Domain of Functions JEE 2024: Domain vs Differentiability

What does the domain of functions JEE 2024 question actually ask?

The supplied answer to this domain of functions JEE 2024 question is C: 150, but its worked solution finds differentiability, not the ordinary domain. The task asks for the two boundary values of the permitted outer intervals, then requires: $α^{2} + β^{3}$

The supplied question prints:

f (x) = \frac{\sqrt{x^{2} - 25}}{4 - x^{2}} + \log_{10} (x^{2}) + 2 x - 15

The supplied worked solution instead puts the entire quadratic inside the logarithm and finds where the function is differentiable, not merely defined. The steps below follow that interpretation and retain C: 150.

The record is tagged 2024, but its date field reads 2023-01-24, meaning 24 January 2023, with slot Morning. This is not a verified 2024 sitting; Hard and 180 seconds are question-bank labels, not measured student performance.

How do the square root and denominator restrict the intervals?

A numerator square root permits a zero radicand for domain. Under the supplied differentiability interpretation, the radicand must be strictly positive here because the radical has no finite derivative at its endpoints.

Square-root domain: x^{2} - 25 \geq 0

Supplied differentiability condition: x^{2} - 25 > 0

Factor first, following the supplied method. The product is positive outside its roots:

x^{2} - 25 = (x - 5) (x + 5) > 0

x < - 5 or x > 5

On these open intervals:

\frac{d}{d x} \sqrt{x^{2} - 25} = \frac{x}{\sqrt{x^{2} - 25}}

At both endpoints, this derivative has no finite one-sided limit from within the permitted set. The denominator is nonzero there, so it does not cancel the square-root singularity: $4 - (\pm 5)^{2} = - 21$

Next, exclude denominator zeros:

4 - x^{2} \neq 0 ⟹ x \neq - 2, 2

Neither excluded value lies in the outer intervals already obtained. The denominator therefore removes nothing further once: $| x | > 5$

How do you solve the logarithm inequality and intersect the intervals?

The grouped logarithm permits inputs below negative five or above three. Intersecting this with the square-root differentiability condition leaves inputs below negative five or above five.

Use the solution’s declared grouping: $\log_{10} (x^{2} + 2 x - 15)$

A real logarithm cannot accept zero or a negative argument. Therefore: $x^{2} + 2 x - 15 > 0$

The quadratic formula gives:

x = \frac{- 2 \pm \sqrt{4 + 60}}{2} = \frac{- 2 \pm 8}{2}

$x = - 5, 3$ The factorisation confirms both roots. Since the quadratic opens upwards, it is positive outside them:

x^{2} + 2 x - 15 = (x + 5) (x - 3)

x \in (- \infty, - 5) \cup (3, \infty)

Intersect the restrictions because every component must be valid simultaneously:

[(- \infty, - 5) \cup (3, \infty)] \cap [(- \infty, - 5) \cup (5, \infty)]

Left interval: both restrictions permit inputs below negative five.
Right interval: above five is the stronger restriction and automatically satisfies the logarithm’s requirement of above three.

The differentiability set is: $(- \infty, - 5) \cup (5, \infty)$

Why is the answer 150, and do the endpoints belong to the domain?

The boundary values give 150, regardless of endpoint inclusion. For the grouped expression, positive five belongs to the domain but is excluded from the differentiability set.

Read off and calculate:

α = - 5, β = 5

α^{2} = 25, β^{3} = 125

α^{2} + β^{3} = 25 + 125 = 150

This gives option C. Check each endpoint separately:

Grouped expression at negative five: the logarithm argument is zero, so the function is undefined.

(- 5)^{2} + 2 (- 5) - 15 = 0

Grouped expression at positive five: the radical is zero, the denominator is negative twenty-one, and the logarithm argument is twenty. Every component is defined.

\sqrt{5^{2} - 25} = 0, 4 - 5^{2} = - 21, 5^{2} + 2 (5) - 15 = 20

The ordinary real domain of the grouped expression is therefore: $(- \infty, - 5) \cup [5, \infty)$

For the literally printed expression, only the squared input is inside the logarithm. Both endpoints give a positive logarithm argument and a nonzero denominator:

(\pm 5)^{2} = 25, 4 - (\pm 5)^{2} = - 21

Its ordinary domain is: $(- \infty, - 5] \cup [5, \infty)$

The supplied open-interval domain is not the ordinary domain of either transcription. The supplied answer, 150, follows under the declared differentiability interpretation; domain alone does not justify excluding every square-root endpoint.

