# The the value of 'a' for which $f$ is onto.

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}$$\boldsymbol{And} \boldsymbol{Trial\Longrightarrow}Taking the values of a from the options and calculating the f(x) and predicting the answer on behalf of these results. • So what is it's minimum? – Vim Oct 6 '17 at 7:29 • @Vim minimum value of 'a' should be 1 – Kislay Tripathi Oct 6 '17 at 7:33 ## 3 Answers If function f is onto then (6,\infty) must be its image. But does there exist any quadratic function with this image? No, hence d) is the correct answer here. The image of a quadratic function has shape [c,\infty) or (-\infty,c] where c\in\mathbb R denotes a constant. Can you find out why yourself? Just solve the following equation:$$f\left(\frac{a-3}{2}\right)=6,$$but in all case you'll get$[6,+\infty)$, which says that the answer is$(d)$. • but how can we fix the x at some value when the function is supposed to be onto for whatever x we may choose. – Kislay Tripathi Oct 6 '17 at 7:49 • @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. – Michael Rozenberg Oct 6 '17 at 8:10 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.$x^2-(a-3)x+a+6=6x^2-(a-3)x+a=0\Delta=(a-3)^2-4a = a^2-10a+9 = (a-1)(a-9) \ge 0$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$a=0$] • But$6\notin(6,\infty)\$. – drhab Oct 6 '17 at 7:48
• What's the evidence for your last statement "f isn't a function"? – Vim Oct 6 '17 at 8:09