If 1^+((x - 1)(6 + lambda(x - 1)) + mu(1 - x)/(x - 1)3) = -1, where lambda, mu R, then lambda + mu is equal to

Year	2026	2025	2024	Total
Questions	12	24	16	52

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

is

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

If

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

If x → 1⁺((x - 1)(6 + λ (x - 1)) + μ (1 - x))/((x - 1)³) = -1$\lim_{x \to 1^{+}}\frac{(x - 1)(6 + \lambda\cos(x - 1)) + \mu\sin(1 - x)}{(x - 1)^3} = -1$, where λ, μ in R$\lambda, \mu \in \mathbb{R}$, then \lambda + \mu is equal to

Solution & Explanation

Related Formula

Core Logic

Step 1: Match Coefficients for Existence

Step 2: Solve System of Equations

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

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If $x \to 1⁺((x - 1)(6 + λ (x - 1)) + μ (1 - x))/((x - 1)³) = -1$ , where $λ, μ in R$ $\lambda, \mu \in \mathbb{R}$, then \lambda + \mu is equal to