Applications of Derivatives JEE 2024: Derivative Sum 2890

What must we find in this two-input derivative question?

The answer is 2890 because reversing the two inputs forces equality. This applications of derivatives jee 2024 problem asks for an exact derivative sum, not a function chosen to fit a lower bound.

A differentiable, real-valued function on the positive real numbers has a change between any two inputs at least as large as the corresponding change in the natural logarithm plus the input:

f (x) - f (y) \geq \ln (\frac{x}{y}) + x - y, \forall x, y > 0 .

Find:

\sum_{n = 1}^{20} f^{'} (\frac{1}{n^{2}}) .

This is a numerical-answer question, with no supplied options. The bank labels it 2024 but supplies the conflicting date 2023-01-24; neither a paper date nor a shift is verified here.

How does subtracting the known function force equality?

Subtracting the logarithmic and linear terms leaves an auxiliary function whose value must be at least its value at every other input. Reversing the inputs gives the opposite inequality. Together, these statements force the auxiliary function to be constant, not merely non-decreasing.

Because both inputs are positive, the logarithm expands as:

\ln (\frac{x}{y}) = \ln x - \ln y .

Rearrange the given condition:

f (x) - \ln x - x \geq f (y) - \ln y - y .

Following the supplied solution’s auxiliary-function method, define:

g (t) = f (t) - \ln t - t .

The condition becomes:

g (x) \geq g (y), \forall x, y > 0 .

There is no restriction that the first input must exceed the second. The statement holds for every ordered pair, so exchanging the inputs gives: $g (y) \geq g (x) .$

Combining the two inequalities:

g (x) = g (y), \forall x, y > 0 .

Thus, the auxiliary function is constant throughout the domain. A derivative lower bound alone would not justify choosing one function and declaring its derivative unique; the reversed-pair argument supplies that missing justification.

Why is the derivative exact rather than just a lower bound?

The original function must equal the natural logarithm plus the input plus a constant. Only that constant remains free, and differentiation removes it. Every admissible function therefore has the same derivative.

From the constant auxiliary function:

g (x) = C ⟹ f (x) = \ln x + x + C .

Differentiate every term before substituting an input:

\frac{d}{d x} \ln x = \frac{1}{x}, \frac{d}{d x} x = 1, \frac{d}{d x} C = 0 .

The result is an exact derivative, rather than just a lower bound:

f^{'} (x) = \frac{1}{x} + 1 .

Substitute the reciprocal square to evaluate a general summand:

f^{'} (\frac{1}{n^{2}}) = \frac{1}{1 / n^{2}} + 1 = n^{2} + 1 .

Check the forced function in the original condition. The constant cancels, leaving:

f (x) - f (y) & = (\ln x + x + C) - (\ln y + y + C) & = \ln (\frac{x}{y}) + x - y .

The inequality is satisfied with equality for every allowed pair. This verifies the function’s form against the full hypothesis, not just a derivative condition.

How do the twenty derivative values add to 2890?

The sum contains the first twenty squares and twenty copies of one. Calculate those contributions separately: the squares contribute 2870, while the constant derivative contributes another 20.

Split the sum:

\sum_{n = 1}^{20} (n^{2} + 1) = \sum_{n = 1}^{20} n^{2} + \sum_{n = 1}^{20} 1 .

Use the sum-of-squares formula:

\sum_{n = 1}^{m} n^{2} = \frac{m (m + 1) (2 m + 1)}{6} .

Substitute twenty for the upper limit:

\sum_{n = 1}^{20} n^{2} & = \frac{20 \times 21 \times 41}{6} & = \frac{20 \times 7 \times 41}{2} & = 10 \times 7 \times 41 & = 2870 .

There are twenty terms, so:

\sum_{n = 1}^{20} 1 = 20 .

Therefore:

2870 + 20 = 2890 .

Numerical answer: 2890. Check both ends of the index range: the first summand is 2 and the twentieth is 401.

f^{'} (1) = 2, f^{'} (\frac{1}{400}) = 401 .

