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Select problems from exercise 1.3

solutions to selected NCERT problems
In this article we will look at solutions of a few selected problems from exercise 1.3 of NCERT.
Problem 1:
Show that f:[1,1]R given by f(x)=xx+2 is one-one. Find the inverse of the function f:[1,1]Range f.
(Hint: For yRange f, y=f(x)=xx+2, for some x in [1,1], i.e. x=2y(1y) )
Let's first manipulate the function into a form more easier to read.
Note by this manipulation, we now have x at only one place instead of two in the expression.
Now, we will show that f is one-one. Let x1 and x2 be two different inputs to f having the same output.
12x1+2=12x2+22x1+2=2x2+2for x12 and x22 we can writex1+2=x2+2x1=x2
This means the two inputs are the same and our initial assumption is false. For the same output, we cannot have two different inputs. f is one-one.
Let us now find the inverse of f. The answer is already given in the question hint but we will look at it in a bit more detail here.
try it out
The first question is, what is the range of f? Remember f:[1,1]R, f(x)=12x+2.
Choose 1 answer:

So our function is f:[1,1][1,13].
Let y=f(x)=12x+2.
To find the inverse function, we simply express x in terms of y.
That is it. f1(y)=2y1y.
We had f:[1,1][1,13]. Can you show that for 1y13, 12y1y1 ?
Problem 2:
Consider f:R+[5,) given by f(x)=9x2+6x5. Show that f is invertible with f1(y)=y+613 .
Let's understand the behaviour of f(x) better by writing it in a perfect square form.
Our domain is R+ or (0,). See that as 0<x<+, 5<f(x)<+. Here is the graph.
Clearly f(x) is one-one and onto. So, f is invertible.
Let y=f(x)=(3x+1)26. To find the inverse, we simply express x in terms of y.
In the calculation above, we took only the positive square root, because 3x+1 is positive as xR+.
Finally, f1(y)=y+613 .

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