Let f:Rarrow R be defined as If f is continuous everywhere in R and m is the number of points where f is NOT…

Year	2026	2025	2024	Total
Questions	12	24	16	52

Year

2026

2025

2024

Total

Questions

Solution & Explanation

Related Formula

For a function to be continuous at a boundary point $x = x₀$ $x = x_0$, the left-hand limit, right-hand limit, and exact function value must all match:

x \to x₀^- f(x) = x \to x₀^+ f(x) = f(x₀)

Core Logic

Let's enforce continuity at the critical boundaries, $x = 1$ $x = 1$ and $x = 0$ $x = 0$:

Continuity at $x = 1$ $x = 1$:

f(1^-) = f(1) = 1² + c(1) + 2 = 3 + c

f(1^+) = 2(1) + 1 = 3

Equating both configurations: $3 + c = 3 c = 0$ $3 + c = 3 \implies c = 0$.

Continuity at $x = 0$ $x = 0$:

f(0^+) = f(0) = 0² + 0 + 2 = 2

f(0^-) = h \to 0 (a - b (2h))/(h²)

Using the Taylor expansion $(2h) = 1 - (4h²)/(2!) + (16h⁴)/(4!) - = 1 - 2h² + (2)/(3)h⁴ -$

h \to 0 (a - b(1 - 2h² + (2)/(3)h⁴ - ))/(h²) = h \to 0 ((a-b) + 2bh² - (2)/(3)bh⁴ + )/(h²)

For the limit to exist and remain finite, the constant term must vanish: $a - b = 0 a = b$ $a - b = 0 \implies a = b$. The value of the limit is then equal to $2b$ $2b$. To satisfy continuity: $2b = 2 b = 1 a = 1$ $2b = 2 \implies b = 1 \implies a = 1$.

Step 1: Checking Differentiability at x = 0

Evaluating the Left-Hand Derivative (LHD) at $x = 0$ $x = 0$ using values $a=1, b=1$ $a=1, b=1$:

LHD = h \to 0 (f(-h) - f(0))/(-h) = h \to 0 ((1 - (2h))/(h²) - 2)/(-h)

LHD = h \to 0 ((2 - (2)/(3)h² + ) - 2)/(-h) = h \to 0 (2)/(3)h = 0

Evaluating the Right-Hand Derivative (RHD) at $x = 0$ $x = 0$:

RHD = h \to 0 (f(h) - f(0))/(h) = h \to 0 ((h² + 2) - 2)/(h) = h \to 0 h = 0

Since $LHD = RHD = 0$ $\text{LHD} = \text{RHD} = 0$, the function is fully differentiable at $x = 0$ $x = 0$.

Step 2: Checking Differentiability at x = 1

Evaluating derivatives at $x = 1$ $x = 1$ with parameter $c = 0$ $c = 0$:

For $0 \leq x \leq 1$ $0 \le x \le 1$, $f(x) = x² + 2 f'(x) = 2x f'(1^-) = 2$ .
For $x > 1$ $x > 1$, $f(x) = 2x + 1 f'(x) = 2 f'(1^+) = 2$ .

Since the left derivative equals the right derivative at $x = 1$ $x = 1$, the function is differentiable at $x = 1$ $x = 1$.

Thus, the function is differentiable everywhere, giving $m = 0$ $m = 0$ points of non-differentiability.

Step 3: Finding the Requested Evaluation Sum

Now substitute the values $m=0, a=1, b=1, c=0$ $m=0, a=1, b=1, c=0$ into the target equation:

m + a + b + c = 0 + 1 + 1 + 0 = 2

Pattern Recognition

Sees: Continuity conditions paired with rational surd trigonometric expansion. Shortcut: When tracking indeterminate limits like $(a-b 2x)/(x²)$ $\frac{a-b\cos 2x}{x^2}$, matching expansions row by row prevents typical computation errors encountered with standard L'Hopital differentiation loops.

Chapter Mix

Class 12 Mathematics: Limits, Continuity and Differentiability

More Limits, Continuity and Differentiability Previous-Year Questions — Page 10

Q20 jee_main_2024_30_jan_morning Limits

Let

f:[-(π)/(2),(π)/(2)] \to R

be a differentiable function such that

f(0) = (1)/(2)

. If the

x \to 0 x \int₀x f(t) dtex² - 1 = α

, then

8α²

is equal to:

A. $16$ $16$
B. $2$ $2$
C. $1$ $1$
D. $4$ $4$

Solution

Related Formula

y \to 0 (e^y - 1)/(y) = 1

Leibniz Integral Rule:

(d)/(dx) \int₀^x f(t) dt = f(x)

Core Logic

Given limit is:

α = x \to 0 x \int₀x f(t) dtex² - 1

Multiply and divide the denominator by $x²$ $x^2$ to use standard exponential limit:

