Question

f(x) and g(x) be two functions such that f(g(x))=x then slope of line of symmetry of f(x) and g(x) is.​

Answer 1

Step-by-step explanation:

f(x+y)=f(x)+f(y)

Consider f(x)=e

x

+e

−x

Now f(−x)=e

−x

+e

x

=f(x)

Since

f(−x)=f(x), hence it is an even function.

Therefore it is symmetric about the y axis.

However f(x) intersects the y axis at y=1.

Hence the above f(x) is symmetric about the point (0,1).

Therefore it is not symmetric about the origin.

Now consider

f(x)=ln(x)

As x→0

f(x)→−∞

Furthermore the domain of f(x) is x>0.

The graph of f(x)=ln(x) decreases steeply in the region (0,1) as compared to the interval x>0.

Hence the graph f(x)=lnx is not symmetric about the origin.

Thirdly consider

f(x+y)=f(x)+f(y)

Then f(x) is of the form f(x)=λx where λ is a constant.

Now this represents a straight line passing through the origin, having a slope λ.

Hence it is symmetric about the origin.

Answer 2

Answer:

Correct option is

C

f(x+y)=f(x)+f(y)

Consider f(x)=ex+e−x

Now f(−x)=e−x+ex=f(x)

Since

f(−x)=f(x), hence it is an even function.

Therefore it is symmetric about the y axis.

However f(x) intersects the y axis at y=1.

Hence the above f(x) is symmetric about the point (0,1).

Therefore it is not symmetric about the origin.

Now consider

f(x)=ln(x)

As x→0

f(x)→−∞

Furthermore the domain of f(x) is x>0.

The graph of f(x)=ln(x) decreases steeply in the region (0,1) as compared to the interval x>0.

Hence the graph f(x)=lnx is not symmetric about the origin.

Thirdly consider

f(x+y)=f(x)+f(y)

Then f(x) is of the form f(x)=λx where λ is a constant.

Now this represents a straight line passing through the origin, having a slope λ.

Hence it is symmetric about the origin.

Step-by-step explanation:

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