Question

Match the following:

	List-I		List-II
(P)	If the normal to the curve $y=f(x)$ at $x=0$ be given by the equation $3x-y+3=0$, then the value of $\lim_{x \to 0} x^2 \{f(x^2)-5f(4x^2) +4f(7x^2)\}^{-1}$	(1)	20
(Q)	Slope of tangent to the $y = \int_{x^2}^{x^3} \frac{dt}{\sqrt{1+t^2}}$ at $x=1$ is	(2)	$-\frac{1}{3}$
(R)	Slope of common tangent $y=x^2-x+1$ and $y=x^2-3x+1$ is	(3)	$\frac{1}{\sqrt{2}}$
(S)	Coordinate of the point on $y^2=8x$ which is nearest to the circle $x^2+(y+6)^2=1$ is $(a,b)$ then $a^2+b^2$ is	(4)	-2

• A. If the normal to the curve $y=f(x)$ at $x=0$ be given by the equation $3x-y+3=0$, then the value of $\lim_{x \to 0} x^2 \{f(x^2)-5f(4x^2) +4f(7x^2)\}^{-1}$
• B. Slope of tangent to the $y = \int_{x^2}^{x^3} \frac{dt}{\sqrt{1+t^2}}$ at $x=1$ is
• C. Slope of common tangent $y=x^2-x+1$ and $y=x^2-3x+1$ is
• D. Coordinate of the point on $y^2=8x$ which is nearest to the circle $x^2+(y+6)^2=1$ is $(a,b)$ then $a^2+b^2$ is

Answer 1

[P-2] [Q-3] [R-4] [S-1]

Answer 2

The problem requires us to match four calculus problems (P, Q, R, S) with their respective numerical answers (1, 2, 3, 4).

Part (P): We are given that the normal to the curve $y=f(x)$ at $x=0$ is $3x-y+3=0$. The equation of the normal can be rewritten as $y = 3x+3$. The slope of the normal at $x=0$ is $m_N = 3$. The slope of the tangent at $x=0$ is $f'(0)$. Since the tangent is perpendicular to the normal, their slopes satisfy $m_T \cdot m_N = -1$. So, $f'(0) \cdot 3 = -1 \Rightarrow f'(0) = -\frac{1}{3}$. Also, the point $(0, f(0))$ lies on the normal. Substituting $x=0$ into the normal equation $3(0)-y+3=0$, we get $y=3$. Therefore, $f(0)=3$.

We need to evaluate the limit: $L = \lim_{x \to 0} x^2 \{f(x^2)-5f(4x^2) +4f(7x^2)\}^{-1}$. This can be written as $L = \lim_{x \to 0} \frac{x^2}{f(x^2)-5f(4x^2) +4f(7x^2)}$. Let $u=x^2$. As $x \to 0$, $u \to 0$. The limit becomes: $L = \lim_{u \to 0} \frac{u}{f(u)-5f(4u) +4f(7u)}$. Substituting $u=0$, we get $\frac{0}{f(0)-5f(0)+4f(0)} = \frac{0}{3-5(3)+4(3)} = \frac{0}{3-15+12} = \frac{0}{0}$ form. We can apply L'Hopital's Rule. Differentiating the numerator and the denominator with respect to $u$: $L = \lim_{u \to 0} \frac{\frac{d}{du}(u)}{\frac{d}{du}(f(u)-5f(4u) +4f(7u))}$ $L = \lim_{u \to 0} \frac{1}{f'(u) - 5f'(4u) \cdot 4 + 4f'(7u) \cdot 7}$ $L = \lim_{u \to 0} \frac{1}{f'(u) - 20f'(4u) + 28f'(7u)}$ Now, substitute $u=0$: $L = \frac{1}{f'(0) - 20f'(0) + 28f'(0)}$ $L = \frac{1}{(1-20+28)f'(0)}$ $L = \frac{1}{9f'(0)}$ Substitute $f'(0) = -\frac{1}{3}$: $L = \frac{1}{9(-\frac{1}{3})} = \frac{1}{-3} = -\frac{1}{3}$. So, (P) matches with (2).

Part (Q): We need to find the slope of the tangent to $y = \int_{x^2}^{x^3} \frac{dt}{\sqrt{1+t^2}}$ at $x=1$. The slope of the tangent is $\frac{dy}{dx}$. We use Leibniz integral rule: $\frac{d}{dx} \int_{a(x)}^{b(x)} F(t) dt = F(b(x)) \cdot b'(x) - F(a(x)) \cdot a'(x)$. Here, $F(t) = \frac{1}{\sqrt{1+t^2}}$, $a(x) = x^2$, $b(x) = x^3$. $\frac{dy}{dx} = \frac{1}{\sqrt{1+(x^3)^2}} \cdot \frac{d}{dx}(x^3) - \frac{1}{\sqrt{1+(x^2)^2}} \cdot \frac{d}{dx}(x^2)$ $\frac{dy}{dx} = \frac{1}{\sqrt{1+x^6}} \cdot (3x^2) - \frac{1}{\sqrt{1+x^4}} \cdot (2x)$ Now, evaluate at $x=1$: $\left. \frac{dy}{dx} \right|_{x=1} = \frac{1}{\sqrt{1+1^6}} \cdot (3 \cdot 1^2) - \frac{1}{\sqrt{1+1^4}} \cdot (2 \cdot 1)$ $= \frac{1}{\sqrt{2}} \cdot 3 - \frac{1}{\sqrt{2}} \cdot 2$ $= \frac{3}{\sqrt{2}} - \frac{2}{\sqrt{2}}$ $= \frac{1}{\sqrt{2}}$. So, (Q) matches with (3).

