Ncert Solutions Maths class 12th

Let f, g : R → R be two real valued function defined as f(x) = ${\begin{matrix} - | x + 3 |, & x < 0 \\ e^{x}, & x \geq 0 \end{matrix}$ and $g (x) = {\begin{matrix} x^{2} + k_{1} x, & x < 0 \\ 4 x + k_{2}, & x \geq 0 \end{matrix}$ , where k₁ and k₂ are real constants. If (gof) is differentiable at x = 0, then (gof)(-4) is equal to

0 Follower 3 Views

Payal Gupta

Contributor-Level 10

GOF is differentiable at x = 0

So R.H.D = L.H.D.

$\frac{d}{d x} (4 e^{x} + k_{2}) = \frac{d}{d x} ({(- | x + 3 |)}^{2} - k_{1} | x + 3 |)$

⇒ 4 = 6 – k₁ Þ k₁ = 2

Now g (f (-4) + g (f (4)

= 2 (2e⁴ – 1)

0 0

New answer posted

3 months ago

If the lines $\frac{x - 3}{- 3} = \frac{y - 2}{2 k} = \frac{z - 3}{2} & \frac{x - 1}{3 k} = \frac{y - 1}{1} = \frac{z - 6}{- 5}$ are perpendicular, find the value of k

0 Follower 2 Views

Vishal Baghel

Contributor-Level 10

The direction of ratios of the lines, $\frac{x - 3}{- 3} = \frac{y - 2}{2 k} = \frac{z - 3}{2} & \frac{x - 1}{3 k} = \frac{y - 1}{1} = \frac{z - 6}{- 5}$ , are $- 3, 2 k, 2 a n d 3 k, 1, - 5$ respectively.

It is known that two lines with direction ratios, $a_{1}, b_{1}, c_{1} a n d a_{2}, b_{2}, c_{2}$ , are perpendicular, if $a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2} = 0$

$\begin{array}{l} ∴ - 3 (3 k) + 2 k * 1 + 2 (- 5) = 0 \\ \Rightarrow - 9 k + 2 k - 10 = 0 \\ \Rightarrow 7 k = - 10 \\ \Rightarrow k = \frac{- 10}{7} \end{array}$

Therefore, for k= -10/7, the given lines are perpendicular to each other.

0 0

New answer posted

3 months ago

Kindly Consider the following

0 Follower 1 View

Vishal Baghel

Contributor-Level 10

Kindly go through the solution

0 0

New answer posted

3 months ago

Ncert Solutions Maths class 12th Limits and Derivatives

Find gof and fog , if:

(i) f(x) = |x| and g(x) = |5x - 2|

(ii) f(x) = 8x³ and g(x) = x^1/3

0 Follower 3 Views

Vishal Baghel

Contributor-Level 10

Kindly go through the solution

0 0

New answer posted

3 months ago

17. Kindly consider the following

0 Follower 2 Views

Contributor-Level 10

17. Kindly go through the solution

$=$ $= \underset{x \to 0}{l i m} \frac{2 s i n^{2} x}{2 s i n^{2} \frac{x}{2}}$

$= \frac{\underset{x \to 0}{l i m} {(\frac{s i n x}{x})}^{2} * x^{2}}{\underset{x \to 0}{l i m} {(\frac{s i n^{} \frac{x}{2}}{\frac{x}{2}})}^{2} {\frac{x}{2}}^{2}}$