Why does the supplied explanation of option B fail?

The supplied material identifies B: 175, but does not support a valid error-to-B derivation. Changing endpoint inclusion cannot produce it, and the supplied incorrect roots do not produce it either.

Allowing a zero numerator radicand is correct for domain, but insufficient for differentiability: $x^{2} - 25 \geq 0$

Changing endpoint inclusion leaves the boundary values unchanged. It cannot turn 150 into 175:

α = - 5, β = 5 ⟹ α^{2} + β^{3} = 150

The supplied erroneous roots have the correct sum but the wrong product:

- 7 + 5 = - 2, (- 7) (5) = - 35 \neq - 15

Even using those incorrect boundary values gives:

α = - 7, β = 5

α^{2} + β^{3} = (- 7)^{2} + 5^{3} = 49 + 125 = 174 \neq 175

There is no supported route to B in the supplied explanation. Verify both the sum and product of quadratic roots, then recompute the requested expression independently.

Which three practice questions test domain and differentiability?

These original practice questions, not additional verified PYQs, test numerator endpoints, denominator endpoints and finite derivatives. Each has a compact worked answer immediately after the question.

What is the domain of the radical quotient in Question 1?

The numerator permits zero, but the denominator does not. Find the domain of:

g (x) = \frac{\sqrt{x^{2} - 9}}{x - 4}

Apply both restrictions:

x^{2} - 9 \geq 0, x \neq 4

x \leq - 3 or x \geq 3

Removing four gives:

(- \infty, - 3] \cup [3, 4) \cup (4, \infty)

What changes when the radical is in the denominator in Question 2?

The radicand must now be strictly positive because the denominator cannot be zero. Find the domain of:

g (x) = \frac{\log_{10} (x^{2} + x - 6)}{\sqrt{x^{2} - 16}}

The logarithm also requires a positive argument. Factor and solve:

x^{2} + x - 6 = (x + 3) (x - 2) > 0 ⟹ x < - 3 or x > 2

x^{2} - 16 > 0 ⟹ x < - 4 or x > 4

Their intersection is:

(- \infty, - 4) \cup (4, \infty)

Does the function in Question 3 have a finite right derivative at two?

No finite right derivative exists, although the endpoint belongs to the domain. Find the domain and check the right derivative for: $h (x) = \sqrt{x - 2}$

The domain condition gives:

x - 2 \geq 0 ⟹ [2, \infty)

The endpoint difference quotient diverges:

\frac{h (2 + t) - h (2)}{t} = \frac{\sqrt{t}}{t} = \frac{1}{\sqrt{t}} ⟶ + \infty (t \to 0^{+})

Use Limits Practice Questions JEE: 6 Drills With Solutions to practise one-sided limits. Check the endpoint value first, then the difference quotient.

Frequently asked questions

What is the domain of the function as printed?

For the printed expression, with only x² inside the logarithm, the ordinary real domain is (-∞, -5] ∪ [5, ∞). Both endpoints are allowed because the square root is defined, the denominator is nonzero and the logarithm argument is positive.

What changes if x² + 2x - 15 is inside the logarithm?

The ordinary real domain becomes (-∞, -5) ∪ [5, ∞). At x = -5, the logarithm argument is zero, so that endpoint is excluded. At x = 5, every component is defined, but the square-root term has no finite right derivative, so the differentiability set is (-∞, -5) ∪ (5, ∞).

Why is the supplied answer 150?

The boundary values are α = -5 and β = 5, giving α² + β³ = 25 + 125 = 150, which is option C. Changing endpoint inclusion does not change these boundary values. The supplied open intervals describe differentiability under the grouped-logarithm interpretation, not the ordinary domain.

Is this a verified JEE 2024 question?

No, the supplied record does not establish a verified 2024 sitting. It is tagged 2024, but its date field reads 2023-01-24 with the slot listed as Morning.