Let $f (x) = x^{2} - 3 x + 4$ , the value(s) of $x$ which satisfies $f (1) + f (x) = f (1) f (x)$ are:

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- 2

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Solution

The correct option is D $1$ or $2$
$f (x) = x^{2} - 3 x + 4$
Now
$f (1) = 1 - 3 + 4 = 5 - 3$
$= 2$ .
Hence
$f (1) + f (x) = f (1) f (x)$ implies
$2 + (x^{2} - 3 x + 4) = 2 (x^{2} - 3 x + 4)$
$x^{2} - 3 x + 6 = 2 x^{2} - 6 x + 8$
$0 = 2 x^{2} - x^{2} - 6 x + 3 x + 8 - 6$
$0 = x^{2} - 3 x + 2$
$x^{2} - 3 x + 2 = 0$
$(x - 2) (x - 1) = 0$
$x = 1, 2$
Hence the values of $x$ are $1$ or $2$ .

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Let f(x)=x2−3x+4 , the value(s) of x which satisfies f(1)+f(x)=f(1)f(x) are:

The correct option is D 1 or 2f(x)=x2−3x+4Now f(1)=1−3+4=5−3 =2.Hence f(1)+f(x)=f(1)f(x) implies 2+(x2−3x+4)=2(x2−3x+4)x2−3x+6=2x2−6x+80=2x2−x2−6x+3x+8−60=x2−3x+2x2−3x+2=0(x−2)(x−1)=0x=1,2Hence the values of x are 1 or 2.

Let $f (x) = x^{2} - 3 x + 4$ , the value(s) of $x$ which satisfies $f (1) + f (x) = f (1) f (x)$ are: