Solution & Explanation
### Related Formula
The greatest integer function [u]$[u]$ changes value and experiences a step discontinuity at any point where its inner argument u$u$ takes on an integer value.
### Core Logic
Analyze the potential points where either component function argument changes into an integer within the interval x in [0, 4]$x \in [0, 4]$.
1. For left[fracx^22right]$\left[\frac{x^2}{2}\right]$:
fracx^22$\frac{x^2}{2}$ can range from frac02 = 0$\frac{0}{2} = 0$ up to frac162 = 8$\frac{16}{2} = 8$.
Integer values are reached at fracx^22 = 0, 1, 2, 3, 4, 5, 6, 7, 8$\frac{x^2}{2} = 0, 1, 2, 3, 4, 5, 6, 7, 8$, which means critical test locations are:
x = 0, \, sqrt2, \, 2, \, sqrt6, \, sqrt8, \, sqrt10, \, sqrt12, \, sqrt14, \, 4$$x = 0, \, \sqrt{2}, \, 2, \, \sqrt{6}, \, \sqrt{8}, \, \sqrt{10}, \, \sqrt{12}, \, \sqrt{14}, \, 4$$
2. For [sqrtx]$[\sqrt{x}]$:
sqrtx$\sqrt{x}$ can range from sqrt0 = 0$\sqrt{0} = 0$ to sqrt4 = 2$\sqrt{4} = 2$.
Integer values are reached at sqrtx = 0, 1, 2$\sqrt{x} = 0, 1, 2$, which means critical test locations are:
x = 0, \, 1, \, 4$$x = 0, \, 1, \, 4$$
### Step 1: Audit Each Critical Point
Combine the set of test points within domain boundaries (0, 4)$(0, 4)$:
x in \1, \, sqrt2, \, 2, \, sqrt6, \, sqrt8, \, sqrt10, \, sqrt12, \, sqrt14\$$x \in \{1, \, \sqrt{2}, \, 2, \, \sqrt{6}, \, \sqrt{8}, \, \sqrt{10}, \, \sqrt{12}, \, \sqrt{14}\}$$
Let's evaluate the left and right hand limits at these specific values:
- At x = 1$x = 1$: [sqrtx]$[\sqrt{x}]$ steps up while left[fracx^22right]$\left[\frac{x^2}{2}\right]$ is constant implies$\implies$ Discontinuous.
- At x = sqrt2$x = \sqrt{2}$: left[fracx^22right]$\left[\frac{x^2}{2}\right]$ steps up while [sqrtx]$[\sqrt{x}]$ is constant implies$\implies$ Discontinuous.
- At x = 2$x = 2$: Both functions experience an simultaneous integer step. Let's inspect:
- f(2) = [2] - [sqrt2] = 2 - 1 = 1$f(2) = [2] - [\sqrt{2}] = 2 - 1 = 1$
- f(2^-) = [1.99] - [1.41] = 1 - 1 = 0$f(2^-) = [1.99] - [1.41] = 1 - 1 = 0$
Since LHL neq$\neq$ value at point, it is Discontinuous.
Continuing this verification down the full combined list confirms that none of the step jumps cancel each other out.
### Step 2: Sum the Discontinuity Points
Counting all isolated inner points within (0, 4)$(0, 4)$ yields exactly 8$8$ locations:
textTotal Points = 8$$\text{Total Points} = 8$$
### Pattern Recognition
When two greatest integer functions drop steps simultaneously at the same point (like at x=2$x=2$), always write out the explicit left and right limits manually, as simultaneous steps occasionally step in matching directions and maintain unexpected continuity.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Mathematics: Limits, Continuity and Differentiability
More Limits, Continuity and Differentiability Previous-Year Questions
Q10
jee_main_2026_21_jan_morning
1^infinity Limit Form with L'Hopital's Rule
Let
f: R to (0, infty)$f: R \to (0, \infty)$ be a
twice differentiable function such that
f(3) = 18$f(3) = 18$ ,
f'(3) = 0$f'(3) = 0$ and
f''(3) = 4$f''(3) = 4$ .
Then
lim_x to 1 left( log_e left( fracf(2 + x)f(3) right)^frac18(x-1)^2 right)$\lim_{x \to 1} \left( \log_{e} \left( \frac{f(2 + x)}{f(3)} \right)^{\frac{18}{(x-1)^2}} \right)$ is equal to:
Solution
### Related Formula
For a limit of 1^infty$1^{\infty}$ form, lim_x to a [g(x)]^h(x)$\lim_{x \to a} [g(x)]^{h(x)}$ equals:
e^lim_x to a h(x)[g(x) - 1]$$e^{\lim_{x \to a} h(x)[g(x) - 1]}$$
### Core Logic
Let T = lim_x to 1 left( fracf(x + 2)f(3) right)^frac18(x - 1)^2$T = \lim_{x \to 1} \left( \frac{f(x + 2)}{f(3)} \right)^{\frac{18}{(x - 1)^2}}$.