α = x \to 0 x \int₀x f(t) dt( ex² - 1x²) \cdot x²

Since $x\to 0 ex² - 1x² = 1$ , the expression simplifies to:

α = x \to 0 x \int₀x f(t) dt1 \cdot x² = x \to 0 \int₀x f(t) dtx

Step 1: Applying L'Hôpital's Rule

This is a $0/0$ $0/0$ form. Apply L'Hôpital's Rule by differentiating numerator and denominator w.r.t $x$ $x$:

α = x \to 0 (d)/(dx) \int₀x f(t) dt(d)/(dx)(x) = x \to 0 (f(x))/(1)

By continuity of differentiable function $f$ $f$ at $0$ $0$: $α = f(0)$ $\alpha = f(0)$

Step 2: Final Calculation

We are given $f(0) = (1)/(2)$ $f(0) = \frac{1}{2}$, so $α = (1)/(2)$ $\alpha = \frac{1}{2}$. We need to find $8α²$ $8\alpha^2$:

8α² = 8 ((1)/(2))² = 8 ((1)/(4)) = 2

Pattern Recognition

Standard expansion/limits on isolated terms in denominators immediately reduce the power of $x$ $x$, setting up a trivial Leibniz derivative application.

Chapter Mix

Class 11 Maths: Limits and Derivatives Class 12 Maths: Integrals

Q29 jee_main_2024_30_jan_morning Differentiability

If the function

f(x) = cases (1)/(|x|) & ,|x| \geq 2 ax² + 2b & ,|x| < 2 cases

is differentiable on

R

, then

48 (a + b)

is equal to

Numerical Answer. Answer: 15 to 15

Solution

Related Formula

Continuity at x=c: x \to c^- f(x) = x \to c^+ f(x)

Differentiability at x=c: x \to c^- f'(x) = x \to c^+ f'(x)

Core Logic

Rewrite the piecewise function without absolute values:

f(x) = cases (1)/(x) & , x \geq 2 ax² + 2b & , -2 < x < 2 -(1)/(x) & , x \leq -2 cases

Step 1: Applying Continuity

For $f(x)$ $f(x)$ to be continuous at $x = 2$ $x = 2$:

x \to 2^- (ax² + 2b) = x \to 2^+ (1)/(x)

a(2)² + 2b = (1)/(2) \Rightarrow 4a + 2b = (1)/(2) (1)

Because the function is even, continuity at $x = -2$ $x = -2$ yields the exact same equation: $4a + 2b = 1/2$ $4a + 2b = 1/2$.

Step 2: Applying Differentiability

Find the derivative $f'(x)$ $f'(x)$ for piecewise sections:

f'(x) = cases -(1)/(x²) & , x > 2 2ax & , -2 < x < 2 (1)/(x²) & , x < -2 cases

For $f(x)$ $f(x)$ to be differentiable at $x = 2$ $x = 2$:

x \to 2^- (2ax) = x \to 2^+ (-(1)/(x²))

4a = -(1)/(4) \Rightarrow a = -(1)/(16)

Step 3: Finding variables and final target

Substitute $a$ $a$ back into equation (1):

4(-(1)/(16)) + 2b = (1)/(2)

-(1)/(4) + 2b = (1)/(2) \Rightarrow 2b = (3)/(4) \Rightarrow b = (3)/(8)

We need to evaluate $48(a + b)$ $48(a + b)$:

48(-(1)/(16) + (3)/(8)) = 48((-1 + 6)/(16)) = 48((5)/(16)) = 3 \times 5 = 15

Pattern Recognition

Piecewise differentiability forces simultaneous linear equations matching function values and their first derivatives at boundary limits.

Chapter Mix

Class 12 Maths: Continuity and Differentiability

Q10 jee_main_2024_31_jan_evening Limits of Functions

Let

f: R \to (0, \infty)

be strictly increasing function such that

x \to \infty (f(7x))/(f(x)) = 1

. Then, the value of

x arrow \infty [ (f(5x))/(f(x)) - 1 ]

is equal to

A. $4$ $4$
B. $0$ $0$
C. $7/5$ $7/5$
D. $1$ $1$

Solution

Related Formula

Sandwich / Squeeze Theorem: If g(x) \leq h(x) \leq k(x) and g(x) = k(x) = L, then h(x) = L

Core Logic

Since $f$ $f$ is a strictly increasing function mapping to $(0,\infty)$ $(0,\infty)$: For $x > 0$ $x > 0$, we have $x < 5x < 7x$ $x < 5x < 7x$. Thus, $f(x) < f(5x) < f(7x)$ $f(x) < f(5x) < f(7x)$. Divide everything by $f(x)$ $f(x)$ (which is strictly positive):

1 < (f(5x))/(f(x)) < (f(7x))/(f(x))

Take the limit as $x \to \infty$ $x \to \infty$:

x\to\infty 1 \leq x\to\infty (f(5x))/(f(x)) \leq x\to\infty (f(7x))/(f(x))

