Continuity & DifferentiabilitymediumFree

Where Is f(2/(x-2)) Discontinuous for f(x) = 1/(x^2 - 17x + 66) | JEE

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Question
If f(x)=1x2−17x+66f(x) = \dfrac{1}{x^{2}-17x+66}, then f(2x−2)f\left(\dfrac{2}{x-2}\right) is discontinuous at
Ax=2, 73, 2511x = 2,\ \dfrac73,\ \dfrac{25}{11}
Bx=2, 83, 2411x = 2,\ \dfrac83,\ \dfrac{24}{11}
Cx=2, 73, 2411x = 2,\ \dfrac73,\ \dfrac{24}{11}
Dno where in its domaincorrect
Solution
Step 1: Substitute u=2x−2u = \dfrac{2}{x-2} into ff.
f(2x−2)=14(x−2)2−34x−2+66.f\left(\frac{2}{x-2}\right) = \frac{1}{\dfrac{4}{(x-2)^{2}} - \dfrac{34}{x-2} + 66} .
 Step 2: Clear the inner fractions by multiplying top and bottom by (x−2)2(x-2)^{2}.
=(x−2)24−34(x−2)+66(x−2)2.= \frac{(x-2)^{2}}{4 - 34(x-2) + 66(x-2)^{2}} .
 Step 3: Expand the denominator.
66(x2−4x+4)−34x+68+4=66x2−264x+264−34x+72=66x2−298x+336.66\left(x^{2}-4x+4\right) - 34x + 68 + 4 = 66x^{2} - 264x + 264 - 34x + 72 = 66x^{2} - 298x + 336 .
 Step 4: Factorise it. Taking out 22 and solving 33x2−149x+168=033x^{2}-149x+168 = 0, the discriminant is 1492−4(33)(168)=22201−22176=25149^{2}-4(33)(168) = 22201-22176 = 25, so the roots are 149±566=73, 2411\dfrac{149 \pm 5}{66} = \dfrac73,\ \dfrac{24}{11}:
66x2−298x+336=2(3x−7)(11x−24).66x^{2}-298x+336 = 2(3x-7)(11x-24) .
 Step 5: So the composite function is
f(2x−2)=(x−2)22(3x−7)(11x−24).f\left(\frac{2}{x-2}\right) = \frac{(x-2)^{2}}{2(3x-7)(11x-24)} .
 Step 6: Identify where it is *undefined*: at x=2x = 2 (where the inner function 2x−2\frac{2}{x-2} does not exist) and at x=73x = \dfrac73, x=2411x = \dfrac{24}{11} (where the denominator vanishes). Step 7: Draw the distinction the question is testing. Those three values are **not in the domain** of the composite function at all. Continuity and discontinuity are only defined at points of the domain, so the function is not discontinuous there - it simply is not defined there. Step 8: At every point of its actual domain the function is a quotient of polynomials with non-zero denominator, hence continuous. Answer: (4).
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