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y_2 = (\sin x)^{\cos x}
\text{Apply } \log \text{ on B.S}
\log y_2 = \log (\sin x)^{\cos x}
\log y_2 = \cos x \cdot \log (\sin x)
\text{Diff. w.r.t } x
\frac{d}{dx}(\log y_2) = \frac{d}{dx}(\cos x \cdot \log(\sin x))
\frac{1}{y_2} \cdot \frac{dy_2}{dx} = \cos x \cdot \frac{d}{dx}(\log(\sin x)) + \log(\sin x) \cdo... | |
\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 | |
\textbf{(3)} \text{ S.T the lines joining origin to the points of intersection of the curve } 5x^2 - 2xy + y^2 + 3x + 3y - 2 = 0 \text{ and the st. line } x - y - \sqrt{2} = 0 \text{ are mutually perpendicular.}
x - y - \sqrt{2} = 0
x - y = \sqrt{2}
\frac{x - y}{\sqrt{2}} = t \rightarrow \textcircled{2}
\text{Homog... | |
\textbf{(4)} \text{ Sol:-}
\text{Let } x \text{ be a side of the square and } r \text{ be the radius of circle}
\text{Given length of the wire } = l
4x + 2\pi r = l
2\pi r = l - 4x
r = \frac{l - 4x}{2\pi} \rightarrow \textcircled{1}
\text{Sum of Areas } (A) = x^2 + \pi r^2
A = x^2 + \pi \left( \frac{l - 4x}{2\pi... | |
\boxed{7x - 7y + 4 = 0}
\textbf{Q.1.2.5}
\textcircled{1} \text{ Points } (5, -4), (7, 6) \text{ ratio } 2:3
\text{Let the point } P(x, y)
\text{Condition: } PA : PB = 2 : 3
\frac{PA}{PB} = \frac{2}{3}
3PA = 2PB
\text{S.O.B.S (Squaring on both sides)}
9PA^2 = 4PB^2
PA = \sqrt{(x-5)^2 + (y+4)^2} \quad PA^2 = (x-... | |
\frac{dy_1}{dx} = ya^{y-1} + a^y \log a \cdot \frac{dy}{dx} \rightarrow \textcircled{3} \text{ eq}^n
\text{Also, } y_2 = y^a
\text{Apply } \log \text{ on b.s}
\log y_1 = \log y^a
\log y_2 = a \log y
\text{Diff. w.r.t } x
\frac{d}{dx}(\log y_2) = \frac{d}{dx}(a \log y)
\frac{1}{y_2} \cdot \frac{dy_2}{dx} = a \cdo... | |
\text{N/4}
\cos \beta = \frac{l + m + n}{\sqrt{3}}, \quad \cos \gamma = \frac{l - m + n}{\sqrt{3}}, \quad \cos \beta = \frac{l + m - n}{\sqrt{3}}
\text{Now:}
\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma + \cos^2 \delta
= \left( \frac{l+m+n}{\sqrt{3}} \right)^2 + \left( \frac{-l+m+n}{\sqrt{3}} \right)^2 + \left( \fr... | |
2x^2 + y^2 - 6 = 0
\textbf{2)} \text{ When the origin is shifted to } (-1, 2) \text{ by the translation of axes, find the transformed equation of } x^2 + y^2 + 2x - 4y + 1 = 0
\text{Sol: Given, } O'(h, k) = (-1, 2)
\text{\& the original equation is}
x^2 + y^2 + 2x - 4y + 1 = 0 \rightarrow \textcircled{1}
\text{Let... | |
\text{Eq}^n \text{ of } AB: \quad y - y_1 = m(x - x_1)
y - 3 = 2(x + 1)
y - 3 = 2x + 2
2x - y + 2 + 3 = 0
2x - y + 5 = 0 \rightarrow \textcircled{5}
\text{Solving } \textcircled{4}, \textcircled{5} \text{ eq}^n \quad \text{(MRLM - Matrix Row Linear Method)}
\begin{array}{ccc|c}
-3 & -5 & & -3 \\
\times & \times &... | |
\text{The required plane is parallel to } \textcircled{1} \text{ is of the form } x + 2y + 3z + k = 0 \rightarrow \textcircled{2}
\textcircled{2} \text{ P.T.P } (1, 1, 1)
1 + 2(1) + 3(1) + k = 0
k + 6 = 0
k = -6
\text{Sub } k = -6 \text{ in } \textcircled{2}
\boxed{x + 2y + 3z - 6 = 0}
\textbf{11)} \text{ Find t... | |
\ast \text{ If } x^{\log y} = \log x, \text{ then S.T } \frac{dy}{dx} = \frac{y}{x} \left[ \frac{1 - \log x \cdot \log y}{\log^2 x} \right]
\text{Sol)} \quad \frac{dy}{dx} = \frac{y}{x} \left[ \frac{1 - \log x \cdot \log y}{\log x} \right]
x^{\log y} = \log x
\text{Apply } \log \text{ on b.s}
\log x^{\log y} = \log... | |
\ast \text{ If } x^{2/3} + y^{2/3} = a^{2/3}, \text{ then } \frac{dy}{dx} = -\sqrt[3]{\frac{y}{x}}
\text{Sol)} \quad x^{2/3} + y^{2/3} = a^{2/3}
\text{d.w.r.t } x \text{ on b.s}
\frac{d}{dx}(x^{2/3}) + \frac{d}{dx}(y^{2/3}) = \frac{d}{dx}(a^{2/3}) \quad \left( \because \frac{d}{dx}(x^n) = n \cdot x^{n-1} \right)
\f... | |
\textbf{Q.4.2} \quad y = \tan^{-1} \left( \frac{\sqrt{1+x^2} + \sqrt{1-x^2}}{\sqrt{1+x^2} - \sqrt{1-x^2}} \right)
\text{Sol)} \text{ Let } x^2 = \cos 2\theta \Rightarrow 2\theta = \cos^{-1}(x^2)
\Rightarrow \theta = \frac{1}{2} \cos^{-1}(x^2)
y = \tan^{-1} \left( \frac{\sqrt{1+\cos 2\theta} + \sqrt{1-\cos 2\theta}}{... | |
\textbf{4)} \text{ Show that direction cosines of two lines which are connected by the relations } l + m + n = 0 \text{ and } 2mn + 3nl - 5lm = 0 \text{ are perpendicular to each other.}
\text{Sol)} \quad l + m + n = 0 \rightarrow \textcircled{1}, \quad 2mn + 3nl - 5lm = 0 \rightarrow \textcircled{2}
\text{From } \te... | |
= \frac{1}{2} \left| \frac{n^2 \sqrt{(l_1 m_2 + l_2 m_1)^2 - 4l_1 l_2 m_1 m_2}}{l_1 l_2 m^2 - (l_1 m_2 + l_2 m_1)lm + m_1 m_2 l^2} \right|
= \frac{1}{2} \left| \frac{n^2 \sqrt{4h^2 - 4ab}}{am^2 - 2hlm + bl^2} \right|
= \frac{1}{2} \left| \frac{n^2 \sqrt{4(h^2 - ab)}}{am^2 - 2hlm + bl^2} \right|
= \frac{1}{2} \cdot \... | |
\textbf{(5)} \text{ A window is in the shape of a rectangle surmounted by a semicircle. If the perimeter of the window is } 20 \text{ ft. Find the maximum area.}
\text{Sol)} \text{ Let } 2x \text{ be the length and } y \text{ be the breadth of the rectangle, then } x \text{ is the radius of semi-circle}
\text{Given p... |
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