{"id":35371,"date":"2019-09-26T17:51:50","date_gmt":"2019-09-26T12:21:50","guid":{"rendered":"https:\/\/cracku.in\/blog\/?p=35371"},"modified":"2019-09-26T17:51:50","modified_gmt":"2019-09-26T12:21:50","slug":"functions-and-graphs-questions-for-cat-set-3-pdf","status":"publish","type":"post","link":"https:\/\/cracku.in\/blog\/functions-and-graphs-questions-for-cat-set-3-pdf\/","title":{"rendered":"Functions and Graphs Questions for CAT Set-3 PDF"},"content":{"rendered":"

Functions and Graphs Questions for CAT Set-3 PDF<\/strong><\/span><\/h2>\n

Download important CAT Functions and Graphs Set-3 Questions with Solutions PDF based on previously asked questions in CAT exam. Practice Functions and Graphs Set-3 Questions with Solutions for CAT exam.<\/p>\n

Download Functions and Graphs Questions for CAT Set-3 PDF<\/a><\/p>\n

Crash course to Cracku CAT<\/a><\/p>\n

Download CAT Quant Questions PDF<\/a><\/p>\n

Take Free Mock Test for CAT<\/a><\/p>\n

Question 1:\u00a0<\/b>Let $f(x) = ax^2 + bx + c$, where a, b and c are certain constants and $a \\neq 0$ ?It is known that $f(5) = – 3f(2)$. and that 3 is a root of $f(x) = 0$.What is the other root of f(x) = 0?[CAT 2008]<\/p>\n

a)\u00a0-7<\/p>\n

b)\u00a0– 4<\/p>\n

c)\u00a02<\/p>\n

d)\u00a06<\/p>\n

e)\u00a0cannot be determined<\/p>\n

Question 2:\u00a0<\/b>Let $f(x)\\neq0$ for any ‘x’ be a function satisfying $f(x)f(y) = f(xy)$ for all real x, y. If $f(2) = 4$, then what is the value of $f(\\frac{1}{2})$?<\/p>\n

a)\u00a00<\/p>\n

b)\u00a01\/4<\/p>\n

c)\u00a01\/2<\/p>\n

d)\u00a01<\/p>\n

e)\u00a0cannot be determined<\/p>\n

Question 3:\u00a0<\/b>The function f(x) = |x – 2| + |2.5 – x| + |3.6 – x|, where x is a real number, attains a minimum at<\/p>\n

a)\u00a0x = 2.3<\/p>\n

b)\u00a0x = 2.5<\/p>\n

c)\u00a0x = 2.7<\/p>\n

d)\u00a0None of the above<\/p>\n

Question 4:\u00a0<\/b>If $f(x)=x^3-4x+p$ , and f(0) and f(1) are of opposite signs, then which of the following is necessarily true[CAT 2004]<\/p>\n

a)\u00a0-1 < p < 2<\/p>\n

b)\u00a00 < p < 3<\/p>\n

c)\u00a0-2 < p < 1<\/p>\n

d)\u00a0-3 < p < 0<\/p>\n

Question 5:\u00a0<\/b>For all non-negative integers x and y, f(x, y) is defined as below:
\nf(0, y) = y + 1
\nf(x + 1, 0) = f(x, 1)
\nf(x+ 1, y+ 1)= f(x, f(x+ 1, y))
\nThen, what is the value of f(1,2)?<\/p>\n

a)\u00a0Two<\/p>\n

b)\u00a0Four<\/p>\n

c)\u00a0Three<\/p>\n

d)\u00a0Cannot be determined<\/p>\n

Take free mock tests for CAT<\/a><\/p>\n

Instructions<\/b><\/p>\n

DIRECTIONS for the following questions: These questions are based on the situation given below: In each of the questions\u00a0a pair of graphs F(x) and F1(x) is given. These are composed of straight-line segments, shown as solid lines, in the domain $x\\epsilon (-2, 2)$.<\/p>\n

