Q.f(x) = 5 - |x - 1|. Find the maximum value of f(x), also find the value of x for which f(x) is maximum.

Differentiate $y=\dfrac{\log_e x}{x}$, set $y'=0$ to get $x=e$, and evaluate; the maximum value is $\dfrac1e$.

Maximising a function using the first derivative is a standard NCERT/CBSE Class 12 application of derivatives problem.

Differentiate with the quotient rule:

$y' = \frac{\tfrac{1}{x}\cdot x - \log_e x \cdot 1}{x^2} = \frac{1-\log_e x}{x^2}.$

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Both statements are correct, so the answer is (c).

These are qualitative NCERT/CBSE Class 12 maxima–minima facts.

Statement-I: $|x-3| \ge 0$ for all $x$, with equality only at $x=3$. Hence $f(x)=3+|x-3| \ge 3$, and the least value $3$ is achieved at $x=3$. So it has a local (indeed global) minimum value $3$. True.

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$(x-1)^2 \ge 0$ always, so $5-(x-1)^2$ is largest exactly when the squared term is $0$.

For any real $x$, $(x-1)^2 \ge 0$, with equality exactly when $x=1$. So $5-(x-1)^2 \le 5$ for all $x$, and this maximum value $5$ is attained at $x=1$.

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Set the derivative to zero to find the critical point, then confirm it's a maximum with the second derivative.

Given $f(x)=4x-x^2-3$.

First derivative:

$f'(x) = 4-2x$

Set $f'(x)=0$:

$4-2x=0 \Rightarrow x=2$

Second derivative test: …