Composite functions
Q1–10 of 41If f(x) = √(x³), then f(3x) will be equal to
Given $f(x) = \sqrt{x^3}$.
$f(3x) = \sqrt{(3x)^3} = \sqrt{27x^3} = \sqrt{9 \times 3x^3} = 3\sqrt{3x^3}$.
Correct Option: (c)
Direction for Question 2 to 4: Read the instruction below and solve.
f(x) = f(x – 2) – f(x – 1), x is natural number f(1) = 0, f(2) = 1
Q. The value of f(8) is
Given $f(x) = f(x-2) - f(x-1)$ with $f(1) = 0$ and $f(2) = 1$. Calculating terms sequentially:
$f(3) = f(1) - f(2) = 0 - 1 = -1$
$f(4) = f(2) - f(3) = 1 - (-1) = 2$
$f(5) = f(3) - f(4) = -1 - 2 = -3$
$f(6) = f(4) - f(5) = 2 - (-3) = 5$
$f(7) = f(5) - f(6) = -3 - 5 = -8$
$f(8) = f(6) - f(7) = 5 - (-8) = 13$
$f(9) = f(7) - f(8) = -8 - 13 = -21$
Q2.
$f(8) = 13$.
Correct Option: (b)
Direction for Question 2 to 4: Read the instruction below and solve.
f(x) = f(x – 2) – f(x – 1), x is natural number f(1) = 0, f(2) = 1
Q. The value of f(7) + f(4) is
Given $f(x) = f(x-2) - f(x-1)$ with $f(1) = 0$ and $f(2) = 1$. Calculating terms sequentially:
$f(3) = f(1) - f(2) = 0 - 1 = -1$
$f(4) = f(2) - f(3) = 1 - (-1) = 2$
$f(5) = f(3) - f(4) = -1 - 2 = -3$
$f(6) = f(4) - f(5) = 2 - (-3) = 5$
$f(7) = f(5) - f(6) = -3 - 5 = -8$
$f(8) = f(6) - f(7) = 5 - (-8) = 13$
$f(9) = f(7) - f(8) = -8 - 13 = -21$
Q3.
$f(7) + f(4) = -8 + 2 = -6$.
Correct Option: (b)
What will be the value of ∑f(n) (where n= 1 to 9) ?
Given $f(x) = f(x-2) - f(x-1)$ with $f(1) = 0$ and $f(2) = 1$. Calculating terms sequentially:
$f(3) = f(1) - f(2) = 0 - 1 = -1$
$f(4) = f(2) - f(3) = 1 - (-1) = 2$
$f(5) = f(3) - f(4) = -1 - 2 = -3$
$f(6) = f(4) - f(5) = 2 - (-3) = 5$
$f(7) = f(5) - f(6) = -3 - 5 = -8$
$f(8) = f(6) - f(7) = 5 - (-8) = 13$
$f(9) = f(7) - f(8) = -8 - 13 = -21$
Q4.
$\sum_{n=1}^{9} f(n) = 0 + 1 - 1 + 2 - 3 + 5 - 8 + 13 - 21 = -12$.
Correct Option: (a)
Directions for Questions 5 to 9: Define the following functions:
(i) a @ b = (a + b) / 2
(ii) a # b = a² – b²
(iii) (a ! b) = (a – b) / 2
Q. Find the value of {[(3@4)!(3#2)] @ [(4!3)@(2#3)]}.
$3@4 = 3.5$, $3\#2 = 9 - 4 = 5$ $\implies (3@4)!(3\#2) = 3.5!5 = \frac{3.5-5}{2} = -0.75$.
$4!3 = \frac{4-3}{2} = 0.5$, $2\#3 = 4 - 9 = -5$ $\implies (4!3)@(2\#3) = 0.5@-5 = \frac{0.5-5}{2} = -2.25$.
Combining: $-0.75 @ -2.25 = \frac{-0.75 + (-2.25)}{2} = \frac{-3}{2} = -1.5$.
Correct Option: (c)
Directions for Questions 5 to 9: Define the following functions:
(i) a @ b = (a + b) / 2
(ii) a # b = a² – b²
(iii) (a ! b) = (a – b) / 2
Q. Find the value of (4#3)@(2!3).
$4\#3 = 16 - 9 = 7$.
$2!3 = \frac{2-3}{2} = -0.5$.
$7 @ (-0.5) = \frac{7 + (-0.5)}{2} = \frac{6.5}{2} = 3.25$.
Correct Option: (a)
Directions for Questions 5 to 9: Define the following functions:
(i) a @ b = (a + b) / 2
(ii) a # b = a² – b²
(iii) (a ! b) = (a – b) / 2
Q. Which of the following has a value of 0.25 for a = 0 and b = 0.5?
For $a=0, b=0.5$: $a@b = \frac{0 + 0.5}{2} = 0.25$.
Correct Option: (a)
Directions for Questions 5 to 9: Define the following functions:
(i) a @ b = (a + b) / 2
(ii) a # b = a² – b²
(iii) (a ! b) = (a – b) / 2
Q. Which of the following expressions has a value of 4 for a = 5 and b = 3?
For $a=5, b=3$:
$a!b = 1$, $a\#b = 16$, $a@b = 4$.
Option (b): $(a!b)(a@b) = 1 \times 4 = 4$.
Option (c): $\frac{a\#b}{(a!b)(a@b)} = \frac{16}{1 \times 4} = 4$.
Correct Option: (d) (Both b and c)
Directions for Questions 5 to 9: Define the following functions:
(i) a @ b = (a + b) / 2
(ii) a # b = a² – b²
(iii) (a ! b) = (a – b) / 2
Q. If we define a$b as a³ – b³, then for integers a, b > 2 and a > b which of the following will always be true?
A function F(n) is defined as
F(n – 1) = 1/(2 – F(n))
for all natural numbers ‘n’. If F(1) = 3, then what is the value of [F(1)] + [F(2)] + … + [F(1000)]?
(Here, [x] is equal to the greatest integer less than or equal to x)
Given $F(n-1) = \frac{1}{2 - F(n)} \implies F(n) = 2 - \frac{1}{F(n-1)}$.
With $F(1) = 3$:
$F(2) = 2 - \frac{1}{3} = \frac{5}{3}$
$F(3) = 2 - \frac{3}{5} = \frac{7}{5}$
In general, $F(n) = \frac{2n+1}{2n-1}$.
For $n \ge 2$, $1 < F(n) < 2 \implies [F(n)] = 1$.
Sum $= [F(1)] + [F(2)] + \dots + [F(1000)] = 3 + 1 \times 999 = 1002$.
Correct Option: (b)