JEE Advance - Mathematics (2017 - Paper 1 Offline - No. 16)
Which of the following options is the only INCORRECT combination?
(I) (iii) (P)
(II) (iv) (Q)
(II) (ii) (P)
(III) (i) (R)
Explanation
$$f(x) = x + \ln x - x\ln x$$
$$f(1) = 1 > 0$$
$$f({e^2}) = {e^2} + 2 - 2{e^2} = 2 - {e^2} < 0$$
$$ \Rightarrow f(x) = 0$$ for some $$x \in (1,\,{e^2})$$
$$ \therefore $$ I is correct.
$$f'(x) = 1 + {1 \over x} - \ln x - 1$$
$$ = {1 \over x} - \ln x$$
$$f'(x) > 0$$ for (0, 1)
$$f'(x) < 0$$ for $$(e,\infty )$$
$$ \therefore $$ P and Q are correct, II is correct, III is incorrect.
$$f''(x) = {{ - 1} \over {{x^2}}} - {1 \over x}$$
$$f''(x) < 0$$ for $$(0,\infty )$$
$$ \therefore $$ S, is correct, R is incorrect.
IV is incorrect.
$$\mathop {\lim }\limits_{x \to \infty } f(x) = - \infty $$
$$\mathop {\lim }\limits_{x \to \infty } f'(x) = - \infty $$
$$\mathop {\lim }\limits_{x \to \infty } f''(x) = 0$$
$$ \therefore $$ ii, iii, iv are correct.
$$f(1) = 1 > 0$$
$$f({e^2}) = {e^2} + 2 - 2{e^2} = 2 - {e^2} < 0$$
$$ \Rightarrow f(x) = 0$$ for some $$x \in (1,\,{e^2})$$
$$ \therefore $$ I is correct.
$$f'(x) = 1 + {1 \over x} - \ln x - 1$$
$$ = {1 \over x} - \ln x$$
$$f'(x) > 0$$ for (0, 1)
$$f'(x) < 0$$ for $$(e,\infty )$$
$$ \therefore $$ P and Q are correct, II is correct, III is incorrect.
$$f''(x) = {{ - 1} \over {{x^2}}} - {1 \over x}$$
$$f''(x) < 0$$ for $$(0,\infty )$$
$$ \therefore $$ S, is correct, R is incorrect.
IV is incorrect.
$$\mathop {\lim }\limits_{x \to \infty } f(x) = - \infty $$
$$\mathop {\lim }\limits_{x \to \infty } f'(x) = - \infty $$
$$\mathop {\lim }\limits_{x \to \infty } f''(x) = 0$$
$$ \therefore $$ ii, iii, iv are correct.
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