)e/2 in [–3, 0]. 69. f(x) = 4 − x2 in [– 2, 2]. 70. Discuss the applicability of Rolle’s theorem on the function given by x2 1, if0 x 1()fx .3, f1 x 2xi71. Find the points on the curve y= (cosx– 1) in [0, 2π], where the tangent is parallel to x-axis. 72. Using Rolle’s theorem, find the point on the curve y= x(x– 4), x∈ [0, 4], where the tangent is parallel to x-axis. Verify mean value theorem for each of the functions given Exercises 73 to 76. 1 73. f(x) = in [1, 4].41x−74. f(x) = x3 – 2x2 – x+ 3 in [0, 1]. 75. f(x) = sinx– sin2xin [0, π]. 76. f(x) = 25− x2 in [1, 5]. 77. Find a point on the curve y= (x– 3)2, where the tangent is parallel to the chord joining the points (3, 0) and (4, 1). 78. Using mean value theorem, prove that there is a point on the curve y= 2x2 – 5x+ 3 between the points A(1, 0) and B (2, 1), where tangent is parallel to the chord AB. Also, find that point. Long Answer(L.A.) 79. Find the values of pand qso that ⎧x2 +3x +,⎪ pif x ≤1fx() =⎨ ⎪qx +2, if x >1⎩ is differentiable at x = 1. 80. If xm.yn = (x + y)m+n, prove that 2dyy dy(i) = and (ii) 2 =0.dxx dx 2dy dy 281. If x = sint and y = sin pt, prove that (1–x2) 2 – x +py =0.dxdx dy82. Find , if y = xtanx +dx Objective Type Questions Choose the correct answers from the given four options in each of the Exercises 83 to 96. 2x83. If f (x) = 2x and g (x) = +1 , then which of the following can be a discontinuous2 function (A) f (x) + g (x) (B) f (x) – g (x) ()gx (C) f (x) . g (x) (D) f ()x 4−x2 84. The function f (x) = is4x −x3 (A) discontinuous at only one point (B) discontinuous at exactly two points (C) discontinuous at exactly three points (D) none of these 85. The set of points where the function f given by f (x) = 21 sinx is differentiable isx −1⎧⎫(A) R (B) R – ⎨⎬2⎩⎭ (C) (0, ∞) (D) none of these 86. The function f (x) = cot x is discontinuous on the set (A) {xn : nZ (B) {x =2n : n=π∈} π∈Z} ⎧π⎫ ⎧ nπ⎫(C) ⎨x =(2n +1); n ∈Z⎬ (iv) ⎨x = ; n ∈Z⎬⎩ 2 ⎭⎩ 2 ⎭ x87. The function f (x) = e is (A) continuous everywhere but not differentiable at x = 0 (B) continuous and differentiable everywhere (C) not continuous at x = 0 (D) none of these. 2188. If f (x) = x sin , where x ≠ 0, then the value of the function f at x = 0, so thatxthe function is continuous at x = 0, is (A) 0 (B) – 1 (C) 1 (D) none of these ⎧π mx+1,if x ≤⎪2 πfx89. If ()= ⎨ , is continuous at x = , thenπ 2⎪sin +, xxnif >⎪2⎩nπ(A) m = 1, n = 0 (B) m = + 12 mπ π(C) n = (D) m = n = 2 2 90. Let f (x) = |sin x|. Then (A) f is everywhere differentiable (B) f is everywhere continuous but not differentiable at x = nπ, n ∈ Z. π(C) f is everywhere continuous but not differentiable at x = (2n + 1) ,(D)none of these n ∈ Z. ⎛1−x2 ⎞ dy91. If y = log ⎜⎟, then is equal to⎝1+x2 ⎠dx34x 4x− (A) 41− x (B) 41 x− 1 34x− (C) 44− x (D) 41 x− 92. If y = sin x + y , then dy dx is equal to cos x cos x (A) 2 1y − (B) 12y− sin x sin x (C) 12y− (D) 2 1y − 93. The derivative of cos–1 (2x2 – 1) w.r.t. cos–1x is 1− (A) 2 (B) 221 x− 2 (C) x (D) 1 – x2 94. If x = t2, y = t3, then 2 2 dy dx is 3 3 (A) 2 (B) 4t 3 3 (C) 2t (D) 2t 95. The value of c in Rolle’s theorem for the function f (x) = x3 – 3x in the interval [0, 3] is (A) 1 (B) – 1 3 1(C) (D)2 3 196. For the function f (x) = x + , x ∈ [1, 3], the value of c for mean value theorem is x (A) 1 (B) 3(C) 2 (D) none of these Fill in the blanks in each of the Exercises 97 to 101: 97. An example of a function which is continuous everywhere but fails to be differentiable exactly at two points is __________ . 98. Derivative of x2 w.r.t. x3 is _________. 99. If f (x) = |cosx|, then f ′ = _______ .100. If f (x) = |cosx – sinx | , then f ′ = _______.101. For the curve xy 1, at ⎜ ,⎟ is __________.43dy ⎛ 11 ⎞dx ⎝ 44 ⎠State True or False for the statements in each of the Exercises 102 to 106. 102. Rolle’s theorem is applicable for the function f (x) = |x – 1| in [0, 2]. 103. If f is continuous on its domain D, then | f | is also continuous on D. 104. The composition of two continuous function is a continuous function. 105. Trigonometric and inverse - trigonometric functions are differentiable in their respective domain. 106. If f . g is continuous at x = a, then f and g are separately continuous at x = a.

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