Functions - Vikas Saini

• Q. If f(n) = 14 + 24 + 34 ... n4, then  14 + 34+54 ... (2n – 1)4 be expressed as ?

(a) f(2n-1) – 16 x f(n)                     (b) f(2n-1) – 8 x f(n)

(c) f(2n) – 16 x f(n)                         (d) f(2n) – 8 x f(n)

Solution :-

Method 1

Let’s  take n = 2.

f(n) = f(2) = 17.

F(2n-1) = f(3) = 98.

f(2n) = f(4) = 354.

Value of 14+34 (n=2) = 82.

(a) negative value

(b) negative value

(c)82.

(d)218.

Hence option C.

Method 2

f(n) = 14+24+34………………….n4.

f(2n) = 14+24+34………………….(2n)4.

14+24+34………………….(2n-1)4

= [14+24+34………………….(2n)4 ] – [(24+44+64………………(2n)4]

= f(2n) – 16 [14+24+34…………….n4]

= f(2n) – 16 f(n)

Hence option (c)

Q. ‘ f ’ is a real function such that f(x+y) = f(xy) for all real values of x &y. If f(-7) = 7,then the value f(-40)+f(40) .

Solution :-

f(0-7) = f(0 x -7)

f(-7) = f(0) = 7.

f(0-40) = f(0 x -40)

f(-40) = f(0) = 7.

f(0 + 40) = f(0 x 40)

f(40) = f(0) = 7.

F(-40)+f(40) = 7 + 7 = 14.

Q. If f(x) = (sec x + cosec x) (tan x – cot x) and 45 degree < x  < 90 degree,

Then f(x) lies in which range ?

Solution :-

At x = 45 degree

f(x) = 0.

At x = 90 degree

f(x) = infinitive

Range (0,infinitive)

Q. Let f(x) = ax2+bx+c, where a,b and c are real numbers. If f(x) attains its maximum value at x = 2, then what is the sum of the roots of f(x)=0 ?

Solution :-

f(x) = ax2+bx+c

differentiate with respect to ‘x’

f’(x) = 2ax + b

f’(x) = 0.

2ax + b = 0.

x = -b / 2a.

max value of x given = 2.

2 = -b / 2a

-b/a  = 4.

Sum of roots = -b/a = 4.

Q. Let f(x) = mx3-100x2+3n, where m and n are positive integers. For how many ordered pairs (m,n) will (x-2) be a factor of f(x) ?

Solution :-

If (x-2) is a factor of f(x)

f(2) = m x 23 – 100 x 22 + 3n.

0 = 8m+3n – 400.

8m + 3n = 400.

Ordered pairs = [400 / LCM (8,3)]

= 400 / 24

= 16

Q. F(n – 1) = 1 / ( 2 – F(n) ) for all natural numbers ‘n’. If F(1) = 2, then what is the value [ F(1)+F(2)………..F(50)] ?

Solution :-

F(1) = 2.

2 – f(n) = 1 / f(n-1)

2 – f(2) = 1 / f(1).

f(2) = 2 – ½ = 3/2.

2 – f(3) = 2/3.

f(3) = 2 – 2/3 = 4/3.

2 – f(4) = 3 / 4.

f (4) = 2 – 3/4 = 5/4.

Similarly F(50) = 2 – 49/50 = 51/50.

[f(1)+f(2)……..f(50)]

= [2+3/2+4/3…………………51/50.]

= 51.

Q.  f(x + y) = f(x) + f(y), for all real values of x and y. What is the value of f(2/3) ?

(a) 2/3 f(1)               (b) 2/3

(c) 2/3f(0)                (d) 0.

Solution :-

f( 2/3 + 1/3) = f(1/3)  +  f(2/3)

f(1)  = f(1/3) + f(2/3)…….(1)

f(x + y) = f(x) + f(y)

let x = y,

f(x+x) = f(x) + f(x)

f(2x) = 2 f(x)

f(2 x 1/3) = 2 f(1/3)

f(2/3) = 2f(1/3)

f(1/3) = ½ f(2/3)

putting this value at equation (1)

f(1) = 3/2 f(2/3)

f(2/3)  = 2/3 f(1).

