The number of real solution(s) of the equation x2+3x+2=|x-3|, |x+2| is:

Year	2026	2025	2024	Total
Questions	12	24	16	52

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 $f(x)$ $f(x)$ to be continuous at $x = -(3)/(2)$ $x = -\frac{3}{2}$, the limit as $x \to -(3)/(2)$ $x \to -\frac{3}{2}$ must exist and equal $f(-(3)/(2)) = b$ $f\left(-\frac{3}{2}\right) = b$.

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

Since the denominator is zero at $x = -(3)/(2)$ $x = -\frac{3}{2}$, for the limit to exist, the numerator must also be zero at $x = -(3)/(2)$ $x = -\frac{3}{2}$.

Step 1: Determine 'a'

Set the numerator to $0$ $0$ at $x = -(3)/(2)$ $x = -\frac{3}{2}$:

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 $a = 4$ $a = 4$ into $f(x)$ $f(x)$ for $x \neq -(3)/(2), (1)/(2)$ $x \neq -\frac{3}{2}, \frac{1}{2}$:

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 $x \neq -(3)/(2)$ $x \neq -\frac{3}{2}$.

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

Evaluate $fof(x)$ $fof(x)$:

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 $(7)/(5)$ $\frac{7}{5}$:

(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 $0/0$ $0/0$ form to extract missing variables before addressing composite functions.

Chapter Mix

Class 12 Maths: Limits, Continuity and Differentiability

Q1 jee_main_2026_23_january_evening Continuity of a Function

If

f(x) = cases (a | x | + x² - 2( | x |)( | x |))/(x), & x \neq 0 b, & x = 0 cases

is continuous at

x = 0

, then

a + b

is equal to :

A. $1$ $1$
B. $2$ $2$
C. $0$ $0$
D. $4$ $4$

Solution

Related Formula

For a function to be continuous at $x=0$ $x=0$:

x\to0⁻f(x) = x\to0⁺f(x) = f(0)

Core Logic

For continuity at $x=0$ $x=0$, evaluate the left-hand limit (LHL) and right-hand limit (RHL).

LHL:

x\to0⁻ a|x|+x²-2 |x| |x|x = h\to0 ah+h²-2( ) -h

$= -a + 2$ $= -a + 2$

RHL:

x\to0⁺ a|x|+x²-2 |x| |x|x = h\to0 ah+h²-2( ) h

$= a - 2$ $= a - 2$

Equating both limits to $f(0) = b$ $f(0) = b$: $-a+2 = a-2 = b$ $-a+2 = a-2 = b$

Step 1: Final Calculation

From the above equations:

2a = 4 a = 2

Substitute $a=2$ $a=2$ to find $b$ $b$: $b = 2 - 2 = 0$ $b = 2 - 2 = 0$ Therefore, $a + b = 2 + 0 = 2$ $a + b = 2 + 0 = 2$.

Pattern Recognition

Since $|x|$ $|x|$ behaves differently on left and right, LHL and RHL will have opposite signs for the $|x|/x$ $|x|/x$ term. This immediately forces $a$ $a$ to balance out the remaining expansion limits.

Chapter Mix

Class 12 Maths: Limits, Continuity and Differentiability

Q1 jee_main_2026_24_january_morning Continuity and L'Hospital's Rule

If the function

f(x) = e^x(ex - x - 1) + ₑ( x + x) - x x - x

is Continuous at

x = 0

, then the value of

f(0)

is equal to

A. $2$ $2$
B. $(2)/(3)$ $\frac{2}{3}$
C. $(1)/(2)$ $\frac{1}{2}$
D. $(3)/(2)$ $\frac{3}{2}$

Solution

Related Formula

f(0) = x \to 0 f(x)

x \to 0 ( x - x)/(x³) = (1)/(3)

Core Logic

f(0) = x \to 0 ex - e^x + ln( x + x) - x x - x

Applying L'Hospital's rule:

Step 1: Differentiation

\Rightarrow f(0) = x \to 0 ex \cdot ² x - e^x + x - 1 ² x - 1

\Rightarrow f(0) = x \to 0 ex ( ² x - 1) + (ex - e^x) + x - 1 ² x

Step 2: Limit Evaluation

\Rightarrow f(0) = x \to 0 ( ex + e^x (ex - x - 1) ² x + (1)/( x + 1) )

\Rightarrow f(0) = 1 + 0 + (1)/(2) = (3)/(2)

Pattern Recognition

When expanding or using L'Hospital's, breaking the numerator into standard limits like $(e^t - 1)/t$ $(e^t - 1)/t$ and observing secant/tangent expansions simplifies the process instantly.

