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4.) (\sin x)^{\log x} + x^{\sin x}
Let y = (\sin x)^{\log x} + x^{\sin x}
y_{1} = (\sin x)^{\log x} \quad y_{2} = x^{\sin x}
y = y_{1} + y_{2}
Apply \frac{d}{dx} on B \cdot S
d \cdot w \cdot r to x
\frac{dy}{dx} = \frac{dy_{1}}{dx} + \frac{dy_{2}}{dx} \longrightarrow ①
Now y_{1} = (\sin x)^{\log x}
Apply on B \cd... | |
\frac{1}{y_{1}} \frac{d y_{1}}{d x} = \log x \cdot \frac{d}{d x} \log (\sin x) + \log \sin x \cdot \frac{d}{d x} \log x
\frac{1}{y_{1}} \frac{d y_{1}}{d x} = \log x \cdot \frac{1}{\sin x} \cdot \frac{d}{d x} \sin x + \log (\sin x) \cdot \frac{1}{x}
\frac{1}{y_{1}} \frac{d y_{1}}{d x} = \begin{matrix} \cot x \cdot \lo... | |
⑤ $x^y$
$\frac{d}{dx} \log y_2 = \frac{d}{dx} (\overset{U}{\sin x} \cdot \overset{V}{\log x})$
$\frac{1}{y_2} \frac{dy_2}{dx} = \sin x \frac{d}{dx} \log x + \log x \frac{d}{dx} \sin x$
$\frac{1}{y_2} \frac{dy_2}{dx} = (\sin x \cdot \frac{1}{x} + \overset{\cos x \cdot \log x}{\log x \cdot \cos x})$
$\frac{dy_2}{dx} ... | |
(5) $x^y + y^x = a^b \text{ then } S \cdot T \frac{dy}{dx} = - \left[ \frac{y x^{y-1} + y^x \log y}{x^y \log x + x y^{x-1}} \right]$
$y_1 = x^y \quad y_2 = y^x$
$y_1 + y_2 = a^b$
$Apply \frac{d}{dx} \text{ on } B \cdot S$
$\frac{dy_1}{dx} + \frac{dy_2}{dx} = \frac{d}{dx} ( a^b )$
$\frac{dy_1}{dx} + \frac{dy_2}{dx}... | |
$$\frac{1}{y_{1}} \cdot \frac{dy_{1}}{dx} = y \frac{d}{dx} \log x + \log x \frac{dy}{dx}$$
$$\frac{1}{y_{1}} \frac{dy_{1}}{dx} = y \frac{1}{x} + \log x \frac{dy}{dx}$$
$$\frac{dy_{1}}{dx} = y_{1} ( y \cdot x^{-1} + \log x \frac{dy}{dx} )$$
$$\frac{dy_{1}}{dx} = x^{y} ( y \cdot x^{-1} + \log x \frac{dy}{dx} )$$
$$\frac{... | |
\log y_2 = \log y^x
\log y_2 = x \log y
Apply \frac{d}{dx} on B S
d. w. r to x
\frac{d}{dx} \log y_2 = \frac{d}{dx} (\stackrel{u}{x} \cdot \stackrel{v}{\log y})
\frac{1}{y_2} \frac{dy_2}{dx} = x \frac{d}{dx} \log y + \log y \frac{d}{dx} x
\frac{1}{y_2} \frac{dy_2}{dx} = x \cdot \frac{1}{y} \frac{dy}{dx} + \log y
\frac{... | |
\frac{dy_1}{dx} + \frac{dy_2}{dx}
y \cdot x^{y-1} + x^y \log x \frac{dy}{dx} + x \cdot y^{x-1} \frac{dy}{dx} + y^x \log y
= 0
(x^y \log x + xy^{x-1}) \frac{dy}{dx} = - y \cdot x^{y-1} - y^x \log y
(x^y \log x + xy^{x-1}) \frac{dy}{dx} = -(y \cdot x^{y-1} + y^x \log y)
\frac{dy}{dx} = - \left( \frac{y \cdot x^{y-1} ... | |
\frac{dy}{dx} = \frac{dy_1}{dx} + \frac{dy_2}{dx} \longrightarrow (1)
y_1 = x^{\tan x}
Apply \log on BS
\log y_1 = \log x^{\tan x}
\log y_1 = \tan x \cdot \log x
d \cdot w \cdot r \text{ to } x
\frac{d}{dx} \log y_1 = \frac{d}{dx} (\tan x \cdot \log x)
\frac{1}{y_1} \frac{dy_1}{dx} = \tan x \frac{d}{dx} \log x + \log ... | |
y_{2}=(\sin x)^{\cos x}
\text{Apply log on B.S}
\log y_{2} = \log (\sin x)^{\cos x}
\log y_{2} = \cos x \cdot \log (\sin x)
d \cdot w \cdot r \text{ to } x
\frac{d}{dx} \log y_{2} = \frac{d}{dx} (\cos x \cdot \log (\sin x))
\frac{1}{y_{2}} \frac{d y_{2}}{d x} = \cos x \frac{d}{dx} \log (\sin x)
+ \log (\sin x) \frac{d}... | |
e
$(\tan x)$
$(\sin x)$
$\frac{d}{dx} \cos x$
$\sin X$
$(\log x)$
$\frac{dy^{2}}{dx} = y^{2} (\text{cat } x \cdot \text{cot } x - \sin x \cdot \text{Log } \sin x)$
$\frac{dy^{2}}{dx} = (\sin x)^{\cos x} (\cot x \cdot \text{Cot } x - \sin x \cdot \log \sin x) \longrightarrow \textcircled{3}$
$\text{Sub eqn } \textcir... | |
Sol:-
(and part $\log \ y \cdot \frac{d}{dx} \log \ x + \log \ x \cdot \frac{d}{dx} \log \ y = \frac{1}{\log \ x} \cdot \frac{d}{dx}$
$\log \ x$
$\log \ y \cdot \frac{1}{x} + \log \ x \cdot \frac{1}{y} \cdot \frac{dy}{dx} = \frac{1}{\log \ x} \cdot \frac{1}{x}... | |
Sol:-
Given f(x) = sin^{-1} \sqrt{\frac{x - \beta}{1 - \beta}}
sin f(x) = \sqrt{\frac{x - \beta}{2 - \beta}}
S \cdot O \cdot b \cdot S
sin^{2} f(x) = \frac{x - \beta}{1 - \beta}
W \cdot K \cdot T
cos^{2} f(x) = 1 - sin^{2} f(x)
= 1 - \left( \frac{x - \beta}{1 - \beta} \right)
= \frac{1 - \beta - x + \beta}{1 - \beta... | |
Also
$\tan^{2}f(x) = \frac{\sin^{2}f(x)}{\cos^{2}f(x)}$
