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Provide solution for RD Sharma maths class12 Chapter Functions exercise 2.1 question 3

Answers (1)

f:N \rightarrow N, given byf(x)=x^2+x+1is one-one but not onto.


Let x and y be any two elements in the domain (N), such that f(x)=f(y)

So,f is one-one.


f:N \rightarrow N,  defined byf(x)=x^2+x+1

Let us prove that the given function is one-one.


We havef(x)=x^2+x+1                                                                                                                 … (i)

Calculate f(y)=y^2+x+1                                                                                                              … (ii)

Now equate equation (i) and (ii)

  \begin{aligned} &f(x)=f(y) \\ &x^{2}+x+1=y^{2}+y+1 \\ &x^{2}+x-y^{2}-y=0 \end{aligned}

\begin{array}{r} x^{2}-y^{2}+x-y=0 \\ {\left[\because a^{2}-b^{2}=(a+b)(a-b)\right]} \\ (x+y)(x-y)+x-y=0 \\ (x-y)[(x+y+1)]=0 \end{array}

From (iii) equation, we can write

x-y=0, x+y+1\neq 0

x+y+1\neq 0 , because if we substitutex=1, y= 1


Hence, x+y+1\neq 0 for anyx\in N

Where as,

x-y =0


So,f is a one-one function.

But, f(x)=x^3+x+1\geq 3 for all x\in N

So,f(x) doesn’t assume values of1 and 2

\therefore f is not an onto function.

Hence, we proved that, f(x)=x^2+x+1 is one-one function not an onto function.

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