Download Graphs Questions for XAT<\/a><\/p>\n Enroll to XAT 2023 Crash Course<\/a><\/p>\n Question 1:\u00a0<\/b>Let r be a real number and $f(x) = \\begin{cases}2x -r & ifx \\geq r\\\\ r &ifx < r\\end{cases}$. Then, the equation $f(x) = f(f(x))$ holds for all real values of $x$ where<\/p>\n a)\u00a0$x > r$<\/p>\n b)\u00a0$x \\leq r$<\/p>\n c)\u00a0$x \\neq r$<\/p>\n d)\u00a0$x \\geq r$<\/p>\n 1)\u00a0Answer\u00a0(B)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n When x< r<\/p>\n f(x) = r<\/p>\n f(x) =\u00a0f(f(x))<\/p>\n r = f(r)<\/p>\n r= 2r-r<\/p>\n r=r<\/p>\n When x>=r<\/p>\n f(x) = 2x-r<\/p>\n f(x) = f(f(x))<\/p>\n 2x-r = f(2x-r)<\/p>\n 2x-r = 2(2x-r) – r<\/p>\n 2x-r = 4x-3r<\/p>\n or, x=r<\/p>\n Therefore x<= r<\/p>\n Question 2:\u00a0<\/b>Suppose for all integers x, there are two functions f and g such that $f(x) + f (x – 1) – 1 = 0$ 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>\n 2)\u00a0Answer:\u00a012<\/b><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n Given,<\/p>\n $f\\left(x\\right)+f\\left(x-1\\right)=1$ …… (1)<\/p>\n $f\\left(x^2-x\\right)=5$ ……\u00a0 (2)<\/p>\n $g\\left(x\\right)=x^2$<\/p>\n Substituting x = 1 in (1) and (2), we get<\/p>\n f(0) = 5<\/p>\n f(1) + f(0) = 1<\/p>\n f(1) = 1 – 5 = -4<\/p>\n f(2) + f(1) = 1<\/p>\n f(2) = 1 + 4 = 5<\/p>\n f(n) = 5 if n is even and f(n) = -4 if n is odd<\/p>\n f(g(5)) + g(f(5)) = f(25) + g(-4) = -4 + 16 = 12<\/p>\n Question 3:\u00a0<\/b>Let $f(x)$ be a quadratic polynomial in $x$ such that $f(x) \\geq 0$ for all real numbers $x$. If f(2) = 0 and f( 4) = 6, then f(-2) is equal to<\/p>\n a)\u00a012<\/p>\n b)\u00a024<\/p>\n c)\u00a06<\/p>\n d)\u00a036<\/p>\n 3)\u00a0Answer\u00a0(B)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n $f(x) \\geq 0$for all real numbers $x$, so D<=0<\/p>\n Since f(2)=0 therefore x=2 is a root of f(x)<\/p>\n Since the discriminant of f(x) is less than equal to 0 and 2 is a root so we can conclude that D=0<\/p>\n Therefore f(x) =\u00a0$a\\left(x-2\\right)^2$<\/p>\n f(4)=6<\/p>\n or, 6 =\u00a0$a\\left(x-2\\right)^2$<\/p>\n a= 3\/2<\/p>\n $f\\left(-2\\right)=\\ -\\frac{3}{2}\\left(-4\\right)^2=24$<\/p>\n Question 4:\u00a0<\/b>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>\n 4)\u00a0Answer:\u00a044<\/b><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n It is given,<\/p>\n $\\Sigma_{n=1}^N\\ \\left[\\frac{1}{5}+\\frac{n}{25}\\right]=25$<\/p>\n $\\Sigma_{n=1}^N\\ \\left[\\frac{5+n}{25}\\right]=25$<\/p>\n For n = 1 to n = 19, value of function is zero.<\/p>\n For n = 20 to n = 44, value of function will be 1.<\/p>\n 44 = 20 + n – 1<\/p>\n n = 25 which is equal to given value.