If f(x)= begincases displaystyle lim t arrow ∞(6t+10t)(1/t), x=0 x2+4 x+ C , x ≠ 0 endcases is a continuous function, then minimum value of f(x) is

Question

If f(x)= begincases displaystyle lim t arrow ∞(6t+10t)(1/t), x=0 x2+4 x+ C , x ≠ 0 endcases is a continuous function, then minimum value of f(x) is (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

f(0)=10 displaystyle lim t arrow ∞(((6/10))t+1)(1/t)=10 displaystyle lim x arrow 0 f(x)=x2+4 x+ c=c c=10 f(x)=x2+4 x+10=(x+2)2+6 minimum value =6

Q. If $f (x) = ⎩ ⎨ ⎧ t \to \infty lim (6^{t} + 1 0^{t})^{\frac{1}{t}}, x^{2} + 4 x + C x = 0, x \neq = 0$ is a continuous function, then minimum value of $f (x)$ is

$f (0) = 10 t \to \infty lim ((\frac{6}{10})^{t} + 1)^{\frac{1}{t}} = 10$
$x \to 0 lim f (x) = x^{2} + 4 x + c = c$
$c = 10$
$f (x) = x^{2} + 4 x + 10 = (x + 2)^{2} + 6$
minimum value $= 6$

Q. If f(x)=⎩⎨⎧​t→∞lim​(6t+10t)t1​,x2+4x+C​x=0,x=0​ is a continuous function, then minimum value of f(x) is

f(0)=10t→∞lim​((106​)t+1)t1​=10 x→0lim​f(x)=x2+4x+c=c c=10 f(x)=x2+4x+10=(x+2)2+6 minimum value =6

Q. If $f (x) = ⎩ ⎨ ⎧ t \to \infty lim (6^{t} + 1 0^{t})^{\frac{1}{t}}, x^{2} + 4 x + C x = 0, x \neq = 0$ is a continuous function, then minimum value of $f (x)$ is

$f (0) = 10 t \to \infty lim ((\frac{6}{10})^{t} + 1)^{\frac{1}{t}} = 10$
$x \to 0 lim f (x) = x^{2} + 4 x + c = c$
$c = 10$
$f (x) = x^{2} + 4 x + 10 = (x + 2)^{2} + 6$
minimum value $= 6$