Official Quant thread for CAT 2013

vbhvgupta · March 19, 2013, 7:02am

The expression f(x) is cubic in x in which the coefficient of x^3 is 1. If f(1) = 5 and f(2) = 8 and f(3) = 11 find f(4)?

rnishant231 · March 19, 2013, 7:03am

@Subhashdec2 said:
area of the triangle is 1/2 *10sqrt(2)*10sqrt(2)=100it is the same as area of a square with sides 10draw a square with x=0 y=0 x=10 and y=10no of integral points will be 11*11=121

Is it true for all the cases..
I mean if area of the figure comes out to be non integer and we can't transform the figure into a square ??

subhashdec2 · March 19, 2013, 7:08am

@rnishant231 said:
Is it true for all the cases..I mean if area of the figure comes out to be non integer and we can't transform the figure into a square ??

then try a rectangle....

i use square and rectangle and such figures coz it is easy that way

there would be a formula or something but i dont know it...

amresh_maverick · March 19, 2013, 7:08am

@vbhvgupta said:
The expression f(x) is cubic in x in which the coefficient of x^3 is 1. If f(1) = 5 and f(2) = 8 and f(3) = 11 find f(4)?

f(x) =x^3 -6x^2 +14x -4

f(4)=20

2013ismine · March 19, 2013, 7:09am

@vbhvgupta said:
The expression f(x) is cubic in x in which the coefficient of x^3 is 1. If f(1) = 5 and f(2) = 8 and f(3) = 11 find f(4)?

x^3+bx^2+cx+d

b+c+d=4 from 1

4b+2c+d=0 from 2

9b+3c+d=-16 from 3

subtarcting 1 and 2

3b + c =-4 ....4

subtracting 2 and 3

5b +c =-16.........5

subtracting 4 and 5

2b=-12...b=-6

c=14

d=-4

so f(x)= x^3-6x^2+14x-4

f(4)= 64-96+56-4=64-40-4=20??

vj12 · March 19, 2013, 7:09am

@vbhvgupta

@vbhvgupta said:
The expression f(x) is cubic in x in which the coefficient of x^3 is 1. If f(1) = 5 and f(2) = 8 and f(3) = 11 find f(4)?

20

the_loser · March 19, 2013, 7:11am

@albiesriram said:

FIGURE IS TRIANGLE (as given)

base line would entail 21 points ( -10 to +10)

a line above this base line(into the given triangle) would entail 19 points(-9 to + 9)

.

..

so on

So here we just need to find the summation of 1st 11 odd numbers = 1+3+5+.... +21

= 121

subhashdec2 · March 19, 2013, 7:12am

@vbhvgupta said:
The expression f(x) is cubic in x in which the coefficient of x^3 is 1. If f(1) = 5 and f(2) = 8 and f(3) = 11 find f(4)?

b+c+d=1

8+4b+2c+d=8

4b+2c+d=0

27+9b+3c+d=11

9b+3c+d=-16

5b+c=-16

3b+c=-1

2b=-15=>b=-15/2

c=-1+45/2

c=43/2

d=1+15/2-43/2 = -13

64-15/2 *16 +43/2 *4 -13

64-120 +86-13

17?

vbhvgupta · March 19, 2013, 7:14am

@Subhashdec2 said:
b+c+d=18+4b+2c+d=84b+2c+d=027+9b+3c+d=119b+3c+d=-165b+c=-163b+c=-12b=-15=>b=-15/2c=-1+45/2c=43/2d=1+15/2-43/2 = -1364-15/2 *16 +43/2 *4 -1364-120 +86-1317?

