Let $f(x) = \begin{cases} \frac{ax^2+2ax+3}{4x^2+4x-3}, & x \neq -\frac{3}{2}, \frac{1}{2} \\ b, & x = -\frac{3}{2}, \frac{1}{2} \end{cases}$ be continuous at $x = -\frac{3}{2}$. If $f \circ f(x) = \frac{7}{5}$, then $x$ is equal to: Options: (A) 4 (B) 0 (C) 2 (D) 1

### Analyzing the Setup The function is defined as $f(x) = \frac{ax^2+2ax+3}{4x^2+4x-3}$ for $x \neq -\frac{3}{2}, \frac{1}{2}$, with $f(x) = b$ at the points of interest. In the context of JEE Advanced, **continuity** at $x = -\frac{3}{2}$ implies that the function must not have a jump or a vertical asymptote at that point. Since the denominator vanishes at $x = -\frac{3}{2}$, the numerator must also vanish to create a $\frac{0}{0}$ indeterminate form, which can then be resolved via limits. ### The Algebraic Surgery We begin by factoring the denominator $D(x) = 4x^2+4x-3$. By splitting the middle term, we write: $$D(x) = 4x^2+6x-2x-3 = (2x+3)(2x-1)$$ This confirms that the singularity at $x = -\frac{3}{2}$ is caused by the factor $(2x+3)$. For the function to be continuous, the numerator $N(x) = ax^2+2ax+3$ must also contain the factor $(2x+3)$. We set $N(-\frac{3}{2}) = 0$: $$a\left(-\frac{3}{2}\right)^2 + 2a\left(-\frac{3}{2}\right) + 3 = 0$$ Simplifying this expression: $$a\left(\frac{9}{4}\right) - 3a + 3 = 0$$ Multiplying the entire equation by $4$ to clear the fraction yields $9a - 12a + 12 = 0$, which simplifies to $-3a = -12$. Thus, we find **$a = 4$**. ### The Composite Dance With $a=4$, the numerator becomes $4x^2+8x+3$, which factors into $(2x+3)(2x+1)$. The function simplifies as follows: $$f(x) = \frac{(2x+3)(2x+1)}{(2x+3)(2x-1)} = \frac{2x+1}{2x-1}$$ We now evaluate the composite function $f(f(x)) = \frac{7}{5}$. We substitute $f(x)$ into itself: $$f(f(x)) = \frac{2\left(\frac{2x+1}{2x-1}\right) + 1}{2\left(\frac{2x+1}{2x-1}\right) - 1}$$ ### The Final Resolution To simplify the composite expression, we find a common denominator for both the numerator and the denominator: $$f(f(x)) = \frac{\frac{2(2x+1) + (2x-1)}{2x-1}}{\frac{2(2x+1) - (2x-1)}{2x-1}} = \frac{4x+2+2x-1}{4x+2-2x+1} = \frac{6x+1}{2x+3}$$ We set this result equal to the given value: $$\frac{6x+1}{2x+3} = \frac{7}{5}$$ Cross-multiplying gives $5(6x+1) = 7(2x+3)$, which expands to $30x + 5 = 14x + 21$. Solving for $x$: $$16x = 16 \Rightarrow **x = 1**$$

JEE Main 2026 (23 January Shift 1)

LEVELJEE Main

Animated Solution for Mathematics - Limits, Continuity and Differentiability: Let $f (x) = {\frac{a x ^{2} + 2 a x + 3}{4 x ^{2} + 4 x - 3}, b, x \neq = - \frac{3}{2}, \frac{1}{2} x = - \frac{3}{2}, \frac{1}{2}$ be continuous at $x = - \frac{3}{2}$ . If $f \circ f (x) = \frac{7}{5}$ , then $x$ is equal to:

Q: Wait, why do we set the numerator to zero? Can't we just plug in the value?

That is a great question! If you just plug in $x = -\frac{3}{2}$ into the original function, you get a $\frac{0}{0}$ form, which is indeterminate. In the context of continuity, for the limit to exist at a point where the denominator is zero, the numerator must also be zero. This allows us to 'remove' the singularity by canceling the common factor, which is exactly what we need for the function to be continuous.

