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Q.

The range of a function y = f (x) is the set of all possible output values f (x) corresponding to every input x in the domain of f and is denoted as f (A) if A is the domain. For finding the range of a function y = f (x), first of all, find the domain of f.If the domain consists of finite number of points, then the range consists of set of corresponding f (x) values. If the domain consists of whole real line or real line minus some finite points, then express x in terms of y as x = g(y). The values of y for which g is defined is the required range.If the domain is a finite interval, find the intervals in which f (x) increases/decreases and then find the extreme values of the function in those intervals. The union of those intervals is the required range.If ex + ef(x) = e, then range of the function f isThe range of the function f(x)=sin⁡πx2+1x4+1. where [⋅] denotes the greatest integer function, isRange of values of f (x) = 1 + sinx + sin3x + sin5x+….., x ∈ −π2,π2 is

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a

(–∞, 1]

b

(–∞, 1)

c

(1, ∞)

d

[1, ∞)

e

[0, 1]

f

[–1, 1]

g

{0}

h

None of these

i

(0, 1)

j

(– ∞, ∞)

k

(– 2, 2)

l

None of these

answer is , , .

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Detailed Solution

We have,ex + ef(x) = e ⇒ ef(x) = e – ex⇒ f (x) = log (e – ex)For f (x) to be defined, e – ex > 0⇒ e1 > ex ⇒ x < 1∴ Domain of f = (– ∞, 1)Let y = log (e – ex) ⇒ ey = e – ex⇒ ex = e – ey⇒ x = log (e – ey)For x to be real, e – ey > 0⇒ e1 > ey ⇒ y < 1∴ Range of f = (– ∞, 1)Since [x2 + 1] is an integer,∴sin⁡πx2+1=0⇒f(x)=sin⁡πx2+1x4+1=0Hence, Range of f = Rf = {0}We have, f(x)=1+sin⁡xcos2⁡x⇒f′(x)=cos2⁡x(cos⁡x)+sin⁡x(2cos⁡xsin⁡x)cos4⁡x=cos⁡xcos2⁡x+2sin2⁡xcos4⁡x=1+sin2⁡xcos3⁡x⇒ f ′(x) > 0.∴ f (x) is increasing function.limx→−π2 1+sin⁡xcos2⁡x=−∞ and, limx→π2 1+sin⁡xcos2⁡x=∞∴ Range = (–∞, ∞)
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