Question

1
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1. Problem: If f:R\{0} → R is defined by f(x) = x+- then prove that
(f (x)) = f (x?) + f (1).​

Answer 1

Answer:

I hope it will help you

Explanation:

[−12,12][-12,12]

(−1,1)−{0}(-1,1)-{0}

R−[−12,12]R-[-12,12]

R−[−1,1]R-[-1,1]

Answer :

A

Solution :

We have,f(x)=x1+x2,x∈Rf(x)=x1+x2,x∈R

Ist Method f(x) is an odd function and maximum occur at x = 1

From the graph it is clear that range of f(x) is

[−12,12][-12,12]

IInd Method f(x)=1x+1xf(x)=1x+1x

If x>0x>0, then by AM≥GMAM≥GM, we get x+1x≥2x+1x≥2

⇒1x+1x≤12⇒0<f(x)≤12⇒1x+1x≤12⇒0<f(x)≤12

If x<0x<0, then by AM≥GMAM≥GM, we get x+1x≤−2x+1x≤-2

⇒1x+1x≥−12⇒−12≤f(x)≤0⇒1x+1x≥-12⇒-12≤f(x)≤0

If x = 0, then f(x)=01+0=0f(x)=01+0=0

Thus, −12≤f(x)≤12-12≤f(x)≤12

Hence, f(x)∈[−12,12]f(x)∈[-12,12]

IIIrd Method

Let y=x1+x2⇒yx2−x+y=0y=x1+x2⇒yx2-x+y=0

∵x∈R, so D≥0∵x∈R, so D≥0

⇒1−4y2≥0⇒1-4y2≥0

⇒(1−2y)(1+2y)≥0⇒y∈[−12,12]⇒(1-2y)(1+2y)≥0⇒y∈[-12,12]

So, range is [−12,12][-12,12].