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

64. Kindly consider the following

$e^{x} + {e^{x}}^{2} + \dots + {e^{x}}^{5}$

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alok kumar singh | Contributor-Level 10

4 months ago

64. Let y = e^x + e^x² + … + e^x⁵.

$∴ \frac{d y}{d x} = \frac{d}{d x}$ (e^x + e^x² + e^x³ + e^x⁴ + e^x⁵).

$= \frac{d e^{x}}{d x} + \frac{d e^{x^{2}}}{d x} + \frac{d e^{x^{3}}}{d x} + \frac{d}{d x} e^{x^{4}} + \frac{d}{d x} e^{x^{5 .}}$

$= e^{x} + e^{x^{2}} \frac{d x^{2}}{d x} + e^{x^{3}} \frac{d x^{3}}{d x} + e^{x^{4}} \frac{d x^{4}}{d x} + e^{x^{5}} \frac{d x^{5}}{d x} .$

= e^x + 2x e^x² + 3x²e^x³ + 4x³e^x⁴ + 5ee^x⁴

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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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64. Kindly consider the following

$e^{x} + {e^{x}}^{2} + \dots + {e^{x}}^{5}$