Let the function f(x) = (x2 - 1)|x2 - ax + 2| + |x| be not differentiable at the two points x = alpha = 2 and x = beta. Then the distance of the point (alpha, beta) from the line 12x + 5y + 10 = 0 is equal to:

The expression |x| is everywhere differentiable. Thus, non-differentiability relies completely on the modulus function containing the quadratic factor, i.e., |x2 - ax + 2|. Non-differentiability points generally happen where: Given one of the roots is alpha = 2: Substituting a=3 gives the other root beta = 1. However, evaluating differentiability at x=1 reveals properties that invalidate standard options. Due to a technical contradiction in the configuration of the differentiable constraints at x=1, this question was officially dropped by NTA. If a quadratic inside a modulus has distinct real roots, it normally creates non-differentiable sharp turns unless a repeated multiplying factor outside cancels it out. Class 12 Mathematics: Limits, Continuity and Differentiability

Let the function f(x) = (x2 - 1)|x2 - ax + 2| + |x| be not differentiable at the two points x = alpha = 2 and x = beta. Then the distance of the point (alpha, beta) from the line 12x + 5y + 10 = 0 is

The expression |x| is everywhere differentiable. Thus, non-differentiability relies completely on the modulus function containing the quadratic factor, i.e., |x2 - ax + 2|. Non-differentiability points generally happen where: Given one of the roots is alpha = 2: Substituting a=3 gives the other root beta = 1. However, evaluating differentiability at x=1 reveals properties that invalidate standard options. Due to a technical contradiction in the configuration of the differentiable constraints at

Let the function f(x) = (x2 - 1)|x2 - ax + 2| + |x| be not differentiable at the two points x = alpha = 2 and x = beta. Then the distance of the point (alpha, beta) from the line 12x + 5y + 10 = 0 is equal to:

The expression |x| is everywhere differentiable. Thus, non-differentiability relies completely on the modulus function containing the quadratic factor, i.e., |x2 - ax + 2|. Non-differentiability points generally happen where: Given one of the roots is alpha = 2: Substituting a=3 gives the other root beta = 1. However, evaluating differentiability at x=1 reveals properties that invalidate standard options. Due to a technical contradiction in the configuration of the differentiable constraints at x=1, this question was officially dropped by NTA. If a quadratic inside a modulus has distinct real roots, it normally creates non-differentiable sharp turns unless a repeated multiplying factor outside cancels it out. Class 12 Mathematics: Limits, Continuity and Differentiability

Let the function f(x) = (x2 - 1)|x2 - ax + 2| + |x| be not differentiable at the two points x = alpha = 2 and x = beta. Then the distance of the point (alpha, beta) from the line 12x + 5y + 10 = 0 is

The expression |x| is everywhere differentiable. Thus, non-differentiability relies completely on the modulus function containing the quadratic factor, i.e., |x2 - ax + 2|. Non-differentiability points generally happen where: Given one of the roots is alpha = 2: Substituting a=3 gives the other root beta = 1. However, evaluating differentiability at x=1 reveals properties that invalidate standard options. Due to a technical contradiction in the configuration of the differentiable constraints at

Let the function f(x) = (x2 - 1)|x2 - ax + 2| + |x| be not differentiable at the two points x = alpha = 2 and x = beta…

Year	2026	2025	2024	Total
Questions	11	24	16	51

Q10 jee_main_2026_21_jan_morning 1^infinity Limit Form with L'Hopital's Rule

Let

f: R \to (0, \infty)

be a twice differentiable function such that

f(3) = 18

,

f'(3) = 0

and

f''(3) = 4

. Then

x \to 1 ( ₑ ( (f(2 + x))/(f(3)) )(18)/((x-1)²) )

is equal to:

A. 1
B. 9
C. 2
D. 18

Solution

### Related Formula For a limit of

1\infty

form,

x \to a [g(x)]h(x)

equals:

e^ x \to a h(x)[g(x) - 1]

### Core Logic Let

T = x \to 1 ( (f(x + 2))/(f(3)) )(18)/((x - 1)²)

. As

x \to 1

,

(f(x+2))/(f(3)) \to (f(3))/(f(3)) = 1

. The exponent goes to

\infty

. This is a standard

1\infty

form.

T = e^ x \to 1 (18)/((x - 1)²) ( (f(x + 2) - f(3))/(f(3)) )

### Step 1: Simplify Exponent Limit Given

f(3) = 18

:

T = e^ x \to 1 (18)/((x - 1)²) ( (f(x + 2) - f(3))/(18) )

T = e^ x \to 1 (f(x + 2) - f(3))/((x - 1)²)

This is a

(0)/(0)

form limit. ### Step 2: Apply L'Hopital's Rule Differentiate numerator and denominator w.r.t

x

:

T = e^ x \to 1 (f'(x + 2))/(2(x - 1))

This is still a

(0)/(0)

form since

f'(3) = 0

. Apply L'Hopital's Rule again:

T = e^ x \to 1 (f''(x + 2))/(2)

Substitute

x = 1

:

T = e(f''(3))/(2)

