Let f-x- - -0-x-3u-2 - 2u - 2-du- where x satisfies the inequality log-2 -1-6x - x-2 - 5- - 0- If minimum and maximum values of f-x- are -m- and -M- respectively then M-m-4 is-

6x x^2 5 nbsp;⩾ 0 ⇒ x ϵ [1, 5] f(x) = [u^3 + u^2 + 2u]_0^x = x^3 + x^2 + 2x f'(x) = 3x^2 + 2x + 2 a gt; 0 and D lt; 0 ⇒ f(x) is increasing in [1, 5 ...

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Byju's Answer

Standard XII

Mathematics

Local Maxima

Let f(x) = ∫_...

Question

Let $f (x) = \int_{0}^{x} (3 u^{2} + 2 u + 2) d u,$ where x satisfies the inequality $l o g_{2} (1 + \sqrt{6 x - x^{2} - 5}) ⩾ 0.$ If minimum and maximum values of f(x) are 'm' and 'M' respectively then $\frac{M + m}{4}$ is

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Solution

$6 x - x^{2} - 5 ⩾ 0$
$\Rightarrow x ϵ [1, 5]$
$f (x) = [u^{3} + u^{2} + 2 u]_{0}^{x} = x^{3} + x^{2} + 2 x$
$f^{'} (x) = 3 x^{2} + 2 x + 2$
a > 0 and D < 0 $\Rightarrow$ f(x) is increasing in [1, 5]
Hence, max. value of f(5) = 160 = M
min. value of f(1) = 4 = m
$\frac{m + M}{4} = \frac{164}{4} = 41$

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Let f(x)=∫x0(3u2+2u+2)du, where x satisfies the inequality log2(1+√6x−x2−5)⩾0. If minimum and maximum values of f(x) are 'm' and 'M' respectively then M+m4 is

6x−x2−5⩾0 ⇒x ϵ [1,5] f(x)=[u3+u2+2u]x0=x3+x2+2x f′(x)=3x2+2x+2 a > 0 and D < 0 ⇒ f(x) is increasing in [1, 5] Hence, max. value of f(5) = 160 = M min. value of f(1) = 4 = m m+M4=1644=41

Let $f (x) = \int_{0}^{x} (3 u^{2} + 2 u + 2) d u,$ where x satisfies the inequality $l o g_{2} (1 + \sqrt{6 x - x^{2} - 5}) ⩾ 0.$ If minimum and maximum values of f(x) are 'm' and 'M' respectively then $\frac{M + m}{4}$ is