Functions  Vikas Saini

Q. If f(n) = 1^{4} + 2^{4} + 3^{4 }... n^{4}, then 1^{4} + 3^{4}+5^{4 }... (2n – 1)^{4 }be expressed as ?
(a) f(2n1) – 16 x f(n) (b) f(2n1) – 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(2n1) = f(3) = 98.
f(2n) = f(4) = 354.
Value of 1^{4}+3^{4} (n=2) = 82.
(a) negative value
(b) negative value
(c)82.
(d)218.
Hence option C.
Method 2
f(n) = 1^{4}+2^{4}+3^{4}………………….n^{4}.
f(2n) = 1^{4}+2^{4}+3^{4}………………….(2n)^{4}.
1^{4}+2^{4}+3^{4}………………….(2n1)^{4}
= [1^{4}+2^{4}+3^{4}………………….(2n)^{4} ] – [(2^{4}+4^{4}+6^{4}………………(2n)^{4}]
= f(2n) – 16 [1^{4}+2^{4}+3^{4}…………….n^{4}]
= 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(07) = f(0 x 7)
f(7) = f(0) = 7.
f(040) = 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) = ax^{2}+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) = ax^{2}+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) = mx^{3}100x^{2}+3n, where m and n are positive integers. For how many ordered pairs (m,n) will (x2) be a factor of f(x) ?
Solution :
If (x2) is a factor of f(x)
f(2) = m x 2^{3 }– 100 x 2^{2} + 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(n1)
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 coprime 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(3b2) = 3b – 2.
3b 2 = a2
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) = 4^{x} / (4^{x} + 2)
Then find the value of f(1/100)+f(2/100) ... f(99/100).
Solution :
f (x) = 4^{x} / 4^{x} + 2.
f ( 1 – x ) = 4^{(1 – x) }/ 4^{(1 – x )} + 2.
= 4 / 4 + 2.4^{x}
= 2 / 2 + 4^{x}.
f(x) + f(1 – x) = [4^{x } / 4^{x }+ 2] + [ 2 / 4^{x} + 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 ax^{2}+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) = ax^{2}+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) = 2x^{2}5x+4.
f(5) = 2(5)^{2} – 5(5)+4 = 29.
Q. If f(x) = x^{4} + x^{3} + x^{2} + x + 1, where x is a positive integer greater than 1.What will be the remainder if f(x^{5}) is divided by f(x) ?
Solution :
f(x) = x^{4} + x^{3} + x^{2} + x + 1.
f(x^{5}) = x^{20} + x^{15} + x^{10} + x^{5} + 1.
Let’s take x = 2.
f (x) = f(2) = 31.
f (x^{5}) = f(32) = 2^{20} + 2^{15} + 2^{10} + 2^{5}+1.
f(32) mod f(2) = 2^{20} + 2^{15} + 2^{10} + 2^{5}+1 mod 31
= (2^{5})^{4 }+ (2^{5})^{3}+(2^{5})^{2 }+(2^{5} ) + 1 mod 31
= 1+1+1+1+1
= 5.
Q . If x is real, then find smallest value of the expression 3x^{2} – 4x +7.
Solution :
Suppose f(x) = 3x^{2}4x+7
f’(x) = 6x – 4
6x – 4 = 0
x = 2/3.
f(2/3) = 17/3.