Select Answer:

Progressive Hints

Hint 1

Hint 1: The Spark. You are dealing with a piecewise function that is continuous at

x = - \frac{3}{2}

. This means the limit as

x

approaches

- \frac{3}{2}

must exist and equal the function value. Since the denominator

4 x^{2} + 4 x - 3

becomes zero at

x = - \frac{3}{2}

, the numerator

a x^{2} + 2 a x + 3

must also be zero at that point to avoid an undefined singularity. This is your gateway to finding

a

Hint 2

Hint 2: The Arsenal. Use the factor theorem. Since

x = - \frac{3}{2}

is a root of the denominator

4 x^{2} + 4 x - 3

(2 x + 3)

must be a factor. Factorize the denominator completely. Then, apply the same logic to the numerator: since it must also be zero at

x = - \frac{3}{2}

(2 x + 3)

must be a factor of the numerator as well. This will allow you to cancel the singularity.

Hint 3

Hint 3: The Execution. Once you find

a = 4

, substitute it back into

f (x)

and simplify the expression by canceling the common factor

(2 x + 3)

. You will be left with

f (x) = \frac{2 x + 1}{2 x - 1}

. Now, to solve

f (f (x)) = \frac{7}{5}

, substitute the simplified

f (x)

into itself:

f (f (x)) = \frac{2 ( \frac{2 x + 1}{2 x - 1} ) + 1}{2 ( \frac{2 x + 1}{2 x - 1} ) - 1}

. Simplify this complex fraction and equate it to

\frac{7}{5}

to solve for

x

Visualized Solution

Analyze the Continuity Condition

Given function $f (x) = {\frac{a x ^{2} + 2 a x + 3}{4 x ^{2} + 4 x - 3}, b, x \neq = - \frac{3}{2}, \frac{1}{2} x = - \frac{3}{2}, \frac{1}{2}$
For $f (x)$ to be continuous at $x = - \frac{3}{2}$ , the limit $lim_{x \to - \frac{3}{2}} f (x)$ must exist and be finite.
Since the denominator $4 x^{2} + 4 x - 3$ is zero at $x = - \frac{3}{2}$ , the numerator $a x^{2} + 2 a x + 3$ must also be zero at $x = - \frac{3}{2}$ .

Factorize the Denominator

Denominator: $D (x) = 4 x^{2} + 4 x - 3$
Splitting the middle term: $4 x^{2} + 6 x - 2 x - 3$
Factoring: $2 x (2 x + 3) - 1 (2 x + 3) = (2 x + 3) (2 x - 1)$

Apply the Vanishing Numerator Condition

Numerator $N (x) = a x^{2} + 2 a x + 3$
Condition: $N (- \frac{3}{2}) = 0$
Substitution: $a (- \frac{3}{2})^{2} + 2 a (- \frac{3}{2}) + 3 = 0$

Solve for Coefficient $a$

$a (\frac{9}{4}) - 3 a + 3 = 0$
Multiply by $4$ : $9 a - 12 a + 12 = 0$
$- 3 a = - 12 ⟹ a = 4$

Simplify the Function $f (x)$

Substitute $a = 4$ into $f (x)$ : $f (x) = \frac{4 x ^{2} + 8 x + 3}{4 x ^{2} + 4 x - 3}$
Factorize numerator: $4 x^{2} + 8 x + 3 = 4 x^{2} + 6 x + 2 x + 3 = (2 x + 3) (2 x + 1)$
$f (x) = \frac{( 2 x + 3 ) ( 2 x + 1 )}{( 2 x + 3 ) ( 2 x - 1 )}$

Final Expression for $f (x)$

For $x \neq = - \frac{3}{2}$ , cancel the common factor $(2 x + 3)$
Simplified $f (x) = \frac{2 x + 1}{2 x - 1}$

Set up Composite Function $f (f (x))$

Given condition: $f (f (x)) = \frac{7}{5}$
Substitute $f (x)$ into itself: $f (f (x)) = \frac{2 ( f ( x )) + 1}{2 ( f ( x )) - 1}$

Substitute $f (x)$ Expression

$f (f (x)) = \frac{2 ( \frac{2 x + 1}{2 x - 1} ) + 1}{2 ( \frac{2 x + 1}{2 x - 1} ) - 1}$