20 bhai.

subhashdec2 · March 19, 2013, 7:15am

@vbhvgupta said:
20 bhai.

koi calc mistake ho gae hogi yaar....

method yahi h

tusharrr · March 19, 2013, 7:17am

Bhai ye karo Na

ABC is a triangle with ∠BAC=60∘,AB=5 and AC=25. D is a point on the internal angle bisector of ∠BAC such that BD=DC. What is AD^2?
Details and assumptions
It is not stated that D lies on BC. This assumption is not necessarily true.

vj12 · March 19, 2013, 8:19am

@vbhvgupta

@vbhvgupta said:
If one of the root of the equation x^2-10x+16=0 is half of one of the root of the equation x^2-4Rx+16=0 Find R such that both the equation have integral roots

Edit:

One root of the 'first' equation is half of one root of the second equation not the other way. Just wrote it so that noone else makes the same mistake.

adiflexs007 · March 19, 2013, 8:25am

@Tusharrr 300 ?

Cosine rule...(Hint).

bodhi_vriksha · March 19, 2013, 8:30am

@Tusharrr said:
Bhai ye karo NaABC is a triangle with ∠BAC=60∘,AB=5 and AC=25. D is a point on the internal angle bisector of ∠BAC such that BD=DC. What is AD^2?Details and assumptionsIt is not stated that D lies on BC. This assumption is not necessarily true.

Using cosine rule in triangle ABD and ACD and equating BD² and CD², we get

25 + AD² - 10*AD*cos30 = 625 + AD² - 50*AD*cos30

i.e. 40*AD*cos30 = 600

i.e. AD = 30/rt(3) = 10rt(3)

i.e. AD² = 300.

I hope there is no calculational blunder. Pls check :)

Team BV - Kamal Lohia

rnishant231 · March 19, 2013, 8:33am

@vbhvgupta said:
If one of the root of the equation x^2-10x+16=0 is half of one of the root of the equation x^2-4Rx+16=0 Find R such that both the equation have integral roots

roots of 1st eq = 8,2

so second eqn should have roots either 16 or 4

a+b =4R
ab =16

16 = 1*16 or 2*8 or 4*4

only 4*4 satisfies both condition

so (x-4)^2

R =2 ??

chirpibird · March 19, 2013, 8:45am

@Tusharrr said:
Bhai ye karo NaABC is a triangle with ∠BAC=60∘,AB=5 and AC=25. D is a point on the internal angle bisector of ∠BAC such that BD=DC. What is AD^2?Details and assumptionsIt is not stated that D lies on BC. This assumption is not necessarily true.

300

mailtoankit · March 19, 2013, 8:48am

@vbhvgupta said:
If one of the root of the equation x^2-10x+16=0 is half of one of the root of the equation x^2-4Rx+16=0 Find R such that both the equation have integral roots

R = 2 ?

x^2 -10x + 16 = 0

x = 2,8

a + b = 4R

ab = 16

a = 4....so b = 4

R = 2...satisfies x^2 - 4*2x + 16 = 0 ---> x^2 -8x + 16 = 0

chirpibird · March 19, 2013, 8:50am

@vbhvgupta said:
If one of the root of the equation x^2-10x+16=0 is half of one of the root of the equation x^2-4Rx+16=0 Find R such that both the equation have integral roots

R=2?

chirpibird · March 19, 2013, 8:54am

@vbhvgupta said:
The expression f(x) is cubic in x in which the coefficient of x^3 is 1. If f(1) = 5 and f(2) = 8 and f(3) = 11 find f(4)?

20

mailtoankit · March 19, 2013, 9:01am

@vbhvgupta said:
The expression f(x) is cubic in x in which the coefficient of x^3 is 1. If f(1) = 5 and f(2) = 8 and f(3) = 11 find f(4)?

20 ?

f(x) = x^3 + bx^2 + cx + d

f(1) = 1 + b + c + d = 5....i

f(2) = 8 + 4b + 2c + d = 8 -----> 4b + 2c + d = 0....ii

f(3) = 27 + 9b + 3c + d = 11 ---> 9b + 3c + d = -16....iii

solve for b,c and d

b = -6 , c = 14 , d = -4

f(x) = x^3 - 6x^2 + 14x - 4 ---> f(4) = 4^3 - 6(4)^2 + 14*4 - 4 = 20