As x to 1$x \to 1$, fracf(x+2)f(3) to fracf(3)f(3) = 1$\frac{f(x+2)}{f(3)} \to \frac{f(3)}{f(3)} = 1$. The exponent goes to infty$\infty$.
This is a standard 1^infty$1^{\infty}$ form.
T = e^lim_x to 1 frac18(x - 1)^2 left( fracf(x + 2) - f(3)f(3) right)$$T = e^{\lim_{x \to 1} \frac{18}{(x - 1)^2} \left( \frac{f(x + 2) - f(3)}{f(3)} \right)}$$
### Step 1: Simplify Exponent Limit
Given f(3) = 18$f(3) = 18$:
T = e^lim_x to 1 frac18(x - 1)^2 left( fracf(x + 2) - f(3)18 right)$$T = e^{\lim_{x \to 1} \frac{18}{(x - 1)^2} \left( \frac{f(x + 2) - f(3)}{18} \right)}$$
T = e^lim_x to 1 fracf(x + 2) - f(3)(x - 1)^2$$T = e^{\lim_{x \to 1} \frac{f(x + 2) - f(3)}{(x - 1)^2}}$$
This is a frac00$\frac{0}{0}$ form limit.
### Step 2: Apply L'Hopital's Rule
Differentiate numerator and denominator w.r.t x$x$:
T = e^lim_x to 1 fracf'(x + 2)2(x - 1)$$T = e^{\lim_{x \to 1} \frac{f'(x + 2)}{2(x - 1)}}$$
This is still a frac00$\frac{0}{0}$ form since f'(3) = 0$f'(3) = 0$.
Apply L'Hopital's Rule again:
T = e^lim_x to 1 fracf''(x + 2)2$$T = e^{\lim_{x \to 1} \frac{f''(x + 2)}{2}}$$
Substitute x = 1$x = 1$:
T = e^fracf''(3)2$$T = e^{\frac{f''(3)}{2}}$$
### Step 3: Final Calculation
Given f''(3) = 4$f''(3) = 4$:
T = e^frac42 = e^2$$T = e^{\frac{4}{2}} = e^2$$
The question asks for log_e(T)$\log_e(T)$:
log_e(T) = log_e(e^2) = 2$$\log_e(T) = \log_e(e^2) = 2$$
### Pattern Recognition
When expanding f(x)$f(x)$ around an extrema (f'(a)=0$f'(a)=0$) inside a 1^infty$1^{\infty}$ limit form, double L'Hopital isolates the second derivative divided by the Taylor factorial (2! = 2$2! = 2$). The limit elegantly shrinks directly to f''(a)/2$f''(a)/2$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Maths: Limits and Derivatives
Class 12 Maths: Limits, Continuity and Differentiability
Q21
jee_main_2026_21_jan_evening
Limits of Sum
Let
[cdot]$[\cdot]$ denote the
greatest integer function and
f(x)=lim_ntoinftyfrac1n^3sum_k=1^nleft[frack^23^xright]$f(x)=\lim_{n\to\infty}\frac{1}{n^{3}}\sum_{k=1}^{n}\left[\frac{k^{2}}{3^{x}}\right]$. Then
12sum_j=1^inftyf(j)$12\sum_{j=1}^{\infty}f(j)$ is equal to
Numerical Answer. Answer: 2 to 2
Solution
### Related Formula
textSandwich Theorem for greatest integer function: x - 1 < [x] leq x$$\text{Sandwich Theorem for greatest integer function: } x - 1 < [x] \leq x$$
sum_k=1^n k^2 = fracn(n+1)(2n+1)6$$\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}$$
### Core Logic
Evaluate
f(x)$f(x)$ using Squeeze Theorem on the summation bounds due to the
Greatest Integer Function.
