Question

Let $f(x) = \begin{cases} x^2-1, & 0 < x < 2{} \\[2ex] 2x+3, & 2 \le x < 3{} \end{cases}$, the quadratic equation whose roots are $\displaystyle \lim_{x \to 2^-}f(x)$ and $\displaystyle \lim_{x \to 2^+}f(x)$ is

• A. $x^2 -6x +9=0$
• B. $x^2 -7x +8=0$
• C. $x^2 -14x +49=0$
• D. $x^2 -10x +21=0$

Answer 1

The Correct Option is D

\(\displaystyle \lim_{x \to 2^-}f(x)\) \(=\displaystyle \lim_{h \to 0}(2-h)^2-1\)

 \(=\displaystyle \lim_{h \to 0}4+h^2-4h-1=3\) \(\displaystyle \lim_{x \to 2^+}f(x)\)\(=\displaystyle \lim_{h \to 0}2(2+h)+3\) 

\(=\displaystyle \lim_{h \to 0}4+2h+3=7\)
 \(\therefore\) Required quadratic equation is \(x^2 - ( 3 + 7)x + ( 3 \times 7) = 0\)
 \(\Rightarrow x^2 -10x+21=0\)

Answer 2

We are given a piecewise function:  
f(x) = {

• \(x^2 - 1\), for \(0 < x < 2\)
• \(2x + 3\), for \(2 \leq x < 3\)

We are to find the limits:

• \(\lim_{x \to 2^-} f(x)\) (from the left)
• \(\lim_{x \to 2^+} f(x)\) (from the right)

Step 1: Left-hand limit
Since \(x \to 2^-\), use \(f(x) = x^2 - 1\): 
\[ \lim_{x \to 2^-} f(x) = 2^2 - 1 = 4 - 1 = 3 \]

Step 2: Right-hand limit
Since \(x \to 2^+\), use \(f(x) = 2x + 3\): 
\[ \lim_{x \to 2^+} f(x) = 2(2) + 3 = 4 + 3 = 7 \]

Now we construct the quadratic equation whose roots are 3 and 7. 
The general form is: \[ x^2 - (\text{sum of roots})x + (\text{product of roots}) = 0 \] 
Sum = \(3 + 7 = 10\), Product = \(3 \times 7 = 21\) 
\[ \Rightarrow x^2 - 10x + 21 = 0 \]

 

Correct answer: \(x^2 - 10x + 21 = 0\)

Answer 3

First, we need to find the limits of f(x) as x approaches 2 from the left and from the right.

1. Limit as x approaches 2 from the left (x -> 2^-):

For x < 2, f(x) = x² - 1

lim_x->2^- f(x) = lim_x->2^- (x² - 1) = (2)² - 1 = 4 - 1 = 3

2. Limit as x approaches 2 from the right (x -> 2⁺):

For x ≥ 2, f(x) = 2x + 3

lim_x->2⁺ f(x) = lim_x->2⁺ (2x + 3) = 2(2) + 3 = 4 + 3 = 7

So, the roots of the quadratic equation are 3 and 7.

A quadratic equation with roots α and β can be written as:

x² - (α + β)x + αβ = 0

In this case, α = 3 and β = 7.

α + β = 3 + 7 = 10

αβ = 3 * 7 = 21

Therefore, the quadratic equation is:

x² - 10x + 21 = 0

Answer:

X²-10x+21 = 0