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f: $\mathbb{R}\rightarrow$(6,$\infty$) , $f(x) = x^{2} -(a-3)x +a+6,$ then the value of 'a' for which function is onto

(a) $(1,9)$

(b) $[1,9]$

(c) $\{1,9\}$

(d) None of these

$\boldsymbol{My}$$\boldsymbol{Approach}$$\Longrightarrow$ Using hit and trial method I can say (d) is correct.

$\boldsymbol{My}$$\boldsymbol{Question}$$\Longrightarrow$What is the proper algebraic way to tackle these kind of question.??

If using Hit andd trial is ideal method ??

$\boldsymbol{Hit}$$\boldsymbol{And}$ $\boldsymbol{Trial\Longrightarrow}$Taking the values of a from the options and calculating the f(x) and predicting the answer on behalf of these results.

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  • $\begingroup$ So what is it's minimum? $\endgroup$ – Vim Oct 6 '17 at 7:29
  • $\begingroup$ @Vim minimum value of 'a' should be 1 $\endgroup$ – Kislay Tripathi Oct 6 '17 at 7:33
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If function $f$ is onto then $(6,\infty)$ must be its image.

But does there exist any quadratic function with this image?

No, hence d) is the correct answer here.

The image of a quadratic function has shape $[c,\infty)$ or $(-\infty,c]$ where $c\in\mathbb R$ denotes a constant.

Can you find out why yourself?

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Just solve the following equation: $$f\left(\frac{a-3}{2}\right)=6,$$ but in all case you'll get $[6,+\infty)$, which says that the answer is $(d)$.

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  • $\begingroup$ but how can we fix the x at some value when the function is supposed to be onto for whatever x we may choose. $\endgroup$ – Kislay Tripathi Oct 6 '17 at 7:49
  • $\begingroup$ @Kislay Tripathi Because your $f$ works so $f:\mathbb R\rightarrow[k,+\infty)$ for some real $k$. If you wish to get $(6,+\infty)$ then it's impossible. $\endgroup$ – Michael Rozenberg Oct 6 '17 at 8:10
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As $f(x)$ opens upward, it suffices to find $a$ such that the minimum of the function is $6$, i.e. $f(x)=6$ has root.

$x^2-(a-3)x+a+6=6$

$x^2-(a-3)x+a=0$

$\Delta=(a-3)^2-4a = a^2-10a+9 = (a-1)(a-9) \ge 0$

Therefore, $a \le 1$ or $a \ge 9$. Rejecting $a<1$ and $a>9$ gives us (c).

[Note that your question is bs as $f$ isn't even a function when $a=0$]

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  • $\begingroup$ But $6\notin(6,\infty)$. $\endgroup$ – drhab Oct 6 '17 at 7:48
  • $\begingroup$ What's the evidence for your last statement "f isn't a function"? $\endgroup$ – Vim Oct 6 '17 at 8:09

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