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Critical Point

If A >0, c , ...

Question

If $A > 0, c, d, u, v$ are non-zero constants, and the graphs of $f (x) = | A x + c | + d$ and $g (x) = - | A x + u | + v$ intersect exactly at 2 points (1, 4) and (3, 1) then the value of $\frac{u + c}{A}$ equals to

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Solution

The correct option is B
-4

$f (x) = | A x + c | + d g (x) = - | A x + u | + v$
which are sides of parallelogram and the diagonals bisect each other
$∴ (- \frac{u}{A}) + (- \frac{c}{A}) = 3 + 1 \Rightarrow \frac{u + c}{A} = - 4$

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If A>0,c,d,u,v are non-zero constants, and the graphs of f(x)=|Ax+c|+d and g(x)=−|Ax+u|+v intersect exactly at 2 points (1, 4) and (3, 1) then the value of u+cA equals to

The correct option is B -4 f(x)=|Ax+c|+dg(x)=−|Ax+u|+v which are sides of parallelogram and the diagonals bisect each other ∴(−uA)+(−cA)=3+1⇒u+cA=−4

If $A > 0, c, d, u, v$ are non-zero constants, and the graphs of $f (x) = | A x + c | + d$ and $g (x) = - | A x + u | + v$ intersect exactly at 2 points (1, 4) and (3, 1) then the value of $\frac{u + c}{A}$ equals to

The correct option is B
-4

$f (x) = | A x + c | + d g (x) = - | A x + u | + v$
which are sides of parallelogram and the diagonals bisect each other
$∴ (- \frac{u}{A}) + (- \frac{c}{A}) = 3 + 1 \Rightarrow \frac{u + c}{A} = - 4$