EJC 2026 H3 Mathematics - Functions - Notes, Tutorial, and Solutions to Tutorial

1 H3 MATHEMATICS (9820) FUNCTIONS 1 Even and Odd Functions If f( ) f( )xx−= for every x in the domain, then f is called an even function. The graph of an even function is symmetric about the y-axis. The following functions are examples of even functions: • x • 2nx , where n∈  • cos x The sum of any two even functions is also even, and the product of any two even functions is even. Example 1 Given that f( )i x are even functions of x, show that 1 f () f () n i i xx = =∑ is even. Solution: 1 1 f () f () f() f( ) (shown) n i i n i i xx x x = = −= − = = ∑ ∑ ( f ( ) are even functions, i.e. f ( ) f ( )i iix xx−= ) If f( ) f( )xx−= − for every x in the domain, then f is called an odd function. The graph of an odd function is symmetric about the origin. i.e. we can obtain the entire graph of f( )yx= from itself by rotating through 180  about the origin. The following functions are examples of odd functions: • 21nx + , where n∈  • sin x The sum of any two odd functions is odd. However, products work slightly differently: the product of two odd functions is even, and the product of an odd function and an even function is odd.

2 Example 2 [2019/NJC/Prelim/Q5 (modified)] A function f: → is ( ) ( ) ( ) ( ) an odd function, if f f an even function, if f f xx xx −= − −= for any real value of x. (a) It is given that f: → is an odd function. Determine with justification, whether each of the following functions is even or odd: (i) the composite function ff, [2] (ii) the derivative of f, given that f is differentiable everywhere on  . [3] (b) Prove that for every function g: → , there exists a unique decomposition into the sum of an odd function and an even function. (Hint: consider ( ) ( )gg 2 xx+− .) [4] (c) Let the decomposition of ex be ( )sinh x and ( )cosh x where sinh and cosh are odd and even functions respectively. Show that sinh coshy A xB x= + is a solution to the differential equation 2 2 d d y yx = . [3]

3 2 Convex and Concave Functions Before we dive into convex and concave functions, a brief note on increasing and decreasing functions. The definition of increasing (and decreasing) functions is not based on their derivatives. Non- differentiable functions can also be increasing. We can determine if a function is increasing by comparing its values, but there are two differing conventions: A function f is increasing on an interval I if: • (Convention A): 12f( ) f( )xx ≤ for all 12xx< in I • (Convention B): 12f( ) f( )xx < for all 12xx< in I Proponents of Convention A tend to call Convention B “strictly increasing”, while proponents of Convention B tend to call Convention A “non-decreasing”. Always look at the definition or context to determine which convention is being used. It is not true that 12f( ) f( )xx < for all 12xx< in I (i.e. Convention B) corresponds to “the derivative is always positive”. Here are two canonical counterexamples: Example 3 The function 1 3g( )xx= is strictly increasing in x∈  . g (0)′ is undefined. The function 3h( )xx= is strictly increasing in x∈  . h (0) 0′ = . A function f is convex if for any two points 1x , 2x in the domain of the function and 01 t≤≤ , we have ( ) ( ) ( ) ( )1 21 2f (1 ) f 1 ftx t x t x t x+− ≤ +− Note that equality holds when 12 .xx= A function f is strictly convex if the inequality in the previous definition is strict. Graphically, if f is convex, then the line segment joining two points on the curve lies above or on the curve. The inequality ( ) ( ) ( ) ( )1 21 2f (1 ) f 1 ftx t x t x t x+− ≤ +− is sometimes called Jensen’s Inequality. O y x ( )( )11,fxx 12 (1 )tx t x+− ( ) ( ) ( )12f 1ftx t x +− ( )( )22,fxx( )12f (1 )tx t x+− t 1 t−

4 A function f is concave if f− is convex. Again, convex functions need not be differentiable. However, a sufficient condition for a function to be convex is that its second derivative is non-negative. Example 4 [2019/NYJC/Prelim/Q4] The tangent at the point ( ),xy on the curve with equation ( )fyx= passes through the point ( ),ax , where a is a positive constant and x < a. (i) Show that ( ) d d yax xy x−= − . [2] (ii) By making the substitution ( )y a xz= − , obtain a differential equation in x and z. [3] (iii) Solve this differential equation and, given that the curve ( )fyx= passes through the origin, show that ( ) ( )f ln axx ax x a −= −+  . [6] (iv) What can you say about the concavity of the curve ( )fyx= ? Justify your answer. [3] (v) Given that ( )f xa→ as xa→ , explain why   2 0 1f d 2 a xx a  . [2]

