Function: CAT Previous Year Questions

Ans: A

Explanation:To find the domain of $h(x) = f(g(x)) + g(f(x))$, we need to consider the domains of $f(x)$, $g(x)$, $f(g(x))$, and $g(f(x))$.

The domain of $f(x) = \frac{x}{(2x-1)}$ is all real numbers except where the denominator is zero, i.e., $2x-1 = 0 \implies x = \frac{1}{2}$.
So, Domain of $f = \mathbb{R} \setminus \{\frac{1}{2}\}$.

The domain of $g(x) = \frac{x}{(x-1)}$ is all real numbers except where the denominator is zero, i.e., $x-1 = 0 \implies x = 1$.
So, Domain of $g = \mathbb{R} \setminus \{1\}$.

Now, let’s find the expression for $f(g(x))$:
$f(g(x)) = f\left(\frac{x}{x-1}\right) = \frac{\frac{x}{x-1}}{2\left(\frac{x}{x-1}\right)-1}$
$f(g(x)) = \frac{\frac{x}{x-1}}{\frac{2x}{x-1}-1} = \frac{\frac{x}{x-1}}{\frac{2x – (x-1)}{x-1}} = \frac{\frac{x}{x-1}}{\frac{x+1}{x-1}} = \frac{x}{x+1}$

For $f(g(x))$ to be defined:
1. $g(x)$ must be defined. So, $x \neq 1$.
2. The denominator of $f(g(x))$ cannot be zero. So, $x+1 \neq 0 \implies x \neq -1$.
Therefore, the domain of $f(g(x))$ is $\mathbb{R} \setminus \{1, -1\}$.

Next, let’s find the expression for $g(f(x))$:
$g(f(x)) = g\left(\frac{x}{2x-1}\right) = \frac{\frac{x}{2x-1}}{\frac{x}{2x-1}-1}$
$g(f(x)) = \frac{\frac{x}{2x-1}}{\frac{x – (2x-1)}{2x-1}} = \frac{\frac{x}{2x-1}}{\frac{-x+1}{2x-1}} = \frac{x}{-x+1}$

For $g(f(x))$ to be defined:
1. $f(x)$ must be defined. So, $x \neq \frac{1}{2}$.
2. The denominator of $g(f(x))$ cannot be zero. So, $-x+1 \neq 0 \implies x \neq 1$.
Therefore, the domain of $g(f(x))$ is $\mathbb{R} \setminus \{\frac{1}{2}, 1\}$.

The domain of $h(x) = f(g(x)) + g(f(x))$ is the intersection of the domains of $f(g(x))$ and $g(f(x))$.
This means $x$ must not be in the set $\{1, -1\}$ (from $f(g(x))$) and $x$ must not be in the set $\{\frac{1}{2}, 1\}$ (from $g(f(x))$).

Combining these restrictions, $x$ cannot be $1$, $-1$, or $\frac{1}{2}$.
Thus, the domain of $h(x)$ is all real numbers except $-1, \frac{1}{2}, 1$.

Comparing this with the given options:
Option A: $-1, \frac{1}{2}, \text{and } 1$

Correct_Option:A

Ans: C

We are given the expression $y= \dfrac{2x-3}{2x^{2}+4x-6}$.
Rearranging this equation to express it in terms of $x$ yields:
$2yx^2 + 4yx – 6y = 2x – 3$
This can be further simplified to:
$2yx^2 + (4y-2)x – 6y + 3 = 0$
Let’s refer to this as Equation (1).

Equation (1) represents a quadratic equation in the variable $x$, and since $x$ is a real number, the discriminant of this quadratic must be non-negative.
Applying the discriminant condition, we get:
$(4y-2)^2 + 4(2y)(-(6y-3)) \geq 0$
Expanding and simplifying the inequality:
$(16y^2 – 16y + 4) + 8y(3-6y) \geq 0$
$16y^2 – 16y + 4 + 24y – 48y^2 \geq 0$
$-32y^2 + 8y + 4 \geq 0$
Dividing by -4 and reversing the inequality sign:
$8y^2 – 2y – 1 \leq 0$

To find the roots of the quadratic $8y^2 – 2y – 1 = 0$, we use the quadratic formula:
$y = \dfrac{-(-2) \pm \sqrt{(-2)^2 – 4(8)(-1)}}{2(8)}$
$y = \dfrac{2 \pm \sqrt{4 + 32}}{16}$
$y = \dfrac{2 \pm \sqrt{36}}{16}$
$y = \dfrac{2 \pm 6}{16}$
The roots are $y = \dfrac{2+6}{16} = \dfrac{8}{16} = \dfrac{1}{2}$ and $y = \dfrac{2-6}{16} = \dfrac{-4}{16} = -\dfrac{1}{4}$.

Since the coefficient of $y^2$ in $8y^2 – 2y – 1$ is positive (8), the quadratic expression is less than or equal to zero between its roots.
Therefore, the possible values of $y$, which represent the range of the function $f(x)$, are:
$\left[-\dfrac{1}{4}, \dfrac{1}{2}\right]$

Option C is the correct answer.

Ans: D

Explanation:The problem asks for the number of surjective functions from set A to set B.
Set A has elements $A = \{2, 3, 5, 7, 11, 13\}$. The number of elements in A is $|A| = 6$.
Set B has elements $B = \{1, 8, 27\}$. The number of elements in B is $|B| = 3$.
A function f from A to B is surjective if for every element b in B, there is at least one element a in A such that f(a) = b. This means that the range of the function must be equal to the codomain B.

The total number of functions from A to B without any restrictions is $|B|^{|A|} = 3^6$.

We can use the Principle of Inclusion-Exclusion to find the number of surjective functions.
The formula for the number of surjective functions from a set of size n to a set of size m is:
$m! S(n, m) = \sum_{k=0}^{m} (-1)^k \binom{m}{k} (m-k)^n$
where $S(n, m)$ is the Stirling number of the second kind.

In this case, $n = |A| = 6$ and $m = |B| = 3$.
The number of surjective functions is:
$3! S(6, 3) = \binom{3}{0}(3-0)^6 – \binom{3}{1}(3-1)^6 + \binom{3}{2}(3-2)^6 – \binom{3}{3}(3-3)^6$
$= \binom{3}{0}3^6 – \binom{3}{1}2^6 + \binom{3}{2}1^6 – \binom{3}{3}0^6$
$= 1 \times 729 – 3 \times 64 + 3 \times 1 – 1 \times 0$
$= 729 – 192 + 3 – 0$
$= 729 – 192 + 3$
$= 537 + 3$
$= 540$

Alternatively, we can think of partitioning the 6 elements of A into 3 non-empty subsets, and then assigning each of these subsets to one of the 3 elements of B.
The number of ways to partition a set of 6 elements into 3 non-empty subsets is given by the Stirling number of the second kind, $S(6, 3)$.
$S(n, k) = \frac{1}{k!} \sum_{j=0}^{k} (-1)^{k-j} \binom{k}{j} j^n$
$S(6, 3) = \frac{1}{3!} \sum_{j=0}^{3} (-1)^{3-j} \binom{3}{j} j^6$
$S(6, 3) = \frac{1}{6} \left( (-1)^3 \binom{3}{0} 0^6 + (-1)^2 \binom{3}{1} 1^6 + (-1)^1 \binom{3}{2} 2^6 + (-1)^0 \binom{3}{3} 3^6 \right)$
$S(6, 3) = \frac{1}{6} \left( -1 \times 1 \times 0 + 1 \times 3 \times 1 + (-1) \times 3 \times 64 + 1 \times 1 \times 729 \right)$
$S(6, 3) = \frac{1}{6} \left( 0 + 3 – 192 + 729 \right)$
$S(6, 3) = \frac{1}{6} \left( 540 \right)$
$S(6, 3) = 90$

This means there are 90 ways to partition the 6 elements of A into 3 non-empty groups.
Once we have these 3 non-empty groups, we need to map them to the 3 distinct elements of B. Since the function must be surjective, each group must be mapped to a different element of B. There are $3!$ ways to assign these 3 groups to the 3 elements of B.
So, the total number of surjective functions is $S(6, 3) \times 3! = 90 \times 6 = 540$.