### Step 3: Final Calculation Given

f''(3) = 4

:

T = e(4)/(2) = e²

The question asks for

ₑ(T)

:

ₑ(T) = ₑ(e²) = 2

### Pattern Recognition When expanding

f(x)

around an extrema (

f'(a)=0

) inside a

1\infty

limit form, double L'Hopital isolates the second derivative divided by the Taylor factorial (

2! = 2

). The limit elegantly shrinks directly to

f''(a)/2

. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Limits and Derivatives Class 12 Maths: Limits, Continuity and Differentiability

Q21 jee_main_2026_21_jan_evening Limits of Sum

Let

[\cdot]

denote the greatest integer function and

f(x)= n\to\infty 1n³Σk=1ⁿ[ k²3x]

. Then

12Σj=1\inftyf(j)

is equal to

Numerical Answer. Answer: 2 to 2

Solution

### Related Formula

Sandwich Theorem for greatest integer function : x - 1 < [x] \leq x

Σk=1ⁿ k² = (n(n+1)(2n+1))/(6)

### Core Logic Evaluate

f(x)

using Squeeze Theorem on the summation bounds due to the Greatest Integer Function.

(k²)/(3^x) - 1 < [(k²)/(3^x)] \leq (k²)/(3^x)

### Step 1: Evaluate Limit f(x) Sum over bounds:

Σk=1ⁿ ( (k²)/(3^x) - 1 ) < Σk=1ⁿ [ (k²)/(3^x) ] \leq Σk=1ⁿ (k²)/(3^x)

(1)/(3^x) (n(n+1)(2n+1))/(6) - n < Σk=1ⁿ [ (k²)/(3^x) ] \leq (1)/(3^x) (n(n+1)(2n+1))/(6)

Divide by

n³

and apply limit

n \to \infty

:

n \to \infty ( (2n³ + 3n² + n)/(6n³ \cdot 3^x) - (1)/(n²) ) < f(x) \leq n \to \infty (2n³ + 3n² + n)/(6n³ \cdot 3^x)

f(x) = (2)/(6 \cdot 3^x) = (1)/(3 \cdot 3^x) = 13x+1

### Step 2: Evaluate Final Summation We need

12Σj=1\infty f(j)

:

12 Σj=1\infty 13j+1 = 12 ( (1)/(3²) + (1)/(3³) + \dots )

This is an infinite geometric progression with

a = (1)/(9)

and

r = (1)/(3)

.

Sum = (a)/(1 - r) = (1/9)/(1 - 1/3) = (1/9)/(2/3) = (1)/(6)

Finally,

12 \times (1)/(6) = 2

. ### Pattern Recognition When evaluating infinite limits over greatest integer sums

n \to \infty 1np+1 Σ [k^p/C]

, the G.I.F brackets can be safely replaced by exact values due to Sandwich theorem bounding the

-1

residual to

0

. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Limits, Continuity and Differentiability Class 11 Maths: Sequence and Series

Q5 jee_main_2026_22_january_evening Limits using Expansion

If

x \to 0 e(a-1)x + 2 bx + (c-2)e-xx x - ₑ(1+x) = 2

, then

a² + b² + c²

is equal to:

A. 5
B. 3
C. 7
D. 9

Solution

### Related Formula Standard Taylor series expansions:

e^x = 1 + x + (x²)/(2!) + , x = 1 - (x²)/(2!) + , ₑ(1+x) = x - (x²)/(2) +

### Core Logic Expand numerator and denominator around

x = 0

: Denominator:

x(1 - (x²)/(2)) - (x - (x²)/(2)) = (x²)/(2) + O(x³)

. For limit to be finite, coefficients of

x⁰

and

x¹

in numerator must be zero: - Coefficient of

x⁰

:

1 + 2 + c - 2 = 0 c = -1

- Coefficient of

x¹

:

(a-1) - (c-2) = 0 a - 1 + 3 = 0 a = -2

### Step 1: Coefficient of x^2 Numerator coefficient of

x²

is

((a-1)²)/(2) - b² + (c-2)/(2)

. Given limit value is

2

:

(((a-1)²)/(2) - b² + (c-2)/(2))/(1/2) = 2 (9)/(2) - b² - (3)/(2) = 1 b² = 2

### Step 2: Final Calculation

a² + b² + c² = (-2)² + 2 + (-1)² = 4 + 2 + 1 = 7

### Pattern Recognition Match powers of

x

in Taylor series to resolve indeterminate limit form

(0)/(0)

. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Limits, Continuity and Differentiability

Q15 jee_main_2026_22_january_evening Points of Discontinuity and Min Function

Let

[\cdot]

denote the greatest integer function, and let

f(x) = \sqrt(2)x, x²

. Let

S = x in (-2,2) : the function g(x) = |x|[x²] is discontinuous at x

. Then

Σx in S f(x)

equals:

A. $2 - \sqrt(2)$ $2 - \sqrt{2}$
B. $2\sqrt(6) - 3\sqrt(2)$ $2\sqrt{6} - 3\sqrt{2}$
C. $1 - \sqrt(2)$ $1 - \sqrt{2}$
D. $\sqrt(6) - 2\sqrt(2)$ $\sqrt{6} - 2\sqrt{2}$

