Set A={2,3,5,7,11,13} so |A|=6Set B={1, 8, 27} so |B|=3Without any restrictions, each element in A can map to any of the 3 elements in B. Thus, the total number of functions is: $3^6=729$ Excluding Functions That Miss One Element in B:&nbsp;If a function does not map to an element in B, there are 2 elements in B left for mapping. The total number of such functions (for each specific element not mapped) is: $2^6=64$Since there are 3 elements in B, the total number of such functions is: 3x64=192Adding Back Functions That Miss Two Elements in B: If a function misses two elements in B, there is only 1 element left for mapping. The total number of such functions is: $1^6=1$&nbsp;Since there are&nbsp;$^3C_2$ ways to choose which two elements are missed, the total number of such functions is: 3Using the inclusion-exclusion principle, the number of functions where all elements of B are mapped by at least one element of A is:729-192+3=540.&nbsp;

Looking at the additional information about the prime numbers should make one realise that they are the key to solving the question.&nbsp;f(16000) can be written as&nbsp;$f\left(2^8\times\ 5^4\right)$Now, we can try to find these individual values:For any prime p: f(p)=1$f\left(p^2\right)=f\left(p\right)f\left(p\right)+f\left(p\right)+f\left(p\right)=1+1+1=3$$f\left(p^3\right)=f\left(p^2\right)f\left(p\right)+f\left(p^2\right)+f\left(p\right)=3+3+1=7$This way, we can find the function output for any prime number raised to a power.&nbsp;We can see that each new exponent is twice the previous output +1, solving this way till prime raised to power 8$f\left(p^4\right)=7+7+1=15$$f\left(p^5\right)=15+15+1=31$$f\left(p^6\right)=31+31+1=63$$f\left(p^7\right)=63+63+1=127$$f\left(p^8\right)=127+127+1=255$Using these values in the original expression of&nbsp;$f\left(2^8\times\ 5^4\right)=f\left(2^8\right)f\left(5^4\right)+f\left(2^8\right)+f\left(5^4\right)$ we get$f\left(2^8\times\ 5^4\right)=\left(255\times\ 15\right)+255+15=4095$Therefore, Option A is the correct answer.&nbsp;

The expression on the right-hand side will have two critical points: 2 and -2For any value of x greater than equal to 2, the equation changes to x+2-(x-2) = 4the value of min{x,2} would be 2, so we would want max{x,2} to be 4+2 = 6Therefore, x=6 works.&nbsp;For any value of x less than equal to -2,&nbsp;the equation changes to -(x+2)+(x-2) = -4the value of max{x,2} would be 2, so we would want min{x,2} to be 6 again; this cannot be the case.In general, subtracting the smaller (min) of the two values from the bigger (max) can not lead to a negative number. The max it can lead to is a 0When x lies between -2 and 2,&nbsp; the equation becomes 2xThe maximum function will give back 2, and the minimum function will give back x, with the right-hand side giving 2xSolving this we would get 2-x = 2x which is x=2/3Therefore, there are two real values of x for which the given equation holds.&nbsp;Hence, 2 is the correct answer.The graph of the function on the right hand side can be visualised as:<img src="https://lightyellow-stork-291267.hostingersite.com/wp-content/uploads/2025/11/new.image_mPxCmPT.png">Hence, 2 is the correct answer.&nbsp;

We are given,&nbsp;$f(x) + 2f \left(\cfrac{1}{x}\right) = 3x$Substituting&nbsp;$\frac{1}{x}\ for\ x$$f\left(\dfrac{1}{x}\right)+2f\left(x\right)=\dfrac{3}{x}$Multiplying the second equation by 2 we will have&nbsp;$2f\left(\dfrac{1}{x}\right)+4f\left(x\right)=\dfrac{6}{x}$Subtracting the first equation from the second equation we have,&nbsp;$3f\left(x\right)=\frac{6}{x}-3x$$f\left(x\right)=\frac{2}{x}-x$We want the sum of values when this function equals 3,&nbsp;$\frac{2}{x}-x=3$$x^2+3x-2=0$Since the discriminant is greater than zero, both values of x will be real, and we can directly take the sum of values of $x$ to be,&nbsp;$-\frac{3}{1}$Answer is -3.&nbsp;

Given that f(3x + 2y, 2x - 5y) = 19x.Let us assume the function f(a, b) is a linear combination of a and b.=&gt; f(3x+2y, 2x-5y) = m(3x+2y) + n(2x-5y) = 19x=&gt; 3m + 2n = 19 and 2m - 5n = 0Solving we get m = 5 and n = 2=&gt; f(a, b) = 5a+2b=&gt; f(x, 2x) = 5x + 2(2x) = 9x = 27 =&gt; x = 3.