$= \frac{\frac{x - \beta}{\alpha - \beta}}{\frac{\alpha - x}{\alpha - \beta}}$
$\tan^{2} \cdot f(x) = \frac{x - \beta}{\alpha - x}$
$\tan f(x) = \sqrt{\frac{x - \beta}{\alpha - x}}$
$f(x) = \tan^{-1} \sqrt{\frac{x - \beta}{\alpha - x}}$
$f(x) = g(x) (\because give... | |
Continuity $(G \to Sem 0)$
$\rule{4cm}{0.4pt}$
$\fbox{* lim \to limit}$
(1) $\lim_{x \to 0} \frac{\sin x}{x} = 1$ $\qquad \qquad$ (2) $\lim_{x \to 0} \frac{\sin ax}{x} = a$
(3) $\lim_{x \to 0} \frac{\tan x}{x} = 1$ $\qquad \qquad$ (4) $\lim_{x \to 0} \frac{\tan ax}{x} = a$
Def :- If $f(x)$ is continous at $x=a$
$\qq... | |
Sol:-
\lim_{x \to 0} f(x) = \lim_{x \to 0} \frac{\cos ax - \cos bx}{x^2}
\cos C - \cos P = -2 \sin \left( \frac{C+P}{2} \right) \sin \left( \frac{C-P}{2} \right)
C = ax, P = bx
= \lim_{x \to 0} \frac{-2 \sin \left( \frac{ax + bx}{2} \right) \sin \left( \frac{ax - bx}{2} \right)}{x^2}
\neq -2 \lim_{x \to 0} \frac{\s... | |
\lim_{x \to 0} \frac{\sin ax}{x} = a
= -2 \left( \frac{a+b}{2} \right) \left( \frac{a-b}{2} \right)
= -\frac{(a^{2} - b^{2})}{2}
= -\frac{a^{2} - b^{2}}{2}
= \frac{1}{2} (b^{2} - a^{2})
\lim_{x \to 0} f(x) = f(0)
\therefore f(x) \text{ is continuous at } x = 0 | |
② if f defined by
f(x) = \begin{cases} \frac{\sin 2x}{x} , \text{ if } x \neq 0 \\ 1 , \text{ if } x = 0 \end{cases} continuous at
h=0
\lim_{x \to 0} f(x) = \lim_{x \to 0} \frac{\sin 2x}{x}
= 2
[\because \lim_{x \to} \frac{\sin ax}{x} = a]
and . f(0) = 1
\lim_{x \to 0} f(x) \neq f(0)
f(x) is not continuous
at x=0 | |
\neq 0
0
ous at
h=0
\frac{x}{x}
Since x=a
uous
x=0
(5) f(x) = \left\{ \begin{array}{ll} (x^2-9)/(x^2-2x-3) & \text{if } 0 \le x < 5 \\ & \text{and} \\ & x \neq 3 \\ 1.5, & \text{if } x=3 \\ & \text{at the} \\ & \text{point} \\ & x=3 \end{array} \right.
Sol :-
\lim_{x \to 3} f(x) = \lim_{x \to 3} \frac{x^2-9}{x^2-2x-3... | |
(4) Check the continuity of the
following functions at $x=2$
$f(x) = \begin{cases} \frac{1}{2}(x^2-4) & \text{if } 0 < x < 2 \\ 0 & \text{if } x=0 \\ 2-8x^{-3} & \text{if } x > 2 \end{cases}$ (5)
$L \cdot H \cdot L = \lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} \frac{1}{2}(x^2-4)$
$= \frac{1}{2}(2^2-4) = \frac{1}{2}(0)$
$... | |
since $\lim_{x \to 2^{-}} f(x) \neq \lim_{x \to 2^{+}} f(x)$
$f(x)$ is not conti . at $x=2$
(5)
$f(x) = \begin{cases} k^2x-k. & \text{if } x > 1 \\ 2 & \text{if } x < 1 \end{cases}$
at $x=1$
$\rule{2cm}{0.4pt}$
$L \cdot H \cdot L = \lim_{x \to 1^{-}} f(x) = \lim_{x \to 1^{-}} (2) = 2$
$R \cdot H \cdot L = \lim_{x \... | |
$$(k-2) \quad (k+1) = 0$$
$$k-2=0 \quad | \quad K+1=0$$
$$k=2 \quad \quad | \quad K=-1$$
$$\therefore K=2 \text{ (or) } = -1$$
$$6. \text{ Find the const } a, b \text{ show}$$
$$\text{that the } f^{n} \text{ f given by}$$
$$f(x) \begin{cases} \sin x, x \le 0 \\ x^{2}+a, \text{ if } 0 < x < 1 \\ bx+3, \text{ if } 1 \le... | |
at @ x=3
L . H . L = \lim_{x \to 3^-} f(x) = \lim_{x \to 3^-} (bx+3)
= b(3)+3
= 3b+3
R \cdot H \cdot L = \lim_{x \to 3^+} f(x) = \lim_{x \to 3^+} (-3) = -3
f(x) is + cont at x=3
L \cdot H \cdot L = R \cdot H \cdot L
3b + 3 = -3
3b = -3-3
3b = -6
b = -2 | |
I - 120
whose
eqns
$m^2 = 0$
$\rightarrow$ ②
$l^2 + m^2 + 2 ml$
Case I
Put in $m = 0$ in ③
$l = -0 - n = -n$
$l : m : n = -n : 0 : n$
$l : m : n = \frac{-n}{n} : \frac{0}{n} : \frac{n}{n}$
$l : m : n = -1 : 0 : 1$
D.r's are (-1, 0, 1)
Divide with $\sqrt{a^2 + b^2 + c^2}$
$= \sqrt{(-1)^2 + 0^2 + 1^2} = \sqrt{1 + 0 + 1... | |
Case - II
sub l + m = -n
l = -(-n) - n = n - n = 0
l : m : n = 0 : -n : n
l : m : n = \frac{0}{n} : \frac{-n}{n} : \frac{n}{n}
l : m : n = 0 : -1 : 1
D.R's are (0, -1, 1)
Divide with \sqrt{l^2 + m^2 + n^2}
= \sqrt{0^2 + (-1)^2 + 1^2} = \sqrt{0+1+1}
= \sqrt{2}
D.C's of second line are
(l_2, m_2, n_2) = (\frac{0}{\sqrt{2... | |
then . \cos \theta = | l_1 l_2 + m_1 m_2 + n_1 n_2 |
\cos \theta = | ( \frac{-1}{\sqrt{2}} ) (0) + (0) ( \frac{-1}{\sqrt{2}} )
+ ( \frac{1}{\sqrt{2}} ) ( \frac{1}{\sqrt{2}} ) |