<\/p>\n This implies N = 44<\/p>\n Question 5:\u00a0<\/b>Let $0 \\leq a \\leq x \\leq 100$ and $f(x) = \\mid x – a \\mid + \\mid x – 100 \\mid + \\mid x – a – 50\\mid$. Then the maximum value of f(x) becomes 100 when a is equal to<\/p>\n a)\u00a025<\/p>\n b)\u00a0100<\/p>\n c)\u00a050<\/p>\n d)\u00a00<\/p>\n 5)\u00a0Answer\u00a0(C)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n x>=a, so |x-a| = x-a<\/p>\n x<100, so |x-100| = 100-x<\/p>\n f(x) = (x-a) + (100-x) + |x-a-50| =100<\/p>\n or, |x-a-50| = a<\/p>\n From the graph we can can see that when x=a then<\/p>\n |x-a-50|=a<\/p>\n or, a= 50<\/p>\n Similarly when x=a+100<\/p>\n |x-a-50|=a<\/p>\n or, a= 50<\/p>\n So value of a is 50 when f(x) is 100.<\/p>\n Take\u00a0 SNAP mock tests here<\/strong><\/span><\/a><\/p>\n Enrol to 10 SNAP Latest Mocks For Just Rs. 499<\/a><\/strong><\/span><\/p>\n Question 6:\u00a0<\/b>Given below are two statements: a)\u00a0Both Statement I and Statement II are correct<\/p>\n b)\u00a0Both Statement I and Statement II are incorrect<\/p>\n c)\u00a0Statement I is correct but Statement II is incorrect<\/p>\n d)\u00a0Statement I is incorrect but Statement II is correct<\/p>\n 6)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n Statement I:<\/p>\n Given arithmetic mean of\u00a0(7,8,9,a,b,10,8,7) is 9 and mode is 8<\/p>\n 7+8+9+a+b+10+8+7 =\u00a0$9\\times\\ 8$<\/p>\n 49+a+b = 72<\/p>\n a+b = 23<\/p>\n It is mentioned that mode is 8, 8 occurred twice and 7 also occurred twice. Therefore, a or b should be 8<\/p>\n If a = 8, b = 15<\/p>\n If b = 8, a = 15<\/p>\n ab = 120<\/p>\n Statement I is true<\/p>\n Statement II:<\/p>\n a + b + ab = 84<\/p>\n 1 + a + b + ab = 85<\/p>\n (1+a)(1+b) = 85<\/p>\n 85 = 1\u00a0$\\times\\ $ 85 or 5\u00a0$\\times\\ $ 17<\/p>\n It cannot be 1$\\times\\ $85, as it is mentioned both a and b are positive integers<\/p>\n If 1+a = 5, 1+b = 17<\/p>\n a =4 and b = 16<\/p>\n If 1+a = 17, 1+b = 5<\/p>\n a = 16 and b = 4<\/p>\n In both cases, a+b = 20<\/p>\n Statement II is true<\/p>\n Answer is option A.<\/p>\n Question 7:\u00a0<\/b>The following observations are arranged in ascending order: 26, 29, 42, 53, X, X+2, X+5, 75, 82, 93: If the median is 65; Find the value of X.<\/p>\n a)\u00a070<\/p>\n b)\u00a055<\/p>\n c)\u00a064<\/p>\n d)\u00a058<\/p>\n 7)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n There are even number of observations in the given data. Therefore, median is the average of X and X+2. Answer is option C.<\/p>\n Question 8:\u00a0<\/b>$f(x) = \\frac{2x + 2}{2x – 2}$, where y = f(x). Find the ratio of x to f(y).<\/p>\n a)\u00a0$\\sqrt{x} : \\sqrt{y}$<\/p>\n b)\u00a0$x^3 : y^3$<\/p>\n c)\u00a01 : 2<\/p>\n d)\u00a01 : 1<\/p>\n 8)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n $f\\left(y\\right)=f\\left(f\\left(x\\right)\\right)$<\/p>\n or,\u00a0$f\\left(y\\right)=f\\left(\\frac{2x+2}{2x-2}\\right)$<\/p>\n $f\\left(y\\right)=\\frac{2\\left(\\frac{2x+2}{2x-2}\\right)+2}{2\\left(\\frac{2x+2}{2x-2}\\right)-2}$<\/p>\n or,\u00a0$f\\left(y\\right)=\\frac{8x}{8}=x$<\/p>\n Therefore x:f(y) = 1:1<\/p>\n Question 9:\u00a0<\/b>If $f(x)=x^{2}-7x$ and $g(x)=x+3$, then the minimum value of $f(g(x))-3x$ is:<\/p>\n a)\u00a0-20<\/p>\n b)\u00a0-12<\/p>\n c)\u00a0-15<\/p>\n d)\u00a0-16<\/p>\n 