Let fx-x2-1-0 - x - 2 2x-3-2 - x - 3- The quadratic equation whose roots are limx- 2-fx and limx- 2-fx is

The correct option is A x^2 10x+21=0 For 0 lt; x lt; 2 , f(x) = nbsp;x^2 1 lim_ x →2^ nbsp; f(x) = 3 For, 2 lt; x l ...

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Byju's Answer

Standard XII

Mathematics

Right Hand Limit

Let fx=x2-1...

Question

Let $f (x) = {\begin{matrix} x^{2} - 1, 0 < x < 2 2 x + 3, 2 \leq x < 3 \end{matrix},$
The quadratic equation whose roots are $lim x \to 2^{-} f (x)$ and $lim x \to 2^{+} f (x)$ is

x^{2} - 10 x + 21 = 0

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x^{2} - 6 x + 9 = 0

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x^{2} - 14 x + 49 = 0

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x^{2} + 6 x + 9 = 0

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Solution

The correct option is A $x^{2} - 10 x + 21 = 0$
For $0 < x < 2$ ,
$f (x) = x^{2} - 1$
${lim}_{x \to 2^{-}} f (x) = 3$
For, $2 < x < 3$
$f (x) = 2 x + 3$
${lim}_{x \to 2^{+}} f (x) = 7$
Thus the equation with roots, 3 and 7 is
$x^{2} - 10 x + 21 = 0$

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Similar questions

Let f(x) = $x^{2}$ -1, 0<x<2
and 2x+3, 2 $\leq$ x<3, The quadratic equation whose roots are ${lim}_{x \to 2^{-}} f (x), a n d {lim}_{x \to 2^{+}} f (x)$

Q. Let

f (x) = (\begin{matrix} x^{2} - 1, if 0 < x < 2 2 x + 3, if 2 \leq x < 3 \end{matrix}

, a quadratic equation whose roots are

lim x \to 2^{-} f (x) and lim x \to 2^{+} f (x)

Q. Let

f (x) = {\begin{matrix} x^{2} - 1; 0 < x < 2 2 x + 3; 2 \leq x < 3 \end{matrix}

The quadratic equation whose roots are

lim x \to 2^{-} f (x)

and

lim x \to 2^{+} f (x)

I f f (x) = {\begin{matrix} x^{2} - 3, & 2 < x < 3 2 x + 5, & 3 < x < 4 \end{matrix}

, the equation whose roots are

lim x \to 3 - f (x)

and

lim x \to 3 + f (x) i s

Q. If

f (x) = \frac{x^{2} + 6 x}{sin x}

, then

lim x \to 0^{-} f (x) =

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Let f(x)={x2−1,0<x<22x+3,2≤x<3, The quadratic equation whose roots are limx→2−f(x) and limx→2+f(x) is

The correct option is A x2−10x+21=0For 0<x<2,f(x)=x2−1limx→2−f(x)=3For, 2<x<3 f(x)=2x+3limx→2+f(x)=7Thus the equation with roots, 3 and 7 isx2−10x+21=0

Let $f (x) = {\begin{matrix} x^{2} - 1, 0 < x < 2 2 x + 3, 2 \leq x < 3 \end{matrix},$
The quadratic equation whose roots are $lim x \to 2^{-} f (x)$ and $lim x \to 2^{+} f (x)$ is

The correct option is A $x^{2} - 10 x + 21 = 0$
For $0 < x < 2$ ,
$f (x) = x^{2} - 1$
${lim}_{x \to 2^{-}} f (x) = 3$
For, $2 < x < 3$
$f (x) = 2 x + 3$
${lim}_{x \to 2^{+}} f (x) = 7$
Thus the equation with roots, 3 and 7 is
$x^{2} - 10 x + 21 = 0$