The calculation using the inclusion-exclusion principle directly gives the number of surjective functions, which is:
Number of surjective functions = $3^6 – \binom{3}{1} 2^6 + \binom{3}{2} 1^6$
$= 729 – 3 \times 64 + 3 \times 1$
$= 729 – 192 + 3$
$= 537 + 3 = 540$.

The options provided are:
A: 665
B: 667
C: 537
D: 540

Our calculated value is 540, which matches option D.

The final answer is $\boxed{540}$.
Correct_Option:D

Ans: A

Explanation:The given functional equation is $f(xy) = f(x)f(y) + f(x) + f(y)$.
We can add 1 to both sides of the equation to get:
$f(xy) + 1 = f(x)f(y) + f(x) + f(y) + 1$
$f(xy) + 1 = (f(x) + 1)(f(y) + 1)$

Let $g(x) = f(x) + 1$. Then the functional equation becomes:
$g(xy) = g(x)g(y)$
This means that $g$ is a completely multiplicative function.

We are given that $f(p) = 1$ for every prime number $p$.
So, $g(p) = f(p) + 1 = 1 + 1 = 2$ for every prime number $p$.

Now we need to find the value of $f(160000)$.
First, let’s find the prime factorization of 160000.
$160000 = 16 \times 10000 = 2^4 \times (10^2)^2 = 2^4 \times 10^4 = 2^4 \times (2 \times 5)^4 = 2^4 \times 2^4 \times 5^4 = 2^8 \times 5^4$.

Now we can find $g(160000)$:
Since $g$ is completely multiplicative, $g(160000) = g(2^8 \times 5^4) = g(2^8) \times g(5^4)$.
Also, $g(a^n) = (g(a))^n$.
So, $g(2^8) = (g(2))^8$ and $g(5^4) = (g(5))^4$.

Since 2 and 5 are prime numbers, we have $g(2) = 2$ and $g(5) = 2$.
Therefore, $g(2^8) = 2^8$ and $g(5^4) = 2^4$.

Now, $g(160000) = g(2^8) \times g(5^4) = 2^8 \times 2^4 = 2^{8+4} = 2^{12}$.

We have $g(160000) = 2^{12}$.
We know that $g(x) = f(x) + 1$, so $f(x) = g(x) – 1$.
Therefore, $f(160000) = g(160000) – 1 = 2^{12} – 1$.

Let’s calculate $2^{12}$:
$2^{10} = 1024$
$2^{11} = 1024 \times 2 = 2048$
$2^{12} = 2048 \times 2 = 4096$

So, $f(160000) = 4096 – 1 = 4095$.

Comparing this value with the given options:
Option A: 4095
Option B: 8191
Option C: 2047
Option D: 1023

The calculated value matches Option A.

Correct_Option:A

Ans: C

Explanation:We are given the functional equation $f(x) + 2f \left(\cfrac{1}{x}\right) = 3x$ for any non-zero real number $x$.
We are asked to find the sum of all possible values of $x$ for which $f(x) = 3$.

First, let’s find an expression for $f(x)$.
From the given equation:
1) $f(x) + 2f \left(\cfrac{1}{x}\right) = 3x$

Now, let’s replace $x$ with $\frac{1}{x}$ in the given equation. Since $x$ is a non-zero real number, $\frac{1}{x}$ is also a non-zero real number.
$f \left(\cfrac{1}{x}\right) + 2f \left(\cfrac{1}{\frac{1}{x}}\right) = 3 \left(\cfrac{1}{x}\right)$
$f \left(\cfrac{1}{x}\right) + 2f(x) = \cfrac{3}{x}$

Now we have a system of two linear equations with $f(x)$ and $f \left(\cfrac{1}{x}\right)$:
1) $f(x) + 2f \left(\cfrac{1}{x}\right) = 3x$
2) $2f(x) + f \left(\cfrac{1}{x}\right) = \cfrac{3}{x}$

To eliminate $f \left(\cfrac{1}{x}\right)$, we multiply equation (2) by 2:
$4f(x) + 2f \left(\cfrac{1}{x}\right) = \cfrac{6}{x}$ (Equation 3)

Now, subtract equation (1) from equation (3):
$(4f(x) + 2f \left(\cfrac{1}{x}\right)) – (f(x) + 2f \left(\cfrac{1}{x}\right)) = \cfrac{6}{x} – 3x$
$3f(x) = \cfrac{6}{x} – 3x$

Divide by 3:
$f(x) = \cfrac{2}{x} – x$

Now we need to find the values of $x$ for which $f(x) = 3$.
Substitute the expression for $f(x)$:
$\cfrac{2}{x} – x = 3$

To solve for $x$, we can multiply the entire equation by $x$ (since $x$ is non-zero):
$2 – x^2 = 3x$

Rearrange the terms to form a quadratic equation:
$x^2 + 3x – 2 = 0$

We need to find the sum of all possible values of $x$ for which this quadratic equation holds. For a quadratic equation of the form $ax^2 + bx + c = 0$, the sum of the roots is given by $-\frac{b}{a}$.

In our equation, $a = 1$, $b = 3$, and $c = -2$.
The sum of the roots is $-\frac{3}{1} = -3$.

Let’s check if the roots are real and non-zero. The discriminant of the quadratic equation $x^2 + 3x – 2 = 0$ is $\Delta = b^2 – 4ac = 3^2 – 4(1)(-2) = 9 + 8 = 17$. Since the discriminant is positive, there are two distinct real roots.
The roots are $x = \frac{-3 \pm \sqrt{17}}{2}$. Both of these roots are non-zero.
Therefore, the sum of all possible values of $x$ for which $f(x) = 3$ is -3.

Correct_Option: C

Ans: 3

Explanation:
Let the given equation be
f(3x + 2y, 2x – 5y) = 19x (1)

We are asked to find the value of x for which f(x, 2x) = 27.
This means we need to find values of ‘a’ and ‘b’ such that the arguments of f in equation (1) become x and 2x respectively.
Let 3x + 2y = x (2)
And 2x – 5y = 2x (3)

From equation (3), we have 2x – 5y = 2x.
Subtracting 2x from both sides gives -5y = 0, which implies y = 0.

Now substitute y = 0 into equation (2):
3x + 2(0) = x
3x = x
Subtracting x from both sides gives 2x = 0, which implies x = 0.

This substitution method means that if we set the arguments of f to be x and 2x, the corresponding values of the original x and y must be 0 and 0. However, this does not help us find the value of x for which f(x, 2x) = 27.

Instead, let’s try to express the arguments of f in equation (1) in terms of the arguments of f we are interested in, which are x and 2x.
Let u = 3x + 2y and v = 2x – 5y.
We want to find the values of the original x and y that produce the arguments x and 2x.
So, we set:
3x’ + 2y’ = x (4)
2x’ – 5y’ = 2x (5)
where x’ and y’ are the original variables in the definition of f. We need to solve for x’ and y’ in terms of x.

Multiply equation (4) by 5 and equation (5) by 2 to eliminate y’:
5 * (3x’ + 2y’) = 5 * x => 15x’ + 10y’ = 5x
2 * (2x’ – 5y’) = 2 * 2x => 4x’ – 10y’ = 4x

Add the two modified equations:
(15x’ + 10y’) + (4x’ – 10y’) = 5x + 4x
19x’ = 9x
x’ = (9/19)x

Now, substitute the value of x’ back into equation (4):
3 * ((9/19)x) + 2y’ = x
(27/19)x + 2y’ = x
2y’ = x – (27/19)x
2y’ = (19x – 27x) / 19
2y’ = -8x / 19
y’ = -4x / 19

Now we use the original given equation with these new expressions for x’ and y’:
f(3x’ + 2y’, 2x’ – 5y’) = 19x’
Substitute the values of x’ and y’ we found:
f(x, 2x) = 19 * ((9/19)x)
f(x, 2x) = 9x

We are given that f(x, 2x) = 27.
So, we set 9x equal to 27:
9x = 27
x = 27 / 9
x = 3

Final_Answer:3

Ans: C

If the discriminant ($b^2 – 4ac$) is negative, the quadratic function $f(x) = ax^2 + bx + c$ will maintain a constant sign for all real values of x. Specifically, if the leading coefficient (a) is positive, then $f(x) > 0$ for all x. Conversely, if the leading coefficient (a) is negative, then $f(x) < 0$ for all x.

Given that $f(m) < 0$ and m is an integer, we can infer the following about the set of possible integer values for m:

* If the leading coefficient of $x^2$ is positive, then there are no integer values of m for which $f(m) < 0$. The set of such integers is thus empty.
* If the leading coefficient of $x^2$ is negative, then $f(x) < 0$ for all real numbers x, and consequently for all integers m. The set of possible integer values for m is therefore the set of all integers.

Ans: B

Explanation:
Let the quadratic polynomial be $f(x) = ax^2 + bx + c$.
We are given that $f(x) \geq 0$ for all real numbers $x$. This implies that the parabola opens upwards, so $a > 0$, and the discriminant is less than or equal to zero, i.e., $b^2 – 4ac \leq 0$.
We are also given that $f(2) = 0$. Since $f(x) \geq 0$ for all real numbers $x$, $x=2$ must be the only root of the quadratic equation $f(x)=0$. This means that the quadratic has a repeated root at $x=2$.
Therefore, the quadratic polynomial can be written in the form $f(x) = k(x-2)^2$ for some constant $k$.
Since $f(x) \geq 0$ for all real numbers $x$, and $(x-2)^2 \geq 0$, we must have $k \geq 0$.
We are given that $f(4) = 6$. Substituting this into the expression for $f(x)$:
$f(4) = k(4-2)^2 = k(2)^2 = 4k$.
So, $4k = 6$, which gives $k = \frac{6}{4} = \frac{3}{2}$.
Thus, the quadratic polynomial is $f(x) = \frac{3}{2}(x-2)^2$.
We need to find $f(-2)$.
$f(-2) = \frac{3}{2}(-2-2)^2 = \frac{3}{2}(-4)^2 = \frac{3}{2}(16) = 3 \times 8 = 24$.

Correct_Option: B

Ans: 12

Explanation:The first given condition is $f(x) + f(x – 1) – 1 = 0$.
This can be rewritten as $f(x) = 1 – f(x – 1)$.

Let’s find the values of f for a few integers:
If $x=1$, $f(1) = 1 – f(0)$.
If $x=2$, $f(2) = 1 – f(1) = 1 – (1 – f(0)) = f(0)$.
If $x=3$, $f(3) = 1 – f(2) = 1 – f(0)$.
If $x=4$, $f(4) = 1 – f(3) = 1 – (1 – f(0)) = f(0)$.

We can observe a pattern here:
$f(x) = f(0)$ if $x$ is even.
$f(x) = 1 – f(0)$ if $x$ is odd.

The second given condition is $g(x) = x^2$.
We are also given $f(x^2 – x) = 5$.
Let’s analyze the argument of f: $x^2 – x = x(x-1)$.
The product of two consecutive integers is always even.
So, $x^2 – x$ is always an even number for any integer $x$.

Since $x^2 – x$ is always even, according to our derived pattern for $f(x)$, we must have $f(x^2 – x) = f(0)$.
We are given $f(x^2 – x) = 5$.
Therefore, $f(0) = 5$.

Now we can determine the general form of $f(x)$:
If $x$ is even, $f(x) = f(0) = 5$.
If $x$ is odd, $f(x) = 1 – f(0) = 1 – 5 = -4$.

We need to find the value of $f(g(5)) + g(f(5))$.

First, let’s find $g(5)$:
$g(5) = 5^2 = 25$.
So, $f(g(5)) = f(25)$.
Since 25 is an odd number, $f(25) = -4$.

Next, let’s find $f(5)$:
Since 5 is an odd number, $f(5) = -4$.
So, $g(f(5)) = g(-4)$.
$g(-4) = (-4)^2 = 16$.

Finally, we need to calculate $f(g(5)) + g(f(5))$:
$f(g(5)) + g(f(5)) = f(25) + g(-4) = -4 + 16 = 12$.

Final_Answer:12

Ans: B

Explanation:We are given the function $f(x) = \begin{cases}2x -r & ifx \geq r\\ r &ifx < r\end{cases}$ and we need to find the values of $x$ for which the equation $f(x) = f(f(x))$ holds.

We consider two cases for $x$:

Case 1: $x \geq r$
In this case, $f(x) = 2x – r$.
Now we need to evaluate $f(f(x)) = f(2x – r)$.
We need to compare $2x – r$ with $r$.

Subcase 1.1: $2x – r \geq r$
This implies $2x \geq 2r$, which means $x \geq r$. This condition is already satisfied in Case 1.
So, if $x \geq r$, then $f(2x – r) = 2(2x – r) – r = 4x – 2r – r = 4x – 3r$.
The equation $f(x) = f(f(x))$ becomes $2x – r = 4x – 3r$.
$2r = 2x$, which means $x = r$.
This solution $x=r$ is consistent with the condition $x \geq r$.

Subcase 1.2: $2x – r < r$
This implies $2x < 2r$, which means $x < r$.
This condition contradicts the initial assumption of Case 1 ($x \geq r$). So, there are no solutions in this subcase.

Case 2: $x < r$
In this case, $f(x) = r$.
Now we need to evaluate $f(f(x)) = f(r)$.
Since $r \geq r$ is true, we use the first part of the function definition for $f(r)$: $f(r) = 2r – r = r$.
The equation $f(x) = f(f(x))$ becomes $r = r$.
This equation is true for all $x$ such that $x < r$.