9)\u00a0Answer\u00a0(D)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n Now we have : Question 10:\u00a0<\/b>Two sets AA and BB are defined as follows:- a)\u00a0n(AA) > n(BB)<\/p>\n b)\u00a0n(AA) < n(BB)<\/p>\n c)\u00a0AA = BB<\/p>\n d)\u00a0n(AA) $\\leq$ n(BB)<\/p>\n 10)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n AA= {1, 2, 3, 4}; n(AA)= 4<\/p>\n BB= {1, 2, 3, 4}; n(BB)=4<\/p>\n So, AA=BB<\/p>\n Question 11:\u00a0<\/b>Let $g(x)+g(\\frac{1}{x})=1+3x$. Find the value of $g(3)$.<\/p>\n a)\u00a0$-4$<\/p>\n b)\u00a0$-\\frac{1}{2}$<\/p>\n c)\u00a0$\\frac{1}{2}$<\/p>\n d)\u00a0$3$<\/p>\n 11)\u00a0Answer\u00a0(E)<\/strong><\/p>\n Solution:<\/b><\/p>\n The question appeared in CMAT 2021 Slot 2 and the question is incorrect . Marks were awarded to students.<\/p>\n Question 12:\u00a0<\/b>Which of the following equations best describes the graph given below? a)\u00a0$\\mid x+y\\mid-\\mid x-y\\mid=6$<\/p>\n b)\u00a0$\\mid x-y\\mid+\\mid x+y\\mid=10$<\/p>\n c)\u00a0$\\mid x\\mid-\\mid y\\mid=6$<\/p>\n d)\u00a0$\\mid x+y\\mid-\\mid x-y\\mid=0$<\/p>\n 12)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n In these type of questions Question 13:\u00a0<\/b>Which of the following graphs represent the function of x? Choose the correct answer from the options given below :<\/p>\n a)\u00a0(A), (B) and (C) only<\/p>\n b)\u00a0(A), (B) and (D) only<\/p>\n c)\u00a0(A) and (D) only<\/p>\n d)\u00a0(A) and (C) only<\/p>\n 13)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n Graph A and D represent the function of x, because a vertical line is drawn in (a) meets the graph at only one point i.e., for one x, in the domain there exist only one f(x) in the codomain.<\/p>\n Question 14:\u00a0<\/b>In how many ways can a pair of integers (x , a) be chosen such that $x^{2}-2\\mid x\\mid+\\mid a-2\\mid=0$ ?<\/p>\n a)\u00a06<\/p>\n b)\u00a05<\/p>\n c)\u00a04<\/p>\n d)\u00a07<\/p>\n 14)\u00a0Answer\u00a0(D)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n $x^{2}-2\\mid x\\mid+\\mid a-2\\mid=0$<\/p>\n where x>= 0 and x>=2<\/p>\n $x^2-2x+a-2\\ =0$ Using quadratic equation we have $x=\\ 1+\\sqrt{\\ 3-a}\\ and\\ x=1-\\sqrt{\\ 3-a}$ Only two integer values are possible<\/p>\n a=2 and a=3. So corresponding “x” values are x=1 and a=3, x=2 and a=2, x=0 and a=2<\/p>\n where x>=0 and x<2<\/p>\n Applying the above process we get x=1 and a=1<\/p>\n where x<0 and x>=2 we get a=3 and x=-1 , a=2 and x=-2<\/p>\n where x<0 and x<2 we get a=1 and x=-1<\/p>\n Hence there are total 7 values possible<\/p>\n Question 15:\u00a0<\/b>In a group of 10 students, the mean of the lowest 9 scores is 42 while the mean of the\u00a0highest 9 scores is 47. For the entire group of 10 students, the maximum possible\u00a0mean exceeds the minimum possible mean by<\/p>\n a)\u00a05<\/p>\n b)\u00a04<\/p>\n c)\u00a03<\/p>\n d)\u00a06<\/p>\n 15)\u00a0Answer\u00a0(B)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n 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 that<\/p>\n x(2)+x(3)+x(4)+……..x(10)= 47*9=423……………….