Solution

### Related Formula Greatest integer function

[x²]

is discontinuous where

x²

takes integer values, except possibly where

|x| = 0

. ### Core Logic In

(-2, 2)

,

x² in [0, 4)

. Integer values occur at

x = 0, \pm 1, \pm\sqrt(2), \pm\sqrt(3)

. At

x = 0

,

g(0) = 0

and

x \to 0 g(x) = 0

, so

g(x)

is continuous at

x = 0

. Points of discontinuity:

S = -1, 1, -\sqrt(2), \sqrt(2), -\sqrt(3), \sqrt(3)

. ### Step 1: Evaluate f(x) for x in S For

f(x) = \sqrt(2)x, x²

: -

f(-1) = -\sqrt(2), 1 = -\sqrt(2)

-

f(1) = \sqrt(2), 1 = 1

-

f(-\sqrt(2)) = -2, 2 = -2

-

f(\sqrt(2)) = 2, 2 = 2

-

f(-\sqrt(3)) = -\sqrt(6), 3 = -\sqrt(6)

-

f(\sqrt(3)) = \sqrt(6), 3 = \sqrt(6)

### Step 2: Summation

Σx in S f(x) = -\sqrt(2) + 1 - 2 + 2 - \sqrt(6) + \sqrt(6) = 1 - \sqrt(2)

### Pattern Recognition Check origin continuity explicitly for

|x|[x²]

; evaluate

\sqrt(2)x, x²

case-by-case on set

S

. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Limits, Continuity and Differentiability

Q9 jee_main_2026_23_january_morning Continuity

Let

f(x) = cases (ax² + 2ax + 3)/(4x² + 4x - 3), & x \neq -(3)/(2), (1)/(2) b, & x = -(3)/(2), (1)/(2) cases

be continuous at

x = -(3)/(2)

. If

fof(x) = (7)/(5)

, then

x

is equal to:

A. $2$ $2$
B. $1$ $1$
C. $0$ $0$
D. $1.4$ $1.4$

Solution

### Core Logic For

f(x)

to be continuous at

x = -(3)/(2)

, the limit as

x \to -(3)/(2)

must exist and equal

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

.

x \to -3/2 (ax² + 2ax + 3)/((2x - 1)(2x + 3))

Since the denominator is zero at

x = -(3)/(2)

, for the limit to exist, the numerator must also be zero at

x = -(3)/(2)

. ### Step 1: Determine 'a' Set the numerator to

0

at

x = -(3)/(2)

:

a(-(3)/(2))² + 2a(-(3)/(2)) + 3 = 0

(9a)/(4) - 3a + 3 = 0

(-3a)/(4) + 3 = 0 \Rightarrow (3a)/(4) = 3 \Rightarrow a = 4

### Step 2: Simplify f(x) Substitute

a = 4

into

f(x)

for

x \neq -(3)/(2), (1)/(2)

:

f(x) = (4x² + 8x + 3)/((2x - 1)(2x + 3))

Factorizing the numerator:

4x² + 8x + 3 = (2x + 1)(2x + 3)

Thus,

f(x) = ((2x + 1)(2x + 3))/((2x - 1)(2x + 3)) = (2x + 1)/(2x - 1)

for

x \neq -(3)/(2)

. ### Step 3: Solve f(f(x)) = 7/5 Evaluate

fof(x)

:

f(f(x)) = f((2x + 1)/(2x - 1)) = (2((2x + 1)/(2x - 1)) + 1)/(2((2x + 1)/(2x - 1)) - 1)

= (2(2x + 1) + (2x - 1))/(2(2x + 1) - (2x - 1)) = (4x + 2 + 2x - 1)/(4x + 2 - 2x + 1) = (6x + 1)/(2x + 3)

Equate to

(7)/(5)

:

(6x + 1)/(2x + 3) = (7)/(5) \Rightarrow 5(6x + 1) = 7(2x + 3)

30x + 5 = 14x + 21 \Rightarrow 16x = 16 \Rightarrow x = 1

### Pattern Recognition Indeterminate forms at points of continuity explicitly lock polynomial coefficients. Always resolve the

0/0

form to extract missing variables before addressing composite functions. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Limits, Continuity and Differentiability

Let the function $f(x) = (x² - 1)|x² - ax + 2| + |x|$ be not differentiable at the two points $x = α = 2$ $x = \alpha = 2$ and $x = β$ $x = \beta$. Then the distance of the point $(α, β)$ $(\alpha, \beta)$ from the line $12x + 5y + 10 = 0$ $12x + 5y + 10 = 0$ is equal to:

Solution & Explanation

Core Logic

Step 1: Evaluation of Roots

Step 2: Conclusion

Pattern Recognition

Chapter Mix

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Solution & Explanation

Core Logic

Step 1: Evaluation of Roots

Step 2: Conclusion

Pattern Recognition

Chapter Mix

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