frack^23^x - 1 < left[frack^23^xright] leq frack^23^x$$\frac{k^2}{3^x} - 1 < \left[\frac{k^2}{3^x}\right] \leq \frac{k^2}{3^x}$$
### Step 1: Evaluate Limit f(x)
Sum over bounds:
sum_k=1^n left( frack^23^x - 1 right) < sum_k=1^n left[ frack^23^x right] leq sum_k=1^n frack^23^x$$\sum_{k=1}^n \left( \frac{k^2}{3^x} - 1 \right) < \sum_{k=1}^n \left[ \frac{k^2}{3^x} \right] \leq \sum_{k=1}^n \frac{k^2}{3^x}$$
frac13^x fracn(n+1)(2n+1)6 - n < sum_k=1^n left[ frack^23^x right] leq frac13^x fracn(n+1)(2n+1)6$$\frac{1}{3^x} \frac{n(n+1)(2n+1)}{6} - n < \sum_{k=1}^n \left[ \frac{k^2}{3^x} \right] \leq \frac{1}{3^x} \frac{n(n+1)(2n+1)}{6}$$
Divide by
n^3$n^3$ and apply limit
n to infty$n \to \infty$:
lim_n to infty left( frac2n^3 + 3n^2 + n6n^3 cdot 3^x - frac1n^2 right) < f(x) leq lim_n to infty frac2n^3 + 3n^2 + n6n^3 cdot 3^x$$\lim_{n \to \infty} \left( \frac{2n^3 + 3n^2 + n}{6n^3 \cdot 3^x} - \frac{1}{n^2} \right) < f(x) \leq \lim_{n \to \infty} \frac{2n^3 + 3n^2 + n}{6n^3 \cdot 3^x}$$
f(x) = frac26 cdot 3^x = frac13 cdot 3^x = frac13^x+1$$f(x) = \frac{2}{6 \cdot 3^x} = \frac{1}{3 \cdot 3^x} = \frac{1}{3^{x+1}}$$
### Step 2: Evaluate Final Summation
We need
12sum_j=1^infty f(j)$12\sum_{j=1}^{\infty} f(j)$:
12 sum_j=1^infty frac13^j+1 = 12 left( frac13^2 + frac13^3 + ldots right)$$12 \sum_{j=1}^{\infty} \frac{1}{3^{j+1}} = 12 \left( \frac{1}{3^2} + \frac{1}{3^3} + \ldots \right)$$
This is an infinite geometric progression with
a = frac19$a = \frac{1}{9}$ and
r = frac13$r = \frac{1}{3}$.
textSum = fraca1 - r = frac1/91 - 1/3 = frac1/92/3 = frac16$$\text{Sum} = \frac{a}{1 - r} = \frac{1/9}{1 - 1/3} = \frac{1/9}{2/3} = \frac{1}{6}$$
Finally,
12 times frac16 = 2$12 \times \frac{1}{6} = 2$.
### Pattern Recognition
When evaluating infinite limits over greatest integer sums
lim_n to infty frac1n^p+1 sum [k^p/C]$\lim_{n \to \infty} \frac{1}{n^{p+1}} \sum [k^p/C]$, the G.I.F brackets can be safely replaced by exact values due to Sandwich theorem bounding the
-1$-1$ residual to
0$0$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Maths: Limits, Continuity and Differentiability
Class 11 Maths: Sequence and Series
Q5
jee_main_2026_22_january_evening
Limits using Expansion
If lim_x to 0 frace^(a-1)x + 2 cos bx + (c-2)e^-xx cos x - log_e(1+x) = 2$\lim_{x \to 0} \frac{e^{(a-1)x} + 2 \cos bx + (c-2)e^{-x}}{x \cos x - \log_e(1+x)} = 2$, then a^2 + b^2 + c^2$a^2 + b^2 + c^2$ is equal to:
Solution
### Related Formula
Standard Taylor series expansions:
e^x = 1 + x + fracx^22! + dots, quad cos x = 1 - fracx^22! + dots, quad log_e(1+x) = x - fracx^22 + dots$$e^x = 1 + x + \frac{x^2}{2!} + \dots, \quad \cos x = 1 - \frac{x^2}{2!} + \dots, \quad \log_e(1+x) = x - \frac{x^2}{2} + \dots$$
### Core Logic
Expand numerator and denominator around x = 0$x = 0$:
Denominator: xleft(1 - fracx^22right) - left(x - fracx^22right) = fracx^22 + O(x^3)$x\left(1 - \frac{x^2}{2}\right) - \left(x - \frac{x^2}{2}\right) = \frac{x^2}{2} + O(x^3)$.
For limit to be finite, coefficients of x^0$x^0$ and x^1$x^1$ in numerator must be zero:
- Coefficient of x^0$x^0$: 1 + 2 + c - 2 = 0 implies c = -1$1 + 2 + c - 2 = 0 \implies c = -1$
- Coefficient of x^1$x^1$: (a-1) - (c-2) = 0 implies a - 1 + 3 = 0 implies a = -2$(a-1) - (c-2) = 0 \implies a - 1 + 3 = 0 \implies a = -2$
### Step 1: Coefficient of x^2
Numerator coefficient of x^2$x^2$ is frac(a-1)^22 - b^2 + fracc-22$\frac{(a-1)^2}{2} - b^2 + \frac{c-2}{2}$.