Why would dropping the linear term give the wrong answer 2870?

The response 2870 results if you keep the logarithm’s derivative but drop the linear term’s derivative. The input itself is a variable term, not a constant. Its derivative is one; only the arbitrary constant differentiates to zero.

The incorrect route is:

f (x) = \ln x + x + C \overset{incorrect differentiation}{\to} f^{'} (x) = \frac{1}{x} ⟹ \sum_{n = 1}^{20} n^{2} = 2870 .

Omitting one at each of twenty evaluation points lowers the answer by exactly twenty:

\sum_{n = 1}^{20} [(n^{2} + 1) - n^{2}] = 20 .

The repair is to differentiate all terms first, then substitute the input. Separate the sum into its square and constant contributions so neither disappears.

Here, 2870 is an illustrative incorrect numerical response, not an official distractor. The supplied question has no answer options.

Which two practice questions use the same reversed-input argument?

Both questions below are original related practice, not verified JEE PYQs. Each asks you to subtract a known expression and exchange the inputs to prove that the remainder is constant. The first then requires a derivative sum; the second requires a limit.

Question 1. A differentiable function on the positive real numbers satisfies:

h (x) - h (y) \geq 2 \ln (\frac{x}{y}) + 3 (x - y), \forall x, y > 0 .

Find:

\sum_{k = 1}^{5} h^{'} (\frac{1}{k}) .

Worked check: exchanging the inputs forces the following expression to be constant: $h (t) - 2 \ln t - 3 t = C .$

Differentiate, evaluate, and add:

h^{'} (t) = \frac{2}{t} + 3, h^{'} (\frac{1}{k}) = 2 k + 3 .

\sum_{k = 1}^{5} (2 k + 3) = 2 (1 + 2 + 3 + 4 + 5) + 3 \times 5 = 30 + 15 = 45 .

Question 2. A differentiable function on the real numbers satisfies:

p (x) - p (y) \geq x^{2} - y^{2}, \forall x, y \in R .

Evaluate:

{lim}_{h \to 0} \frac{p (3 + h) - p (3)}{h} .

Worked check: reversing the inputs again gives equality, so:

p (t) - t^{2} = C, p (t) = t^{2} + C .

Substitute into the difference quotient. The constant cancels before division, and division is valid away from zero:

\frac{p (3 + h) - p (3)}{h} = \frac{(3 + h)^{2} - 9}{h} = \frac{6 h + h^{2}}{h} = 6 + h, h \neq 0 .

Therefore:

{lim}_{h \to 0} (6 + h) = 6 .

When an inequality holds for every pair of inputs, test the reversed pair before differentiating.

Next step: photograph a doubt on JEEnius and photograph any question you are stuck on and get a step-by-step solution, with a free-body diagram when the question needs one (20 free).

Frequently asked questions

Why does reversing x and y turn the inequality into equality?

Define g(t) = f(t) - ln t - t, so the given condition becomes g(x) ≥ g(y) for all positive x and y. Since every ordered pair is allowed, swapping the inputs also gives g(y) ≥ g(x). Both inequalities can hold only if g(x) = g(y), making g constant.

How do we get f'(x) = 1/x + 1 from the inequality?

Reversing the inputs proves that f(x) - ln x - x is constant, so f(x) = ln x + x + C. Differentiating gives f'(x) = 1/x + 1. This is an exact derivative, not merely a lower bound.

Why is the derivative sum 2890 and not 2870?

Since f'(x) = 1/x + 1, substituting x = 1/n² gives f'(1/n²) = n² + 1. The first twenty squares sum to 2870, and the twenty copies of 1 add another 20, giving 2890. The incorrect answer 2870 drops the derivative of the linear term x.

Which JEE 2024 shift is this derivative question from?

The paper date and shift are not verified in the supplied material. The question bank labels it 2024 but also supplies the conflicting date 2023-01-24, so the article does not assign it to a specific shift.