\cos \theta = | 0 + 0 + \frac{1}{2} | = \frac{1}{2}
\cos \theta = \cos 60
\theta = 60^\circ = \frac{\pi}{3}
③ Find the angle b/w the lines
... | |
\begin{array}{l}
\underline{Case} \\
\\
sub\ l,.\ m\ in\ ② \\
\\
6(-3l-5n)n-2nl+5l(-3l-5n)=0 \\
\\
-\ 18\ ln-30n^{2}-2nl-15l^{2}-25ln=0 \\
\\
-\ 15\ l^{2}-45\ ln-30n^{2}=0 \\
\\
-\ 15\ (l^{2}+3ln+2n^{2})=0 \\
\\
\cdot\ l^{2}+3ln+2n^{2}=0 \\
\\
\cdot\ l^{2}+2ln+ln+2n^{2}=0 \\
\\
l(l+2n)+n(l+2n)=0 \\
\\
(l+2n)\ (l+n)=0 \... | |
-5n)=0
Case-I
25n=0
put II l = -2n in ③
m = -3(-2n) - 5n
m = 6n - 5n
m = n
l : m : n = -2n : n : n
l : m : n = \frac{-2n}{n} : \frac{n}{n} : \frac{n}{n}
l : m : n = -2 : 1 : 1
Dr's are (-2, 1, 1)
Divide with \sqrt{a^2+b^2+c^2}
= \sqrt{(-2)^2+1^2+1^2} = \sqrt{4+9+1} = \sqrt{6}
Dc's of first line are
(l_1, m_1, n_1) = ... | |
Case -II
Sub \cdot II l = -n in (3)
m = -3(-n) - 5n
m = 3n - 5n
m = -2n
l : m : n = -n : -2n : n
l : m : n = \frac{-n}{n} : \frac{-2n}{n} : \frac{n}{n}
l : m : n = -1 : -2 : 1
Dr's are (-1, -2, 1)
Divide with \sqrt{a^2 + b^2 + c^2}
= \sqrt{(-1)^2 + (-2)^2 + 1^2} = \sqrt{1 + 4 + 1} = \sqrt{6}
DC's of second line are | |
-2:1
1+4+1
= 6
-\sqrt{6}
(l_2, m_2, n_2) = \left( \frac{-1}{\sqrt{6}}, \frac{-2}{\sqrt{6}}, \frac{1}{\sqrt{6}} \right)
If \theta is the angle b/w two
lines
whose d.c's are
(l_1, m_1, n_1) and (l_2, m_2, n_2)
\cos \theta = | l_1 l_2 + m_1 m_2 + n_1 n_2 |
\cos \theta = | \left( \frac{-2}{\sqrt{6}} \right) \left( \frac... | |
6
the
$y^{2}$
-1
0
(
$(1)^{2}=0$
$y$
$(2y)^{2}=0$
$+4y^{2}$
$2 k^{2} x^{2} - 2 k^{2} x y + 3 k^{2} y^{2} + 2 k x^{2} + 4 k x y - k x y$
$- 2 k y^{2} - x^{2} - 4 y^{2} - 4 x y = 0$
___________________________
$k^{2}$
$(2 k^{2} + 2 k - 1) x^{2} + (- 2 k^{2} + 4 k - k - 4) x y$
$+ (3 k^{2} - 2 k - 4) y^{2}$
$= 0$
Given t... | |
⑤ S.T the lines joining origin to
the points of intersection of the
curve 3x^2 - xy + y^2 + 3x + 3y - 2 = 0
and the st. line x - y - \sqrt{2} = 0
are mutually perpendicular.
x - y - \sqrt{2} = 0
x - y = \sqrt{2}
\frac{x - y}{\sqrt{2}} = 1 \longrightarrow ②
Homogenising ① with the help of ②.
x^2 - xy + y^2 + 3x (1) +... | |
Th 1 (05) 6 :-
① let the eq $ax^2 + 2hxy + by^2 = 0$ represent
a pair of st. lines then the angle $\theta$.
b/w the lines is given by $cos \theta = \frac{|a+b|}{\sqrt{(a-b)^2 + (2h)^2}}$
Sol :- Let $ax^2 + 2hxy + by^2 = 0$ rep
a pair of lines.
$l_1 x + m_1 y = 0 \rightarrow ① \quad l_2 x + m_2 y = 0 \rightarrow ②$
$a... | |
present
Q.
[cot a]
-> 0
- m_2 b
+ m_1 m_2^2
xy
2
y=lx
z=0
\cos \theta = \frac{|l_1 l_2 + m_1 m_2|}{\sqrt{(l_1^2 + m_1^2) \cdot (l_2^2 + m_2^2)}} \quad 2
= \frac{| a + b |}{\sqrt{l_1^2 + l_2^2 + l_1^2 m_2^2 + l_2^2 m_1^2 + m_1^2 m_2^2}}
= \frac{| a + b |}{\sqrt{(l_1 l_2)^2 + (m_1 m_2)^2 + (l_1 m_2)^2 + (l_2 m_1)^2}}
... | |
② show that the product of the
perpendicular distances from
a point (\alpha, \beta) to the pair of
st lines ax^2 + 2hxy + by^2 = 0
is \frac{|a\alpha^2 + 2h\alpha\beta + b\beta^2|}{\sqrt{(a-b)^2 + 4h^2}}
_________________________________
Sol :- Let ax^2 + 2hxy + by^2 = 0 rep a pair of lines
l_1x + m_1y = 0 \implies ① l_... | |
= \frac{|(l_1 \alpha + m_1 \beta)(l_2 \alpha + m_2 \beta)|}{\sqrt{(l_1^2 + m_1^2) \cdot (l_2^2 + m_2^2)}}
= \frac{|l_1 l_2 \alpha^2 + l_1 m_2 \alpha \beta + l_2 m_1 \alpha \beta + m_1 m_2 \beta^2|}{\sqrt{l_1^2 l_2^2 + l_1^2 m_2^2 + l_2^2 m_1^2 + m_1^2 m_2^2}}
= \frac{|l_1 l_2 \alpha^2 + (l_1 m_2 + l_2 m_1) \alpha \be... | |
$\alpha\beta$
$- m_1, m_2 \beta^2$
$m_1^2 m_2^2$
$+ m_1, m_2 y^2$
$+ (l_2 m_1)^2$
$m_1 m_2$
$(l_1 m_2)^2$
$l_2 m_1)^2$
$2 l_1 l_2 m_1 m_2$
$| a\alpha^2 + 2h\alpha\beta + b\beta^2 |$
$= \frac{| a\alpha^2 + 2h\alpha\beta + b\beta^2 |}{\sqrt{(l_1 l_2 - m_1 m_2)^2 + (l_1 m_2 + l_2 m_1)^2}}$
$= \frac{| a\alpha^2 + 2h\alpha\... | |
Comparing like term
$l_1 l_2 = a, l_1 m_2 + l_2 m_1 = 2h, m_1 m_2 = b$
given line $lx + my + n = 0 \longrightarrow \textcircled{3}$
sol \textcircled{1}, \textcircled{2} is = (0,0)
sol \textcircled{1}, \textcircled{3} .