Given limit value is 2$2$:
fracfrac(a-1)^22 - b^2 + fracc-221/2 = 2 implies frac92 - b^2 - frac32 = 1 implies b^2 = 2$$\frac{\frac{(a-1)^2}{2} - b^2 + \frac{c-2}{2}}{1/2} = 2 \implies \frac{9}{2} - b^2 - \frac{3}{2} = 1 \implies b^2 = 2$$
### Step 2: Final Calculation
a^2 + b^2 + c^2 = (-2)^2 + 2 + (-1)^2 = 4 + 2 + 1 = 7$$a^2 + b^2 + c^2 = (-2)^2 + 2 + (-1)^2 = 4 + 2 + 1 = 7$$
### Pattern Recognition
Match powers of x$x$ in Taylor series to resolve indeterminate limit form frac00$\frac{0}{0}$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Maths: Limits, Continuity and Differentiability
Q15
jee_main_2026_22_january_evening
Points of Discontinuity and Min Function
Let [cdot]$[\cdot]$ denote the greatest integer function, and let f(x) = min\sqrt2x, x^2\$f(x) = \min\{\sqrt{2}x, x^2\}$. Let S = \x in (-2,2) : textthe function g(x) = |x|[x^2] text is discontinuous at x\$S = \{x \in (-2,2) : \text{the function } g(x) = |x|[x^2] \text{ is discontinuous at } x\}$. Then sum_x in S f(x)$\sum_{x \in S} f(x)$ equals:
- A. 2 - sqrt2$2 - \sqrt{2}$
- B. 2sqrt6 - 3sqrt2$2\sqrt{6} - 3\sqrt{2}$
- C. 1 - sqrt2$1 - \sqrt{2}$
- D. sqrt6 - 2sqrt2$\sqrt{6} - 2\sqrt{2}$
Solution
### Related Formula
Greatest integer function [x^2]$[x^2]$ is discontinuous where x^2$x^2$ takes integer values, except possibly where |x| = 0$|x| = 0$.
### Core Logic
In (-2, 2)$(-2, 2)$, x^2 in [0, 4)$x^2 \in [0, 4)$. Integer values occur at x = 0, pm 1, pmsqrt2, pmsqrt3$x = 0, \pm 1, \pm\sqrt{2}, \pm\sqrt{3}$.
At x = 0$x = 0$, g(0) = 0$g(0) = 0$ and lim_x to 0 g(x) = 0$\lim_{x \to 0} g(x) = 0$, so g(x)$g(x)$ is continuous at x = 0$x = 0$.
Points of discontinuity: S = \-1, 1, -sqrt2, sqrt2, -sqrt3, sqrt3\$S = \{-1, 1, -\sqrt{2}, \sqrt{2}, -\sqrt{3}, \sqrt{3}\}$.
### Step 1: Evaluate f(x) for x in S
For f(x) = min\sqrt2x, x^2\$f(x) = \min\{\sqrt{2}x, x^2\}$:
- f(-1) = min\-sqrt2, 1\ = -sqrt2$f(-1) = \min\{-\sqrt{2}, 1\} = -\sqrt{2}$
- f(1) = min\sqrt2, 1\ = 1$f(1) = \min\{\sqrt{2}, 1\} = 1$
- f(-sqrt2) = min\-2, 2\ = -2$f(-\sqrt{2}) = \min\{-2, 2\} = -2$
- f(sqrt2) = min\2, 2\ = 2$f(\sqrt{2}) = \min\{2, 2\} = 2$
- f(-sqrt3) = min\-sqrt6, 3\ = -sqrt6$f(-\sqrt{3}) = \min\{-\sqrt{6}, 3\} = -\sqrt{6}$
- f(sqrt3) = min\sqrt6, 3\ = sqrt6$f(\sqrt{3}) = \min\{\sqrt{6}, 3\} = \sqrt{6}$
### Step 2: Summation
sum_x in S f(x) = -sqrt2 + 1 - 2 + 2 - sqrt6 + sqrt6 = 1 - sqrt2$$\sum_{x \in S} f(x) = -\sqrt{2} + 1 - 2 + 2 - \sqrt{6} + \sqrt{6} = 1 - \sqrt{2}$$
### Pattern Recognition
Check origin continuity explicitly for |x|[x^2]$|x|[x^2]$; evaluate min\sqrt2x, x^2\$\min\{\sqrt{2}x, x^2\}$ case-by-case on set S$S$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Maths: Limits, Continuity and Differentiability
Q9
jee_main_2026_23_january_morning
Continuity
Let f(x) = begincases fracax^2 + 2ax + 34x^2 + 4x - 3, & x neq -frac32, frac12 \\ b, & x = -frac32, frac12 endcases$f(x) = \begin{cases} \frac{ax^2 + 2ax + 3}{4x^2 + 4x - 3}, & x \neq -\frac{3}{2}, \frac{1}{2} \\ b, & x = -\frac{3}{2}, \frac{1}{2} \end{cases}$ be continuous at x = -frac32$x = -\frac{3}{2}$. If fof(x) = frac75$fof(x) = \frac{7}{5}$, then x$x$ is equal to:
- A. 2$2$
- B. 1$1$
- C. 0$0$
- D. 1.4$1.4$
Solution
### Core Logic
For f(x)$f(x)$ to be continuous at x = -frac32$x = -\frac{3}{2}$, the limit as x to -frac32$x \to -\frac{3}{2}$ must exist and equal fleft(-frac32right) = b$f\left(-\frac{3}{2}\right) = b$.
