IIT JEE , AIEEE Forum - Mathematics , Differential Calculus

tarun_bits

find the range of following functions........

1.... f(x)= / sin x / + / cos x /.......... x = [0, pie ]

2...... $f(x) = {sin}^{2}x - 5sinx - 6$

yaar show the whole method and....ya....every effort gets a rate..

:)

tarun_bits

note that............/ gf / denotes mode of gf....

allamraju

Is the ans to the 2nd question [-10,0].If correct,I will explain.

And the ans for 1st question is [1, $\sqrt{2}$ ].Is it correct?

risin

1. f(x)=sinx+cosx= $\sqrt{2}sin(\Pi/4 \;+x) \;for\;xE[0,\Pi/2]=>1<=f(x)<=\sqrt{2}\\f(x)=sinx-cosx=>\sqrt{2}(sin(x-\Pi/4)\;for\;xE[\Pi/2,\Pi]=>-1<=f(x)<=1$

or 1<=f(x)<=root(2)

allamraju

@risin,in second quadrant also,sinx is +ve,so,sin(x-pi/4) cant take the value of -1/rt2 for x in[pi/2,pi].So,I think the ans is [1,rt2]

spideyunlimited

2. f(x) = sin^2 x - 5 sinx - 6

f(x) = sin^2 x - 5 sinx - 6
= sin^2 x - 6 sinx + sinx - 6
= (sin x +1) (sin x - 6)
max value is when sin x = -1 ..... ie. 0
min. value is when sin x = 1 ....ie. -10

so range is [-10, 0]

risin

@allamraju:sorry i telling 1 to myself and typed -1..thanks.

rahul1993

Re:help in range of functions

oneyeartogo

tarun_bits

Answers are
for
1......[ 1 , root of 2 ]
2......[ -10 , + 10 ]

@risin how you got root 2 ( sin pi/4 + x)....i am getting pi/4 as @....
and then
r00t 2 <=root 2 sin( x + @) <= root 2

please tell where am i wrong

tarun_bits

for second ques me getting same but answer is differnet

allamraju

For 2nd question [-10,10] can't be the ans.(put the expression=10 then we don't get any soln.)

For 1st question,split the interval into [0,pi/2] and [pi/2,pi]

In [0,pi/2],f(x)=sinx+cosx= $\sqrt{2}$ sin(x+pi/4).Now,0 $\le$ x $\le$ pi/2 $\Rightarrow$ pi/4 $\le$ x+pi/4 $\le$ 3pi/4 $\Rightarrow$ 1 $\le$ sin(x+pi/4) $\le$ $\sqrt{2}$ .Similarly in [pi/2,pi],f(x)=sinx-cosx= $\sqrt{2}$ sin(x-pi/4).Now,pi/2 $\le$ x $\le$ pi $\Rightarrow$ pi/4 $\le$ x-pi/4 $\le$ 3pi/4 $\Rightarrow$ 1 $\le$ x $\le$ $\sqrt{2}$ .Hence the range is [1, $\sqrt{2}$ ]

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Is the ans to the 2nd question [-10,0].If correct,I will explain.

And the ans for 1st question is [1,].Is it correct?

@risin,in second quadrant also,sinx is +ve,so,sin(x-pi/4) cant take the value of -1/rt2 for x in[pi/2,pi].So,I think the ans is [1,rt2]

For 2nd question [-10,10] can't be the ans.(put the expression=10 then we don't get any soln.)

For 1st question,split the interval into [0,pi/2] and [pi/2,pi]

In [0,pi/2],f(x)=sinx+cosx=sin(x+pi/4).Now,0xpi/2pi/4x+pi/43pi/41sin(x+pi/4).Similarly in [pi/2,pi],f(x)=sinx-cosx=sin(x-pi/4).Now,pi/2xpipi/4x-pi/43pi/41x.Hence the range is [1,]

And the ans for 1st question is [1, $\sqrt{2}$ ].Is it correct?