Combining the results from both cases:
From Case 1, we found that $f(x) = f(f(x))$ holds for $x = r$.
From Case 2, we found that $f(x) = f(f(x))$ holds for all $x < r$.

Therefore, the equation $f(x) = f(f(x))$ holds for all real values of $x$ such that $x < r$ or $x = r$. This means the equation holds for all $x \leq r$.

Let’s check the given options:
Option A: $x > r$ – We found a solution only at $x=r$ in this range, but the equation $2x-r = 4x-3r$ is only satisfied when $x=r$. For $x>r$, $2x-r > r$, so $f(x) = 2x-r$ and $f(f(x)) = 2(2x-r)-r = 4x-3r$. $2x-r = 4x-3r \implies 2x = 2r \implies x=r$. This contradicts $x>r$. So, this option is incorrect.

Option B: $x \leq r$ – This includes $x < r$ and $x = r$. We found that for $x < r$, $f(x) = r$ and $f(f(x)) = f(r) = r$, so $f(x) = f(f(x))$. For $x = r$, $f(x) = 2r - r = r$ and $f(f(x)) = f(r) = r$, so $f(x) = f(f(x))$. Thus, the equation holds for all $x \leq r$.

Option C: $x \neq r$ – This includes $x < r$ and $x > r$. We know it holds for $x < r$, but not for $x > r$. So, this option is incorrect.

Option D: $x \geq r$ – This includes $x = r$ and $x > r$. We found it holds for $x=r$, but not for $x > r$. So, this option is incorrect.

Therefore, the equation $f(x) = f(f(x))$ holds for all real values of $x$ where $x \leq r$.

The final answer is $\boxed{x \leq r}$.

Correct_Option:B

Ans: B

Explanation:Let the given expression be $y = \frac{x^{2} – 6x + 10}{3-x}$.
We are given the condition $x < 3$. This implies $3-x > 0$.

We can rewrite the numerator by completing the square:
$x^2 – 6x + 10 = (x^2 – 6x + 9) + 1 = (x-3)^2 + 1$.

So, $y = \frac{(x-3)^2 + 1}{3-x}$.
Since $3-x > 0$, we can write $(x-3)^2 = (-(3-x))^2 = (3-x)^2$.
Therefore, $y = \frac{(3-x)^2 + 1}{3-x}$.

Let $u = 3-x$. Since $x < 3$, we have $u > 0$.
The expression becomes $y = \frac{u^2 + 1}{u} = \frac{u^2}{u} + \frac{1}{u} = u + \frac{1}{u}$.

We need to find the minimum possible value of $u + \frac{1}{u}$ for $u > 0$.
By the AM-GM inequality, for positive numbers $a$ and $b$, we have $\frac{a+b}{2} \ge \sqrt{ab}$, which implies $a+b \ge 2\sqrt{ab}$.
In our case, $a=u$ and $b=\frac{1}{u}$. Both are positive since $u > 0$.
So, $u + \frac{1}{u} \ge 2\sqrt{u \cdot \frac{1}{u}} = 2\sqrt{1} = 2$.

The minimum value of $u + \frac{1}{u}$ is 2.
This minimum value occurs when $u = \frac{1}{u}$, which means $u^2 = 1$. Since $u > 0$, we have $u=1$.
When $u=1$, $3-x = 1$, so $x = 3-1 = 2$.
This value of $x=2$ satisfies the condition $x < 3$.
Thus, the minimum possible value of the expression is 2.

We can also use calculus to find the minimum.
Let $f(x) = \frac{x^{2} – 6x + 10}{3-x}$.
We find the derivative $f'(x)$ using the quotient rule:
$f'(x) = \frac{(2x-6)(3-x) – (x^2-6x+10)(-1)}{(3-x)^2}$
$f'(x) = \frac{(6x – 2x^2 – 18 + 6x) + (x^2 – 6x + 10)}{(3-x)^2}$
$f'(x) = \frac{-2x^2 + 12x – 18 + x^2 – 6x + 10}{(3-x)^2}$
$f'(x) = \frac{-x^2 + 6x – 8}{(3-x)^2}$

To find critical points, we set $f'(x) = 0$:
$-x^2 + 6x – 8 = 0$
$x^2 – 6x + 8 = 0$
$(x-2)(x-4) = 0$
The critical points are $x=2$ and $x=4$.

We are given $x < 3$. So, we only consider $x=2$.
Now we check the second derivative or the sign of the first derivative around $x=2$.
For $x < 2$ (and $x < 3$), let's take $x=1$.
$f'(1) = \frac{-(1)^2 + 6(1) – 8}{(3-1)^2} = \frac{-1+6-8}{4} = \frac{-3}{4} < 0$.
So, $f(x)$ is decreasing for $x < 2$.

For $2 < x < 3$, let's take $x=2.5$.
$f'(2.5) = \frac{-(2.5)^2 + 6(2.5) – 8}{(3-2.5)^2} = \frac{-6.25 + 15 – 8}{(0.5)^2} = \frac{0.75}{0.25} = 3 > 0$.
So, $f(x)$ is increasing for $2 < x < 3$.

This means that $x=2$ is a local minimum.
The value of the expression at $x=2$ is:
$y = \frac{(2)^2 – 6(2) + 10}{3-2} = \frac{4 – 12 + 10}{1} = \frac{2}{1} = 2$.

As $x \to 3^-$, $3-x \to 0^+$. The numerator approaches $(3)^2 – 6(3) + 10 = 9 – 18 + 10 = 1$.
So, as $x \to 3^-$, $y \to \frac{1}{0^+} \to +\infty$.
Thus, the minimum value is indeed 2.

Correct_Option: B

Ans: D

$f(x)=\frac{x^{2}+2x+4}{2x^{2}+4x+9}$
By observing the proportional coefficients of the $x^2$ and $x$ terms in the numerator and denominator (specifically the 1:2 ratio), we can infer a strategy to simplify the expression by manipulating the numerator.

$f(x)=\frac{x^2+2x+4}{2x^2+4x+9}=\frac{x^2+2x+4.5-0.5}{2x^2+4x+9}$
This can be rewritten as:
$ \frac{x^2+2x+4.5}{2x^2+4x+9}-\frac{0.5}{2x^2+4x+9} $
Noticing that $2x^2+4x+9 = 2(x^2+2x+4.5)$, the first term simplifies to $\frac{1}{2}$.
Therefore, the expression becomes:
$ \frac{1}{2}-\frac{0.5}{2x^2+4x+9} $
Now, the variable $x$ is solely present in the denominator.

To find the maximum value of the function, we consider the case where the denominator $2x^2+4x+9$ approaches infinity. In this scenario, the fraction $\frac{0.5}{2x^2+4x+9}$ tends to zero, and the entire expression approaches $\frac{1}{2}$. However, since the denominator can never truly be infinite, the maximum value is not inclusively $\frac{1}{2}$, hence $\frac{1}{2})$.

To find the minimum value, we need to determine the minimum value of the quadratic expression in the denominator, $2x^2+4x+9$. This occurs at the vertex of the parabola. We find the derivative with respect to $x$ and set it to zero:
$ \frac{d}{dx}(2x^2+4x+9) = 4x+4 $
Setting $4x+4=0$ yields $x=-1$.

