Let the tangent at any point P(x, y) on a cur | vedclass

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(A) Let the tangent at $P(x, y)$ intersect the $x$-axis at $A(\alpha, 0)$ and the $y$-axis at $B(0, \beta)$.The equation of the tangent is $Y - y = \f

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(A) Let the tangent at $P(x, y)$ intersect the $x$-axis at $A(\alpha, 0)$ and the $y$-axis at $B(0, \beta)$.The equation of the tangent is $Y - y = \frac{dy}{dx}(X - x)$.For $A$,$Y = 0 \implies -y = \frac{dy}{dx}(\alpha - x) \implies \alpha = x - y \frac{dx}{dy}$.For $B$,$X = 0 \implies Y - y = \frac{dy}{dx}(-x) \implies Y = y - x \frac{dy}{dx}$.Given $PA: PB = 1: k$,by section formula,$x = \frac{k \cdot \alpha + 1 \cdot 0}{k + 1} = \frac{k \alpha}{k + 1} \implies \alpha = \frac{k + 1}{k} x$.Substituting $\alpha$: $\frac{k + 1}{k} x = x - y \frac{dx}{dy} \implies \frac{x}{k} = -y \frac{dx}{dy} \implies \frac{dy}{dx} = -\frac{ky}{x}$.Integrating: $\int \frac{dy}{y} = -k \int \frac{dx}{x} \implies \ln y = -k \ln x + C \implies y x^k = C$.Passing through $(1, 1) \implies C = 1$. Passing through $(\frac{1}{10}, 100) \implies 100 \cdot (\frac{1}{10})^k = 1 \implies 10^2 \cdot 10^{-k} = 10^0 \implies 2 - k = 0 \implies k = 2$.The differential equation is $e^{\frac{dy}{dx}} = 2x + 1 \implies \frac{dy}{dx} = \ln(2x + 1)$.Integrating: $y = \int \ln(2x + 1) dx = \frac{1}{2} (2x + 1) \ln(2x + 1) - x + C$.Using $y(0) = 2$: $2 = \frac{1}{2}(1)(0) - 0 + C \implies C = 2$.So,$y(x) = \frac{2x + 1}{2} \ln(2x + 1) - x + 2$.$y(1) = \frac{3}{2} \ln 3 - 1 + 2 = \frac{3}{2} \ln 3 + 1$.$4y(1) - 5 \ln 3 = 4(\frac{3}{2} \ln 3 + 1) - 5 \ln 3 = 6 \ln 3 + 4 - 5 \ln 3 = \ln 3 + 4$.

Column $I$	Column $II$
$(A)$ The number of solutions of the equation $x e^{\sin x}-\cos x=0$ in the interval $(0, \frac{\pi}{2})$	$(p)$ $1$
$(B)$ Value$(s)$ of $k$ for which the planes $k x+4 y+z=0, 4 x+k y+2 z=0$ and $2 x+2 y+z=0$ intersect in a straight line	$(q)$ $2$
$(C)$ Value$(s)$ of $k$ for which $\|x-1\|+\|x-2\|+\|x+1\|+\|x+2\|=4 k$ has integer solution$(s)$	$(r)$ $3$
$(D)$ If $y^{\prime}=y+1$ and $y(0)=1$,then value$(s)$ of $y(\ln 2)$	$(s)$ $4$
	$(t)$ $5$

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