$x$ $y$ $1$
$m_1$ $0$ $l_1$ $m_1$
$m$ $n$ $l$ $m$
$\frac{x}{m_1 n - 0} = \frac{y}{0 - nl_1} = \frac{1... | |
2=b
B = \left( \frac{m_2 n}{l_2 m - l m_2} , \frac{- n l_2}{l_2 m - l m_2} \right)
the\ area\ at\ \Delta O AB = \frac{1}{2} (x_1 y_2 - x_2 y_1)
= \frac{1}{2} \left| \left( \frac{m_1 n}{l_1 m - l m_1} \right) \left( \frac{- n l_2}{l_2 m - l m_2} \right) \right.
\qquad - \left( \frac{m_2 n}{l_2 m - l m_2} \right) \le... | |
(3)
$= \frac{1}{2} \left| \frac{n^2 \sqrt{(l_1 m_2 + l_2 m_1)^2 - 4 l_1 l_2 m_1 m_2}}{l_1 l_2 m^2 - (l_1 m_2 + l_2 m_1) l m + m_1 m_2 l^2} \right|$
$= \frac{1}{2} \left| \frac{n^2 \sqrt{4(h^2 - ab)}}{am^2 - 2 h l m + b l^2} \right|$
$= \frac{1}{2} \cdot \left| \frac{n^2 \sqrt{4(h^2 - ab)}}{am^2 - 2 h l m + b l^2} \r... | |
m_1 m_2
③ Orthocentre :- Steps in orthocentre
Step 1:- Given points are A, B, C
Let \overline{AP}, \overline{BE} are
altitudes drawn A
from A and B to the E
sides BC and AC B D C
Step 2:- To find eqn of \overline{AP}, slope of BC (m)
... | |
$P(x_1, y_1) \text{ to } y - y_1 = -\frac{1}{m}(x - x_1)$
$\rightarrow (2)$
Step (2) :- sol (1) and (2)
$O = ( , )$
(3) find the orthocentre of the triangle
with the vertices $(-2, -1), (6, -1)$
$(2, 5)$
Sol :- Given points are $A, B, C$.
Let $\overline{AD}, \overline{BF}$ are
altitude drawn
from $A$ and $B$ to the ... | |
slope of AD = \frac{-1}{m} = \frac{-1}{\frac{-3}{2}} = \frac{2}{3}
The eqn of AD P.T. P (Passing through
the Point)
A(-2, -1)
y - y_{1} = \frac{-1}{m} \cdot (x - x_{1})
y + 1 = \frac{2}{3} (x + 2)
3y + 3 = 2x + 4
2x + 4 - 3y - 3 = 0
slope of AC (m) = \frac{5 + 1}{2 + 2}
m = \frac{6}{4} = \frac{3}{2}
BE \perp AC
slope ... | |
$y+1 = -\frac{2}{3} (x-6)$
$3y + 3 = -2x+12$
$3y + 3 + 2x - 12 = 0$
$2x + 3y - 9 = 0 \rightarrow (2)$
$Sol \text{ } (1), (2)$
$\quad \quad x \quad \quad y \quad \quad 1$
$-3 \quad \text{ } 1 \quad \text{ } 2 \quad -3$
$3 \quad -9 \quad \text{ } 2 \quad \text{ } 3$
$\frac{x}{27-3} = \frac{y}{2+18} = \frac{1}{6+6}$... | |
orthocentre (O) = ( 2, \frac{5}{3} )
④ A ( -5, -7 )
B ( 13, 2 )
C ( -5, 6 )
Given points are A, B, C
Let AD , BE are
altitude drawn
from A and B to the sides
BC and AC
A
E
B P C
Slope of BC (m) = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}
m = \frac{6 - 2}{-5 - 13} = \frac{4}{-18} = \frac{-2}{9}
AD \per... | |
The eqn of \overline{AD} P-T-P
A(-5, -7) is
y - y_{1} = -\frac{1}{m} (x - x_{1})
2y + 14 = 9x + 45
9x + 45 - 2y - 14 = 0
9x - 2y + 31 = 0 \longrightarrow ①
slope of AC(m) = \frac{6 + 7}{-5 + 5}
m = \frac{13}{0}
BE \perp AC
slope of BE = -\frac{1}{m} = \frac{-1}{\frac{13}{0}} = \frac{0}{13} = 0
The eqn of \overline{BE} ... | |
y-y_{1}=\frac{-1}{m}(x-x_{1}) \textcircled{1}
y-2=0 (-x. 13)
y-2=0
y=2
Sub in y=2 in \textcircled{1}
9x-2(2)+31=0
9x-4+31=0
9x+27=0
9x=-27
x=\frac{-27}{9}=-3
Centhan (-3, 2) | |
⑪ $\cdot Q(h, k) \cdot f(x, y) \cdot$
$ax + by + c = 0$ then
$\frac{h - x_1}{a} = \frac{k - y_1}{b} = \frac{-(ax_1 + by_1 + c)}{a^2 + b^2}$
sol :- Given line $ax + by + c = 0 \longrightarrow$ ①
& the points are $P(x_1, y_1) \& Q(h, k)$
slope of $PQ (m_2) = \frac{k - y_1}{h - x_1}$
since $PQ \perp$ to ①
slope of ... | |
\frac{h - x_1}{a} = t \quad \frac{k - y_1}{b} = t
h - x_1 = at \quad k - y_1 = bt
h = x_1 + at \quad k = y_1 + bt - ③
Since \odot (h, k) lie on ①
ah + bk + c = 0
a(x_1 + at) + b(y_1 + bt) + c = 0
ax_1 + a^2 t + by_1 + b^2 t + c = 0
t(a^2 + b^2) = -ax_1 - by_1 - c
t(a^2 + b^2) = -(ax_1 + by_1 + c)
t = \frac{-(ax_1 ... | |
(iv) let $\Theta(h, k)$ is the foot of the
$\perp$ from $P(-1, 3)$ on the line $5x-y-18=0$
$\frac{h - x_1}{a} = \frac{k - y_1}{b} = \frac{-(ax_1 + by_1 + C)}{a^2 + b^2}$
$\frac{h + 1}{5} = \frac{k - 3}{-1} = \frac{-(5(-1) - 3 - 18)}{25 + 1}$
$= \frac{-(-5 - 3 - 18)}{26}$
$= \frac{26}{26} = 1$
$\frac{h + 1}{5} = 1$ |... | |
the
-5-18-A
C)
Image theorem
If Q (h, k) is the image of
point - P(x_1, y_1) with respective to the
st. line ax + by + c = 0 then P.T
\frac{h - x_1}{a} = \frac{k - y_1}{b} = \frac{-2(ax_1 + by_1 + c)}{a^2 + b^2}
and find the image of (1, -2) w.r.t
to st. line 2x - 3y + 5 = 0
sol:- Given line ax + by + c = 0 ①
& the po... | |
\frac{k - y_1}{b} = \frac{h - x_1}{a}
\frac{h - x_1}{a} = \frac{k - y_1}{b} = t \text{ (let) } ②
\frac{h - x_1}{a} = t \quad \mid \quad \frac{k - y_1}{b} = t
h - x_1 = at \quad \mid \quad k - y_1 = bt
h = x_1 + at \quad \mid \quad k = y_1 + bt - ③
let R be the mid point of PQ
R = \left( \frac{x_1 + h}{2}, \frac{y_1 ... | |
(8) (0,6) and (6,0)
Sol:- A(0,6) ; B(6,0)
Let P(x,y) be a point on the locus.