Substituting $x=-1$ into the quadratic expression:
$ 2(-1)^2 + 4(-1) + 9 = 2 – 4 + 9 = 7 $
So, the minimum value of the denominator is 7.

Now, substituting this minimum denominator value back into the simplified function:
$ \frac{1}{2}-\frac{0.5}{7} = \frac{1}{2}-\frac{1/2}{7} = \frac{1}{2}-\frac{1}{14} = \frac{7}{14}-\frac{1}{14} = \frac{6}{14} = \frac{3}{7} $
Thus, the range of the expression is $[\frac{3}{7},\frac{1}{2})$.

Ans: D

Explanation:We are given two functions:
$f(x) = x^2 – 7x$
$g(x) = x + 3$

We need to find the minimum value of $f(g(x)) – 3x$.

First, let’s find the expression for $f(g(x))$.
Substitute $g(x)$ into $f(x)$:
$f(g(x)) = f(x+3)$
Replace $x$ in $f(x)$ with $(x+3)$:
$f(x+3) = (x+3)^2 – 7(x+3)$

Expand the expression:
$(x+3)^2 = x^2 + 2(x)(3) + 3^2 = x^2 + 6x + 9$
$-7(x+3) = -7x – 21$

So, $f(g(x)) = (x^2 + 6x + 9) + (-7x – 21)$
$f(g(x)) = x^2 + 6x + 9 – 7x – 21$
$f(g(x)) = x^2 – x – 12$

Now, we need to find the minimum value of $f(g(x)) – 3x$.
Substitute the expression for $f(g(x))$:
$f(g(x)) – 3x = (x^2 – x – 12) – 3x$
$f(g(x)) – 3x = x^2 – x – 3x – 12$
$f(g(x)) – 3x = x^2 – 4x – 12$

Let $h(x) = x^2 – 4x – 12$. This is a quadratic function in the form $ax^2 + bx + c$, where $a=1$, $b=-4$, and $c=-12$.
Since the coefficient of the $x^2$ term ($a=1$) is positive, the parabola opens upwards, which means it has a minimum value.

The minimum value of a quadratic function $ax^2 + bx + c$ occurs at the vertex, where $x = -\frac{b}{2a}$.
In our case, $a=1$ and $b=-4$.
The x-coordinate of the vertex is $x = -\frac{-4}{2(1)} = \frac{4}{2} = 2$.

Now, substitute this value of $x$ back into the expression for $h(x)$ to find the minimum value:
Minimum value = $h(2) = (2)^2 – 4(2) – 12$
Minimum value = $4 – 8 – 12$
Minimum value = $-4 – 12$
Minimum value = $-16$

Alternatively, we can complete the square for $h(x) = x^2 – 4x – 12$:
$h(x) = (x^2 – 4x + 4) – 4 – 12$
$h(x) = (x-2)^2 – 16$
The term $(x-2)^2$ is always greater than or equal to 0. The minimum value of $(x-2)^2$ is 0, which occurs when $x=2$.
Therefore, the minimum value of $h(x)$ is $0 – 16 = -16$.

Comparing this with the given options:
Option A: -20
Option B: -12
Option C: -15
Option D: -16

The minimum value is -16.

Correct_Option: D

Ans: D

Explanation:The condition $f(5+x)=f(5-x)$ implies that the function $f(x)$ is symmetric about the line $x=5$.
Let the roots of $f(x)=0$ be $r_1, r_2, r_3, r_4$.
Since the roots are distinct, we have $r_1 \neq r_2 \neq r_3 \neq r_4$.
The symmetry of the function about $x=5$ means that if $r$ is a root, then $10-r$ must also be a root. This is because if $f(r) = 0$, then $f(5+(r-5)) = f(5-(r-5))$, which simplifies to $f(r) = f(10-r)$. If $f(r)=0$, then $f(10-r)=0$, meaning $10-r$ is also a root.

Since there are four distinct real roots, we can pair them up based on this symmetry. Let the roots be denoted as $r_1, r_2, r_3, r_4$.
We can group these roots into pairs such that the sum of each pair is 10.
Suppose $r_1$ is one root. Then $10-r_1$ must be another root.
Suppose $r_2$ is a different root. Then $10-r_2$ must be a third root.

Since there are four distinct roots, these pairs must cover all the roots.
Let the roots be $r_1$, $10-r_1$, $r_2$, and $10-r_2$.
For these roots to be distinct, we must have:
1. $r_1 \neq 10-r_1 \implies 2r_1 \neq 10 \implies r_1 \neq 5$.
2. $r_2 \neq 10-r_2 \implies 2r_2 \neq 10 \implies r_2 \neq 5$.
3. $r_1 \neq r_2$.
4. $r_1 \neq 10-r_2 \implies r_1+r_2 \neq 10$.
5. $10-r_1 \neq r_2 \implies 10 \neq r_1+r_2$.
6. $10-r_1 \neq 10-r_2 \implies r_1 \neq r_2$.

The set of four distinct roots is $\{r_1, 10-r_1, r_2, 10-r_2\}$.
The sum of these roots is $r_1 + (10-r_1) + r_2 + (10-r_2) = r_1 + 10 – r_1 + r_2 + 10 – r_2 = 20$.

Alternatively, consider the roots as $a, b, c, d$.
Since $f(5+x) = f(5-x)$, if $r$ is a root, then $10-r$ is also a root.
Let the four distinct roots be $r_1, r_2, r_3, r_4$.
We can pair them as follows:
If $r_1$ is a root, then $10-r_1$ is also a root. Let $r_2 = 10-r_1$.
If $r_3$ is a root (and $r_3 \neq r_1$ and $r_3 \neq r_2$), then $10-r_3$ is also a root. Let $r_4 = 10-r_3$.
For the roots to be distinct, $r_1 \neq 5$ and $r_3 \neq 5$. Also $r_1 \neq r_3$ and $r_1 \neq 10-r_3$.

The sum of the roots is $r_1 + r_2 + r_3 + r_4 = r_1 + (10-r_1) + r_3 + (10-r_3) = 10 + 10 = 20$.

The final answer is $\boxed{20}$.

Correct_Option:D

Ans: B

Explanation:Given the quadratic function $f(x) = x^2 + ax + b$.
The condition $f(x) \geq 0$ for all real x means that the parabola opens upwards (which is true since the coefficient of $x^2$ is 1, which is positive) and it either touches the x-axis at one point or does not intersect the x-axis. This implies that the discriminant of the quadratic equation $f(x) = 0$ must be less than or equal to zero.
The discriminant is $\Delta = a^2 – 4(1)(b) = a^2 – 4b$.
So, $a^2 – 4b \leq 0$, which means $a^2 \leq 4b$.