$= \frac{{(1)}^{2} * x^{2} * 4}{{(1)}^{2} * x^{2}}$

= 4

0 0

New answer posted

3 months ago

Kindly consider the following

$\int e^{- 3 x} c o s^{3} x d x$

0 Follower 1 View

Contributor-Level 10

This is a Long Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t ? ? ? ? I = ? \underset{I I}{e^{? 3 x}} \underset{I}{c o s^{3} x} d x \\ ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? = c o s^{3} x . ? e^{? 3 x} ? d x ? ? (D (c o s^{3} x) . ? e^{? 3 x} ? d x) d x \\ ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? = c o s^{3} x . \frac{e^{? 3 x}}{? 3} ? ? (3 c o s^{2} x (? s i n x) . \frac{e^{? 3 x}}{? 3}) d x \\ ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? = ? \frac{1}{3} e^{? 3 x} c o s^{3} x ? ? c o s^{2} x s i n x . e^{? 3 x} d x \\ ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? = ? \frac{1}{3} e^{? 3 x} c o s^{3} x ? ? (1 ? s i n^{2} x) s i n x . e^{? 3 x} d x \\ ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? = ? \frac{1}{3} e^{? 3 x} c o s^{3} x ? ? s i n x . e^{? 3 x} d x + ? \underset{I}{s i n^{3} x} . \underset{I I}{e^{? 3 x}} d x \\ ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? = ? \frac{1}{3} e^{? 3 x} c o s^{3} x ? ? s i n x . e^{? 3 x} d x + s i n^{3} x ? e^{? 3 x} d x ? ? (D (s i n^{3} x) . ? e^{? 3 x} ? d x) d x \\ ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? = ? \frac{1}{3} e^{? 3 x} c o s^{3} x ? ? s i n x . e^{? 3 x} d x + s i n^{3} x . \frac{e^{? 3 x}}{? 3} ? ? (3 s i n^{2} x . c o s x . \frac{e^{? 3 x}}{? 3}) d x \\ ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? = ? \frac{1}{3} e^{? 3 x} c o s^{3} x ? ? s i n x . e^{? 3 x} d x ? \frac{1}{3} e^{? 3 x} s i n^{3} x + ? s i n^{2} x c o s x . e^{? 3 x} d x \\ ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? = ? \frac{1}{3} e^{? 3 x} c o s^{3} x ? ? s i n x . e^{? 3 x} d x ? \frac{1}{3} e^{? 3 x} s i n^{3} x + ? (1 ? c o s^{2} x) c o s x . e^{? 3 x} d x \\ ? ? ? ? ? ? ? ? ? ? ? ? ? I = ? \frac{1}{3} e^{? 3 x} c o s^{3} x ? [s i n x . \frac{e^{? 3 x}}{? 3} ? ? c o s x . \frac{e^{? 3 x}}{? 3} d x] ? \frac{1}{3} e^{? 3 x} s i n^{3} x + ? c o s x . e^{? 3 x} d x ? ? c o s^{3} x . e^{? 3 x} d x \\ ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? = ? \frac{1}{3} e^{? 3 x} c o s^{3} x + s i n x . \frac{e^{? 3 x}}{3} ? ? c o s x . \frac{e^{? 3 x}}{? 3} d x ? \frac{1}{3} e^{? 3 x} s i n^{3} x + ? c o s x . e^{? 3 x} d x ? I \\ ? ? ? ? ? ? ? ? ? 2 I = \frac{e^{? 3 x}}{? 3} [c o s^{3} x + s i n^{3} x] ? [s i n x . \frac{e^{? 3 x}}{3} ? ? c o s x . \frac{e^{? 3 x}}{? 3} d x] + ? c o s x . e^{? 3 x} d x \\ ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? = \frac{e^{? 3 x}}{? 3} [c o s^{3} x + s i n^{3} x] + \frac{1}{3} s i n x . e^{? 3 x} ? \frac{1}{3} ? c o s x . e^{? 3 x} d x + ? c o s x . e^{? 3 x} d x \\ ? ? ? ? ? ? ? ? 2 I = \frac{e^{? 3 x}}{? 3} [c o s^{3} x + s i n^{3} x] + \frac{1}{3} s i n x . e^{? 3 x} + \frac{2}{3} ? c o s x . e^{? 3 x} d x \\ N o w, ? I_{1} = \frac{2}{3} ? \underset{I}{c o s x} . \underset{I I}{e^{? 3 x}} d x \\ ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? = \frac{2}{3} [c o s x . ? e^{? 3 x} d x ? ? (D (c o s x) . ? e^{? 3 x} ? d x) d x] \\ ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? = \frac{2}{3} [c o s x . \frac{e^{? 3 x}}{? 3} ? ? ? s i n x . \frac{e^{? 3 x}}{? 3} d x] \\ ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? = \frac{2}{3} [c o s x . \frac{e^{? 3 x}}{? 3} ? \frac{1}{3} ? s i n x . e^{? 3 x} d x] \end{array}$

