Official Quant thread for CAT 2013

MBA CAT 2022
1381
@sowmyanarayanan said:
@krum@sauravd2001@rkshtsurana@techsurgeSolution says y = 2^sinx + 2^cosxFrom symmetry, y would have minimum value when sin x = cos x = -1/_/2so b)Can anyone explain what the bold line means?
see my solution and attachment as graph of function
1382
@sowmyanarayanan said:
OA says b) share your approach. let us try coming to the answer from there
i drew a graph, generally max hota hai jab equal hoon
pi/4 par


graph mein same par negative happens on pi + pi/4
so lowest case here
1383
If sin x, cos x and tan x are in G.P., then cot^6 x €“cot^2 x =
(a) 0 (b) €“1
(c) 1 (d) Depends on x
no OA
1384
@sowmyanarayanan said:
@krum@sauravd2001@rkshtsurana@techsurgeSolution says y = 2^sinx + 2^cosxFrom symmetry, y would have minimum value when sin x = cos x = -1/_/2so b)Can anyone explain what the bold line means?
bhai ye samajh nahi aa raha to differentitate karke jo solution nikala hai, use refer kar lo
trigonometry and graphs mein main bhi weak hoon
1385
@ASHISH_mantra said:
@billuisback@YouMadFellow@[email protected]@vijay_chandola@onlytjPlease help by Solving this Distance Puzzleiwillanswer.in/1634/solve-this...
54
1386
@catter2011 bhai aa gaya samjh kya galti kar raha tha n iske liye i have a thinking reply if m wrong from 4 quadrants we have only in third quadrant both cos and sin are -ve so simply we could have put 225 m i right or wrong?
1387
@ASHISH_mantra said:
@billuisback@YouMadFellow@[email protected]@vijay_chandola@onlytjPlease help by Solving this Distance Puzzleiwillanswer.in/1634/solve-this...
54 Miles.
1388
@ASHISH_mantra said:
@billuisback@YouMadFellow@[email protected]@vijay_chandola@onlytjPlease help by Solving this Distance Puzzleiwillanswer.in/1634/solve-this...
54 Miles.

Teh bird will travel for the time taken by two friends to meet. So, total distance will be 18* 3, 54.
1389
@sowmyanarayanan said:
@krum@sauravd2001@rkshtsurana@techsurgeSolution says y = 2^sinx + 2^cosxFrom symmetry, y would have minimum value when sin x = cos x = -1/_/2so b)Can anyone explain what the bold line means?
min in 3rd quad when both sin and cos are -ve

=>2^(-1/root)+2^(-1/root2)
=>2/2^(1/root2)
=>2^(1-1/root2)
=>2^(root2*root2-1*root2)/2
=>_/2^(2 - _/2)
1390

The sum of series below is


1/1.2.3 + 1/3.4.5 +1/5.6.7......


a) e^2-1
b) 2log2 -1 [base 10]
c) log2 -1
d) None of these


1391
@sauravd2001 said:
@catter2011 bhai aa gaya samjh kya galti kar raha tha n iske liye i have a thinking reply if m wrong from 4 quadrants we have only in third quadrant both cos and sin are -ve so simply we could have put 225 m i right or wrong?
sahi hein.. 3rd q brings min valur for sinx + cosx and atmost minima at 225 deg
1392
@rkshtsurana said:
If sin x, cos x and tan x are in G.P., then cot^6 x €“cot^2 x =(a) 0 (b) €“1(c) 1 (d) Depends on x no OA
i get the final equation to be 1/sin^2 x - 1/cos x
so answer d) ?
1393
@rkshtsurana said:
If sin x, cos x and tan x are in G.P., then cot^6 x 창€“cot^2 x =(a) 0 (b) 창€“1(c) 1 (d) Depends on x no OA
1

given cos^2 x = sin x * tan x
=>cos^3 x = sin^2 x

cot^6 x 창€“cot^2 x
=>(cos^2 x)/(sin^2 x)*(cos^4 x/sin^4 x -1)
=>1/cos x*(1/cos^2 x -1)
=>1/cos x*(sin^2 x/cos^2x)
=>sin^2 x/cos^3 x
=>1

@sowmyanarayanan
1394

1/2( 1/2 - 1/n(n+1) ]

n tends to infinity
roughly 1/4 = 0.25
options
a) 6.xx
b and c are -ve

so d) none of these ?
@techsurge @techsurge said:The sum of series below is1/1.2.3 + 1/3.4.5 +1/5.6.7......a) e^2-1b) 2log2 -1 [base 10]c) log2 -1d) None of these
1395
@krum said:
1
how?
1396
@krum said:
1
approach..no OA
@sowyanarayanan
1397
@techsurge said:
The sum of series below is1/1.2.3 + 1/3.4.5 +1/5.6.7......a) e^2-1b) 2log2 -1 [base 10]c) log2 -1d) None of these
none??? 1/4
1398
@rkshtsurana said:
approach..no OA@sowyanarayanan
check ^^^, edited earlier
1399
@rkshtsurana said:
If sin x, cos x and tan x are in G.P., then cot^6 x –cot^2 x =(a) 0 (b) –1(c) 1 (d) Depends on x no OA

@rkshtsurana said:
If sin x, cos x and tan x are in G.P., then cot^6 x –cot^2 x =(a) 0 (b) –1(c) 1 (d) Depends on x no OA
1

on solving we get sin^2x =cos^3x

cot^6x-cot^2x=

cos^6x/sin^6x- cos^2x/sin^2x
substitute sin^2x = cos^3x

1/cos^3x-1/cosx
sin^2x/cos^3x=1
1400
@rkshtsurana said:
approach..no OA@sowyanarayanan
=> cos x = _/ (tan x * sin x)
=> cos^3 x = sin^2 x

solving it in cot^6 x - cot^2 x get 1/sin^2 x - 1/cos x