IIT JEE , AIEEE Forum - Mathematics , Differential Calculus - Range

ravikumar

Find the range of f(x) = 1+ sin x + (sinx)³ + (sin x)⁵........

where x belongs to (- pie /2, pie/2)

(Hint: 1+ sin x + (sinx)³ + (sin x)⁵........ in G.P., so we can use sum to infinite terms of a G.P.)

bhuwanaroracorroded

f(x)=sec²x

aditi_g

@ bhuwanaroracorroded how does f(x) come out to be sec^2 x???
f(x) will be 1 + secx tanx

pannaguma

f(x) = a/1-r = 1/1-sin^2x = 1/cos^2x = sec^2x

ayshwarya

d range is (-infinty, infinty )

rishabh29288

y = 1 + sinx/1-sinx^2 = 1 + secxtanx
the derivative = secx^3 + secxtanx^2
which is positive for -pi/2 to pi/2
therfore lowest value at -pi/2 and max at pi/2
the range = (-infinity,infinity)

ash_speed

looking at it logically...
sin x max. value is 1 n min value is -1..
as the power of sin x is always odd ...there is no change in the disturbance in sign..
so logically..
the range must b from -infinity to infinity..

hsbhatt

As pointed out by bhuwanesh, S = sec²x whose range is [1,

)

hsbhatt

As pointed out by bhuwanesh, S = sec²x whose range is [1,

)

master_purav

f(x) = a/1-r = 1/1-sin x

1/f(x) = 1 - sin x

Now sin x is between -1 and 1

1/f(x) will be maximum if sin is minimum. So max val = 1-(-1) = 2

1/f(x) will be minimum if sin is maximum. That is for zero.

0 <= 1/f(x) <= 2

so 1/2 <= f(x) <= infinity

So f(x) lies between [1/2, infinity)

hsbhatt

Sorry for being careless in reading the qn.

f(x) = 1+ sin x + (sinx)³ + (sin x)⁵........ = 1+tanxsecx. No limits.

ash_speed's observation also gives the same result

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