(1)<\/p>\n Similarly x(1)+x(2)+x(3)+x(4)…………….+x(9)= 42*9=378……………(2)<\/p>\n Subtracting both the equations we get x(10)-x(1)=45<\/p>\n Now, the sum of the 10 observations from equation (1) is 423+x(1)<\/p>\n 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=42.5<\/p>\n Maximum value of x(1) is 42. Hence maximum average will be 465\/10=46.5<\/p>\n Hence difference in average will be 46.5-42.5=4 which is the correct answer<\/p>\n Question 16:\u00a0<\/b>Let $f(x)=x^{2}+ax+b$ and $g(x)=f(x+1)-f(x-1)$. If $f(x)\\geq0$ for all real x, and $g(20)=72$. then the smallest possible value of b is<\/p>\n a)\u00a016<\/p>\n b)\u00a04<\/p>\n c)\u00a01<\/p>\n d)\u00a00<\/p>\n 16)\u00a0Answer\u00a0(B)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n $f\\left(x\\right)=\\ x^2+ax+b$<\/p>\n $f\\left(x+1\\right)=x^2+2x+1+ax+a+b$<\/p>\n $f\\left(x-1\\right)=x^2-2x+1+ax-a+b$<\/p>\n $ g(x)=f(x+1)-f(x-1)= 4x+2a$<\/p>\n Now $g(20) = 72$ from this we get $a = -4$\u00a0;\u00a0$f\\left(x\\right)=x^2-4x\\ +b$<\/p>\n For this expression to be greater than zero it has to be a perfect square\u00a0which is possible for $b\\ge\\ 4$<\/p>\n Hence the smallest value of ‘b’<\/strong> is 4.<\/p>\n Question 17:\u00a0<\/b>If $f(x+y)=f(x)f(y)$ and $f(5)=4$, then $f(10)-f(-10)$ is equal to<\/p>\n a)\u00a014.0625<\/p>\n b)\u00a00<\/p>\n c)\u00a015.9375<\/p>\n d)\u00a03<\/p>\n 17)\u00a0Answer\u00a0(C)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n The given function is equivalent to f(x) = $a^x$<\/p>\n Given, f(5) = 4<\/p>\n => $a^5=4=>\\ a=4^{\\frac{1}{5}}$<\/p>\n => f(x) = $4^{\\frac{x}{5}}$<\/p>\n f(10) – f(-10) = 16 – 1\/16 = 15.9375<\/p>\n Question 18:\u00a0<\/b>The area, in sq. units, enclosed by the lines $x=2,y=\\mid x-2\\mid+4$, the X-axis and the Y-axis is equal to<\/p>\n a)\u00a010<\/p>\n b)\u00a06<\/p>\n c)\u00a08<\/p>\n d)\u00a012<\/p>\n 18)\u00a0Answer\u00a0(A)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n The required figure is a trapezium with vertices A(0,0), B(2,0), C(2,4) and D(0,6)<\/p>\n AB = 2 BC = 4 and AD = 6<\/p>\n Area of trapezium =\u00a0$\\frac{1}{2}\\left(sum\\ of\\ the\\ opposite\\ sides\\right)\\cdot height$ =\u00a0$\\frac{1}{2}\\left(4+6\\right)\\cdot2\\ =\\ 10$<\/p>\n Question 19:\u00a0<\/b>If $f(5+x)=f(5-x)$ for every real x, and $f(x)=0$ has four distinct real roots, then the sum of these roots is<\/p>\n a)\u00a00<\/p>\n b)\u00a040<\/p>\n c)\u00a010<\/p>\n d)\u00a020<\/p>\n 19)\u00a0Answer\u00a0(D)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n Let ‘r’ be the root of the function. It follows that f(r) = 0. We can represent this as $f\\left(r\\right)=f\\left\\{5-\\left(5-r\\right)\\right\\}$<\/p>\n Based on the relation: $f\\left(5-x\\right)=f\\left(5+x\\right)$; $f\\left(r\\right)=f\\left\\{5-\\left(5-r\\right)\\right\\}=f\\left\\{5+\\left(5-r\\right)\\right\\}$<\/p>\n $\\therefore\\ f\\left(r\\right)=f\\left(10-r\\right)$<\/p>\n Thus, every root ‘r’ is associated with another root ‘(10-r)’ [these form a pair]. For even distinct roots, in this case four, let us assume the roots to be as follows: $r_1,\\ \\left(10-r_1\\right),\\ r_2,\\ \\left(10-r_2\\right)$<\/p>\n The sum of these roots = $r_1\\ +\\left(10-r_1\\right)+\\ r_2+\\ \\left(10-r_2\\right)\\ =\\ 20$<\/p>\n Hence, Option D is the correct answer.