Chapter Mix

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

Q16 jee_main_2026_24_january_morning Differentiability of Piecewise Functions

Let

α, β in R

be such that the function

f (x) = cases 2 α (x² - 2) + 2 β x & , x < 1 (α + 3) x + (α - β) & , x \geq 1 cases

be differentiable at all

x in R

. Then

34(α + β)

is equal to

A. $84$ $84$
B. $48$ $48$
C. $36$ $36$
D. $24$ $24$

Solution

Related Formula

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

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

Core Logic

Since $f(x)$ $f(x)$ is differentiable at $x=1$ $x=1$, it must be continuous at $x=1$ $x=1$. $f(x) = cases 2α x² + 2β x - 4α & ; x < 1 (α + 3)x + α - β & ; x \geq 1 cases$

Step 1: Continuity Check

f(1^-) = -2α + 2β

f(1^+) = (α + 3) + α - β = 2α - β + 3

Equating:

-2α + 2β = 2α - β + 3

4α - 3β + 3 = 0 (1)

Step 2: Differentiability Check

Differentiate both branches:

f'(x) = cases 4α x + 2β & ; x < 1 α + 3 & ; x > 1 cases

Equate at $x=1$ $x=1$:

f'(1^-) = 4α + 2β

f'(1^+) = α + 3

4α + 2β = α + 3 \Rightarrow 3α + 2β - 3 = 0 (2)

Step 3: Solving Equations

From (2), $β = (3 - 3α)/(2)$ $\beta = \frac{3 - 3\alpha}{2}$. Substitute into (1):

4α - 3((3 - 3α)/(2)) + 3 = 0

8α - 9 + 9α + 6 = 0 \Rightarrow 17α - 3 = 0 \Rightarrow α = (3)/(17)

β = (3 - 9/17)/(2) = (42)/(34) = (21)/(17)

Step 4: Evaluate Final Target

34(α + β) = 34( (3)/(17) + (21)/(17) ) = 34 \times (24)/(17) = 48

Pattern Recognition

For piecewise polynomials, standard constraints of LHL=RHL and LHD=RHD form a solvable linear system. Differentiating standard polynomials directly instead of applying first-principle limits saves 2+ minutes.

Chapter Mix

Class 12 Maths: Limits, Continuity and Differentiability

Q20 jee_main_2026_24_january_evening Continuity of Piecewise Functions

Let

[t]

denote the greatest integer less than or equal to

t

. If the function

f(x)= casesb² ((π)/(2)[(π)/(2)( x+ x) x])&,x<0 x-(1)/(2) 2xx³&,x>0a&,x=0 cases

is continuous at

x = 0

, then

a² + b²

is equal to

A. $(5)/(8)$ $$\frac{5}{8}$$
B. $(9)/(16)$ $$\frac{9}{16}$$
C. $(3)/(4)$ $$\frac{3}{4}$$
D. $(1)/(2)$ $$\frac{1}{2}$$

Solution

Related Formula

Condition for continuity at x = 0: x \to 0^- f(x) = x \to 0^+ f(x) = f(0)

Core Logic

Evaluate the function value at $x = 0$ $x = 0$: $f(0) = a$ $f(0) = a$

Step 1: Evaluate RHL (Right Hand Limit)

For $x > 0$ $x > 0$:

RHL = x \to 0^+ ( x - (1)/(2) 2x)/(x³)

= x \to 0^+ ( x - x x)/(x³) = x \to 0^+ ( x (1 - x))/(x³)

= x \to 0^+ ( ( x)/(x) ) ( (1 - x)/(x²) ) = (1) ( (1)/(2) ) = (1)/(2)

For continuity, $a = RHL$ $a = \text{RHL}$, so $a = (1)/(2)$ $a = \frac{1}{2}$.

