If $L = lim x \to - 2 \frac{3 x^{2} + k x + k - 7}{2 x^{2} + x - 6}$ exists and is finite, where $k$ is a real number, then the values of $k$ and $L$ are

k = 5, L = - 1

No worries! We‘ve got your back. Try BYJU‘S free classes today!

k = 5, L = 1

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

k \in R, L = \frac{3}{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

k = - 5, L = 1

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $k = 5, L = 1$
$L = lim x \to - 2 \frac{3 x^{2} + k x + k - 7}{2 x^{2} + x - 6}$
For $\frac{0}{0}$ form and limit to exist, numerator should tend to $0$ as $x \to - 2$
$3 (- 2)^{2} + k (- 2) + k - 7 = 0 \Rightarrow k = 5 L = lim x \to - 2 \frac{3 x^{2} + 5 x - 2}{2 x^{2} + x - 6} = lim x \to - 2 \frac{(3 x - 1) (x + 2)}{(2 x - 3) (x + 2)} = \frac{- 6 - 1}{- 4 - 3} = 1$

Suggest Corrections

Similar questions

L = lim x \to - 2 \frac{3 x^{2} + k x + k - 7}{2 x^{2} + x - 6}

k

k

L

X

\begin{matrix} X & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 P(X) & 0 & k & 2k & 2k & 3k & k^{2} & 2 k^{2} & 7 k^{2} + k \end{matrix}

k

X

\begin{matrix} X & 1 & 2 & 3 & 4 & 5 P (X) & K^{2} & 2 K & K & 2 K & 5 K^{2} \end{matrix}

P (X > 2)

Join BYJU'S Learning Program