5 3 L’Hopital’s Rule How would we find the value of 0 sinlim x x x→ ? The issue here is that both numerator and denominator tend to zero as 0x→ . Fortunately, we have a practical result that helps us in such circumstances. Suppose we want to find f( )lim g( )xa x x→ , and we have either: limf( ) limg( ) 0 xa xa xx →→ = = or: limf( ) andlimg( ) xa xa xx →→ =±∞ =±∞ . Then the limit can be computed using f () f()lim limg () g ()xa xa xx xx→→ ′= ′ . Example 5 Using L’Hopital’s rule, find (i) 0 sinlim x x x→ , (ii) 0 lim ln x xx +→ , (iii) lim e x x x →−∞ . Note: What does this tell you about how polynomial, exponential and logarithmic growth rates compare?

6 4 Improper Integrals In H2 Mathematics, definite integrals are usually taken over finite intervals f( )d b a xx∫ . In H3 Mathematics, you may have to extend this notion to determine infinite integrals: If f( )x is continuous on [ ),a ∞ for some a∈  , then the improper integral of f over [ ),a ∞ is: f( )d lim f( )d t t aa xx xx ∞ → ∞ =∫∫ If f( )x is continuous on ( ],b−∞ for some b∈  , then the improper integral of f over ( ],b−∞ is: f( )d lim f( )d bb t t xx xx ∞→− −∞ =∫∫ If f( )d c xx ∞ − ∫ and f( )d c xx ∞ ∫ are convergent for some c∈  , then the improper integral of f over  is: f( )d f( )d f( )d c c xx xx xx − ∞∞ ∞∞− = +∫∫∫ Limits can similarly be used to determine integrals over discontinuities: If f( )x is continuous on ( ],a b , then the improper integral of f over ( ],a b is: f( )d l i m f( )d bb taat xx xx +→ =∫∫ If f( )x is continuous on [ ),a b , then the improper integral of f over [ ),a b is: f( )d l i m f( )d b tbaa t xx xx −→ =∫∫ Example 6 Find, if possible, (i) 1 1 dxx ∞ ∫ , (ii) 2 1 1 dxx ∞ ∫ , (iii) 2 0 1 dxx ∞ ∫ , (iv) 1 0 1 dx x∫

7 4 Integration by Substitution and other Techniques In H2 Mathematics, we learnt how to integrate by substitution when the substitution is given. How can we recognize the substitution when it is not given? By the chain rule, ( )( ) ( ) ( ) ( ) ( ) ( ) ( )d f ( g ( ) ) fg g fg g d f gd x xx xx x x cx ′′ ′′= ⇒= + ∫ . Equivalently, when ( )gux= , ( ) dg d ux x ′ = ( )( ) ( ) ( ) ( )d s o f gg df dfd d ux x xu xu u x ′′ ′ ′ = =∫ ∫∫ . Example 7 [(ii): 2019/ASRJC/Prelim/Q3a] Find 2 2 cos(i) cos d , (ii) d sin cos xx xx x xx−∫∫

8 In more abstract situations, the functions and limits of integration can provide hints on the intended substitution. Example 8 Prove that f( )d f( )d t bb aa x x abt= +−∫∫ . Hence prove that, if 0 2 πβ< < , then ln cotsin d 2 x x x πβ β βπ −    =  ⌠⌡ .

9 Example 9 [2019/RI/Prelim/Q4] (i) It is given that A, B and C are real numbers. S how that the quadratic function 2h( ) 2t At Bt C=++ is non-negative for all real values of t if and only if 0A> and 2.AC B≥ [2] (ii) Let f and g be continuous functions o n the interval [, ]ab . By considering ( ) 2 f () g () d b a tx x x+∫ , show that ( )( ) ( )( ) ( ) 222 f () d g () d f () g () d bb b aa a xx xx x x x ≥∫∫ ∫ and determine when equality holds. [3] (iii) Hence show that 1 3 0 51d 2xx+<∫ . [3] (iv) Let k be a continuous and differentiable function on the interva