For the function f(x) = x^4(12ln x - 7),match | vedclass

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Question

Given $f(x) = x^4(12\ln x - 7)$.$f'(x) = 4x^3(12\ln x - 7) + x^4(\frac{12}{x}) = 48x^3\ln x - 28x^3 + 12x^3 = 48x^3\ln x - 16x^3 = 16x^3(3\ln x - 1)$.

Vedclass · Accepted Answer

$(A) 	o S, (B) 	o R, (C) 	o P, (D) 	o Q$

Vedclass · Answer

Given $f(x) = x^4(12\ln x - 7)$.$f'(x) = 4x^3(12\ln x - 7) + x^4(\frac{12}{x}) = 48x^3\ln x - 28x^3 + 12x^3 = 48x^3\ln x - 16x^3 = 16x^3(3\ln x - 1)$.$f''(x) = 48x^2(3\ln x - 1) + 16x^3(\frac{3}{x}) = 144x^2\ln x - 48x^2 + 48x^2 = 144x^2\ln x$.$(A)$ Point of inflection occurs where $f''(x) = 0$,so $144x^2\ln x = 0 \implies \ln x = 0 \implies x = 1$. Then $f(1) = 1^4(12\ln 1 - 7) = -7$. So $(a, b) = (1, -7)$. $a - b = 1 - (-7) = 8$. Thus $(A) 	o S$.$(B)$ Local minimum occurs where $f'(x) = 0$ and $f''(x) > 0$. $16x^3(3\ln x - 1) = 0 \implies \ln x = 1/3 \implies x = e^{1/3}$. Here $t = 1/3$. So $12t = 12(1/3) = 4$. Thus $(B) 	o R$.$(C)$ Concave down where $f''(x) $(D)$ Concave up where $f''(x) > 0$,i.e.,$144x^2\ln x > 0 \implies \ln x > 0 \implies x > 1$. So $p = 1$. Thus $(D) 	o Q$.

Column-$I$	Column-$II$
$(A)$ If $(a, b)$ is a point of inflection,then $a - b$ equals	$(P)$ $3$
$(B)$ If $e^t$ is the point of local minimum,then $12t$ equals	$(Q)$ $1$
$(C)$ If the graph is concave down on $(d, e)$,then $d + 3e$ equals	$(R)$ $4$
$(D)$ If the graph is concave up on $(p, \infty)$,then $p$ equals	$(S)$ $8$

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