given condition is \angle APB = 90^\circ
PA^2 + PB^2 = AB^2
(x-0)^2 + (y-6)^2 + (x-6)^2 + (y-0)^2
= (6-0)^2 + (0-6)^2
x^2 + y^2 + 36 - 12y + x^2 + 36 - 12x + y^2
= 36 + 36
2x^2 + 2y^2 - 12x - 12y = 0
2(x^2 + y^... | |
(11) A(4,0) B(0,4)
Let P(x,y) be a point on the given
condition is \angle APB = 90
PA^2 + PB^2 = AB^2
(x-4)^2 + (y-0)^2 + (x-0)^2 + (y-4)^2 = (0-4)^2 + (4-0)^2
x^2 + 16 - 8x + y^2 + x^2 + y^2 + 16 - 8y = 16 + 16
2x^2 + 2y^2 - 8x - 8y = 0
2(x^2 + y^2 - 4x - 4y) = 0
x^2 + y^2 - 4x - 4y = 0
The locus of P(x,y) is
x^2 + ... | |
Let $P (x,y)$ be a point on the locus
given condition is . $OP = 2PA$
$S \cdot O \cdot B \cdot S$
$OP^{2} = 4 PA^{2}$
$(x-0)^{2} + (y-0)^{2} = 4((x-1)^{2} + (y-2)^{2})$
$x^{2} + y^{2} = 4(x^{2} + 1 - 2x + y^{2} + 4 - 4y)$
$x^{2} + y^{2} = 4x^{2} + 4 - 8x + 4y^{2} + 16 - 16y$
$4x^{2} + 20 - 8x + 4y^{2} - 16y - x^{2} - y... | |
2. Transformation of axes
$\rightarrow$ When the origin is shifted to $O' (h, k)$
by transformation of axes and
its new co-ordinates of $(x, y)$
Then $x = X + h$ ; $y = Y + k$
$\rightarrow$ Write the original equation always
in terms of $(x, y)$ only
$\rightarrow$ write the transformed equ always
in terms of $(X, Y)$ o... | |
axes
$O' (h, k)$
and $(X, Y)$
$(x, y)$
$+ k$
always
always
ly
$y - 11 = 0$
$- 11 = 0 \to \textcircled{0}$
Rotation of axes
If the axes are rotated through
an angle $\theta$ without
changing the angle is called
/rotation of axes
| | X | Y |
| :--- | :--- | :--- |
| x | $\cos \theta$ | $-\sin \theta$ |
| y | $\sin \t... | |
\theta = 45
x \cos 45
x \left( \frac{1}{\sqrt{2}} \right)
\frac{x+y}{\sqrt{2}}
- y
\frac{y^2 + 2xy}{2}
= 9
x^2 + 3y^2
+ 6xy
= 9
16x^2 - 4y^2 = 2(9)
2(8x^2 - 2y^2) = 2(9)
\boxed{8x^2 - 2y^2 = 9}
(3) \frac{\pi}{6} \cdot x^2 + 2\sqrt{3}xy - y^2 = 2a^2
Angle of rotation (\theta) = \frac{\pi}{6} = \frac{180}{6} = 30^\circ... | |
x = X \cos \theta - Y \sin \theta \space , \space y = X \sin \theta + Y \cos \theta
x = X \cos 45 - Y \sin 45 \quad | \quad y = X \sin 45 + Y \cos 45
x = X \left( \frac{1}{\sqrt{2}} \right) - Y \left( \frac{1}{\sqrt{2}} \right) \quad | \quad y = X \left( \frac{1}{\sqrt{2}} \right) + Y \left( \frac{1}{\sqrt{2}} \right)... | |
(\frac{\sqrt{3}x-y}{2})^{2} + 2\sqrt{3} \cdot (\frac{\sqrt{3}x-y}{2}) (\frac{x+\sqrt{3}y}{2})
- (\frac{x+\sqrt{3}y}{2})^{2}
= 2a^{2}
\frac{3x^{2} + y^{2} - 2\sqrt{3}xy}{4} + 2\sqrt{3} (\frac{\sqrt{3}x^{2} + 3xy - xy - \sqrt{3}y^{2}}{4})
= 2a^{2}
- (\frac{x^{2} + 3y^{2} + 2\sqrt{3}xy}{4})
= 2a^{2}
3x^{2} + y^{2} - 2\s... | |
\frac{+\sqrt{3}x}{2}
(\frac{x+\sqrt{3}y}{2})^2
= -2a^2
\frac{3xy-xy-\sqrt{3}y^2}{4}
\frac{x^2+2\sqrt{3}xy}{4}
= 2a^2
y^2-x^2
y^2-2\sqrt{3}xy
= a^2
XY
3Y^2 = 4(2a^2)
4 \cdot A \cdot Q
1 St. lines 6 X 7
1 to 5, 10, 11 5 X 4
2 Pair of St lines
1 to 6
1 to 9
3 Dc's and DR's
1 to 7
S.A.Q's
Locus
1 to 4, 7, 8, 11, 14, 15
2 ... | |
(4)
hol:-
x
x
r
Let x be a side of a square and
r be the radius of circle
given length of the wire = l
4x + 2\pi r = l
2\pi r = l - 4x
r = \frac{l - 4x}{2\pi} (1)
Sum-ab Area (A) = x^2 + \pi r^2
A = x^2 + \pi \left( \frac{l - 4x}{2\pi} \right)^2
= x^2 + \pi \left[ \frac{(l - 4x)^2}{4\pi^2} \right] | |
and
$A = x^2 + \frac{(l - 4x)^2}{4 \pi}$
Let $f(x) = x^2 + \frac{(l - 4x)^2}{4 \pi}$
Diff $w \cdot r$ to $x$
$f'(x) = 2x + \frac{1}{4 \pi} \cdot 2(l - 4x)$
$\cdot (0 - 4(1))$
$f'(x) = 2x - \frac{2}{4 \pi} (l - 4x) \cdot (4)$
$f'(x) = 2x - \frac{2}{\pi} (l - 4x)$
again Diff $w.r$ to $x$
$f''(x) = 2(1) - \frac{2}{... | |
$\pi x = l - 4x$
$\pi x + 4x = l$
$(\pi + 4)x = l \Rightarrow x = \frac{l}{\pi + 4}$
at $x = \frac{l}{\pi + 4} \quad , \quad f''(x) = 2 + \frac{8}{\pi} > 0$
$f(x)$ has minimum at $x = \frac{l}{\pi + 4}$
The length of the square part $= 4x = \frac{4l}{\pi + 4}$
The length of the circle part $= l - 4x$
$= l - \frac{4l}{\... | |
3 p - Geometry
______________
1 Find the centroid of the triangle
(5, 4, 6) (1, -1, 3) , (4, 3, 2)
Given vertices are
A(x_1, y_1, z_1) = (5, 4, 6)
B(x_2, y_2, z_2 = (1, -1, 3)
C(x_3, y_3, z_3) = (4, 3, 2)
The centroid of \triangle ABC is
(G) = \left( \frac{x_1 + x_2 + x_3}{3} , \frac{y_1 + y_2 + y_3}{3} \frac{z_1 +... | |
\frac{x-1}{3}=0 \quad | \quad \frac{y+5}{3}=0 \quad | \quad \frac{z+2}{3}=0
x-1=0 \quad | \quad y+5=0 \quad | \quad z+2=0
x=1 \quad | \quad y=-5 \quad | \quad z=-2
C = (1, -5, -2)
3) Find the centroid of the tetrahedral
whose vertices are (2, 3, -4)
(-3, 3, -2) (-1, 4, 2) (3, 5, 1)
Given vertices are
A(x_1, y_1, z_1... | |
2-
find the co-ordinates of the
vertex 'C' of triangle \triangle ABC if
it's centroid is the origin and
the vertices A B are (1,1,1)
(-2, 4, 1) respectively.