Now, let’s find the expression for $g(x)$.
$g(x) = f(x+1) – f(x-1)$
$f(x+1) = (x+1)^2 + a(x+1) + b = (x^2 + 2x + 1) + (ax + a) + b = x^2 + (2+a)x + (1+a+b)$
$f(x-1) = (x-1)^2 + a(x-1) + b = (x^2 – 2x + 1) + (ax – a) + b = x^2 + (-2+a)x + (1-a+b)$

$g(x) = [x^2 + (2+a)x + (1+a+b)] – [x^2 + (-2+a)x + (1-a+b)]$
$g(x) = x^2 + (2+a)x + 1+a+b – x^2 – (-2+a)x – (1-a+b)$
$g(x) = (2+a – (-2+a))x + (1+a+b – (1-a+b))$
$g(x) = (2+a+2-a)x + (1+a+b-1+a-b)$
$g(x) = 4x + 2a$

We are given that $g(20) = 72$.
Substituting x = 20 into the expression for g(x):
$g(20) = 4(20) + 2a = 80 + 2a$
So, $80 + 2a = 72$.
$2a = 72 – 80$
$2a = -8$
$a = -4$

Now we use the condition $a^2 \leq 4b$ with $a = -4$.
$(-4)^2 \leq 4b$
$16 \leq 4b$
Divide both sides by 4:
$4 \leq b$

The smallest possible value of b is 4.

Let’s check if this value is consistent with the options.
Option A: 16
Option B: 4
Option C: 1
Option D: 0

The smallest possible value of b is 4, which corresponds to Option B.

The final answer is $\boxed{4}$.
Explanation:Given $f(x) = x^2 + ax + b$.
The condition $f(x) \geq 0$ for all real $x$ implies that the discriminant of the quadratic equation $x^2 + ax + b = 0$ is less than or equal to zero.
Discriminant $\Delta = a^2 – 4(1)(b) = a^2 – 4b$.
So, $a^2 – 4b \leq 0$, which means $a^2 \leq 4b$.

Now we compute $g(x) = f(x+1) – f(x-1)$.
$f(x+1) = (x+1)^2 + a(x+1) + b = x^2 + 2x + 1 + ax + a + b = x^2 + (a+2)x + (a+b+1)$.
$f(x-1) = (x-1)^2 + a(x-1) + b = x^2 – 2x + 1 + ax – a + b = x^2 + (a-2)x + (-a+b+1)$.

$g(x) = f(x+1) – f(x-1)$
$g(x) = [x^2 + (a+2)x + (a+b+1)] – [x^2 + (a-2)x + (-a+b+1)]$
$g(x) = x^2 + ax + 2x + a + b + 1 – x^2 – ax + 2x – a – b – 1$
$g(x) = (ax + 2x – ax + 2x) + (a + b + 1 – a – b – 1)$
$g(x) = 4x$.

Wait, I made a mistake in the calculation of $g(x)$. Let’s recompute.
$g(x) = [x^2 + (a+2)x + (a+b+1)] – [x^2 + (a-2)x + (-a+b+1)]$
$g(x) = (x^2 – x^2) + ((a+2)x – (a-2)x) + ((a+b+1) – (-a+b+1))$
$g(x) = 0 + (ax + 2x – ax + 2x) + (a+b+1 + a-b-1)$
$g(x) = 4x + 2a$.

Given $g(20) = 72$.
$g(20) = 4(20) + 2a = 80 + 2a$.
So, $80 + 2a = 72$.
$2a = 72 – 80$.
$2a = -8$.
$a = -4$.

Now we use the condition $a^2 \leq 4b$.
Substitute $a = -4$:
$(-4)^2 \leq 4b$.
$16 \leq 4b$.
Divide by 4:
$4 \leq b$.

The smallest possible value of b is 4.

The final answer is $\boxed{4}$.
Correct_Option:B

Ans: D

Explanation:Let the given expression be $f(x) = \frac{x}{\sqrt{1+x^{4}}}$.
We want to find the maximum possible value of $f(x)$ for real $x$.

Case 1: $x < 0$.
If $x < 0$, then $f(x) < 0$. The maximum value of a negative function will be less than any positive value.

Case 2: $x = 0$.
If $x = 0$, then $f(0) = \frac{0}{\sqrt{1+0^4}} = 0$.

Case 3: $x > 0$.
If $x > 0$, then $f(x) > 0$. To find the maximum value, we can consider squaring the expression, as it will preserve the ordering for positive values.
Let $y = f(x)^2 = \left(\frac{x}{\sqrt{1+x^{4}}}\right)^2 = \frac{x^2}{1+x^4}$.
We want to find the maximum value of $y$ for $x > 0$.
Let $u = x^2$. Since $x > 0$, $u > 0$.
Then $y = \frac{u}{1+u^2}$.
To find the maximum value of $y$, we can use calculus or AM-GM inequality.

Using AM-GM inequality:
For $u > 0$, we know that $1+u^2 \ge 2\sqrt{1 \cdot u^2} = 2u$.
Therefore, $\frac{1+u^2}{u} \ge 2$.
Taking the reciprocal, $\frac{u}{1+u^2} \le \frac{1}{2}$.
So, the maximum value of $y = \frac{u}{1+u^2}$ is $\frac{1}{2}$.
This maximum is achieved when $1 = u^2$, which means $u = 1$ (since $u > 0$).
If $u = x^2 = 1$, then $x = 1$ (since $x > 0$).

When $x = 1$, $f(1) = \frac{1}{\sqrt{1+1^4}} = \frac{1}{\sqrt{2}}$.
Since $y = f(x)^2$, the maximum value of $f(x)$ is $\sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}$.

Alternatively, using calculus for $y = \frac{u}{1+u^2}$:
$\frac{dy}{du} = \frac{(1)(1+u^2) – u(2u)}{(1+u^2)^2} = \frac{1+u^2-2u^2}{(1+u^2)^2} = \frac{1-u^2}{(1+u^2)^2}$.
To find critical points, set $\frac{dy}{du} = 0$:
$1-u^2 = 0 \implies u^2 = 1$.
Since $u > 0$, $u = 1$.
We can check the second derivative or the sign of the first derivative.
For $0 < u < 1$, $\frac{dy}{du} > 0$, so $y$ is increasing.
For $u > 1$, $\frac{dy}{du} < 0$, so $y$ is decreasing.
Thus, $u=1$ corresponds to a maximum.
When $u=1$, $y = \frac{1}{1+1^2} = \frac{1}{2}$.
The maximum value of $f(x)^2$ is $\frac{1}{2}$.
The maximum value of $f(x)$ is $\sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}$.

Comparing the values:
For $x < 0$, $f(x) < 0$.
For $x = 0$, $f(x) = 0$.
For $x > 0$, the maximum value is $\frac{1}{\sqrt{2}}$.
The overall maximum possible value of the expression is $\frac{1}{\sqrt{2}}$.

Correct_Option:D

Ans: C

Explanation:The given functional equation is $f(x+y) = f(x)f(y)$. This is a characteristic property of exponential functions.
We are given that $f(5) = 4$.
We need to find the value of $f(10) – f(-10)$.

First, let’s find $f(10)$.
Using the functional equation, we can write:
$f(10) = f(5+5) = f(5)f(5) = (f(5))^2$
Since $f(5) = 4$, we have:
$f(10) = 4^2 = 16$.

Next, let’s find $f(-10)$.
We can use the property $f(x+y) = f(x)f(y)$ for $y = -x$.
$f(x+(-x)) = f(x)f(-x)$
$f(0) = f(x)f(-x)$

To find $f(0)$, we can set $x=0$ and $y=0$ in the given functional equation:
$f(0+0) = f(0)f(0)$
$f(0) = (f(0))^2$
This implies $f(0) – (f(0))^2 = 0$, so $f(0)(1 – f(0)) = 0$.
Thus, $f(0) = 0$ or $f(0) = 1$.

