Functions and Graphs Questions for CAT Set-2

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Functions and Graphs Questions for CAT Set-2:

Practice Functions and Graphs questions and answers for CAT with detailed solutions and explanations. Download Functions and graphs tricks, concepts and formulas for CAT.

Question 1: f(2x+3) = $4x^2+14x+14$. Find the value of f($\frac{1}{x}$).

a) $\frac{2x^2+x+1}{x^2}$
b) $\frac{x^2+x+1}{x^2}$
c) $\frac{x^2+x+2}{x^2}$
d) $\frac{x^2+2x+1}{x^2}$

Question 2: Given that $\frac{5}{x}f(\frac{1}{x}) + 4f(x) = 6$, find the value of f($\frac{1}{4}$). ($x \neq 0$)

a) $\frac{32}{3}$
b) $\frac{28}{3}$
c) $\frac{24}{3}$
d) $\frac{20}{3}$

Question 3: If f(g(x)) =$2x^2+3x$, g(f(x)) =$x^2+4x-4$, which of the following is the possible value of f(-4)?

a) 1
b) -2
c) -1
d) 2

Question 4: What is the minimum value of the function f(x) = max{2x + 1, 4x – 3, 5 – 2x}

a) -7
b) 3
c) 7
d) 11/3

Question 5: Given that f(x+y) + f(x-y) = 2f(x)f(y) and f(x)$\neq0$, which of the following is true?

a) f(y) is an even function
b) f(x) is an odd function
c) f(y) is both even and odd function
d) f(x) is neither even nor odd function

Question 6: If abcd = 1250, find the minimum integral value of $a^2+b^2+c^2+d^2$ if a,b,c,d are greater than zero.

a) 142
b) 175
c) 176
d) 135

Question 7: f(x+y) = f(x)f(y) and f(x) $\neq$ 0 for all x, y. If f(4) = +5, what is the value of f(-8)?

a) 1/16
b) -1/25
c) 1/25
d) None of these

Question 8: What is the equation of the reflection of the curve $y = 5x^2 – 7x + 2$ in the point (3,-3)?

a) $x = 5y^2 – 7y + 2$
b) $x = 5y^2 + 23y + 29$
c) $y = 5x^2 – 37x + 65$
d) $x = 5y^2 – 37y + 65$

Question 9: For real values of Y, consider the two statements and choose the correct option from the choices available.

Statement 1: The maximum value of $\frac{Y^2 – Y +1}{Y^2 + Y + 1}$ is $3$.
Statement 2: The minimum value of $\frac{Y^2 – Y +1}{Y^2 + Y + 1}$ is $\frac{1}{3}$.

a) Only Statement 1 is true
b) Only Statement 2 is true
c) Both the statements are true
d) Neither of the two statements is true

Question 10: Let #a(x) = $[\frac{x}{a}]$ where [x] represents the greatest integer less than or equal to x. If #8(#7(#6(#5(y)))) = 1 then what is the third digit from left in the least possible value of y.

a) 2
b) 4
c) 6
d) 8

Solutions for Functions and Graphs Questions for CAT Set-2:

Solutions:

f(2x+3) = $4x^2+14x+14$
To find f($\frac{1}{x}$), we have to find f(x).
$4x^2+14x+14$ can be written as $(4x^2+12x+9)$ + $(2x+3)$ + 2
=> f(2x+3) = $(2x+3)^2$ + $(2x+3)$ + 2
=> f(x) = $x^2 + x + 2$
=> f($\frac{1}{x}$) = $\frac{2x^2+x+1}{x^2}$

Substitute x = 4.
$\frac{5}{4}f(\frac{1}{4}) + 4f(4) = 6$ => $\frac{25}{4}f(\frac{1}{4}) + 20f(4) = 30$ – Eqn (1)
Now, substitute x = 1/4.
$20f(4) + 4f(\frac{1}{4})$ = 6 – Eqn (2)
Using (1) and (2), we get:
$30 – \frac{25}{4}f(\frac{1}{4})$ = $6 – 4f(\frac{1}{4})$
$24 = \frac{9}{4}f(\frac{1}{4})$
=> $f(\frac{1}{4})$ = $\frac{32}{3}$

f(g(x)) =$2x^2+3x$
Put f(x) in place of x,
f(g(f(x))) =$2f(x)^2+3f(x)$
f($x^2+4x-4$) =$2f(x)^2+3f(x)$
Put $x^2+4x-4$ = -4
x(x+4) = 0
x = 0,-4
Put x = -4 in the above equation,
f(-4) =$2f(-4)^2+3f(-4)$
$2f(-4)^2 = -2f(-4)$
f(-4) = -1

2x + 1, 4x – 3 and 5 – 2x are three straight lines.The minimum value of f(x) will occur at the intersection point of any of the two lines.
When 2x + 1 intersects 4x – 3, 2x + 1 = 4x – 3 => x = 2, then f(x) = 5
When 2x + 1 intersects 5 – 2x, x = 1 and f(x) = 3
When 4x – 3 intersects 5 – 2x, x = 4/3 and f(x) = 11/3.
Thus f(x) = 3 is the minimum value of f(x).

f(x+y) + f(x-y) = 2f(x)f(y) — (1)
Put -y in place of y.
f(x-y) + f(x+y) = 2f(x)f(-y) — (2)
From (1) and (2), we can say that
2f(x)f(y) = 2f(x)f(-y)
=> f(y) = f(-y)
=> f(y) is an even function.

$a^2+b^2+c^2+d^2\geq 4(a^2b^2c^2d^2)^{1/4}$
$a^2+b^2+c^2+d^2\geq 4(1250^2)^{1/4}$
$a^2+b^2+c^2+d^2\geq 4*25(2^2)^{1/4}$
$a^2+b^2+c^2+d^2\geq 141.4$
Hence the minimum value would be 142.

f(x+0) = f(x)f(0)
f(0) = 1
f(4-4) = f(4)f(-4)
f(-4) = 1/5
f(-8) = f(-4-4) = 1/5*1/5 = 1/25

To get the reflection of the curve in the origin, the x and y coordinates have to be interchanged. Since the reflection is in the point (3, -3) and not the origin, the x coordinate has to be replaced with x-3 and y has to be replaced with y+3.
So, the reflection is: $x – 3 = 5*(y +3)^2 – 7(y + 3) + 2$
=> $x – 3 = 5y^2 + 45 + 30y – 7y – 21 + 2$
=> $x – 3 = 5y^2 + 23y + 26$
=> $x = 5y^2 + 23y + 29$

Let us look at the value of the expression (let us call it E) given.
$\frac{Y^2 – Y +1}{Y^2 + Y + 1} = 1 – \frac{2Y}{Y^2+Y+1} = 1 – \frac{2}{Y+1+\frac{1}{Y}}$
The maximum and minimum values of E depend on the value of the new expression (e) $\frac{2}{Y+1+\frac{1}{Y}}$
We know $Y+\frac{1}{Y}$ takes values between $(-\infty,-2]$ and $[2,\infty)$.
Hence the minimum value of e is when $Y+\frac{1}{Y}$ equals -2. The value of expression E is maximum at this point and equals 1-(-2) = 3
Similarly, the maximum value of e is when $Y+\frac{1}{Y}$ equals 2. The value of expression E is minimum at this point and equals $1 – \frac{2}{3} = \frac{1}{3}$
So, both the statements are true.

= #8(#7(#6($[\frac{y}{5}]$)))
= #8(#7($[\frac{y}{5*6}]$))
= #8($[\frac{y}{5*6*7}]$)
= $[\frac{y}{5*6*7*8}]$
= $[\frac{y}{1680}]$