1 \leq x\to\infty (f(5x))/(f(x)) \leq 1

Therefore, $x\to\infty (f(5x))/(f(x)) = 1$ . The required value is:

x arrow \infty [ (f(5x))/(f(x)) - 1 ] = 1 - 1 = 0

Chapter Mix

Class 11 Maths: Limits and Derivatives Class 12 Maths: Continuity and Differentiability

Q14 jee_main_2024_31_jan_evening Differentiability

Consider the function

f:(0,\infty)\to R

defined by

f(x) = e^-| ₑx|

. If

m

and

n

be respectively the number of points at which

f

is not continuous and

f

is not differentiable, then

m + n

A. $0$ $0$
B. $3$ $3$
C. $1$ $1$
D. $2$ $2$

Solution

Core Logic

Differentiability diagram for Q14 - JEE Main 2024 Evening

The function is $f(x) = e-|ln x|$ $f(x) = e^{-|\ln x|}$. Rewrite piecewise for $(0, \infty)$ $(0, \infty)$:

f(x) = cases e-(-ln x) & if 0 < x < 1 e-ln x & if x \geq 1 cases

f(x) = cases eln x = x & if 0 < x < 1 1eln x = (1)/(x) & if x \geq 1 cases

Check continuity at $x = 1$ $x = 1$:

x \to 1^- f(x) = x \to 1^- x = 1

x \to 1^+ f(x) = x \to 1^+ (1)/(x) = 1

$f(1) = 1$ $f(1) = 1$. The function is continuous everywhere on $(0, \infty)$ $(0, \infty)$. Thus, $m = 0$ $m = 0$.

Check differentiability at $x = 1$ $x = 1$:

LHD = x \to 1^- f'(x) = 1

RHD = x \to 1^+ f'(x) = -(1)/(x²)|x=1 = -1

Since $LHD \neq RHD$ $LHD \neq RHD$, the function is not differentiable at $x = 1$ $x = 1$. Thus, $n = 1$ $n = 1$.

Finally, $m + n = 0 + 1 = 1$ $m + n = 0 + 1 = 1$.

Chapter Mix

Class 12 Maths: Continuity and Differentiability

Q27 jee_main_2024_31_jan_evening Maclaurin Series / L'Hopital

x \to 0 ax²e^x - b ₑ(1 + x) + cxe-xx² x = 1

, then

16(a² + b² + c²)

is equal to

Numerical Answer. Answer: 81 to 81

Solution

Related Formula

e^x = 1 + x + (x²)/(2!) +

ln(1+x) = x - (x²)/(2) + (x³)/(3) -

x \approx x x² x \approx x³

Core Logic

Expand the numerator terms using Maclaurin series around $x=0$ $x=0$:

ax² (1 + x + (x²)/(2) + ) - b (x - (x²)/(2) + (x³)/(3) - ) + cx (1 - x + (x²)/(2) - (x³)/(6) + )

Denominator behavior is $x³$ $x^3$. Group by powers of $x$ $x$: Coefficient of $x$ $x$: $-b + c = 0 c = b$ $-b + c = 0 \implies c = b$ Coefficient of $x²$ $x^2$: $a + (b)/(2) - c = 0 a = c - (b)/(2) = (b)/(2)$ Coefficient of $x³$ $x^3$: $a - (b)/(3) + (c)/(2) = 1$ $a - \frac{b}{3} + \frac{c}{2} = 1$

Substitute $a = b/2$ $a = b/2$ and $c = b$ $c = b$ into the $x³$ $x^3$ equation:

(b)/(2) - (b)/(3) + (b)/(2) = 1

b - (b)/(3) = 1 (2b)/(3) = 1 b = (3)/(2)

This gives $c = (3)/(2)$ $c = \frac{3}{2}$ and $a = (3)/(4)$ $a = \frac{3}{4}$.

Calculate the required value:

16(a² + b² + c²) = 16((9)/(16) + (9)/(4) + (9)/(4))

= 9 + 36 + 36 = 81

Chapter Mix

Class 11 Maths: Limits and Derivatives

Solution & Explanation

Related Formula

Core Logic

Step 1: Checking Differentiability at x = 0

Step 2: Checking Differentiability at x = 1

Step 3: Finding the Requested Evaluation Sum

Pattern Recognition

Chapter Mix

Reference Study Guides

More Limits, Continuity and Differentiability Previous-Year Questions — Page 10

Solution

Related Formula

Core Logic

Step 1: Applying L'Hôpital's Rule

Step 2: Final Calculation

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Applying Continuity

Step 2: Applying Differentiability

Step 3: Finding variables and final target

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Chapter Mix

Solution

Core Logic

Chapter Mix

Solution

Related Formula

Core Logic

Chapter Mix

More Limits, Continuity and Differentiability Questions — jee_main_2024_01_february_morning

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