Let fx-x2-42 ln x-1-e x-2 k- k - R- If least value of K for which -fx is defined for all x -0- - is - then - is where-denotes greatest integer function

F'(x) = x lnx e f'(x) gt; 0 → x ∈ (e, ∞) f'(x) lt; 0 → x ∈ (0, e) f(x)_min =f(e) = 2k 3/4 e^2 2k 3/4e^2 ≥ 0 ⇒ k ≥3/8 e^2 So ∝ = 3/8 e^2 So [∝] is = ...

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Byju's Answer

Standard XII

Mathematics

AM,GM,HM Inequality

Let fx=x2/42 ...

Question

Let $f (x) = \frac{x^{2}}{4} (2 l n x - 1) - e x + 2 k, k \in R$ . If least value of K for which $\sqrt{f (x)}$ is defined for all $x \in (0, \infty)$ is $\propto$ then $[\propto]$ is
(where[.]denotes greatest integer function)

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Solution

F'(x) = x lnx - e
$f^{'} (x) > 0 \to x \in (e, \infty)$
$f^{'} (x) < 0 \to x \in (0, e)$
$f (x)_{m i n} = f (e) = 2 k - \frac{3}{4} e^{2}$
$2 k - \frac{3}{4} e^{2} \geq 0 \Rightarrow k \geq \frac{3}{8} e^{2}$
So $\propto= \frac{3}{8} e^{2} S o [\propto]$ is = 2

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Let f(x)=x24(2lnx−1)−ex+2k,k∈R. If least value of K for which √f(x) is defined for all x∈(0,∞) is ∝ then [∝] is (where[.]denotes greatest integer function)

F'(x) = x lnx - e f′(x)>0→x∈(e,∞) f′(x)<0→x∈(0,e) f(x)min=f(e)=2k−34e2 2k−34e2≥0⇒k≥38e2 So ∝=38e2So[∝] is = 2

Let $f (x) = \frac{x^{2}}{4} (2 l n x - 1) - e x + 2 k, k \in R$ . If least value of K for which $\sqrt{f (x)}$ is defined for all $x \in (0, \infty)$ is $\propto$ then $[\propto]$ is
(where[.]denotes greatest integer function)

F'(x) = x lnx - e
$f^{'} (x) > 0 \to x \in (e, \infty)$
$f^{'} (x) < 0 \to x \in (0, e)$
$f (x)_{m i n} = f (e) = 2 k - \frac{3}{4} e^{2}$
$2 k - \frac{3}{4} e^{2} \geq 0 \Rightarrow k \geq \frac{3}{8} e^{2}$
So $\propto= \frac{3}{8} e^{2} S o [\propto]$ is = 2