Limit, Continuity and Differentiability JEE 2013 (Solved)

What Was the JEE Advanced 2013 Paragraph Question on Differential Inequalities?

f must satisfy −∞ < f(x) < 0 on (0,1). The paragraph described a twice differentiable f on [0,1] with f(0) = f(1) = 0 that satisfies the strict inequality $f^{″} (x) - 2 f^{'} (x) + f (x) > e^{x}$ for all x in [0,1]. It asked which statement must hold for the range of f(x) on (0,1): f(x) remains positive yet finite, f(x) stays between -1/2 and 1/2, f(x) stays between -1/4 and 1, or −∞ < f(x) < 0.

How Does the Official Solution Transform the Inequality to Reach −∞ < f(x) < 0?

Define $g (x) = e^{- x} f (x)$ .

Its first derivative is $g^{'} (x) = e^{- x} (f^{'} (x) - f (x))$ .

Its second derivative is $g^{″} (x) = e^{- x} (f^{″} (x) - 2 f^{'} (x) + f (x))$ .

Substitute the given inequality and multiply by the positive quantity $e^{- x}$ to obtain $g^{″} (x) > 1$ on [0,1]. Boundary values are $g (0) = 0$ and $g (1) = 0$ .

Graph on interval [0,1] with x-axis labelled from 0 to 1, vertical axis labelled g(x), endpoints marked (0,0) and (1,0), straight horizontal chord drawn at y=0, and a smooth strictly convex curve passing through both endpoints while staying strictly below the chord everywhere in

Strict convexity (g'' > 0) implies g lies strictly below the chord joining its endpoints, so $g (x) < 0$ for x in (0,1). Since $e^{- x} > 0$ everywhere, f(x) < 0. As f is real-valued, −∞ < f(x) < 0 on (0,1).

What Sign Error with Convexity Leads Students to the Wrong Positive Range?

Claiming that g''(x) > 0 means the graph lies above its chords instead of below reverses the sign to g(x) > 0 on (0,1). After multiplying by positive $e^{- x}$ this produces f(x) > 0, which matches the incorrect choice that f stays positive and finite.

Forgetting to preserve direction when multiplying by positive $e^{- x}$ yields the same wrong positive conclusion. The correction is that g'' > 0 forces the graph below every chord. With zero endpoints the chord is y = 0, so g(x) < 0 inside (0,1) and f(x) inherits the negative sign.

Why Does g''(x) > 1 Force f(x) to Stay Negative on (0,1)?

g''(x) > 1 > 0 guarantees strict convexity on the closed interval [0,1]. Any strictly convex function on [a,b] with g(a) = g(b) must satisfy g(x) < g(a) = 0 inside (a,b). The comparison is with the linear chord y = 0 that joins the endpoints at height zero.

The positivity of $e^{- x}$ transfers the negative sign directly to f(x) without solving the differential equation.

What Two Related JEE-Level Questions Test the Same Auxiliary-Function Technique?

Consider a twice differentiable f on [0,1] with f(0) = f(1) = 0 that satisfies $f^{″} (x) - 4 f^{'} (x) + 4 f (x) > e^{2 x}$ . Determine the sign of f on (0,1). This tests convexity and sign analysis after the same exponential multiplier reduces the left side to g''(x) > 1.

Next, let f be differentiable on [0, ∞) with f(0) = 0 and satisfy f'(x) ≥ f(x) + 1 for x ≥ 0. Construct an auxiliary function to prove f(x) ≥ x on [0, ∞). This tests boundedness and monotonicity via a first-derivative transformation.

You can search every JEE Main paper from 2002 and every Advanced paper from 2007 by year, subject or chapter in the past-paper archive, each with a worked solution.

Why Does This Problem Sit Inside the Limits, Continuity and Differentiability Chapter?

The problem uses differentiability to define g' and g''. Continuity of g on the closed interval [0,1] lets us apply the chord property without gaps at the endpoints. No explicit limit evaluation appears, yet the strict inequality acts like the limiting case of an equality that would give g'' = 1.

It contrasts with simpler L'Hospital or continuity questions in the same chapter because the core demand is inequality manipulation and convexity.

What Quick Checklist Should You Run Before Attempting Similar Problems?

Multiply by positive integrating factor $e^{- x}$ to simplify left side to exact derivative.
Check second derivative sign for convexity or concavity.
Evaluate auxiliary function at boundaries.
Apply chord property only when endpoints are equal.
Transfer sign back using the factor's positivity.

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Keep going with P-Block Elements JEE 2022: Blast Furnace Question Analysis.

Frequently asked questions

What must be true about the range of f(x) in JEE Advanced 2013 differential inequality question?

The range must satisfy −∞ < f(x) < 0 on (0,1). Define g(x) = e^{-x} f(x) to obtain g''(x) > 1 with g(0) = g(1) = 0. Strict convexity forces g(x) below the chord y = 0, so g(x) < 0 on (0,1) and thus f(x) < 0 since e^{-x} > 0.

How do you solve the JEE 2013 question on f''(x) - 2f'(x) + f(x) > e^x?

Define g(x) = e^{-x} f(x). Then g'(x) = e^{-x}(f'(x) - f(x)) and g''(x) = e^{-x}(f''(x) - 2f'(x) + f(x)) > 1 on [0,1]. With g(0) = g(1) = 0 and g'' > 0, g is strictly convex and lies below the chord y = 0, implying g(x) < 0 and f(x) < 0 on (0,1).

Why does g''(x) > 1 imply that f(x) is negative in the 2013 JEE problem?

g''(x) > 1 guarantees strict convexity on [0,1]. Any strictly convex function with g(0) = g(1) = 0 must satisfy g(x) < 0 for x in (0,1) because it lies strictly below the chord joining the endpoints. Since f(x) = e^x g(x) and e^x > 0, f(x) inherits the negative sign.

What sign error leads to choosing positive f(x) in JEE 2013 convexity question?

Students sometimes reverse the convexity property and claim g''(x) > 0 means the graph lies above its chords, producing g(x) > 0 and then f(x) > 0. The correct statement is that a strictly convex function lies below its chords. With zero endpoints the chord is y = 0, forcing g(x) < 0 inside (0,1).

Limit, Continuity and Differentiability JEE 2013: f(x) < 0 on (0,1)