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Continuity and Differentiability Ncert Solutions Maths class 12th Maths Class 12th Continuity and Differentiability

21. Discuss the continuity of the cosine, cosecant, secant and cotangent functions.

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alok kumar singh | Contributor-Level 10

4 months ago

21. For two continuous fxn f(x) and g(x), $\frac{f (x)}{g (x)}, \frac{g (x)}{f (x)},$

$\frac{1}{f (x)} \frac{1}{g (x)}$ are also continuous

Let f(x) = sin x is defined x R.

Let C E R such that x = c + h. so, as x c, h 0

now, f(c) = sin c.

$\underset{x \to c}{l i m}$ f(i) = $\underset{x \to c}{l i m}$ sin x = $\underset{h \to 0}{l i m}$ sin (c + h).

= $\underset{h \to 0}{l i m}$ (sin c cos h + cos c sin h)

= sin c cos 0 + cos c sin 0

= sin c 1 + 0

= sin c

= f(c)

So, f is continuous.

Then, $\frac{1}{f (x)}$ is also continuous

$\Rightarrow \frac{1}{s i n (x)}$ is also continuous

$\Rightarrow$ cosec x is also continuous

Let g(x) = cos x is defined x R.

Then, g(c) = cos c

$\underset{x \to c}{l i m}$ g(x) = $\underset{x \to c}{l i m}$ . cos x

= $\underset{h \to 0}{l i m}$ cos (c + h).

= $\underset{h \to 0}{l i m}$ (cos c cos h sin c sin h.)

= cos c cos h sin c sin h

= cos c.

= g(c)

So, g is continuous

Then,&nb

...more

21. For two continuous fxn f(x) and g(x), $\frac{f (x)}{g (x)}, \frac{g (x)}{f (x)},$

$\frac{1}{f (x)} \frac{1}{g (x)}$ are also continuous

Let f(x) = sin x is defined x R.

Let C E R such that x = c + h. so, as x c, h 0

now, f(c) = sin c.

$\underset{x \to c}{l i m}$ f(i) = $\underset{x \to c}{l i m}$ sin x = $\underset{h \to 0}{l i m}$ sin (c + h).

= $\underset{h \to 0}{l i m}$ (sin c cos h + cos c sin h)

= sin c cos 0 + cos c sin 0

= sin c 1 + 0

= sin c

= f(c)

So, f is continuous.

Then, $\frac{1}{f (x)}$ is also continuous

$\Rightarrow \frac{1}{s i n (x)}$ is also continuous

$\Rightarrow$ cosec x is also continuous

Let g(x) = cos x is defined x R.

Then, g(c) = cos c

$\underset{x \to c}{l i m}$ g(x) = $\underset{x \to c}{l i m}$ . cos x

= $\underset{h \to 0}{l i m}$ cos (c + h).

= $\underset{h \to 0}{l i m}$ (cos c cos h sin c sin h.)

= cos c cos h sin c sin h

= cos c.

= g(c)

So, g is continuous

Then, $\frac{1}{g (x)}$ is also continuous

$\Rightarrow \frac{1}{c o s x}$ is also continuous

$\Rightarrow$ cos x is also continuous.

Hence, $\frac{g (x)}{f (x)}$ is also continuous

$\Rightarrow \frac{c o s x}{s i n x}$ is also continuous

$\Rightarrow$ cot x is also continuous .

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Let f and g be twice differentiable even functions on (-2, 2) such that $f (\frac{1}{4}) = 0, f (\frac{1}{2}) = 0, f (1) = 1$ and $f (\frac{3}{4}) = 0, g (1) = 2$

Then, the minimum number of solutions of $f (x) g " (x) + f' (x) g' (x) = 0 i n (- 2, 2)$ is equal to……….

Vishal Baghel

f (x) is an even function

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So, f (x) has at least four roots in (-2, 2)

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It is same as number of roots of $\frac{d}{d x} (f (x) \cdot g' (x)) = 0$ will have atleast 4 roots in (-2, 2)

Let f be a real valued continuous function on [0, 1] and f(x) = $x + \int_{0}^{1} (x - t) f (t) d t .$ Then, which of the following points (x, y) lies on the curve y = f(x)?

Vishal Baghel

$f (x) = x + x \int_{0}^{1} f (t) d t - \int_{0}^{1} t_{0} f (t) d t$

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option (D) satisfies

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alok kumar singh

$f' (x) = \frac{n_{1} \cdot f (x)}{x - 3} + n_{2} \cdot \frac{f (x)}{x - 5}$

$= \frac{f (x) \cdot (n_{1} + n_{2})}{(x - 3) (x - 5)} (x - \frac{(5 n_{1} + 3 n_{2})}{n_{1} + n_{2}})$

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Let f(x) be non-constant thrice differentiable function defined on $(- \infty, \infty)$ such that f(x) = f(6 – x) and f’(0) = 0 = f’(2) = f’(5). If ‘n’ is the minimum number of roots of (f”(x))² + f(x)f”(x) = 0 in the interval x $\in$ [0, 6] then sum of digits of n equals

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21. Discuss the continuity of the cosine, cosecant, secant and cotangent functions.

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