Simplify the Composite Fraction

Numerator: $\frac{2 ( 2 x + 1 ) + 1 ( 2 x - 1 )}{2 x - 1} = \frac{4 x + 2 + 2 x - 1}{2 x - 1}$
Denominator: $\frac{2 ( 2 x + 1 ) - 1 ( 2 x - 1 )}{2 x - 1} = \frac{4 x + 2 - 2 x + 1}{2 x - 1}$

Final Simplified Composite Function

$f (f (x)) = \frac{6 x + 1}{2 x + 3}$

Equate to Given Value

Set $\frac{6 x + 1}{2 x + 3} = \frac{7}{5}$

Cross-Multiplication

$5 (6 x + 1) = 7 (2 x + 3)$
$30 x + 5 = 14 x + 21$

Solve for $x$

$30 x - 14 x = 21 - 5$
$16 x = 16 ⟹ x = 1$

00:00 / 00:00

The Sigma Insight: Continuity at a Point and in an Interval

Analyzing the Setup

The function is defined as

f (x) = \frac{a x ^{2} + 2 a x + 3}{4 x ^{2} + 4 x - 3}

for $x eq -\frac{3}{2}, \frac{1}{2}$, with

f (x) = b

at the points of interest.

In the context of JEE Advanced, continuity at

x = - \frac{3}{2}

implies that the function must not have a jump or a vertical asymptote at that point. Since the denominator vanishes at

x = - \frac{3}{2}

, the numerator must also vanish to create a

\frac{0}{0}

indeterminate form, which can then be resolved via limits.

The Algebraic Surgery

We begin by factoring the denominator

D (x) = 4 x^{2} + 4 x - 3

. By splitting the middle term, we write:

D (x) = 4 x^{2} + 6 x - 2 x - 3 = (2 x + 3) (2 x - 1)

This confirms that the singularity at

x = - \frac{3}{2}

is caused by the factor

(2 x + 3)

. For the function to be continuous, the numerator

N (x) = a x^{2} + 2 a x + 3

must also contain the factor

(2 x + 3)

We set

N (- \frac{3}{2}) = 0

a (- \frac{3}{2})^{2} + 2 a (- \frac{3}{2}) + 3 = 0

Simplifying this expression:

a (\frac{9}{4}) - 3 a + 3 = 0

Multiplying the entire equation by

4

to clear the fraction yields

9 a - 12 a + 12 = 0

, which simplifies to

- 3 a = - 12

. Thus, we find $a = 4$ .

The Composite Dance

With

a = 4

, the numerator becomes

4 x^{2} + 8 x + 3

, which factors into

(2 x + 3) (2 x + 1)

. The function simplifies as follows:

f (x) = \frac{( 2 x + 3 ) ( 2 x + 1 )}{( 2 x + 3 ) ( 2 x - 1 )} = \frac{2 x + 1}{2 x - 1}

We now evaluate the composite function

f (f (x)) = \frac{7}{5}

. We substitute

f (x)

into itself:

f (f (x)) = \frac{2 ( \frac{2 x + 1}{2 x - 1} ) + 1}{2 ( \frac{2 x + 1}{2 x - 1} ) - 1}

The Final Resolution

To simplify the composite expression, we find a common denominator for both the numerator and the denominator:

f (f (x)) = \frac{\frac{2 ( 2 x + 1 ) + ( 2 x - 1 )}{2 x - 1}}{\frac{2 ( 2 x + 1 ) - ( 2 x - 1 )}{2 x - 1}} = \frac{4 x + 2 + 2 x - 1}{4 x + 2 - 2 x + 1} = \frac{6 x + 1}{2 x + 3}

We set this result equal to the given value:

\frac{6 x + 1}{2 x + 3} = \frac{7}{5}

Cross-multiplying gives

5 (6 x + 1) = 7 (2 x + 3)

, which expands to

30 x + 5 = 14 x + 21

. Solving for

x

16 x = 16 \Rightarrow * * x = 1 * *

Common Traps & FAQs

Wait, why do we set the numerator to zero? Can't we just plug in the value?

That is a great question! If you just plug in

x = - \frac{3}{2}

into the original function, you get a

\frac{0}{0}

form, which is indeterminate. In the context of continuity, for the limit to exist at a point where the denominator is zero, the numerator must also be zero. This allows us to 'remove' the singularity by canceling the common factor, which is exactly what we need for the function to be continuous.