If $f(0) = 0$, then for any $x$, $f(x) = f(x+0) = f(x)f(0) = f(x) \cdot 0 = 0$.
This would mean $f(x) = 0$ for all $x$. However, we are given $f(5) = 4$, which contradicts $f(x) = 0$.
Therefore, $f(0)$ must be $1$.

Now, using $f(0) = f(x)f(-x)$ and $f(0) = 1$, we get:
$1 = f(x)f(-x)$
So, $f(-x) = \frac{1}{f(x)}$.

We need to find $f(-10)$. Using the above relation:
$f(-10) = \frac{1}{f(10)}$
We found $f(10) = 16$.
So, $f(-10) = \frac{1}{16}$.

Now, we can calculate $f(10) – f(-10)$:
$f(10) – f(-10) = 16 – \frac{1}{16}$
To subtract these, we find a common denominator:
$16 – \frac{1}{16} = \frac{16 \cdot 16}{16} – \frac{1}{16} = \frac{256}{16} – \frac{1}{16} = \frac{256 – 1}{16} = \frac{255}{16}$.

Now, we convert the fraction $\frac{255}{16}$ to a decimal.
$255 \div 16$:
$255 = 16 \times 10 + 95$
$95 = 16 \times 5 + 15$
So, $255 = 16 \times 15 + 15$.
$\frac{255}{16} = 15 + \frac{15}{16}$.
To convert $\frac{15}{16}$ to decimal:
$\frac{15}{16} = 15 \times \frac{1}{16}$
$\frac{1}{16} = 0.0625$
So, $\frac{15}{16} = 15 \times 0.0625 = 0.9375$.
Therefore, $\frac{255}{16} = 15 + 0.9375 = 15.9375$.

Comparing this with the given options:
Option A: 14.0625
Option B: 0
Option C: 15.9375
Option D: 3

The calculated value matches Option C.

The final answer is $\boxed{15.9375}$.
Correct_Option:C

Ans: 3

Explanation:The function $f(x + y) = f(x) f(y)$ where $x, y$ are positive integers and $f(1) = 2$ is an exponential function.
Let $x = 1, y = 1$, then $f(2) = f(1) f(1) = 2 \times 2 = 2^2$.
Let $x = 2, y = 1$, then $f(3) = f(2) f(1) = 2^2 \times 2 = 2^3$.
By induction, we can prove that $f(x) = 2^x$ for any positive integer $x$.

We are given the equation $f(a + 1) + f(a + 2) + … + f(a + n) = 16 (2^n – 1)$.
Substituting $f(x) = 2^x$ into the equation, we get:
$2^{a+1} + 2^{a+2} + … + 2^{a+n} = 16 (2^n – 1)$

This is a geometric series with the first term $A = 2^{a+1}$, the common ratio $R = 2$, and the number of terms $N = n$.
The sum of a geometric series is given by $S_N = A \frac{R^N – 1}{R – 1}$.
So, the sum of the left side is:
$2^{a+1} \frac{2^n – 1}{2 – 1} = 2^{a+1} (2^n – 1)$

Now, we equate this to the right side of the given equation:
$2^{a+1} (2^n – 1) = 16 (2^n – 1)$

Since $n$ is a positive integer, $2^n – 1 \neq 0$. We can divide both sides by $(2^n – 1)$:
$2^{a+1} = 16$

We know that $16 = 2^4$.
So, $2^{a+1} = 2^4$.
Equating the exponents, we get:
$a + 1 = 4$
$a = 4 – 1$
$a = 3$

Final_Answer:3

Ans: 12

The given property is that for any two numbers m and n, the function f(mn) is equivalent to the product of f(m) and f(n).

Let’s explore some values:
If we set m=1 and n=1, the property becomes f(1*1) = f(1)*f(1), which simplifies to f(1) = [f(1)]^2. This implies that f(1) must be either 0 or 1.

If we set m=1 and n=2, the property is f(1*2) = f(1)*f(2), meaning f(2) = f(1)*f(2). For this to hold true, f(1) must be 1 (since f(2) is not necessarily zero).

Using the property, we can deduce:
f(4) = f(2*2) = f(2)*f(2) = [f(2)]^2
f(8) = f(4*2) = f(4)*f(2) = [f(2)]^2 * f(2) = [f(2)]^3

We are given that f(24) = 54.
We can express 24 as 8 * 3. Therefore, f(24) = f(8*3) = f(8)*f(3).
Substituting the expression for f(8), we get f(24) = [f(2)]^3 * f(3).
So, we have the equation: [f(2)]^3 * f(3) = 54.

We can factor 54 as 27 * 2, which is also [3]^3 * 2.
Comparing [f(2)]^3 * f(3) with [3]^3 * 2, we can infer that f(2) = 3 and f(3) = 2.

Now we need to determine the value of f(18).
We can express 18 as 2 * 9, and 9 as 3*3.
So, f(18) = f(2 * 3 * 3) = f(2) * f(3) * f(3) = f(2) * [f(3)]^2.
Substituting the values we found for f(2) and f(3):
f(18) = 3 * [2]^2 = 3 * 4 = 12.

Ans: 32

Explanation:Let $f(x) = \min(2x^2, 52-5x)$ for positive real numbers $x$. We want to find the maximum possible value of $f(x)$.

The function $f(x)$ is the minimum of two functions: $g(x) = 2x^2$ and $h(x) = 52-5x$.
$g(x) = 2x^2$ is a parabola opening upwards, and its value increases as $x$ increases for $x>0$.
$h(x) = 52-5x$ is a linear function with a negative slope, and its value decreases as $x$ increases.

The graph of $f(x)$ will follow the graph of $g(x)$ when $g(x) \le h(x)$, and will follow the graph of $h(x)$ when $h(x) \le g(x)$. The maximum value of $f(x)$ will occur at the point where the two functions intersect, or at the peak of the “valley” formed by the two functions.

To find the intersection point, we set $g(x) = h(x)$:
$2x^2 = 52 – 5x$
$2x^2 + 5x – 52 = 0$

We can solve this quadratic equation using the quadratic formula:
$x = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}$
Here, $a=2$, $b=5$, $c=-52$.
$x = \frac{-5 \pm \sqrt{5^2 – 4(2)(-52)}}{2(2)}$
$x = \frac{-5 \pm \sqrt{25 + 416}}{4}$
$x = \frac{-5 \pm \sqrt{441}}{4}$
$x = \frac{-5 \pm 21}{4}$

We have two possible values for $x$:
$x_1 = \frac{-5 + 21}{4} = \frac{16}{4} = 4$
$x_2 = \frac{-5 – 21}{4} = \frac{-26}{4} = -\frac{13}{2}$

Since $x$ is a positive real number, we take $x=4$.

At $x=4$, the value of $g(x)$ is $2(4^2) = 2(16) = 32$.
At $x=4$, the value of $h(x)$ is $52 – 5(4) = 52 – 20 = 32$.
So, at $x=4$, $f(x) = \min(32, 32) = 32$.

