Q.Find the local maximum and local minimum values of $f(x) = 2x^3 - 15x^2 + 36x + 10$ using the first derivative test.

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:

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

Set $y'=0$: $1-\log_e x = 0 \Rightarrow \log_e x = 1 \Rightarrow x = e$.

For $x<e$, $y'>0$ and for $x>e$, $y'<0$, so $x=e$ is a maximum. The value there is

$y = \frac{\log_e e}{e} = \frac{1}{e}.$

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.

Statement-II: $\sin x$ is periodic with period $2\pi$; it reaches its maximum $1$ at $x=\tfrac{\pi}{2}+2n\pi$ and its minimum $-1$ at $x=-\tfrac{\pi}{2}+2n\pi$ for every integer $n$. So it has infinitely many maximum and minimum values. True.

$(x-1)^2 \ge 0$ 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 $x=1$.

Using calculus to confirm: let $f(x)=5-(x-1)^2$. Then $f'(x) = -2(x-1)$, which is $0$ at $x=1$. Since $f''(x) = -2 < 0$, $x=1$ gives a maximum, and $f(1) = 5$.

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:

$f''(x) = -2 < 0$ for all $x$, so $x=2$ gives a maximum (the graph is a downward parabola).