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Question: The function \(y = f\left( x \right)\) is defined by \(x = 2t - \left| t \right|,\;y = {t^2} + t\lef...

The function y=f(x)y = f\left( x \right) is defined by x=2t−∣t∣,  y=t2+t∣t∣,  t∈Rx = 2t - \left| t \right|,\;y = {t^2} + t\left| t \right|,\;t \in R in the interval x∈[−1,1]x \in \left[ { - 1,1} \right] then-
A. f(x) is continuous everywhere
B. f(x) is not continuous at x = 0
C. f(x) is continuous but not derivable at x = 0
D. f(x) is a constant function

Explanation

Solution

Hint: This is a problem of continuity and differentiability. We will first try to define the function at different values of t, and then check their continuity and differentiability at the critical points. The function is continuous at a point if the limit of that function exists at the given point such that the left hand limit is equal to the right hand limit. The function is differentiable if its derivative exists at that particular point.

Complete step-by-step answer:

We have been given the function y=f(x)y = f\left( x \right) is defined by x=2t−∣t∣,  y=t2+t∣t∣,  t∈Rx = 2t - \left| t \right|,\;y = {t^2} + t\left| t \right|,\;t \in R. We will first open the modulus sign for the function using the property that-
∣t∣=t,  t>0\left| t \right| = t,\;t > 0
∣t∣=−t,  t<0\left| t \right| = - t,\;t < 0
Using this property, we can write that-
At;  t>0,   x;=2t−t=t y;=t2+t.t=2t2 At;  t=0, x;=0−0=0 y;=0+0=0 At;  t<0, x;=2t−(−t)=3t y;=t2+t(−t)=0 \begin{aligned} &At;\;t > 0,\; \\\ &x; = 2t - t = t \\\ &y; = {t^2} + t.t = 2{t^2} \\\ &At;\;t = 0, \\\ &x; = 0 - 0 = 0 \\\ &y; = 0 + 0 = 0 \\\ &At;\;t < 0, \\\ &x; = 2t - \left( { - t} \right) = 3t \\\ &y; = {t^2} + t\left( { - t} \right) = 0 \\\ \end{aligned}

Now we will define the function f(x). We know the values of the functions at different ranges of t. So we will write the ranges of the function in terms of x, so we can write that-
\begin{aligned} & f\left( x \right) = \left\\{ {\begin{array}{*{20}{l}} {0,\;t < 0} \\\ {0,\;t = 0\;} \\\ {2{t^2},\;t > 0} \end{array}} \right. \\\ & f\left( x \right) = \left\\{ {\begin{array}{*{20}{l}} {0,\; - 1 \leqslant x < 0} \\\ {0,\;x = 0\;} \\\ {2{t^2},\;1 \geqslant x > 0} \end{array}} \right. \\\ \end{aligned}

Now, this function has three critical points, at x = 0, -1 and 1. The critical point for modulus functions is usually 0. The points -1 and 1 have been chosen because these are the endpoints of the values of x. First, we will check the continuity on these points using the formula-
lim⁡x→af(x)=lim⁡x→a+f(x)=lim⁡x→a−f(x)\mathop {\lim }\limits_{x \to a} f\left( x \right) = \mathop {\lim }\limits_{x \to {a^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {a^ - }} f\left( x \right)
For the point x = -1, we can write that-
lim⁡x→−1f(x)=lim⁡x→−1+f(x)  exists\mathop {\lim }\limits_{x \to - 1} f\left( x \right) = \mathop {\lim }\limits_{x \to - {1^ + }} f\left( x \right)\;exists
lim⁡x→−1(0)=0\mathop {\lim }\limits_{x \to - 1} \left( 0 \right) = 0
The limit of the function exists at x = -1, hence, it is continuous at x = -1. We did not calculate the LHL because it is not in the range of values of x.
For the point x = 0, we can write that-
lim⁡x→0f(x)=lim⁡x→0+f(x)=lim⁡x→0−f(x)\mathop {\lim }\limits_{x \to 0} f\left( x \right) = \mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {0^ - }} f\left( x \right)
lim⁡x→0(0)=lim⁡x→0+(2t2)=lim⁡x→0−(0)\mathop {\lim }\limits_{x \to 0} \left( 0 \right) = \mathop {\lim }\limits_{x \to {0^ + }} \left( {2{t^2}} \right) = \mathop {\lim }\limits_{x \to {0^ - }} \left( 0 \right)
0=2lim⁡t→0(t2)=00 = 2\mathop {\lim }\limits_{t \to 0} \left( {{t^2}} \right) = 0
0=0=00 = 0 = 0
The limit of the function exists at x = 0, as shown. Therefore, it is continuous at x = 0 as well.
For the point x = 1,
lim⁡x→1f(x)=lim⁡x→1−f(x) lim⁡x→1(2t2)=2lim⁡x→1(t2) We;  know  that  x=3t,  for  t>0 2lim⁡t→13(t2)=29 \begin{aligned} &\mathop {\lim }\limits_{x \to 1} f\left( x \right) = \mathop {\lim }\limits_{x \to {1^ - }} f\left( x \right) \\\ &\mathop {\lim }\limits_{x \to 1} \left( {2{t^2}} \right) = 2\mathop {\lim }\limits_{x \to 1} \left( {{t^2}} \right) \\\ &We;\;know\;that\;x = 3t,\;for\;t > 0 \\\ &2\mathop {\lim }\limits_{t \to \dfrac{1}{3}} \left( {{t^2}} \right) = \dfrac{2}{9} \\\ \end{aligned}
The limit of the function exists at x = 1, hence, it is continuous at x = 1. We did not calculate the RHL because it is not in the range of values of x.

Therefore, we can see that the function is continuous for all points for all values of x∈[−1,1]x \in \left[ { - 1,1} \right]. Now, we will differentiate the function at x=0,  0−  and  0+x = 0,\;{0^ - }\;and\;{0^ + }. If the derivatives are equal, then the function is derivable at x = 0.
At;  x→0−, f;(x)=0   f;′(x)=0   Also;  at  x=0,   f;(x)=0   f;′(x)=0   At;  x→0+, f;(x)=2t2 f;′(x)=4t We;  know  that  x=3t, f;′(3t)=4t f;′(0)=0 \begin{aligned} &At;\;x \to {0^ - }, \\\ &f;\left( x \right) = 0\; \\\ &f;'\left( x \right) = 0\; \\\ &Also;\;at\;x = 0,\; \\\ &f;\left( x \right) = 0\; \\\ &f;'\left( x \right) = 0\; \\\ &At;\;x \to {0^ + }, \\\ &f;\left( x \right) = 2{t^2} \\\ &f;'\left( x \right) = 4t \\\ &We;\;know\;that\;x = 3t, \\\ &f;'\left( {3t} \right) = 4t \\\ &f;'\left( 0 \right) = 0 \\\ \end{aligned}
Here we can see that the derivatives are equal in all the three cases, hence f(x) is derivable at x = 0.

We can clearly see that f(x) is continuous and derivable at all values of x, hence the correct option is A.

Note: Students should keep in mind that if a function is continuous, it is not necessary that it will be derivable as well. So, we should always check the derivability of the function whenever required. Also, while checking the continuity at x = 1 and -1, we should not check the RHL and LHL respectively, because they are out of the range for the values of x, which is specified in the question.