choose the answer as<\/p>\n

a. If F1(x) = – F(x)<\/p>\n

b. if F1(x) = F(- x)<\/p>\n

c. if F1(x) = – F(- x)<\/p>\n

d.\u00a0if none of the above is true<\/p>\n

Question 6:\u00a0<\/b>\"\"
\n\"\"<\/p>\n

a)\u00a0a<\/p>\n

b)\u00a0b<\/p>\n

c)\u00a0c<\/p>\n

d)\u00a0d<\/p>\n

Question 7:\u00a0<\/b>\"\"
\n\"\"<\/p>\n

a)\u00a0a<\/p>\n

b)\u00a0b<\/p>\n

c)\u00a0c<\/p>\n

d)\u00a0d<\/p>\n

Question 8:\u00a0<\/b>\"\"
\n\"\"<\/p>\n

a)\u00a0a<\/p>\n

b)\u00a0b<\/p>\n

c)\u00a0c<\/p>\n

d)\u00a0d<\/p>\n

Question 9:\u00a0<\/b>\"\"\"\"<\/p>\n

a)\u00a0a<\/p>\n

b)\u00a0b<\/p>\n

c)\u00a0c<\/p>\n

d)\u00a0d<\/p>\n

Question 10:\u00a0<\/b>The figure below shows the graph of a function f(x). How many solutions does the equation f(f(x)) = 15 have?<\/p>\n

<\/figure>\n

a)\u00a05<\/p>\n

b)\u00a06<\/p>\n

c)\u00a07<\/p>\n

d)\u00a08<\/p>\n

e)\u00a0cannot be determined from the given graph<\/p>\n

Crash course to Cracku CAT<\/a><\/p>\n

Answers & Solutions:<\/strong><\/span><\/p>\n

1)\u00a0Answer\u00a0(B)<\/strong><\/p>\n

f(3) = 9a + 3b + c = 0 f(5) = 25a + 5b + c
\nf(2) = 4a + 2b + c
\nf(5) = -3f(2) => 25a + 5b + c = -12a -6b -3c
\n=> 37a + 11b + 4c = 0 –> (1)
\n4(9a + 3b + c) = 36a + 12b + 4c = 0 –> (2)
\nFrom (1) and (2), a – b = 0 => a = b
\n=> c = -12a
\nThe equation is, therefore, $ax^2 + ax – 12a = 0 => x^2 + x – 12 = 0$
\n=> -4 is a root of the equation.<\/p>\n

2)\u00a0Answer\u00a0(B)<\/strong><\/p>\n

$f(1)^2$ = f(1) => f(1) = 1<\/p>\n

f(2)*(f(1\/2) = f(1) => 4x = 1<\/p>\n

So, f(1\/2) = 1\/4<\/p>\n

3)\u00a0Answer\u00a0(B)<\/strong><\/p>\n

f(x) = |x – 2| + |2.5 – x| + |3.6 – x|<\/p>\n

For x belonging to (-infinity to 2), f(x) = 2-x + 2.5-x + 3.6-x = 8.1-3x<\/p>\n

This attains the minimum value at x=2. Value = 2.1<\/p>\n

For x belonging to (2 to 2.5), f(x) = x-2 + 2.5-x + 3.6-x = 4.1-x<\/p>\n

Attains the minimum value at x = 2.5. Value = 1.6<\/p>\n

For x belonging to (2.5 to 3.6), f(x) = x-2 + x-2.5 + 3.6-x = x-0.9<\/p>\n

Attains the minimum at x=2.5, value = 1.6<\/p>\n

For x > 3.6, f(x) = x-2+x-2.5+x-3.6 = 3x – 8.1<\/p>\n

Attains the minimum at x= 3.6, value = 2.7<\/p>\n

So, min value of the function is 1.6 at x=2.5<\/p>\n

4)\u00a0Answer\u00a0(B)<\/strong><\/p>\n

f(1) = 1-4+p = p-3
\nf(0) = p
\nSince they are of opposite signs, p(p-3) < 0
\n=> 0 < p < 3<\/p>\n

