if x varies as y² and y=z when x=6. Find out the relation between x & y.
Answers
Step-by-step explanation:
x²-3abx+2b²=0
or, {(ax)²-2.ax.3b/2+(3b/2)²}+2b²-9b²/4=0
or, (ax-3b/2)²+(8b²-9b²)/4=0
or, (ax-3b/2)²-(b/2)²=0
or, (ax-3b/2+b/2)(ax-3b/2-b/2)=0
or, (ax-b)(ax-2b)=0
Either, ax-b=0
or, ax=b
Answer:
SOLUTION :–
• Let the function –
\begin{gathered} \\ \implies\bf y = { \tan}^{ - 1} (x) \\ \end{gathered}
⟹y=tan
−1
(x)
• Let the function –
\begin{gathered} \\ \implies\bf x= \tan( \theta) \\ \end{gathered}
⟹x=tan(θ)
• Differentiate with respect to 'θ' –
\begin{gathered} \\\implies\bf \dfrac{dx}{d \theta}= \sec^{2} ( \theta) \\ \end{gathered}
⟹
dθ
dx
=sec
2
(θ)
\begin{gathered} \\\implies\bf \dfrac{dx}{d \theta}= 1 + \tan^{2} ( \theta) \\ \end{gathered}
⟹
dθ
dx
=1+tan
2
(θ)
\begin{gathered} \\\implies\bf \dfrac{dx}{d \theta}= 1 +x^{2} \: \: \: \: \: - - - eq.(1)\\ \end{gathered}
⟹
dθ
dx
=1+x
2
−−−eq.(1)
• Now –
\begin{gathered} \\ \implies\bf y = { \tan}^{ - 1} ( \tan( \theta) ) \\ \end{gathered}
⟹y=tan
−1
(tan(θ))
\begin{gathered} \\ \implies\bf y = \theta\\ \end{gathered}
⟹y=θ
• Differentiate with respect to 'θ' –
\begin{gathered} \\ \implies\bf \dfrac{dy}{d \theta} = \dfrac{d\theta}{d\theta}\\ \end{gathered}
⟹
dθ
dy
=
dθ
dθ
\begin{gathered} \\ \implies\bf \dfrac{dy}{d \theta} =1\\ \end{gathered}
⟹
dθ
dy
=1
• We should write this as –
\begin{gathered} \\ \implies\bf \dfrac{dy}{dx} \times \dfrac{dx}{d \theta} =1\\ \end{gathered}
⟹
dx
dy
×
dθ
dx
=1
• Using eq.(1) –
\begin{gathered} \\ \implies\bf \dfrac{dy}{dx} (1 + {x}^{2})=1\\ \end{gathered}
⟹
dx
dy
(1+x
2
)=1
\begin{gathered} \\ \large \implies{ \boxed{\bf \dfrac{dy}{dx} = \dfrac{1}{(1+{x}^{2})}}}\\ \end{gathered}
⟹
dx
dy
=
(1+x
2
)
1
or, x=b/a
Or, ax-2b=0
Answer:
SOLUTION :–
• Let the function –
\begin{gathered} \\ \implies \bf y = log_{ex}{logx} \\ \end{gathered}
⟹y=log
ex
logx
• Using identity –
\begin{gathered} \\ \implies \bf log_{a}{b} = \dfrac{ log_{e}(b) }{ log_{e}(a) } \\ \end{gathered}
⟹log
a
b=
log
e
(a)
log
e
(b)
• So –
\begin{gathered} \\ \implies \bf y = \dfrac{log_{e}(logx)}{ log_{e}(ex)}\\ \end{gathered}
⟹y=
log
e
(ex)
log
e
(logx)
• Using identity –
\begin{gathered} \\ \implies \bf log_{e}(a.b) = log_{e}(a)+log_{e}(b) \\ \end{gathered}
⟹log
e
(a.b)=log
e
(a)+log
e
(b)
• So –
\begin{gathered} \\ \implies \bf y = \dfrac{log_{e}(logx)}{ log_{e}(e) +log_{e}(x)}\\ \end{gathered}
⟹y=
log
e
(e)+log
e
(x)
log
e
(logx)
\begin{gathered} \\ \implies \bf y = \dfrac{log_{e}(logx)}{ 1 +log_{e}(x)}\\ \end{gathered}
⟹y=
1+log
e
(x)
log
e
(logx)
• Now Differentiate with respect to 'x' –
\begin{gathered} \\ \implies \bf \dfrac{dy}{dx} = \dfrac{[1 +log_{e}(x)] \bigg \{\dfrac{1}{x log_{e}(x)} \bigg \}- [log_{e}(logx)] \bigg( \dfrac{1}{x} \bigg) }{ [1 +log_{e}(x)]^{2} }\\ \end{gathered}
⟹
dx
dy
=
[1+log
e
(x)]
2
[1+log
e
(x)]{
xlog
e
(x)
1
}−[log
e
(logx)](
x
1
)
\begin{gathered} \\ \implies \bf \dfrac{dy}{dx} = \dfrac{\bigg \{\dfrac{1}{x log_{e}(x)}+ \dfrac{1}{x} \bigg\}- \dfrac{log_{e}(logx)}{x}}{ [1 +log_{e}(x)]^{2} }\\ \end{gathered}
⟹
dx
dy
=
[1+log
e
(x)]
2
{
xlog
e
(x)
1
+
x
1
}−
x
log
e
(logx)
\begin{gathered} \\ \implies \bf \dfrac{dy}{dx} = \dfrac{\dfrac{1}{x} \bigg \{\dfrac{1}{ log_{e}(x)}+1 - log_{e}(logx)\bigg\}}{ [1 +log_{e}(x)]^{2} }\\ \end{gathered}
⟹
dx
dy
=
[1+log
e
(x)]
2
x
1
{
log
e
(x)
1
+1−log
e
(logx)}
or, ax=2b
or, x=2b/a
∴, the roots of the given equations are: b/a, 2b/a. Ans.
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