Let f: R (0, infty) be a twice differentiable function such that f(3) = 18 , f'(3) = 0 and f''(3) = 4 . Then 1 ( ( (f(2…

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More Limits, Continuity and Differentiability Previous-Year Questions

Q3 jee_main_2026_21_jan_evening Differentiation

Let

f(x) = x³ + x² f'(1) + 2x f''(2) + f'''(3)

x in R

. Then the value of

f'(5)

is :

A. $(62)/(5)$ $\frac{62}{5}$
B. $(657)/(5)$ $\frac{657}{5}$
C. $(2)/(5)$ $\frac{2}{5}$
D. $(117)/(5)$ $\frac{117}{5}$

Solution

Related Formula

(d)/(dx) (xⁿ) = nxⁿ⁻¹

Core Logic

Differentiate the given polynomial function iteratively to find expressions for $f'(x)$ $f'(x)$ and $f''(x)$ $f''(x)$, treating $f'(1)$ $f'(1)$, $f''(2)$ $f''(2)$, and $f'''(3)$ $f'''(3)$ as constant values.

Step 1: First and Second Derivatives

f'(x) = 3x² + 2x f'(1) + 2f''(2)

f''(x) = 6x + 2f'(1)

Substitute $x = 2$ $x = 2$ into the second derivative to create a relation:

f''(2) = 12 + 2f'(1)

Step 2: Substitute and Solve for Constants

Substitute $f''(2)$ $f''(2)$ back into $f'(x)$ $f'(x)$:

f'(x) = 3x² + 2x f'(1) + 2(12 + 2f'(1))

f'(x) = 3x² + 2(x + 2)f'(1) + 24

Now put $x = 1$ $x = 1$ to solve for $f'(1)$ $f'(1)$:

f'(1) = 3(1)² + 2(1 + 2)f'(1) + 24

f'(1) = 3 + 6f'(1) + 24

-5f'(1) = 27 f'(1) = -(27)/(5)

Step 3: Calculate Required Value

Determine the exact form of $f'(x)$ $f'(x)$:

f'(x) = 3x² + 2(x + 2)(-(27)/(5)) + 24

f'(x) = 3x² - (54)/(5)x - (108)/(5) + (120)/(5)

f'(x) = 3x² - (54)/(5)x + (12)/(5)

Substitute $x = 5$ $x = 5$ to find $f'(5)$ $f'(5)$:

f'(5) = 3(25) - (54)/(5)(5) + (12)/(5)

f'(5) = 75 - 54 + (12)/(5) = 21 + (12)/(5) = (105 + 12)/(5) = (117)/(5)

Pattern Recognition

Treat derivatives evaluated at specific points (like $f'(1)$ $f'(1)$) as fixed scalar constants. Substitute values back sequentially to solve the linear system of constants.

Chapter Mix

Class 12 Maths: Method of Differentiation

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 $f(x)$ $f(x)$ 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 $n³$ $n^3$ and apply limit $n \to \infty$ $n \to \infty$:

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 $12Σj=1\infty f(j)$ $12\sum_{j=1}^{\infty} f(j)$:

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

This is an infinite geometric progression with $a = (1)/(9)$ $a = \frac{1}{9}$ and $r = (1)/(3)$ $r = \frac{1}{3}$.

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

Finally, $12 \times (1)/(6) = 2$ $12 \times \frac{1}{6} = 2$.

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 $-1$ $-1$ residual to $0$ $0$.

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

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 $x = 0$ $x = 0$: Denominator: $x(1 - (x²)/(2)) - (x - (x²)/(2)) = (x²)/(2) + O(x³)$ . For limit to be finite, coefficients of $x⁰$ $x^0$ and $x¹$ $x^1$ in numerator must be zero:

Coefficient of $x⁰$ $x^0$: $1 + 2 + c - 2 = 0 c = -1$ $1 + 2 + c - 2 = 0 \implies c = -1$
Coefficient of $x¹$ $x^1$: $(a-1) - (c-2) = 0 a - 1 + 3 = 0 a = -2$

Step 1: Coefficient of x^2

Numerator coefficient of $x²$ $x^2$ is $((a-1)²)/(2) - b² + (c-2)/(2)$ . Given limit value is $2$ $2$:

(((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 $x$ $x$ in Taylor series to resolve indeterminate limit form $(0)/(0)$ $\frac{0}{0}$.

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 $[x²]$ $[x^2]$ is discontinuous where $x²$ $x^2$ takes integer values, except possibly where $|x| = 0$ $|x| = 0$.

Core Logic

In $(-2, 2)$ $(-2, 2)$, $x² in [0, 4)$ $x^2 \in [0, 4)$. Integer values occur at $x = 0, \pm 1, \pm\sqrt(2), \pm\sqrt(3)$ . At $x = 0$ $x = 0$, $g(0) = 0$ $g(0) = 0$ and $x \to 0 g(x) = 0$ $\lim_{x \to 0} g(x) = 0$, so $g(x)$ $g(x)$ is continuous at $x = 0$ $x = 0$. Points of discontinuity: $S = -1, 1, -\sqrt(2), \sqrt(2), -\sqrt(3), \sqrt(3)$ .

Step 1: Evaluate f(x) for x in S

For $f(x) = \sqrt(2)x, x²$ $f(x) = \min\{\sqrt{2}x, x^2\}$:

$f(-1) = -\sqrt(2), 1 = -\sqrt(2)$
$f(1) = \sqrt(2), 1 = 1$ $f(1) = \min\{\sqrt{2}, 1\} = 1$
$f(-\sqrt(2)) = -2, 2 = -2$ $f(-\sqrt{2}) = \min\{-2, 2\} = -2$
$f(\sqrt(2)) = 2, 2 = 2$ $f(\sqrt{2}) = \min\{2, 2\} = 2$
$f(-\sqrt(3)) = -\sqrt(6), 3 = -\sqrt(6)$
$f(\sqrt(3)) = \sqrt(6), 3 = \sqrt(6)$

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 $|x|[x²]$ $|x|[x^2]$; evaluate $\sqrt(2)x, x²$ $\min\{\sqrt{2}x, x^2\}$ case-by-case on set $S$ $S$.

Chapter Mix

Class 12 Maths: Limits, Continuity and Differentiability

Solution & Explanation

Related Formula

Core Logic

Step 1: Simplify Exponent Limit

Step 2: Apply L'Hopital's Rule

Step 3: Final Calculation

Pattern Recognition

Chapter Mix

Reference Study Guides

More Limits, Continuity and Differentiability Previous-Year Questions

Solution

Related Formula

Core Logic

Step 1: First and Second Derivatives

Step 2: Substitute and Solve for Constants

Step 3: Calculate Required Value

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Evaluate Limit f(x)

Step 2: Evaluate Final Summation

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Coefficient of x^2

Step 2: Final Calculation

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Evaluate f(x) for x in S

Step 2: Summation

Pattern Recognition

Chapter Mix

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