Let g(x)=ax2+bx+c,a,b,c∈N and satisfies ∫01g(x)dx=116. Let f (x) be a continuous and derivable function in (x1,x2). If f(x).f'(x)≥x1−(f(x))4 and limx→x1+(f(x))2=1 and limx→x2−(f(x))2=12, then the minimum value of [x12−x22] is equal to, (where [.] denotes greatest integer function.)

∫01g(x)dx=116⇒∫01(ax2+bx+c)dx=116 ⇒a3+b2+c=116 ⇒2a+3b+6c=11;a,b,c∈N ⇒a=b=c=1 f(x)f'(x)1−(f(x))4≥xIntegration on both sides∫x1x2f(x)f'(x)1−(f(x))4 dx≥∫x1x2x dx 12sin−1((f(x))2)]x1x2≥x22−x122 ⇒π6−π2≥x22−x12 ⇒x12−x22≥π3⇒[x12−x22]min=1

Let g(x)=ax2+bx+c,a,b,c∈N and satisfies ∫01g(x)dx=116. Let f (x) be a continuous and derivable function in (x1,x2). If f(x).f'(x)≥x1−(f(x))4 and limx→x1+(f(x))2=1 and limx→x2−(f(x))2=12, then the minimum value of [x12−x22] is equal to, (where [.] denotes greatest integer function.)

∫01g(x)dx=116⇒∫01(ax2+bx+c)dx=116 ⇒a3+b2+c=116 ⇒2a+3b+6c=11;a,b,c∈N ⇒a=b=c=1 f(x)f'(x)1−(f(x))4≥xIntegration on both sides∫x1x2f(x)f'(x)1−(f(x))4 dx≥∫x1x2x dx 12sin−1((f(x))2)]x1x2≥x22−x122 ⇒π6−π2≥x22−x12 ⇒x12−x22≥π3⇒[x12−x22]min=1

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Let $g (x) = a x^{2} + b x + c, a, b, c \in N$ and satisfies $\int_{0}^{1} g (x) d x = \frac{11}{6} .$ Let f (x) be a continuous and derivable function in $(x_{1}, x_{2}) .$ If $f (x) . f' (x) \geq x \sqrt{1 - {(f (x))}^{4}}$ and $\lim_{x \to x_{1}^{+}} {(f (x))}^{2} = 1$ and $\lim_{x \to x_{2}^{-}} {(f (x))}^{2} = \frac{1}{2},$ then the minimum value of $[x_{1}^{2} - x_{2}^{2}]$ is equal to, (where [.] denotes greatest integer function.)

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$2 - a b c$

$a + b - c$

$b + c - a$

$a - b$

answer is A, B, C.

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Detailed Solution

$\int_{0}^{1} g (x) d x = \frac{11}{6} \Rightarrow \int_{0}^{1} (a x^{2} + b x + c) d x = \frac{11}{6}$
$\Rightarrow \frac{a}{3} + \frac{b}{2} + c = \frac{11}{6}$
$\Rightarrow 2 a + 3 b + 6 c = 11; a, b, c \in N$
$\Rightarrow a = b = c = 1$
$\frac{f (x) f' (x)}{\sqrt{1 - {(f (x))}^{4}}} \geq x$
Integration on both sides
$\int_{x_{1}}^{x_{2}} \frac{f (x) f' (x)}{\sqrt{1 - {(f (x))}^{4}}} d x \geq \int_{x_{1}}^{x_{2}} x d x$
${\frac{1}{2} \sin^{- 1} ({(f (x))}^{2})]}_{x_{1}}^{x_{2}} \geq \frac{x_{2}^{2} - x_{1}^{2}}{2}$
$\Rightarrow \frac{π}{6} - \frac{π}{2} \geq x_{2}^{2} - x_{1}^{2}$
$\Rightarrow x_{1}^{2} - x_{2}^{2} \geq \frac{π}{3} \Rightarrow {[x_{1}^{2} - x_{2}^{2}]}_{\min} = 1$