Given vertices are
A(x_1, y_1, z_1) = (1, 1, 1)
B(x_2, y_2, z_2) = (-2, 4, 1)
& centroid (G) = (0, 0, 0)
Let C(x, y, z) be the third
vertex
(0, 0, 0)
The centr... | |
$= \left( \frac{2-3-1+3}{4} , \frac{3+3+4+5}{4} , \frac{-4-2+2+1}{4} \right)$
$= \left( \frac{1}{4} , \frac{15}{4} , \frac{-3}{4} \right)$
(4) If $(-3, 2, -1), (4, 1, 1), (6, 2, 5)$ the three
vertices and $(4, 2, 2)$ is the
centroid of the tetrahedral find
the forth vertex
Given vertices are
$A(x_1, y_1, z_1) = (-3,... | |
-2+1)
The centroid of the tetrahedral
ABCD is
three
(G) = \left( \frac{x_1 + x_2 + x_3 + x_4}{3}, \frac{y_1 + y_2 + y_3 + y_4}{3}, \frac{z_1 + z_2 + z_3 + z_4}{3} \right)
(4, 2, 2) = \left( \frac{3 + 4 + 6 + x}{4}, \frac{2 + 1 + 2 + y}{4}, \frac{1 + 1 + 5 + z}{4} \right)
fixed
\frac{x + 13}{4} = 4 \quad | \quad \frac{y... | |
7 Find the forth vertex of the
parallelogram whose consecutive
(2, 4, -1) (3, 6, -1) (4, 5, 1)
Given vertices are
A(2, 4, -1) B(3, 6, -1) C(4, 5, 1)
let D(x, y, z) be the vertex are 4th
vertex
midpoint of $\overline{AC} = \text{mid point of } \overline{BD}$
$(\frac{2+4}{2}, \frac{4+5}{2}, \frac{-1+1}{2}) = (\frac{3+x}{... | |
(6) S.T A(1,2,3) B(7,0,1) C(-2,3,5)
find AB, BC, CA
AB = \sqrt{(7-1)^2 + (0-2)^2 + (1-3)^2}
= \sqrt{36+4+4}
= \sqrt{44}
BC = \sqrt{(-2-7)^2 + (3-0)^2 + (4-1)^2}
= \sqrt{81+9+9}
= \sqrt{99}
CA = \sqrt{(-2-1)^2 + (3-2)^2 + (4-3)^2}
= \sqrt{9+1+1} = \sqrt{11}
Now
AB + CA = 2\sqrt{11} + \sqrt{11} = 3\sqrt{11} = BC
\the... | |
(10). S·T the points (1,2,3) (2,3,1)
(3,1,2) form an
equilateral triangle
Given vertices are A(1,2,3)
B(2,3,1)
C(3,1,2)
AB = \sqrt{(2-1)^2 + (3-2)^2 + (1-3)^2}
\sqrt{1+1+4} = \sqrt{6}
BC = \sqrt{(3-2)^2 + (1-3)^2 + (2-1)^2}
BC = \sqrt{1 + (-2)^2 + (1)^2}
BC = \sqrt{1+4+1}
BC = \sqrt{6} | |
(2) $CA = \sqrt{(3-1)^2 + (1-2)^2 + (2-3)^2}$
$CA = \sqrt{4 + 1 + 1} = \sqrt{6}$
$AB = BC = CA$
$\triangle ABC$ is an equilateral $\triangle$le
_______________________________________________________
(15) Find the ratio in which
$XZ$ plane divide the line segment
joining $A(-2, 3, 4) \cdot B(1, 2, 3)$
Given point... | |
3d^2
Differentating Pg 134
Find the derivative of the following
functions from the first
principles
(i) f(x) = \cos^2 x
f(x+h) = \cos^2 (x+h)
al gle
By first principal method
f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}
h
f(x) \lim_{h \to 0} \frac{\cos^2(x+h) - \cos^2 x}{h}
\cos^2 B - \cos^2 A = \sin(A+B) \sin(A-B)... | |
② $I = \lim_{h \to 0} \sin(2x+h) \cdot \lim_{h \to 0} \frac{\sin h}{h} \cdot \lim_{h \to 0}$ $2$
$= \sin(2x+0) \cdot 1 \left\{ \lim_{x \to 0} \frac{\sin x}{x} = 1 \right\} = 2$
$\therefore f'(x) = \sin 2x$
(ii) $f(x) = \sin 2x$
$f(x+h) = \sin 2(x+h) = \sin(2x+2h)$
By First principal method
$f'(x) = \lim_{h \to 0} \fra... | |
h
h
x = 1)
(2x + h
(x)
\lim_{h \to 0} \frac{2 \cos \left( \frac{2x + 2h + 2x}{2} \right) \sin \left( \frac{2x + 2h - 2x}{2} \right)}{h}
= 2 \lim_{h \to 0} \frac{\cos \left( \frac{4x + 2h}{ } \right) \sin h}{h}
= 2 \lim_{h \to 0} \cdot \cos \left( \frac{4x + 2h}{2} \right) \lim_{x \to 0} \cdot \frac{\sin h}{h}
= 2 \cos... | |
(v) $f(x) = \tan 2x$
$f(x+h) = \tan 2(x+h) \stackrel{\therefore}{=} \tan(2x+2h)$
By first principal method
$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$
$\lim_{h \to 0} \frac{\tan(2x+2h) - \tan 2x}{h}$
$= \lim_{h \to 0} \frac{1}{h} \left( \frac{\sin(2x+2h)}{\cos(2x+2h)} - \frac{\sin 2x}{\cos 2x} \right)$
$= \lim... | |
$(v)$
$f$
$= \cdot \lim_{h \to 0} \frac{\text{Cos} \cdot (ax+ah) - \cos ax}{h}$
$(\cos C - \cos D = -2 \sin (\frac{C+D}{2}) \sin (\frac{C-D}{2}))$
$- \lim_{h \to 0} \cdot -2 \sin (\frac{ax+ah+ax}{2}) \cdot \sin \left( \frac{ax+ah-ax}{2} \right)$
$--------------------------------------$
$h$