$\begin{array}{l} = \frac{2}{? 9} c o s x . e^{? 3 x} ? \frac{2}{9} ? s i n x . e^{? 3 x} d x \\ ? ? ? ? ? ? ? ? ? ? ? I_{1} = \frac{2}{? 9} c o s x . e^{? 3 x} ? \frac{2}{9} [s i n x . \frac{e^{? 3 x}}{? 3} ? ? c o s x . \frac{e^{? 3 x}}{? 3} d x] \\ ? ? ? ? ? ? ? ? ? ? ? I_{1} = ? \frac{2}{9} c o s x . e^{? 3 x} + \frac{2}{27} s i n x . e^{? 3 x} ? \frac{2}{27} ? c o s x . e^{? 3 x} d x \\ ? ? ? ? ? ? ? ? ? ? ? I_{1} = ? \frac{2}{9} c o s x . e^{? 3 x} + \frac{2}{27} s i n x . e^{? 3 x} ? \frac{1}{9} . \frac{2}{3} ? c o s x . e^{? 3 x} d x \\ ? ? ? ? ? ? ? ? ? ? ? I_{1} = ? \frac{2}{9} c o s x . e^{? 3 x} + \frac{2}{27} s i n x . e^{? 3 x} ? \frac{1}{9} . I_{1} \\ I_{1} + \frac{1}{9} I_{1} = ? \frac{2}{9} c o s x . e^{? 3 x} + \frac{2}{27} s i n x . e^{? 3 x} \\ ? ? ? ? ? ? ? \frac{10 I_{1}}{9} = ? \frac{2}{9} c o s x . e^{? 3 x} + \frac{2}{27} s i n x . e^{? 3 x} \\ ? ? ? ? ? ? ? ? ? ? ? I_{1} = ? \frac{1}{10} c o s x . e^{? 3 x} + \frac{1}{15} s i n x . e^{? 3 x} \\ S o, ? ? ? ? ? 2 I = ? \frac{1}{3} e^{? 3 x} [s i n^{3} x + c o s^{3} x] + \frac{1}{3} s i n x . e^{? 3 x} ? \frac{1}{10} c o s x . e^{? 3 x} + \frac{1}{15} s i n x . e^{? 3 x} \\ ? ? ? ? ? ? ? ? ? ? ? ? ? I = ? \frac{1}{6} e^{? 3 x} [s i n^{3} x + c o s^{3} x] + \frac{1}{6} s i n x . e^{? 3 x} ? \frac{1}{20} c o s x . e^{? 3 x} + \frac{1}{30} s i n x . e^{? 3 x} \\ ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? = ? \frac{1}{6} e^{? 3 x} [s i n^{3} x + c o s^{3} x] + \frac{1}{5} s i n x . e^{? 3 x} ? \frac{1}{20} c o s x . e^{? 3 x} \\ ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? = \frac{e^{? 3 x}}{24} [s i n 3 x ? c o s 3 x] + \frac{3 e^{? 3 x}}{40} [s i n x \end{array}$

0 0

New answer posted 3 months ago

Kindly consider the following

0 Follower 2 Views

alok kumar singh Contributor-Level 10

This is a Long Answer Type Questions as classified in NCERT Exemplar

0 0

New answer posted 3 months ago

Kindly consider the following

0 Follower 1 View

alok kumar singh Contributor-Level 10

This is a Long Answer Type Questions as classified in NCERT Exemplar

0 0

New answer posted 3 months ago

$\int e^{t a n^{- 1} x} (\frac{1 + x + x^{2}}{1 + x^{2}}) d x$

0 Follower 2 Views

Contributor-Level 10

This is a Long Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t I = \int e^{t a n^{- 1} x} (\frac{1 + x + x^{2}}{1 + x^{2}}) d x \\ P u t t a n^{- 1} x = t \Rightarrow \frac{1}{1 + x^{2}} . d x = d t \\ = \int e^{t} (1 + t a n t + t a n^{2} t) d t = \int e^{t} (s e c^{2} t + t a n t) d t \\ H e r e, f (t) = t a n t \\ ∴ f^{'} (t) = s e c^{2} t \\ = e^{t} . f (t) = e^{t} t a n t = e^{t a n^{- 1} x} . x + C \\ [? \int e^{t} [f (x) + f^{'} (x)] d x = e^{x} f (x) + C] \\ H e n c e, I = e^{t a n^{- 1} x} . x + C \end{array}$

0 0

New answer posted

3 months ago

$\int \frac{2 x - 1}{(x - 1) (x - 2) (x - 3)} d x$

0 Follower 1 View