Part (R): We need to find the slope of the common tangent to $y_1=x^2-x+1$ and $y_2=x^2-3x+1$. Let the slope of the tangent to $y_1$ at $(x_1, y_1)$ be $m_1 = \frac{dy_1}{dx} = 2x-1$. So $m_1 = 2x_1-1$. Let the slope of the tangent to $y_2$ at $(x_2, y_2)$ be $m_2 = \frac{dy_2}{dx} = 2x-3$. So $m_2 = 2x_2-3$. For a common tangent, the slopes must be equal: $m_1 = m_2 = m$. $2x_1-1 = 2x_2-3 \Rightarrow 2x_1 = 2x_2-2 \Rightarrow x_1 = x_2-1$.

The equation of the tangent to $y_1$ at $(x_1, y_1)$ is $Y - (x_1^2-x_1+1) = (2x_1-1)(X-x_1)$. $Y = (2x_1-1)X - x_1(2x_1-1) + x_1^2-x_1+1$ $Y = (2x_1-1)X - 2x_1^2+x_1 + x_1^2-x_1+1$ $Y = (2x_1-1)X - x_1^2+1$.

The equation of the tangent to $y_2$ at $(x_2, y_2)$ is $Y - (x_2^2-3x_2+1) = (2x_2-3)(X-x_2)$. $Y = (2x_2-3)X - x_2(2x_2-3) + x_2^2-3x_2+1$ $Y = (2x_2-3)X - 2x_2^2+3x_2 + x_2^2-3x_2+1$ $Y = (2x_2-3)X - x_2^2+1$.

For these two equations to represent the same common tangent, their constant terms must be equal: $-x_1^2+1 = -x_2^2+1 \Rightarrow x_1^2 = x_2^2$. Substitute $x_1 = x_2-1$: $(x_2-1)^2 = x_2^2$ $x_2^2 - 2x_2 + 1 = x_2^2$ $-2x_2 + 1 = 0 \Rightarrow x_2 = \frac{1}{2}$. Now find $x_1$: $x_1 = x_2-1 = \frac{1}{2}-1 = -\frac{1}{2}$. The slope of the common tangent is $m = 2x_1-1 = 2(-\frac{1}{2})-1 = -1-1 = -2$. (Alternatively, $m = 2x_2-3 = 2(\frac{1}{2})-3 = 1-3 = -2$). So, (R) matches with (4).

Part (S): We need to find the coordinate $(a,b)$ on the parabola $y^2=8x$ which is nearest to the circle $x^2+(y+6)^2=1$. The center of the circle is $C(0, -6)$ and its radius is $r=1$. The point on the parabola nearest to the circle will be such that the line connecting it to the center of the circle is normal to the parabola. Let the point on the parabola be $P(x_0, y_0)$. Since $y^2=8x$, we have $x_0 = \frac{y_0^2}{8}$. The equation of the parabola is $y^2=8x$. Differentiating with respect to $x$: $2y \frac{dy}{dx} = 8 \Rightarrow \frac{dy}{dx} = \frac{4}{y}$. The slope of the tangent at $P(x_0, y_0)$ is $m_T = \frac{4}{y_0}$. The slope of the normal at $P(x_0, y_0)$ is $m_N = -\frac{1}{m_T} = -\frac{y_0}{4}$. The normal line passes through the center of the circle $C(0, -6)$. The slope of the line segment $PC$ is $\frac{y_0 - (-6)}{x_0 - 0} = \frac{y_0+6}{x_0}$. Since $PC$ is normal to the parabola, its slope must be $m_N$: $\frac{y_0+6}{x_0} = -\frac{y_0}{4}$. Substitute $x_0 = \frac{y_0^2}{8}$: $\frac{y_0+6}{\frac{y_0^2}{8}} = -\frac{y_0}{4}$ $\frac{8(y_0+6)}{y_0^2} = -\frac{y_0}{4}$ $32(y_0+6) = -y_0^3$ $32y_0 + 192 = -y_0^3$ $y_0^3 + 32y_0 + 192 = 0$. By inspection, we can test integer factors of 192. If $y_0=-4$: $(-4)^3 + 32(-4) + 192 = -64 - 128 + 192 = -192 + 192 = 0$. So, $y_0=-4$ is a solution. Then $x_0 = \frac{y_0^2}{8} = \frac{(-4)^2}{8} = \frac{16}{8} = 2$. The point $(a,b)$ on the parabola is $(2, -4)$. We need to find $a^2+b^2$. $a^2+b^2 = 2^2 + (-4)^2 = 4 + 16 = 20$. So, (S) matches with (1).

Summary of Matches: (P) $\to$ (2) (Q) $\to$ (3) (R) $\to$ (4) (S) $\to$ (1)

Comparing with the given options, the correct option is [P-2] [Q-3] [R-4] [S-1].