$= -2 \lim_{h \to 0} \fra... | |
$= \lim_{h \to 0} \frac{\sin 2h}{h} \lim_{h \to 0} \frac{1}{\cos(2x+2h) \cos 2x}$
$= 2 \cdot \frac{1}{\cos(2x + 2(0)) \cdot \cos 2x}$
$= \frac{2}{\cos^2 2x}$
$f'(x) = 2 \sec^2 2x$
$(V)$
$f(x) = \sec 3x$
$f(x+h) = \sec 3(x+h) = \sec(3x+3h)$
$[\cdot \text{By first principal method}]$
$f'(x) = \lim_{h \to 0} \frac{f(... | |
$3 \left( \frac{X^2 + Y^2 - 2XY}{2} \right) + 10 \left( \frac{X^2 - Y^2}{2} \right) \quad (5.)$
$\qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad Sol-$
$\quad + 3 \left( \frac{X^2 + Y^2 + 2XY}{2} \right)$
$\qquad \qquad \qquad = 9$
$3x^2 + 3y^2 - 6xy + 10x^2 - 10x^2 + 3x^2$
$\qquad \qquad \qquad \qquad \quad + 3... | |
(2-8x^{-3})
)= 2 - \frac{8}{2^3}
= 2 - \frac{8}{8}
= 2 - 1
= 1
x = 2
x \ge 1
x < 1
= 2
R \cdot H \cdot L = \lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} k^2x - k
= k^2(1) - k
= k^2 - k
Since f(x) is cont on R
\lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x)
2 = k^2 - k
k^2 - k - 2 = 0
k^2 - 2k + k - 2 = 0
k(k - 2) + 1(k - ... | |
$R.H.\widetilde{L} = \lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} (2 - 8/x^3) \qquad R \cdot H$
$= \lim_{x \to 2^+} (2 - \frac{8}{x^3}) = 2 - \frac{8}{2^3}$
$= 2 - \frac{8}{8}$
$= 2 - 1$
$= 1$
and $f(0) = 0$
$f(x)$ is not cont at $x=2$
(5) $f(x) = \begin{cases} K^2x - K & \text{if } x \ge 1 \\ 2 & \text{if } x < 1 \... | |
if x \neq 0
if x = 0
\frac{2x}{x}
cont
at x = 0
x < 5
3
Sol-
\lim_{x \to 3} f(x) = \lim_{x \to 3} \frac{x^2 - 9}{x^2 - 2x - 3} = \lim_{x \to 3}
\frac{x^2 - 3^2}{x^2 + x - 3x - 3}
= \lim_{x \to 3} \frac{(x + 3)(x - 3)}{x(x + 1) - 3(x + 1)} = \lim_{x \to 3} \frac{(x + 3)(x - 3)}{(x + 1)(x - 3)}
= \lim_{x \to 3} \frac{x ... | |
Sol-
② Q
$$f(x) = \begin{cases} \frac{\sin 2x}{x} , & \text{if } x \neq 0 \\ 1 , & \text{if } x = 0 \end{cases}$$
$$\lim_{x \to 0} f(x) = \lim_{x \to 0} \frac{\sin 2x}{x}$$
$$= 2$$
$$\text{and } f(0) = 1$$
$$\lim_{x \to 0} f(x) \neq f(0)$$
$$f(x) \text{ is not cont}$$
$$\text{at } x = 0 \quad ©$$
③ $$f(x) = \begin{ca... | |
lex
, if x \neq 0
if x = 0
Easbx
- \sin \left( \frac{c - b}{2} \right)
\frac{ax - bx}{2}
\left( \frac{a - b}{2} \right) x
\dot{=} -2 \lim_{x \to 0} \frac{\sin \left( \frac{a + b}{2} \right) x}{x} \cdot \lim_{x \to 0} \frac{\sin \left( \frac{a - b}{2} \right) x}{x}
\lim_{x \to 0} \frac{\sin ax}{x} = a
= - 2 \left( \frac... | |
30/10/25 H-W
① f(x) = \left\{ \begin{array}{l} \frac{\cos ax - \cos bx}{x^2} \text{ if } x \neq 0 \\ \frac{1}{2} (b^2 - a^2) \text{ if } x = 0 \end{array} \right. \ge \otimes
Sol:- , \lim_{x \to 0} f(x) = \lim_{x \to 0} \frac{\cos ax - \cos bx}{x^2}
\boxed{\cos C - \cos D = -2 \sin \left( \frac{C+D}{2} \right) \cdot ... | |
10.) Let eqn of the st line be
ax + by + C = 0
①
1 x
Y
P(x_1, y_1)
R ax + by + C = 0 X
Q(h, k)
Let Q(h, k) be the image of P(x_1, y_1)
w.r. to ①
slope of ① = \frac{-a}{b}
slope of PQ = \frac{k - y_1}{h - x_1}
since PQ \perp ①
slope of PQ \times slope of ① = -1
\frac{k - y_1}{h - x_1} \times \frac{-a}{b} = -1
Let \f... | |
\frac{h-x_{1}}{a} = t \quad ; \quad \frac{k-y_{1}}{b} = t
h-x_{1} = at \quad ; \quad k-y_{1} = bt
h = at+x_{1} \quad ; \quad k = bt+y_{1}
\qquad \rightarrow (3) \qquad \qquad \qquad \rightarrow (4)
Let R be the midpoint of PQ
R = \left( \frac{h+x_{1}}{2} \quad , \quad \frac{k+y_{1}}{2} \right)
R lies on (1)
a\lef... | |
17/10/25//
$\star \star \star$ IMP
$I\phi \ y = x\sqrt{a^2+x^2} + a^2 \log(x + \sqrt{a^2+x^2}) \text{ then } P. T. ?$
$\frac{dy}{dx} = 2\sqrt{a^2+x^2}.$
$Sl:- y = x\sqrt{a^2+x^2} + a^2 \log(x + \sqrt{a^2+x^2}).$
$d.w.r.to \ x \ B. S.$
$\frac{d}{dx} y = \frac{d}{dx} \left[ x\sqrt{a^2+x^2} + a^2 \log (x + \sqrt{a^2+x^2})... | |
$22/10/24$ \qquad 9. Differentiation ($15m$)
L.A.Q.