It is given,$\sum_{n=1}^{N} \left[ \frac{1}{5} + \frac{n}{25} \right] = 25$$\sum_{n=1}^{N} \left[ \frac{5 + n}{25} \right] = 25$For n = 1 to n = 19, value of function is zero.For n = 20 to n = 44, value of function will be 1.$44 = 20 + n - 1$n = 25 which is equal to given value.This implies N = 44

x&gt;=a, so |x-a| = x-ax&lt;100, so |x-100| = 100-xf(x) = (x-a) + (100-x) + |x-a-50| =100or, |x-a-50| = a <img src="https://lightyellow-stork-291267.hostingersite.com/wp-content/uploads/2025/11/last.Screenshot_20221212_134752.png">From the graph we can can see that when x=a then|x-a-50|=aor, a= 50Similarly when x=a+100|x-a-50|=aor, a= 50So value of a is 50 when f(x) is 100.

Given,f(x) + f(x−1) = 1 ...... (1)f($x^2$−x) = 5 ...... (2)g(x) = $x^2$Substituting x = 1 in (1) and (2), we getf(0) = 5f(1) + f(0) = 1f(1) = 1 - 5 = -4f(2) + f(1) = 1f(2) = 1 + 4 = 5f(n) = 5 if n is even and f(n) = -4 if n is oddf(g(5)) + g(f(5)) = f(25) + g(-4) = -4 + 16 = 12

$f(x) \ge 0$ for all real numbers $x$, so $D \le 0$Since $f(2) = 0$ therefore $x = 2$ is a root of $f(x)$Since the discriminant of $f(x)$ is less than equal to 0 and 2 is a root so we can conclude that $D = 0$Therefore $f(x) = a(x - 2)^2$$f(4) = 6$or, $6 = a(x - 2)^2$$a = \frac{3}{2}$

When x&lt; rf(x) = rf(x) = f(f(x))r = f(r)r= 2r-rr=rWhen x&gt;=rf(x) = 2x-rf(x) = f(f(x))2x-r = f(2x-r)2x-r = 2(2x-r) - r2x-r = 4x-3ror, x=rTherefore x&lt;= r

Now we have:$f(g(x))−3x$so we get $f(x+3)-3x$ = $(x+3)^2 - 7(x+3) - 3x$= $x^2 - 4x - 12$Now minimum value of expression = $-\frac{D}{4a}\frac{4ac-b^2}{4a}$ We get - $(16+48)/4$ = -16

<img src="https://lightyellow-stork-291267.hostingersite.com/wp-content/uploads/2025/11/6.image_3ehaJXH.png"> The graphs of&nbsp;$2^x+2^{−x} \text{ and } 2−(x−2)^2$&nbsp;never intersect. So, number of solutions=0.Alternate method:We notice that the minimum value of the term in the LHS will be greater than or equal to&nbsp;2 {at x=0;&nbsp;LHS = 2}. However, the term in the RHS is less than or equal to 2 {at x=2; RHS = 2}. The values of x&nbsp;at which both the sides become 2 are distinct; hence, there are zero real-valued solutions to the above equation.

<img src="https://lightyellow-stork-291267.hostingersite.com/wp-content/uploads/2025/11/16.image_D2ggvra.png">The area of the region contained by the lines&nbsp;∣ x ∣ − y ≤ 1, y ≥ 0 and&nbsp;y ≤ 1 is the white region.Total area = Area of rectangle + 2 * Area of triangle =&nbsp;2+($\frac12$ ×&nbsp;2 ×&nbsp;1)&nbsp;=3Hence, 3 is the correct answer.

Let x(1) be the least number and x(10) be the largest number. Now from the condition given in the question , we can say thatx(2)+x(3)+x(4)+........x(10)= 47*9=423...................(1)Similarly x(1)+x(2)+x(3)+x(4)................+x(9)= 42*9=378...............(2)Subtracting both the equations we get x(10)-x(1)=45Now, the sum of the 10 observations from equation (1) is 423+x(1)Now the minimum value of x(10) will be 47 and the minimum value of x(1) will be 2 . Hence minimum average 425/10=42.5Maximum value of x(1) is 42. Hence maximum average will be 465/10=46.5Hence difference in average will be 46.5-42.5=4 which is the correct answer

f(x) =&nbsp;$x^2$ + ax + b$f(x+1)=x^2+2x+1+ax+a+b$$f(x−1)=x^2−2x+1+ax−a+b$g(x)=f(x+1)−f(x−1)=4x+2aNow&nbsp;g(20)=72&nbsp;from this we get&nbsp;a=−4;&nbsp;f(x) = $x^2$ − 4x&nbsp;+ bFor this expression to be greater than zero it has to be a perfect square&nbsp;which is possible for&nbsp;b≥&nbsp;4Hence the smallest value of 'b'&nbsp;is 4.

