JEE Advanced · Mathematics · 13. Parabola

Equation of common tangent of $y=x^2, y=-x^2+4 x-4$ is

A
$y=4(x-1)$
B
$y=0$
C
$y=-4(x-1)$
D
$y=-30 x-50$

Verified Solution

Answer & Solution

Correct Answer

(B)
$y=0$

Step-by-step Solution

Detailed explanation

The equation of tangent to $y=x^2$, be $y=m x-\frac{m^2}{4}$, putting in $y=-x^2+4 x-4$, we should only get one value of $x$ i.e. Discriminate must be zero.
\[
\begin{array}{rrr}
\therefore & m x-\frac{m^2}{4} & =-x^2+4 x-4 \\
\Rightarrow & x^2+x(m-4)+4-\frac{m^2}{4} & =0 \\
\Rightarrow & m(m-4) & =0 \\
& \therefore y=0 & \text { or } m=0,4
\end{array}
\]

\[
\therefore y=0 \text { and } y=4(x-1) \text { are tangents. }
\]

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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A straight line through the vertex $P$ of a $\triangle P Q R$ intersects the side $Q R$ at the point $S$ and the circumcircle of the $\triangle P Q R$ at the point $T$. If $S$ is not the centre of the circumcircle, then

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Let $z_k=\cos \left(\frac{2 k \pi}{10}\right)+i \sin \left(\frac{2 k \pi}{10}\right) ; k=1,2, \ldots, 9$

List - I	List - II
(A) For each $z_k$ there exists a $z_j$ such $z_k \cdot z_j=1$	(P) True
(B) There exists a $k \in\{1,2, \ldots, 9\}$ such that $z_1 \cdot z=z_k$ has no solution z in the set of complex numbers	(Q) False
(C) $\frac{\left\|1-z_1\right\|\left\|1-z_2\right\|\ldots\left\|1-z_9\right\|}{10}$ equals	(R) 1
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From JEE Advanced

Let

E_{1} =

{

x \in ℝ : x \neq 1

and

\frac{x}{x - 1} > 0

} and

E_{2} =

{

x \in E_{1} : \sin^{- 1} (\log_{e} (\frac{x}{x - 1}))

is a real number} (Here the inverse trigonometric function

\sin^{- 1} (x)

assumes values in

[- \frac{π}{2} . \frac{π}{2}]

.)
Let

f : E_{1} \to ℝ

be the function defined by

f (x) = \log_{e} (\frac{x}{x - 1})

And

g : E_{2} \to R

be the function defined by

g (x) = \sin^{- 1} (\log_{e} (\frac{x}{x - 1}))

LIST-I	LIST-II
A. The range of $f$ is	P. $(- \infty, \frac{1}{1 - e}] \cup [\frac{e}{e - 1}, \infty)$
B. The range of $g$ contains	Q. (0, 1)
C. The domain of $f$ contains	R. $[- \frac{1}{2}, \frac{1}{2}]$
D. The domain of $g$ is	S. $(- \infty, 0) \cup (0, \infty)$
	T. $(- \infty, \frac{e}{e - 1}]$
	U. $(- \infty, 0) \cup (\frac{1}{2}, \frac{e}{e - 1}]$

The correct option is:

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	List- $I$		List- II
$(I)$	$2 h + k$	$(P)$	$6$
$(I I)$	$\frac{L e n g t h o f Z W}{L e n g t h o f X Y}$	$(Q)$	$\sqrt{6}$
$(I I I)$	$\frac{A r e a o f t r i a n g l e M Z N}{A r e a o f t r i a n g l e Z M W}$	$(R)$	$\frac{5}{4}$
$(I V)$	$α$	$(S)$	$\frac{21}{5}$
		$(T)$	$2 \sqrt{6}$
		$(U)$	$\frac{10}{3}$

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Equation of common tangent of \(y=x^2, y=-x^2+4 x-4\) is

Answer & Solution

Step-by-step Solution

See the Complete Solution

List - I	List - II
(A) For each \(z_k\) there exists a \(z_j\) such \(z_k \cdot z_j=1\)	(P) True
(B) There exists a \(k \in\{1,2, \ldots, 9\}\) such that \(z_1 \cdot z=z_k\) has no solution z in the set of complex numbers	(Q) False
(C) \(\frac{\left\|1-z_1\right\|\left\|1-z_2\right\|\ldots\left\|1-z_9\right\|}{10}\) equals	(R) 1
(D) \(1-\sum_{k=1}^9 \cos \left(\frac{2 k \pi}{10}\right)\) equals	(S) 2