① If $\sqrt{1-x^2} + \sqrt{1-y^2} = a(x-y)$ then $\frac{dy}{dx} = \sqrt{\frac{1-y^2}{1-x^2}}$
$S_o$: Let , $x = \sin \alpha$ , $y = \sin \beta$
$\sin^{-1} x = \alpha$ , $\sin^{-1} y = \beta$
$\sqrt{1-\sin^2 \alpha} + \sqrt{1-\sin^2 \beta} = a(\sin \alpha - \sin \beta)... | |
LAQ 23/10/25
1 If \sqrt{1-x^{2}} + \sqrt{1-y^{2}} \Rightarrow a(x-y) then \frac{dy}{dx} = \sqrt{\frac{1-y^{2}}{1-x^{2}}}
Sl:- Let, x = \sin \alpha , y = \sin \beta .
\sin^{-1} x = \alpha , \sin^{-1} y = \beta .
\sqrt{1-\sin^{2} \alpha} + \sqrt{1-\sin^{2} \beta} = a(\sin \alpha - \sin \beta)
\sqrt{\cos^{2} \alpha} + \... | |
9.4.4. Find the derivative of (\sin x)^{\log x} + x^{\sin x}
with respective 'x'.
Sol:- Let y = (\sin x)^{\log x}. \surd
Take log on both sides,
\log (y) = \log (\sin x)^{\log x} . (\because \log a^{b} = b. \log a).
\log y = \log x . \log (\sin x)
D.w.r. to 'x' on B.S.
\frac{1}{y} \cdot \frac{dy}{dx} = \frac{1}{x} \log... | |
$10/11/25$ $\underline{\text{Pair of straight lines.}}$ $7051$ $11/11/25$
$5) \ x^{2} + y^{2} = a^{2}$ ① $lx + my = 1$ ② If lines are
"COINCIDE"
Sol:- Homogenise Eqn ① by Eqn ②
$x^{2} + y^{2} = a^{2} (1)^{2}$
$x^{2} + y^{2} = a^{2} (lx + my)^{2}$
$x^{2} + y^{2} = a^{2} (l^{2}x^{2} + m^{2}y^{2} + 2lmxy)$
$x^{2} + y^{... | |
from eqn ③ and ④
\frac{l_1l_2}{-2} = \frac{m_1m_2}{-3} = \frac{n_1n_2}{5} = k \text{ (suppose)}
\frac{l_1l_2}{-2} = k \mid \frac{m_1m_2}{-3} = k \mid \frac{n_1n_2}{5} = k
l_1l_2 = -2k \quad m_1m_2 = -3k \quad n_1n_2 = 5k
angle b/w the lines \cos \theta = l_1l_2 \times m_1m_2 \times n_1n_2
\cos \theta = -2k - 3k + ... | |
$6.1.1 \Rightarrow l + m + n = 0$ ① $2mn + 3nl - 5lm = 0$ ②
Sol :- case ⓘ :-
from eq ①, $l = -m - n$
sub in eq ②
$2mn + 3n(-m - n) - 5m(-m - n) = 0$
$2mn - 3nm - 3n^{2} + 5m^{2} + 5mn = 0$
$5m^{2} + 4mn - 3n^{2} = 0$
Divide by $n^{2}$
$\frac{5m^{2}}{n^{2}} + \frac{4mn}{n^{2}} - \frac{3n^{2}}{n^{2}} = 0$
$5\left(\frac... | |
6.23 Find the direction cosines satisfy the ;
Eqn's $l+m+n=0$ and $mn-2nl-2lm=0$
$Sl:-$ Given eqn of lines $l+m+n=0$ --- ① and
$mn-2nl+2lm=0$ --- ②
from eq①, $l+m+n=0$
$l = -m-n$ sub in eqn ②
$mn-2n(-m-n)-2m(-m-n)=0$
$mn+2nm+2n^2+2m^2+2mn=0$
$2m^2+5mn+2n^2=0$ (Factorise).
$2m^2+4mn+1mn+2n^2=0$
$2m(m+2n)+n(m+2n)=0$
$(m+... | |
* Foot of the perpendicular Theorem + Problems
3.10.1 If $Q(h,k)$ is the foot of the
perpendicular from $P(x_1, y_1)$ on the line
$ax+by+c=0$ then prove that
$\frac{h-x_1}{a} = \frac{k-y_1}{b} = \frac{-(ax_1+by_1+c)}{a^2+b^2}$ also find
foot of $\perp^{er}$ from $(-1,3)$ on the line
$5x-y-18=0$.
Sl :- Proof :- Given ... | |
$$3y+3 = -2x+12$$
$$3y+3+2x-12=0$$
$$2x+3y-9=0 \quad ②$$
solving Eqn ① and ②
$$M \quad \quad R \quad \quad L \quad \quad M$$
$$\quad x \quad \quad \quad \quad y \quad \quad \quad \quad 1$$
$$-3 \quad \searrow \quad \nearrow 1 \quad \searrow \quad \nearrow 2 \quad \searrow \quad \nearrow -3$$
$$3 \quad \nearrow \quad... | |
$\frac{h+1}{5} = 1 \quad , \quad \frac{k-3}{-1} = 1$
$h+1 = 5 \quad , \quad k-3 = -1$
$h = 5-1 \quad k = -1+3$
$h=4 \quad k=2$
$14/11/25$
$\underline{\underline{\text{Image - Theorem + Problem}}}$
3.10.2 :- If $Q(h, k)$ is the image of the point
$P(x_1, y_1)$ w.r.to the st. line $ax+by+c=0$
then $\frac{h-x_1}{a} =... | |
$\frac{k-y_{1}}{b} \times \frac{a}{h-x_{1}} = 1$ $P(x_{1}, y_{1})$
$\frac{k-y_{1}}{b} = \frac{h-x_{1}}{a}$ $ax+by+c=0$
$R$
Let $\frac{k-y_{1}}{b} = \frac{h-x_{1}}{a} = m$... | |
\Rightarrow Circumcentre:-
3.11.2 \quad C.C. \quad A(1, 3) \quad B(-3, 5) \quad C(5, -1)
Let \quad S(x, y) \quad be the \quad C.C.
SA = SB = SC
SA = SB
\Rightarrow (SA)^2 = (SB)^2 \quad (S.O.B.S)
(x - 1)^2 + (y - 3)^2 = (x + 3)^2 + (y - 5)^2
\cancel{x^2} + 1 - 2x + \cancel{y^2} + \cancel{9} - 6y = \cancel{x^2} + ... |
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