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

69. Kindly consider the following

$c o s x . c o s 2 x . c o s 3 x$

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alok kumar singh | Contributor-Level 10

4 months ago

69. Let y = cos x cos 2x cos 3x _____ (i)

Taking log_e on bolk sides.

logy = log (cos x) + log (cos 2x + log (cos 3x)

= log (cos x) + log (cos 2x) + log (cos 3x)

Differentiating w r t ‘x’

$\frac{1}{y} \frac{d y}{d x} = \frac{1}{c o s x} \frac{d}{d x} c o s x + \frac{1}{c o s 2 x} \frac{d}{d x} c o s 2 x + \frac{1}{c o s 3 x} \frac{d}{d x} c o s 3 x .$

$= \frac{(- s i n x)}{c o s x} + \frac{(- s i n 2 x)}{c o s 2 x} \frac{d}{d x} (2 x) + \frac{(- s i n 3 x)}{c o s 3 x} \frac{d (3 x)}{d x}$

= - tan x-2 tan 2x- 3 tan 3x.

$\Rightarrow \frac{d y}{d x}$ = y [- tan x- 2 tan 2x- 3 tan 3x]

Putting value of y from (i) we get,

$= \frac{d y}{d x}$ = - cos x cos 2x cos 3x [tan x + 2 tan 2x + 3 tan 3x]

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Similar Questions for you

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

$f (- \frac{1}{4}) = f (- \frac{1}{2}) = f (\frac{1}{2}) = f (\frac{1}{4}) = 0$

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

$g (\frac{- 3}{4}) = g (\frac{3}{4}) = 0$

So, g (x) has at least two roots in (-2, 2)

now number of roots of f (x) $\cdot g " (x) = f' (x) \cdot g' (x) = 0$

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

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

Let $1 + \int_{0}^{1} f (t) d t = α$

$\int_{0}^{1} t f (t) d t = β$

So, f(x) = x

Now, $α = \int_{0}^{1} f (t) d t + 1$

$α = \int_{0}^{1} (a t - β) d t + 1$

$β = \int_{0}^{1} t \cdot f (t) d t$

$β = \frac{4}{13}, α = \frac{18}{13}$

f(x) = αx – b

$= \frac{18 x - 4}{13}$

option (D) satisfies

Let f : R -> R be a function defined by f(x) = ${(x - 3)}^{n_{1}} {(x - 5)}^{n_{2}}, n_{1}, n_{2} \in N .$ Then which of the following is NOT true?

alok kumar singh

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

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

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

option (C) is incorrect, there will be minima.

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

Vishal Baghel

f (x) = f (6 – x) Þ f' (x) = -f' (6 – x) …. (1)

put x = 0, 2, 5

f' (0) = f' (6) = f' (2) = f' (4) = f' (5) = f' (1) = 0

and from equation (1) we get f' (3) = -f' (3)

$? f' (3) = 0$

So f' (x) = 0 has minimum 7 roots in $x ? [0, 6] ? f'' (x)$ has min 6 roots in $x ? [0, 6]$

h (x) = f' (x) . f' (x)

h' (x) = (f' (x)² + f' (x) f' (x)

h (x) = 0 has 13 roots in x $?$ [0, 6]

h' (x) = 0 has 12 roots in x $?$ [0, 6]

Number of points of non-differentiability of f(x) = $| | x | - 1 | + | c o s π x |;$ -2 < x < 2 is

alok kumar singh

$I n x \in (- 2, 2)$

|x|- 1| is not differentiable at x = -1, 0, 1

|cospx| is not differentiable at x = $\frac{3}{2}, - \frac{1}{2}, \frac{1}{2}, \frac{3}{2}$ -

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69. Kindly consider the following

$c o s x . c o s 2 x . c o s 3 x$