Step 2: Evaluate LHL (Left Hand Limit)

For $x < 0$ $x < 0$ (approaching 0 from negative side):

LHL = x \to 0^- b² ( (π)/(2) [ (π)/(2) ( x + x) x ] )

Look at the inner expression inside the GIF near $x \to 0^-$ $x \to 0^-$: Let $g(x) = (π)/(2) ( x + x) x$ . As $x \to 0^-$ $x \to 0^-$, $x$ $\sin x$ is a small negative number, $x$ $\cos x$ is slightly less than 1.

g(0) = (π)/(2) (0 + 1)(1) = (π)/(2) \approx 1.57

For small negative $x$ $x$, $g(x)$ $g(x)$ will approach $(π)/(2)$ $\frac{\pi}{2}$ but we need to check if it's less than or greater than $(π)/(2)$ $\frac{\pi}{2}$.

g(x) = (π)/(2)( x x + ² x) = (π)/(2)(( 2x)/(2) + (1+ 2x)/(2))

Since $x \to 0^-$ $x \to 0^-$, $2x < 0$ $\sin 2x < 0$ and $2x < 1$ $\cos 2x < 1$. Thus, $g(x)$ $g(x)$ is slightly less than $(π)/(2)$ $\frac{\pi}{2}$ (which is $\approx 1.57$ $\approx 1.57$), so $g(x)$ $g(x)$ is in the interval $(1, 1.57)$ $(1, 1.57)$.

The greatest integer value $[g(x)] = 1$ $[g(x)] = 1$.

LHL = b² ((π)/(2) (1)) = b² ((π)/(2)) = b²

Step 3: Final Calculation

Equate the limits:

LHL = RHL b² = (1)/(2)

Find $a² + b²$ $a^2 + b^2$:

a² + b² = ((1)/(2))² + (1)/(2) = (1)/(4) + (1)/(2) = (3)/(4)

Pattern Recognition

For GIF limits as $x \to 0$ $x \to 0$, expanding into precise Taylor approximations or inspecting trigonometric bounds (e.g., $1.57 - small$ $1.57 - \text{small}$ ) guarantees the exact bounding integer block before taking the final limit step.

Chapter Mix

Class 12 Maths: Limits, Continuity and Differentiability

The number of real solution(s) of the equation x²+3x+2= |x-3|, |x+2|$x^{2}+3x+2=\min\{|x-3|, |x+2|\}$ is:

Solution & Explanation

Related Formula

Core Logic

Step 1: Check Interval x ≤ -2$x \le -2$

Step 2: Check Interval -2 < x ≤ 0.5$-2 < x \le 0.5$

Step 3: Check Interval x > 0.5$x > 0.5$

Pattern Recognition

Chapter Mix

Reference Study Guides

More Limits, Continuity and Differentiability Previous-Year Questions — Page 2

Solution

Core Logic

Step 1: Determine 'a'

Step 2: Simplify f(x)

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

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Final Calculation

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Differentiation

Step 2: Limit Evaluation

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Continuity Check

Step 2: Differentiability Check

Step 3: Solving Equations

Step 4: Evaluate Final Target

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Evaluate RHL (Right Hand Limit)

Step 2: Evaluate LHL (Left Hand Limit)

Step 3: Final Calculation

Pattern Recognition

Chapter Mix

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The number of real solution(s) of the equation $x²+3x+2= |x-3|, |x+2|$ $x^{2}+3x+2=\min\{|x-3|, |x+2|\}$ is:

Step 1: Check Interval $x \leq -2$ $x \le -2$

Step 2: Check Interval $-2 < x \leq 0.5$ $-2 < x \le 0.5$

Step 3: Check Interval $x > 0.5$ $x > 0.5$