Now we need to confirm that this is indeed the maximum value.
Consider values of $x$ slightly less than 4. For example, $x=3$:
$g(3) = 2(3^2) = 2(9) = 18$
$h(3) = 52 – 5(3) = 52 – 15 = 37$
$f(3) = \min(18, 37) = 18$, which is less than 32.

Consider values of $x$ slightly greater than 4. For example, $x=5$:
$g(5) = 2(5^2) = 2(25) = 50$
$h(5) = 52 – 5(5) = 52 – 25 = 27$
$f(5) = \min(50, 27) = 27$, which is less than 32.

The function $f(x)$ increases as $2x^2$ until $x=4$, where it reaches 32. After $x=4$, $f(x)$ follows $52-5x$, which decreases from 32. Therefore, the maximum value of $f(x)$ is 32.

Final Answer:The maximum possible value of $f(x)$ is 32.

Final_Answer:32

Ans: 20

Explanation:The function f(x) is defined as the maximum of two functions: g(x) = 5x and h(x) = 52 – 2x². We are looking for the minimum possible value of f(x) for positive real numbers x.

The minimum value of the maximum of two functions typically occurs when the two functions are equal. Let’s find the point where g(x) = h(x):
5x = 52 – 2x²
2x² + 5x – 52 = 0

We can solve this quadratic equation for x using the quadratic formula:
x = [-b ± sqrt(b² – 4ac)] / 2a
Here, a = 2, b = 5, and c = -52.

x = [-5 ± sqrt(5² – 4 * 2 * -52)] / (2 * 2)
x = [-5 ± sqrt(25 + 416)] / 4
x = [-5 ± sqrt(441)] / 4
x = [-5 ± 21] / 4

We have two possible values for x:
x1 = (-5 + 21) / 4 = 16 / 4 = 4
x2 = (-5 – 21) / 4 = -26 / 4 = -6.5

Since x must be a positive real number, we consider x = 4.

Now, let’s evaluate both functions at x = 4:
g(4) = 5 * 4 = 20
h(4) = 52 – 2 * (4)² = 52 – 2 * 16 = 52 – 32 = 20

At x = 4, both functions are equal to 20. Therefore, f(4) = max(20, 20) = 20.

To confirm that this is a minimum, let’s consider the behavior of the two functions. g(x) = 5x is an increasing linear function. h(x) = 52 – 2x² is a downward-opening parabola.
For x < 4, g(x) < h(x) (e.g., at x=1, g(1)=5, h(1)=50, f(1)=50).
For x > 4, g(x) > h(x) (e.g., at x=5, g(5)=25, h(5)=52-2*25=2, f(5)=25).

So, for x < 4, f(x) = h(x) = 52 - 2x². This is a decreasing function for positive x.
For x > 4, f(x) = g(x) = 5x. This is an increasing function.

The function f(x) decreases until x=4 and then increases. Thus, the minimum value of f(x) occurs at x=4, and the minimum value is 20.

Final_Answer:20

Ans: A

Explanation:We are given two functions $f(x) = \dfrac{5x+2}{3x-5}$ and $g(x) = x^2 – 2x – 1$. We need to find the value of $g(f(f(3)))$.

First, let’s find the value of $f(3)$:
$f(3) = \dfrac{5(3)+2}{3(3)-5} = \dfrac{15+2}{9-5} = \dfrac{17}{4}$

Next, let’s find the value of $f(f(3))$, which is $f\left(\dfrac{17}{4}\right)$:
$f\left(\dfrac{17}{4}\right) = \dfrac{5\left(\dfrac{17}{4}\right)+2}{3\left(\dfrac{17}{4}\right)-5} = \dfrac{\dfrac{85}{4}+2}{\dfrac{51}{4}-5}$
To simplify the numerator and denominator, find a common denominator:
Numerator: $\dfrac{85}{4} + 2 = \dfrac{85}{4} + \dfrac{8}{4} = \dfrac{93}{4}$
Denominator: $\dfrac{51}{4} – 5 = \dfrac{51}{4} – \dfrac{20}{4} = \dfrac{31}{4}$
So, $f\left(\dfrac{17}{4}\right) = \dfrac{\dfrac{93}{4}}{\dfrac{31}{4}} = \dfrac{93}{31} = 3$

Now, we need to find the value of $g(f(f(3)))$, which is $g(3)$:
$g(3) = (3)^2 – 2(3) – 1 = 9 – 6 – 1 = 2$

Thus, $g(f(f(3))) = 2$.

Comparing this result with the given options:
Option A: 2
Option B: $\dfrac{1}{3}$
Option C: 6
Option D: $\dfrac{2}{3}$

The value of $g(f(f(3)))$ is 2, which matches Option A.

The final answer is $\boxed{2}$.
Correct_Option:A

Ans: C

Explanation:We are given two functions, $f(x) = x^2$ and $g(x) = 2^x$. We need to find the value of $f[f(g(x)) + g(f(x))]$ at $x = 1$.

First, let’s evaluate the inner functions at $x=1$:
$g(x) = 2^x$
At $x=1$, $g(1) = 2^1 = 2$.

$f(x) = x^2$
At $x=1$, $f(1) = 1^2 = 1$.

Now, let’s find $f(g(x))$ and $g(f(x))$ at $x=1$.
$f(g(x))$ at $x=1$ means $f(g(1))$. Since $g(1)=2$, we have $f(2)$.
$f(2) = 2^2 = 4$.

$g(f(x))$ at $x=1$ means $g(f(1))$. Since $f(1)=1$, we have $g(1)$.
$g(1) = 2^1 = 2$.

Now, we need to find $f(g(x)) + g(f(x))$ at $x=1$.
This is $f(g(1)) + g(f(1)) = 4 + 2 = 6$.

Finally, we need to find $f[f(g(x)) + g(f(x))]$ at $x = 1$.
This is $f(6)$.
$f(6) = 6^2 = 36$.

Therefore, the value of $f[f(g(x)) + g(f(x))]$ at $x = 1$ is 36.

Let’s double-check the calculations.
Given $f(x) = x^2$ and $g(x) = 2^x$. We want to evaluate $f[f(g(x)) + g(f(x))]$ at $x=1$.

Step 1: Calculate $g(1)$.
$g(1) = 2^1 = 2$.

Step 2: Calculate $f(g(1))$.
$f(g(1)) = f(2) = 2^2 = 4$.

Step 3: Calculate $f(1)$.
$f(1) = 1^2 = 1$.

Step 4: Calculate $g(f(1))$.
$g(f(1)) = g(1) = 2^1 = 2$.

Step 5: Calculate $f(g(1)) + g(f(1))$.
$f(g(1)) + g(f(1)) = 4 + 2 = 6$.

Step 6: Calculate $f[f(g(1)) + g(f(1))]$.
$f[6] = 6^2 = 36$.

The value is 36.

Comparing with the options:
Option A: 16
Option B: 18
Option C: 36
Option D: 40

The calculated value matches Option C.

The final answer is $\boxed{36}$.
Correct_Option:C

Ans: 1

For the expression f(1 * 1) = f(1)f(1), we can simplify this to f(1) = f(1)f(1).
This equation implies that f(1) must be either 0 or 1.
Therefore, the greatest possible value for f(1) is 1.