lim_x to -3/2 fracax^2 + 2ax + 3(2x - 1)(2x + 3)$$\lim_{x \to -3/2} \frac{ax^2 + 2ax + 3}{(2x - 1)(2x + 3)}$$
Since the denominator is zero at x = -frac32$x = -\frac{3}{2}$, for the limit to exist, the numerator must also be zero at x = -frac32$x = -\frac{3}{2}$.
### Step 1: Determine 'a'
Set the numerator to 0$0$ at x = -frac32$x = -\frac{3}{2}$:
aleft(-frac32right)^2 + 2aleft(-frac32right) + 3 = 0$$a\left(-\frac{3}{2}\right)^2 + 2a\left(-\frac{3}{2}\right) + 3 = 0$$
frac9a4 - 3a + 3 = 0$$\frac{9a}{4} - 3a + 3 = 0$$
frac-3a4 + 3 = 0 Rightarrow frac3a4 = 3 Rightarrow a = 4$$\frac{-3a}{4} + 3 = 0 \Rightarrow \frac{3a}{4} = 3 \Rightarrow a = 4$$
### Step 2: Simplify f(x)
Substitute a = 4$a = 4$ into f(x)$f(x)$ for x neq -frac32, frac12$x \neq -\frac{3}{2}, \frac{1}{2}$:
f(x) = frac4x^2 + 8x + 3(2x - 1)(2x + 3)$$f(x) = \frac{4x^2 + 8x + 3}{(2x - 1)(2x + 3)}$$
Factorizing the numerator:
4x^2 + 8x + 3 = (2x + 1)(2x + 3)$$4x^2 + 8x + 3 = (2x + 1)(2x + 3)$$
Thus, f(x) = frac(2x + 1)(2x + 3)(2x - 1)(2x + 3) = frac2x + 12x - 1$f(x) = \frac{(2x + 1)(2x + 3)}{(2x - 1)(2x + 3)} = \frac{2x + 1}{2x - 1}$ for x neq -frac32$x \neq -\frac{3}{2}$.
### Step 3: Solve f(f(x)) = 7/5
Evaluate fof(x)$fof(x)$:
f(f(x)) = fleft(frac2x + 12x - 1right) = frac2left(frac2x + 12x - 1right) + 12left(frac2x + 12x - 1right) - 1$$f(f(x)) = f\left(\frac{2x + 1}{2x - 1}\right) = \frac{2\left(\frac{2x + 1}{2x - 1}\right) + 1}{2\left(\frac{2x + 1}{2x - 1}\right) - 1}$$
= frac2(2x + 1) + (2x - 1)2(2x + 1) - (2x - 1) = frac4x + 2 + 2x - 14x + 2 - 2x + 1 = frac6x + 12x + 3$$= \frac{2(2x + 1) + (2x - 1)}{2(2x + 1) - (2x - 1)} = \frac{4x + 2 + 2x - 1}{4x + 2 - 2x + 1} = \frac{6x + 1}{2x + 3}$$
Equate to frac75$\frac{7}{5}$:
frac6x + 12x + 3 = frac75 Rightarrow 5(6x + 1) = 7(2x + 3)$$\frac{6x + 1}{2x + 3} = \frac{7}{5} \Rightarrow 5(6x + 1) = 7(2x + 3)$$
30x + 5 = 14x + 21 Rightarrow 16x = 16 Rightarrow x = 1$$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$0/0$ form to extract missing variables before addressing composite functions.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Maths: Limits, Continuity and Differentiability
