Functions (XII)

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Question

$N u m b e r o f i n t e g e r a l v a l u e s o f x f o r w h i c h |||x^{2} - x + 4| - 2| - 3| = x^{2} + x - 12 i s$

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Solution

$|||x^{2} - x + 4| - 2| - 3| = x^{2} + x - 12 \Rightarrow ||x^{2} - x + 2| - 3| = x^{2} + x - 12$
$\Rightarrow |x^{2} - x - 1| = x^{2} + x - 12 \Rightarrow - (x^{2} - x - 1) = x^{2} + x - 12 o r (x^{2} - x - 1) = x^{2} + x - 12 \Rightarrow 2 x^{2} = - 13 o r 2 x = 11$
$n o t p o s s i b l e ∴ x = \frac{11}{2} h e n c e o n l y o n e s o l u t i o n$

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Functions (XII)

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Number of integeral values of x for which x2-x+4-2-3 = x2+x-12 is

x2-x+4-2-3 = x2+x-12 ⇒x2-x+2-3= x2+x-12 ⇒x2-x-1= x2+x-12 ⇒-( x2-x-1)= x2+x-12 or (x2-x-1)=x2+x-12 ⇒2x2=-13 or 2x=11not possible ∴x=112 hence only one solution

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$N u m b e r o f i n t e g e r a l v a l u e s o f x f o r w h i c h |||x^{2} - x + 4| - 2| - 3| = x^{2} + x - 12 i s$