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4-2.Quadratic Equations and Inequations
hard

$x$ ની બધી જ વાસ્તવિક કિંમતો માટે $\frac{x}{{{x^2}\, + \,4}}$ ની કિંમતનો વિસ્તાર કેટલો થશે ?

A

$\frac{{ - 1}}{2}\,\, \le \,\,y\,\, \le \,\,\frac{1}{2}$

B

$\frac{{ - 1}}{4}\,\, \le \,\,y\,\, \le \,\,\frac{1}{4}$

C

$\frac{{ - 1}}{6}\,\, \le \,\,y\,\, \le \,\,\frac{1}{6}$

D

આપેલ પૈકી એકપણ નહિ.

Solution

$\,\,\frac{{\rm{x}}}{{{{\rm{x}}^{\rm{2}}}\, + \,\,4}} =y$ લો.

$\,\, \Rightarrow {\rm{ }}{{\rm{x}}^{\rm{2}}}{\rm{y  –  x  +  4y  =  0}}$

હવે $ {\rm{x }} \in {\rm{  R}}$

$ \Rightarrow {\rm{ }}{{\rm{B}}^{\rm{2}}}{\rm{  –  4AC }} \ge {\rm{  0 }}$ 

$ \Rightarrow {\rm{ }}\,{\rm{1  –  4y}}{\rm{. 4y }} \ge \,0$

${\rm{(4y  –  1) (4y  +  1) }} \le {\rm{  0 }}$

$\therefore \,\,\,\frac{{ – 1}}{4}\,\, \le \,\,y\,\, \le \,\,\frac{1}{4}$

Standard 11
Mathematics

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