If S-7-x4x-33- x-3-7- then

S=(7 x)^4(x 3)^3 ⇒dSdx = 4(7 x)^3 (x 3)^3 +3(7 x)^4 (x 3)^2 ⇒dSdx= 0 ⇒ (7 x)^3(x 3)^2 ( 4(x 3) +3(7 x)) = 0 ⇒ 4x +12 +21 3x = 0 ⇒ x = 337 The maximum ...

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Standard XII

Mathematics

First Derivative Test for Local Maximum

If S=7-x4x-33...

Question

$If S = (7 - x)^{4} (x - 3)^{3}, x \in (3, 7), then$

Maximum value of S = \frac{2^{16} \times 3^{3}}{7^{7}}

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Maximum value attained at x = \frac{37}{7}

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Maximum value of S = \frac{2^{18} \times 3^{3}}{7^{7}}

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Maximum value attained at x = \frac{33}{7}

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Solution

The correct option is D $Maximum value attained at x = \frac{33}{7}$
$S = (7 - x)^{4} (x - 3)^{3} \Rightarrow \frac{d S}{d x} = - 4 (7 - x)^{3} (x - 3)^{3} + 3 (7 - x)^{4} (x - 3)^{2} \Rightarrow \frac{d S}{d x} = 0 \Rightarrow (7 - x)^{3} (x - 3)^{2} (- 4 (x - 3) + 3 (7 - x)) = 0 \Rightarrow - 4 x + 12 + 21 - 3 x = 0 \Rightarrow x = \frac{33}{7}$
The maximum value will be
$S = \frac{16^{4} \times 12^{3}}{7^{7}} \Rightarrow S = \frac{2^{22} \times 3^{3}}{7^{7}}$

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If S=(7−x)4(x−3)3, x∈(3,7), then

The correct option is D Maximum value attained at x=337S=(7−x)4(x−3)3⇒dSdx=−4(7−x)3(x−3)3+3(7−x)4(x−3)2⇒dSdx=0⇒(7−x)3(x−3)2(−4(x−3)+3(7−x))=0⇒−4x+12+21−3x=0⇒x=337 The maximum value will be S=164×12377⇒S=222×3377

$If S = (7 - x)^{4} (x - 3)^{3}, x \in (3, 7), then$