# Application of Derivatives - Class 12 - Mathematics

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## Extra Questions - Application of Derivatives | NCERT | Mathematics | Class 12

Find the derivative of $f(x) = \cos x$ at $x = 0$.

**Solution:**

The function given is:
$$
f(x) = \cos x
$$
We need to find its derivative at $x = 0$, denoted as $f'(0)$. According to the definition of the derivative:
$$
f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}
$$
Applying this to find $f'(0)$:
$$
f'(0) = \lim_{h \to 0} \frac{\cos(0+h) - \cos 0}{h} = \lim_{h \to 0} \frac{\cos h - \cos 0}{h}
$$
Since $\cos 0 = 1$, the expression simplifies to:
$$
f'(0) = \lim_{h \to 0} \frac{\cos h - 1}{h}
$$
Using the Taylor series expansion for $\cos h$,
$$
\cos h = 1 - \frac{h^2}{2!} + \frac{h^4}{4!} - \ldots
$$
we substitute in the limit:
$$
f'(0) = \lim_{h \to 0} \frac{1 - \frac{h^2}{2!} + \frac{h^4}{4!} - \ldots - 1}{h} = \lim_{h \to 0} \frac{-\frac{h^2}{2!} + \frac{h^4}{4!} - \ldots}{h}
$$
Factoring out $h$ from the numerator gives:
$$
f'(0) = \lim_{h \to 0} \left(-\frac{h}{2!} + \frac{h^3}{4!} - \ldots\right)
$$
Since this limit results in zero:
$$
\lim_{h \to 0} \left(-\frac{h}{2!} + \frac{h^3}{4!} - \ldots\right) = 0
$$
Therefore,
$$
\boldsymbol{f'(0) = 0}
$$
This conclusion verifies that **the derivative of $\cos x$ at $x = 0$ is indeed zero**, as each non-zero term in the sequence tends to zero as $h$ approaches zero.

Let $y = f(x)$ be a real-valued differentiable function on the set of all real numbers $\mathbb{R}$ such that $f(1) = 1$. If $f(x)$ satisfies $xf^{\prime}(x) = x^2 + f(x) - 2$, then

A. $f(x)$ is an even function.

B. $f(x)$ is an odd function.

C. The minimum value of $f(x)$ is 0.

D. $y = f(x)$ represents a parabola with focus $(1, \frac{5}{4})$.

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Rectangles with perimeter 40 meters will have maximum area if:

A. Length = Breadth

B. Length = 2(Breadth)

C. Length = $\frac{1}{2}$(Breadth)

D. Length = $\frac{1}{4}$(Breadth)