<\/p>\n Question 20:\u00a0<\/b>The area of the region satisfying the inequalities $\\mid x\\mid-y\\leq1,y\\geq0$ and $y\\leq1$ is<\/p>\n 20)\u00a0Answer:\u00a03<\/b><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n The area of the region contained by the lines $\\mid x\\mid-y\\leq1,y\\geq0$ and $y\\leq1$ is the white region.<\/p>\n Total area = Area of rectangle + 2 * Area of triangle =\u00a0$2+\\left(\\frac{1}{2}\\times\\ 2\\times\\ 1\\right)\\ =3$<\/p>\n Hence, 3 is the correct answer.<\/p>\n![](\"https:\/\/cracku.in\/media\/uploads\/Screenshot_20221212_134752.png\")
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\nStatement I : The set of numbers (7,8,9,a,b,10,8,7) has an arithmetic mean of 9 and mode(most frequently occurring number) as 8, Then a x b = 120.
\nStatement II : Let a and b be two positive integers such that a + b + a x b = 84, then a + b =20.
\nIn the light of the above statements, choose the most appropriate answer from the options given below<\/p>\n
\n$\\ \\frac{\\ X+X+2}{2}=65$
\nX + 1 = 65
\nX = 64<\/p>\n
\n$f(g(x))-3x$
\nso we get f(x+3)-3x
\n=\u00a0$\\left(x+3\\right)^2-7\\left(x+3\\right)-3x$
\n=$x^2-4x-12$
\nNow minimum value of expression = $-\\frac{D}{4a}$\u00a0$ \\frac{\\left(4ac-b^2\\right)}{4a}$
\nWe get – (16+48)\/4
\n= -16<\/p>\n
\n$AA = \\left\\{x \\mid x \\in N, x < 5\\right\\}$
\n$AA = \\left\\{x \\mid x \\in N, x2 < 25\\right\\}$, n(XX) = number of elements in XX. Choose the CORRECT option.<\/p>\n
\n![](\"https:\/\/cracku.in\/media\/uploads\/image_4OkcBv9.png\")
<\/p>\n
\nwe go through the options :
\nNow when x=6 or -6 the value of y is 0
\nso let us put value of y=0 in all options and check which option satisfies x=6 and -6
\nPutting x=6 and y=0 in option A
\nwe get |6+0|-|6-0| = 0
\nBut RHS =6 so this option is discarded
\nNow putting in option B
\n|6-0|+|6+0| =12
\nBut RHS is 10
\nso this option is also discarded .
\nNow option C we get
\n|6|-0 =6
\nso we can say graph can be of |x|-|y|=6
\nNow satisfying option D
\nwe get |6+0|-|6-0|=0
\nSo we can say 6,0 and -6,0 satisfies option D as well
\nNow you have to eliminate between C and D
\nIf you observe |x+y|-|x-y|=0 will pass through origin
\nand the graph here is not passing through origin so you can eliminate option D
\nTherefore the graph will be |x|-|y|=6<\/p>\n
\n![](\"https:\/\/cracku.in\/media\/uploads\/364944-a_FH8V4So.png\")
\n![](\"https:\/\/cracku.in\/media\/uploads\/364944-b_wzsv3V8.png\")
\n![](\"https:\/\/cracku.in\/media\/uploads\/364944-c.png\")
\n![](\"https:\/\/cracku.in\/media\/uploads\/364944-d.png\")
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