Let . If f is continuous at , then the value of is: [2026]

Question

Let $f (x) = {\begin{matrix} \frac{1}{3}, & x \leq π / 2 \\ \frac{b (1 - \sin x)}{{(π - 2 x)}^{2}}, & x > π / 2 \end{matrix}$ . 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: [2026]

Smart Achievers · Accepted Answer

(4)

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

$x = π / 2$

$\Rightarrow \frac{1}{3} = \lim_{x \to \frac{π^{+}}{2}} \frac{b (1 - \sin x)}{{(π - 2 x)}^{2}}$

$= \lim_{h \to 0} \frac{b (1 - \cos x)}{4 h^{2}}$

$= \lim_{h \to 0} \frac{b \cdot 2 \cdot \sin^{2} (\frac{h}{2})}{4 h^{2}} = \frac{b}{8}$

$\Rightarrow \frac{1}{3} = \frac{b}{8} \Rightarrow b = \frac{8}{3}$

$I = \int_{0}^{2} | x^{2} + 2 x - 3 | d x$

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

$= \frac{5}{3} + \frac{7}{3} = 4$

Solution

Q.
Let $f (x) = {\begin{matrix} \frac{1}{3}, & x \leq π / 2 \\ \frac{b (1 - \sin x)}{{(π - 2 x)}^{2}}, & x > π / 2 \end{matrix}$ . 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: [2026]

1 5

2 2

3 3

4 4

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Solution

Q. Let f(x)={13,x≤π/2b(1–sinx)(π–2x)2,x>π/2. If f is continuous at x=π/2, then the value of ∫03b–6|x2+2x–3|dx is: [2026]

1 5

2 2

3 3

4 4

Ans. (4) f(x)={1/3,x≤π2b(1–sinx)(π–2x)2,x>π2 limx→π–2f(x)=limx→π+2f(x) ⇒ 13=limx→π+2b(1–sinx)(π–2x)2 =limh→0b(1–cosx)4h2 =limh→0b·2·sin2(h2)4h2=b8 ⇒ 13=b8 ⇒ b=83 I=∫02|x2+2x–3|dx =∫01–(x2+2x–3)dx+∫12(x2+2x–3)dx =53+73=4

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Q.
Let $f (x) = {\begin{matrix} \frac{1}{3}, & x \leq π / 2 \\ \frac{b (1 - \sin x)}{{(π - 2 x)}^{2}}, & x > π / 2 \end{matrix}$ . 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: [2026]