5)\u00a0Answer\u00a0(B)<\/strong><\/p>\n

For f(1,2). First consider x=0 and y=1 and use 3rd given equation, we get f(0,f(1,1)) now for f(1,1) take x=0 and y=0 we get f(0,f(1,0)), for f(1,0) which we use 2nd equation we get f(0,1) whose value is 2. So we have f(0,f(1,0))= f(0,2) whose value is 3 then put this in f(0,f(1,1)) we get f(0,3) we get as 4<\/p>\n

Download CAT Quant Questions PDF<\/a><\/p>\n

Take a free mock test for CAT<\/a><\/p>\n

6)\u00a0Answer\u00a0(D)<\/strong><\/p>\n

The correct relation between the two is:\u00a0F(x) = | F1(x) |<\/p>\n

So, all the three options a), b) and c) can be ruled out. Option d) is the correct answer.<\/p>\n

7)\u00a0Answer\u00a0(B)<\/strong><\/p>\n

The value of F(x) for x < 0 is the same as the value of F1(x) for x > 0.<\/p>\n

So, F1(x) = F(-x)<\/p>\n

Option b) is the correct answer.<\/p>\n

8)\u00a0Answer\u00a0(B)<\/strong><\/p>\n

The value of F(x) for x > 0 is the same as the value of F1(x) for x < 0.<\/p>\n

So, F1(x) = F(-x)<\/p>\n

Option b) is the correct answer.<\/p>\n

9)\u00a0Answer\u00a0(C)<\/strong><\/p>\n

F(0) = 1 ; F1(0) = -1<\/p>\n

F(1) = 0 ; F1(-1) = 0<\/p>\n

F(2) = -1 ; F1(-2) = 1<\/p>\n

=> F1(x) = -F(-x).<\/p>\n

10)\u00a0Answer\u00a0(C)<\/strong><\/p>\n

It has been given that f(f(x)) = 15.
\nFrom the graph, we can see that the value of f(4) = 15 and f(12) = 15
\nTherefore, f(x) can be 4 or 12.<\/p>\n

<\/figure>\n

When f(x) = 4, x can take 4 values
\nWhen f(x) = 12, x can take 3 values.
\nTherefore, there are 4+3 = 7 solutions in total.
\nTherefore, option\u00a0C is the right answer.<\/p>\n

Download CAT Previous Papers PDF<\/a><\/p>\n

Download Free CAT Preparation App<\/a><\/p>\n

We hope this Functions and Graphs Set-3 Questions for CAT with Solutions PDF will be helpful to you.<\/p>\n","protected":false},"excerpt":{"rendered":"

Functions and Graphs Questions for CAT Set-3 PDF Download important CAT Functions and Graphs Set-3 Questions with Solutions PDF based on previously asked questions in CAT exam. Practice Functions and Graphs Set-3 Questions with Solutions for CAT exam. Download CAT Quant Questions PDF Take Free Mock Test for CAT Question 1:\u00a0Let $f(x) = ax^2 + […]<\/p>\n","protected":false},"author":42,"featured_media":35378,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"om_disable_all_campaigns":false,"_mi_skip_tracking":false,"footnotes":""},"categories":[3,169,125,350,362,366],"tags":[1713,2753],"class_list":{"0":"post-35371","1":"post","2":"type-post","3":"status-publish","4":"format-standard","5":"has-post-thumbnail","7":"category-cat","8":"category-downloads","9":"category-featured","10":"category-iift","11":"category-snap","12":"category-xat","13":"tag-cat-2019","14":"tag-functions-and-graphs-set-3-questions-for-cat"},"better_featured_image":{"id":35378,"alt_text":"Functions and Graphs Questions for CAT Set-3 PDF","caption":"Functions and Graphs Questions for CAT Set-3 PDF\n","description":"Functions and Graphs Questions for CAT Set-3 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