Let f(x) = { 1/3 , x pi/2 }. If f is continuous at x=pi/2, then the value of integral from 0 to 3b-6 of |x^2+2x-3| dx is:

The correct answer is Option 4. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Maths Q19 – JEE-MAIN 2026

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Home Maths Limits and DerivativesQuestion #271

Let

f (x) = {\begin{cases} \frac{1}{3} & , x \leq π / 2 \\ \frac{b (1 - \sin x)}{(π - 2 x)^{2}} & , x > π / 2 \end{cases}

. If

f

is continuous at

x = π / 2

, then the value of

\int_{0}^{3 b - 6} | x^{2} + 2 x - 3 | d x

is:

Clue 1 of 3: Continuity Condition

For a function to be continuous at a point $x = a$ , the left-hand limit, right-hand limit, and the function's value at $a$ must all be equal, i.e., $lim_{x \to a^{-}} f (x) = lim_{x \to a^{+}} f (x) = f (a)$ .

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Step 1: Determine the value of 'b' using continuity✦ Active

For $f (x)$ to be continuous at $x = π / 2$ , we must have $lim_{x \to π / 2^{-}} f (x) = lim_{x \to π / 2^{+}} f (x) = f (π / 2)$ . From the definition, $f (π / 2) = \frac{1}{3}$ . The left-hand limit is also $\frac{1}{3}$ . Now, evaluate the right-hand limit:

lim_{x \to π / 2^{+}} f (x) = lim_{x \to π / 2^{+}} \frac{b (1 - \sin x)}{(π - 2 x)^{2}}

Let $x = π / 2 + h$ . As $x \to π / 2^{+}$ , $h \to 0^{+}$ . Substituting this into the limit expression:

lim_{h \to 0^{+}} \frac{b (1 - \sin (π / 2 + h))}{(- π - 2 h)^{2}} = lim_{h \to 0^{+}} \frac{b (1 - \cos h)}{4 h^{2}} = \frac{b}{4} lim_{h \to 0^{+}} \frac{1 - \cos h}{h^{2}} = \frac{b}{4} \cdot \frac{1}{2} = \frac{b}{8}

For continuity, equate this to $f (π / 2)$ :

\frac{b}{8} = \frac{1}{3} ⟹ b = \frac{8}{3}

Step 2: Determine the upper limit of the integral○ Expand

The upper limit of the integral is $3 b - 6$ . Substitute the value of $b$ found in Step 1:

3 b - 6 = 3 (\frac{8}{3}) - 6 = 8 - 6 = 2

So, the integral to evaluate is $\int_{0}^{2} | x^{2} + 2 x - 3 | d x$ .

Step 3: Evaluate the definite integral○ Expand

First, analyze the quadratic expression $x^{2} + 2 x - 3$ . Factorizing it gives $(x + 3) (x - 1)$ . The roots are $x = - 3$ and $x = 1$ . We are integrating from $0$ to $2$ . The sign of the expression changes at $x = 1$ within this interval.

For $x \in [0, 1)$ , $x^{2} + 2 x - 3 < 0$ , so $| x^{2} + 2 x - 3 | = - (x^{2} + 2 x - 3)$ . For $x \in [1, 2]$ , $x^{2} + 2 x - 3 \geq 0$ , so $| x^{2} + 2 x - 3 | = x^{2} + 2 x - 3$ . Split the integral:

\int_{0}^{2} | x^{2} + 2 x - 3 | d x = \int_{0}^{1} - (x^{2} + 2 x - 3) d x + \int_{1}^{2} (x^{2} + 2 x - 3) d x

Evaluate the first part:

\int_{0}^{1} (- x^{2} - 2 x + 3) d x = {[- \frac{x^{3}}{3} - x^{2} + 3 x]}_{0}^{1} = (- \frac{1}{3} - 1 + 3) - (0) = \frac{5}{3}

Evaluate the second part:

\int_{1}^{2} (x^{2} + 2 x - 3) d x = {[\frac{x^{3}}{3} + x^{2} - 3 x]}_{1}^{2} = (\frac{2^{3}}{3} + 2^{2} - 3 (2)) - (\frac{1^{3}}{3} + 1^{2} - 3 (1)) = (\frac{8}{3} + 4 - 6) - (\frac{1}{3} + 1 - 3) = (\frac{8}{3} - 2) - (\frac{1}{3} - 2) = \frac{2}{3} - (- \frac{5}{3}) = \frac{7}{3}

Summing the two parts:

Total integral = \frac{5}{3} + \frac{7}{3} = \frac{12}{3} = 4

💡 Teacher's Secret Hint

Remember to correctly handle the absolute value by splitting the integral at the roots of the quadratic expression within the integration interval.

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Maths Question 19 – JEE-MAIN 2026

Clue 1 of 3: Continuity Condition

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