Option (a)

Q.  5f(x) + 4f( 4x+5 / x – 4)  = 9(2x+1), here x is real number and not equal to 4.

What is the value of f(7) ?

Solution :-

Let’s put x = 7.

5f(7) + 4f(11) = 9 (2 x 7 +1)

5 f(7) + 4 f(11) = 135………….(1)

Now put x = 11.

5 f(11) + 4f(7)  = 9 (2 x 11 +1)

5 f(11) + 4f(7) = 207………..(2)

After solving

f (7)  = -17.

Q. F(a,b) = HCF(a,b) / LCM(a,b)

F(1 / F(a,b) , c)  = 1/12.

a,b,c are distinct positive integers such that any pair of a,b,c are co prime to each other, then what is the sum of a,b and c ?

Solution :-

1 / F(a,b) = LCM(a,b) / HCF (a,b)

Any two co-prime number possible for value of a & here.

F ( any value of a & b, c ) = HCF (a,b,c) / LCM (a,b,c).

HCF(a,b,c) / LCM (a,b,c) = 1/12.

LCM (a,b,c) = 12.

Only possible triplet here is (1,2,4)

a+b+c = 1+2+4 = 8.

Q. Let f(x) = [x]. If ‘a’ and ‘b’ are two real numbers such that f(3b – 2) = a -2 and f(a+2) = b +6, then find the sum of a and b.

Solution :-

Suppose :-  f(x) = [x]

f(3b-2) = 3b – 2.

3b -2 = a-2

a = 3b.

f(a+2) = a+2.

3b+2 = b+6………(2)

2b = 4.

b = 2.

Hence a = 6.

a + b = 8.

Q. If f(x) = 4x / (4x + 2)

Then find the value of f(1/100)+f(2/100) ... f(99/100).

Solution :-

f (x) = 4x / 4x + 2.

f ( 1 – x ) = 4(1 – x) / 4(1 – x ) + 2.

= 4 / 4 + 2.4x

= 2 / 2 + 4x.

f(x) + f(1 – x) = [4x  / 4x + 2] + [ 2 / 4x + 2 ]

= 1.

f(1 / 100) + f(1 – 1/100) = f(1) + f(99/100) = 1.

f (2/100) + f( 1 – 2/100) = f(2) + f(98/100) = 1.

Similarly f(49/100) + f(50/100) = 1.

Hence f(1/100) + f(2/100) + f(3/100)………………………..f(99/100)

= 49.

Q.  The minimum value of ax2+bx+c is 7/8 at x = 5/4. Find the value of expression at x = 5, if the value of the expression at x = 1 is 1.

Solution :-

Let f(x) = ax2+bx+c

f’(x) = 2ax + b.

2ax + b = 0.

x = -b / 2a

At x = 5/4

b = -5, a = 2.

At x = 1,

a(1)2 +b(1)+c = 1.

a+b+c = 1.

2 – 5 +c = 1.

C =4.

f(x) = 2x2-5x+4.

f(5) = 2(5)2 – 5(5)+4 = 29.

Q.  If f(x) = x4 + x3 + x2 + x + 1, where x is a positive integer greater than 1.What will be the remainder if f(x5) is divided by f(x) ?

Solution :-

f(x) = x4 + x3 + x2 + x + 1.

f(x5) = x20 + x15 + x10 + x5 + 1.

Let’s take x = 2.

f (x) = f(2) = 31.

f (x5) = f(32) = 220 + 215 + 210 + 25+1.

f(32) mod  f(2) =  220 + 215 + 210 + 25+1  mod 31

= (25)4 + (25)3+(25)2 +(25 ) + 1 mod 31

= 1+1+1+1+1

= 5.

Q . If x is real, then find  smallest value of the expression 3x2 – 4x +7.

Solution :-

Suppose f(x) = 3x2-4x+7

f’(x) = 6x – 4

6x – 4 = 0

x = 2/3.

f(2/3) = 17/3.

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