Wait, why did we cancel the

(2 x + 3)

term? Isn't that undefined at

x = - \frac{3}{2}

You are absolutely right to be cautious! When we evaluate a limit

lim_{x \to c} f (x)

, we are looking at the behavior of the function as

x

gets arbitrarily close to

c

, but

x

is never actually equal to

c

. Therefore, when we cancel the

(2 x + 3)

term, we are assuming

x \neq = - \frac{3}{2}

. This is perfectly valid in the context of limits and continuity.

Is there a faster way to solve

f (f (x)) = \frac{7}{5}

without full substitution?

While full substitution is the most reliable method, you can sometimes use the property of the function itself. Notice that

f (x) = \frac{2 x + 1}{2 x - 1}

is its own inverse (an involution). If

f (f (x)) = y

, then

f (x) = f^{- 1} (y)

. Since

f (x) = f^{- 1} (x)

here, you could potentially solve

f (x) = f (\frac{7}{5})

or similar, but for JEE, the algebraic substitution is robust and less prone to conceptual errors. Stick to the substitution method to be safe!

Select Answer:

Visualized Solution

Analyze the Continuity Condition

Factorize the Denominator

Apply the Vanishing Numerator Condition

Solve for Coefficient $a$

Simplify the Function $f (x)$

Final Expression for $f (x)$

Set up Composite Function $f (f (x))$

Substitute $f (x)$ Expression

Simplify the Composite Fraction

Final Simplified Composite Function

Equate to Given Value

Cross-Multiplication

Solve for $x$

The Sigma Insight: Continuity at a Point and in an Interval

Analyzing the Setup

The Algebraic Surgery

The Composite Dance

The Final Resolution

Similar Questions

Let $f : R \to R$ be a function defined as $f (x) = ⎩ ⎨ ⎧ \frac{s i n ( a + 1 ) x + s i n 2 x}{2 x}, b, \frac{x + b x ^{3} - x}{b x ^{5/2}}, if x < 0 if x = 0 if x > 0$ If $f$ is continuous at $x = 0$ , then the value of $a + b$ is equal to:

If the function $f (x) = ⎩ ⎨ ⎧ \frac{1}{x} lo g_{e} (\frac{1 + \frac{x}{a}}{1 - \frac{x}{b}}), k, \frac{c o s ^{2} x - s i n ^{2} x - 1}{x ^{2} + 1 - 1}, x < 0 x = 0 x > 0$ is continuous at $x = 0$ , then $\frac{1}{a} + \frac{1}{b} + \frac{4}{k}$ is equal to :

Let $f : (- \frac{π}{4}, \frac{π}{4}) \to R$ be defined as $f (x) = ⎩ ⎨ ⎧ (1 + ∣ sin x ∣)^{\frac{3 a}{∣ s i n x ∣}}, b, e^{c o t 4 x / c o t 2 x}, - \frac{π}{4} < x < 0 x = 0 0 < x < \frac{π}{4}$ If $f$ is continuous at $x = 0$ , then the value of $6 a + b^{2}$ is equal to:

Let $f : R \to R$ be defined as $f (x) = ⎩ ⎨ ⎧ 2 sin (- \frac{π}{2}), a x^{2} + x + b, sin (π x), if x < - 1 if - 1 \leq x \leq 1 if x > 1$ If $f (x)$ is continuous on $R$ , then $a + b$ equals :

For $a, b > 0$ , let $f (x) = ⎩ ⎨ ⎧ \frac{t a n (( a + 1 ) x ) + b t a n x}{x}, 3, \frac{a x + b ^{2} x ^{2} - a x}{b a x x}, x < 0 x = 0 x > 0$ be a continuous function at $x = 0$ . Then $\frac{b}{a}$ is equal to :

If a function $f (x)$ defined by $f (x) = ⎩ ⎨ ⎧ a e^{x} + b e^{- x}, c x^{2} a x^{2} + 2 c x, - 1 < x < 1, 1 \leq x \leq 3 3 < x < 4$ be continuous for some $a, b, c \in R$ and $f^{'} (0) + f^{'} (2) = e$ , then the value of $a$ is :

If $f (x) = ⎩ ⎨ ⎧ \frac{s i n ( a + 2 ) x + s i n x}{x}, b, \frac{( x + 3 x ^{2} ) ^{1/3} - x ^{1/3}}{x ^{4/3}}, x < 0 x = 0 x > 0$ is continuous at $x = 0$ then $a + 2 b$ is equal to:

Function is continuous at $x = 0$ , find $a + 2 b$ . $f (x) = ⎩ ⎨ ⎧ \frac{s i n ( a + 2 ) x + s i n x}{x} b \frac{( x + 3 x ^{2} ) ^{1/3} - x ^{1/3}}{x ^{4/3}}; x < 0; x = 0; x > 0$

If $f (x) = {\frac{a ∣ x ∣ + x ^{2} - 2 ( s i n ∣ x ∣ ) ( c o s ∣ x ∣ )}{x}, b, x \neq = 0 x = 0$ is continuous at $x = 0$ , then $a + b$ is equal to

Select Answer:

Visualized Solution

Analyze the Continuity Condition

Factorize the Denominator

Apply the Vanishing Numerator Condition

Solve for Coefficient .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a

Simplify the Function .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x)

Final Expression for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x)

Set up Composite Function .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(f(x))

Substitute .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x) Expression

Simplify the Composite Fraction

Final Simplified Composite Function

Equate to Given Value

Cross-Multiplication

Solve for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x

The Sigma Insight: Continuity at a Point and in an Interval

Analyzing the Setup

The Algebraic Surgery

The Composite Dance

The Final Resolution

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f:R→R be defined as f(x)=⎩⎨⎧​2sin(−2π​),ax2+x+b,sin(πx),​if x<−1if −1≤x≤1if x>1​ If f(x) is continuous on R, then a+b equals :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } For a,b>0, let f(x)=⎩⎨⎧​xtan((a+1)x)+btanx​,3,bax​xax+b2x2​−ax​​,​x<0x=0x>0​ be a continuous function at x=0. Then ab​ is equal to :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If f(x)=⎩⎨⎧​xsin(a+2)x+sinx​,b,x4/3(x+3x2)1/3−x1/3​,​x<0x=0x>0​ is continuous at x=0 then a+2b is equal to:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Function is continuous at x=0, find a+2b. f(x)=⎩⎨⎧​xsin(a+2)x+sinx​bx4/3(x+3x2)1/3−x1/3​​;x<0;x=0;x>0​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If f(x)={xa∣x∣+x2−2(sin∣x∣)(cos∣x∣)​,b,​x=0x=0​ is continuous at x=0, then a+b is equal to

Solve for Coefficient $a$

Simplify the Function $f (x)$

Final Expression for $f (x)$

Set up Composite Function $f (f (x))$

Substitute $f (x)$ Expression

Solve for $x$

Let $f : R \to R$ be defined as $f (x) = ⎩ ⎨ ⎧ 2 sin (- \frac{π}{2}), a x^{2} + x + b, sin (π x), if x < - 1 if - 1 \leq x \leq 1 if x > 1$ If $f (x)$ is continuous on $R$ , then $a + b$ equals :

For $a, b > 0$ , let $f (x) = ⎩ ⎨ ⎧ \frac{t a n (( a + 1 ) x ) + b t a n x}{x}, 3, \frac{a x + b ^{2} x ^{2} - a x}{b a x x}, x < 0 x = 0 x > 0$ be a continuous function at $x = 0$ . Then $\frac{b}{a}$ is equal to :

If $f (x) = ⎩ ⎨ ⎧ \frac{s i n ( a + 2 ) x + s i n x}{x}, b, \frac{( x + 3 x ^{2} ) ^{1/3} - x ^{1/3}}{x ^{4/3}}, x < 0 x = 0 x > 0$ is continuous at $x = 0$ then $a + 2 b$ is equal to:

Function is continuous at $x = 0$ , find $a + 2 b$ . $f (x) = ⎩ ⎨ ⎧ \frac{s i n ( a + 2 ) x + s i n x}{x} b \frac{( x + 3 x ^{2} ) ^{1/3} - x ^{1/3}}{x ^{4/3}}; x < 0; x = 0; x > 0$

If $f (x) = {\frac{a ∣ x ∣ + x ^{2} - 2 ( s i n ∣ x ∣ ) ( c o s ∣ x ∣ )}{x}, b, x \neq = 0 x = 0$ is continuous at $x = 0$ , then $a + b$ is equal to