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Question Number 58816 by Tawa1 last updated on 30/Apr/19

	Answered by Kunal12588 last updated on 30/Apr/19

	$f r o m l o o k i n g t h e g r a p h w e c a n s a y t h a t$ $m a x i m a o f f u n c t i o n i s a t x = 2$ $f o r A)$ $\frac{d (f (x))}{d x} = 0 [f o r m i n i m a o r m a x i m a]$ $- \frac{1}{3} \times 2 (x - 2) (1) = 0$ $\Rightarrow x = 2$ $o p t i o n A c a n b e c o r r e c t$ $f o r B)$ $\frac{d (f (x))}{d x} = 0$ $- \frac{1}{3} \times 2 (x + 2) (1) = 0$ $\Rightarrow x = - 2$ $o p t i o n B i s n o t c o r r e c t$ $f o r C)$ $\frac{d (f (x))}{d x} = 0$ $\frac{1}{3} \times 2 (x + 2) (1) = 0$ $\Rightarrow x = - 2$ $o p t i o n C i s n o t c o r r e c t$ $f o r D)$ $\frac{d (f (x))}{d x} = 0$ $6 (x - 2) (1) = 0$ $\Rightarrow x = 2$ $∴ o n l y A o r D o p t i o n s c a n b e c o r r e c t$ $n o w l e t s f i n d t h e z e r o e s o f A a n d D$ $A . - \frac{1}{3} {(x - 2)}^{2} + 5 = 0$ $\Rightarrow (x - 2) = \pm \sqrt{- (5) (- 3)}$ $\Rightarrow x = 2 \pm \sqrt{15}$ $D . 3 {(x - 2)}^{2} + 5 = 0$ $\Rightarrow (x - 2) = \pm \sqrt{- \frac{5}{3}}$ $\Rightarrow x = 2 \pm z$ $\Rightarrow x \notin R$ $B u t z e r o e s o f t h e e q u a t i o n i n t h e g r a p h a r e R e a l$ $∴ O n l y A i s c o r r e c t$ $A n s . A . f (x) = - \frac{1}{3} {(x - 2)}^{2} + 5$
	Commented by Kunal12588 last updated on 30/Apr/19

	Commented by tanmay last updated on 30/Apr/19

	$b a h e x c e l l e n t . . . v e r y g o o d . .$
	Commented by Tawa1 last updated on 30/Apr/19

	$God bless you sir$
	Commented by Kunal12588 last updated on 30/Apr/19

	$I t w a s m y p l e a s u r e . I t i s m y f i r s t t r y o f$ $s u c h q u e s t i o n s .$
	Answered by MJS last updated on 30/Apr/19

	${it}^{'} s a hanging parabola \Rightarrow$ $\Rightarrow f (x) = {a x}^{2} + b x + c with a < 0 \Rightarrow$ $\Rightarrow options C, D {don}^{'} t fit$ $A x^{2} - 4 x - 11 = 0 \Rightarrow - \frac{p}{2} = 2 \Rightarrow \max at 2$ $B x^{2} + 4 x - 11 = 0 \Rightarrow - \frac{p}{2} = - 2 \Rightarrow \max at - 2$ $\Rightarrow option A$
	Commented by Tawa1 last updated on 30/Apr/19

	$God bless you sir$
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