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# Find $k$ if the given function is continuous at $x = 0$.$f\left( x \right) = \left\{ {\begin{array}{*{20}{c}} {k\left( {{x^2} - 2} \right)}&{x \leqslant 0} \\ {4x - 1}&{x > 0} \end{array}} \right.$A) $626$B) $1422$C) $1200$D) $1340$

Last updated date: 22nd Jun 2024
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Hint: We will use the basic definition of continuity to solve the given problem. We will take limits from both sides and also verify the functional value at the point of continuity. As the function is given to be continuous, all the values will be the same and we will use the relation formed so far to find the value of unknown.

The defined function is:
$f\left( x \right) = \left\{ {\begin{array}{*{20}{c}} {k\left( {{x^2} - 2} \right)}&{x \leqslant 0} \\ {4x - 1}&{x > 0} \end{array}} \right.$
Also, it is given that the function is continuous at $x = 0$ .
We know that the function $f(x)$ is continuous at any point $x = a$ if the following relation holds:
$\Rightarrow \mathop {\lim }\limits_{x \to {a^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {a^ - }} f\left( x \right) = f\left( a \right)$
That is the left-hand limit, right-hand limit and the functional value are all the same.
In the given example $a = 0$.
Consider the following:
$\Rightarrow \mathop {\lim }\limits_{x \to {0^ + }} f(x) = \mathop {\lim }\limits_{x \to {0^ + }} k\left( {{x^2} - 2} \right)$
It can be observed that the function is easily defined at all points so we can take limit by direct substitution.
$\Rightarrow \mathop {\lim }\limits_{x \to {0^ + }} f(x) = \mathop {\lim }\limits_{x \to {0^ + }} k\left( {{0^2} - 2} \right)$
Therefore, the left-hand limit is:
$\Rightarrow \mathop {\lim }\limits_{x \to {0^ + }} f(x) = - 2k$ ……………………..… (1)
Similarly, we will calculate the right-hand limit.
$\Rightarrow \mathop {\lim }\limits_{x \to {0^ - }} f(x) = \mathop {\lim }\limits_{x \to {0^ - }} \left( {4x - 1} \right)$
It can be observed that the function is easily defined at all points so we can take limit by direct substitution.
$\Rightarrow \mathop {\lim }\limits_{x \to {0^ + }} f(x) = \mathop {\lim }\limits_{x \to {0^ + }} \left( {4(0) - 1} \right)$
Therefore, the left-hand limit is:
$\Rightarrow \mathop {\lim }\limits_{x \to {0^ + }} f(x) = - 1$ …………………...… (2)
Now at $x = 0$ the function is defined as $f\left( x \right) = k\left( {{x^2} - 2} \right)$.
Therefore, $f\left( 0 \right) = - 2k$ ……………………….… (3)
It is given that the function is continuous so all the equations (1) to (3) are equal.
Using this we get,
$\Rightarrow - 2k = - 1$
Simplify the above equation for $k$ and we get,
$\Rightarrow k = \dfrac{1}{2}$
Therefore, the value of $k = \dfrac{1}{2}$.

Hence, the correct option is A.

Note:
Here it was given that the function is continuous so we could use the definition directly. Otherwise, we first need to verify the continuity before going further. Also, the limit was very straightforward to calculate so we directly solved it by simple substitutions.