<img src="https://lightyellow-stork-291267.hostingersite.com/wp-content/uploads/2025/11/image_0fyXIqx.png">The required figure is a trapezium with vertices A(0,0), B(2,0), C(2,4) and D(0,6)AB = 2 BC = 4 and AD = 6Area of trapezium = $\frac12$(sum of opposite sides) x height = $\frac12(4+6) \times 2 = 10$

CAT Functions, Graphs and Statistics

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Question 2.

Question 3.

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Question 6.

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Question 8.

Question 9.

Question 10.

Question 11.

Question 12.

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Free CAT Quant Questions

CAT Functions, Graphs and Statistics

Question 1.

<div><p>Consider two sets $A = \left\{2, 3, 5, 7, 11, 13 \right\}$ and $B = \left\{1, 8, 27 \right\}$. Let f be a function from A to B such that for every element in B, there is at least one element a in A such that $f(a) = b$. Then, the total number of such functions f is</p></div>

Question 2.

<div><p>A function f maps the set of natural numbers to whole numbers, such that f(xy) = f(x)f(y) + f(x) + f(y) for all x, y and f(p) = 1 for every prime number p. Then, the value of f(160000) is</p></div>

Question 3.

<div><p>The number of distinct real values of x, satisfying the equation $max \left\{x, 2\right\} - min\left\{x, 2\right\} = \mid x + 2 \mid - \mid x - 2 \mid$, is</p></div>

Question 4.

<div><p>For any non-zero real number x, let $f(x) + 2f \left(\cfrac{1}{x}\right) = 3x$. Then, the sum of all possible values of x for which $f(x) = 3$, is</p></div>

Question 5.

<div><p>The area of the quadrilateral bounded by the Y -axis, the line x = 5 , and the lines |x &minus; y| &minus; |x &minus; 5| = 2 , is</p></div>

Question 6.

<div><p>Suppose f(x, y) is a real-valued function such that f(3x + 2y, 2x &minus; 5y) = 19x, for all real numbers x and y. The value of x for which f(x, 2x) = 27, is</p></div>

Question 7.

<div><p>For any real number x , let [x] be the largest integer less than or equal to x . If $\sum_{n=1}^N\left[\frac{1}{5}+\frac{n}{25}\right]=25$ then N is</p></div>

Question 8.

<div><p>Let 0 &le; a &le; x &le; 100 and f(x) = |x &minus; a| + |x &minus; 100| + |x &minus; a &minus; 50| . Then the maximum value of f(x) becomes 100 when a is equal to</p></div>

Question 9.

<div><p>Suppose for all integers x, there are two functions f and g such that f(x) + f(x &minus; 1) &minus;1 = 0 and $g(x)=x^2$. If $f\left(x^2-x\right)=5$, then the value of the sum f(g(5)) + g(f(5)) is</p></div>

Question 10.

<div><p>Let f(x) be a quadratic polynomial in x such that f(x) &ge; 0 for all real numbers x. If f(2) = 0 and f(4) = 6, then f(&minus;2) is equal to</p></div>

Question 11.

<div><p>Let r be a real number and $f(x)=\left\{\begin{array}{cl}2 x-rㅤ\text { if } x \geq r \\ rㅤ ㅤ \text { if } x\lt r\end{array}\right.$ Then, the equation f(x) = f(f(x)) holds for all real values of x where</p></div>

Question 12.

<div><p>If $f(x)=x^{2}-7 x$ and g(x) = x + 3 , then the minimum value of f(g(x)) &minus; 3x is</p></div>

Question 13.

<div><p>The number of real-valued solutions of the equation $2^x$ + $2^{-x}$ = 2 - $(x - 2)^2$ is</p></div>

Question 14.

<div><p>The area of the region satisfying the inequalities |x| - y &le; 1, y &ge; 0, and y &le; 1 is</p></div>

Question 15.

<div><p>In a group of 10 students, the mean of the lowest 9 scores is 42 while the mean of the highest 9 scores is 47. For the entire group of 10 students, the maximum possible mean exceeds the minimum possible mean by</p></div>

Question 16.

<div><p>Let f(x) = $x^{2}$ + ax + b and g(x) = f(x + 1) - f(x - 1). If f(x) &ge; 0 for all real x, and g(20) = 72, then the smallest possible value of b is</p></div>

Question 17.

<div><p>The area, in sq. units, enclosed by the lines x = 2, y = |x - 2| + 4, the X-axis and the Y-axis is equal to</p></div>

Free CAT Quant Questions