AP EAMCET · Maths · Application of Derivatives

$V$ is the set of points on the curve $y^3-3 x y+2=0$ where the tangent is vertical then $\mathrm{V}=$

A $\Phi$
B $\{(1,0)\}$
C $\{(1,1)\}$
D $\{(0.0)(1.1)\}$

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Answer & Solution

Correct Answer

(C) $\{(1,1)\}$

Step-by-step Solution

Detailed explanation

We have a curve $y^3-3 x y+2=0$ now differently the given curve w.r.t. $x$ we get \[ \begin{aligned} & 3 y^2 y^1-3(y+x y)=0 \\ & y^2 y^1-y-x y^1=0 \\ & \Rightarrow y^1\left(y^2-x\right)=y \\ & \text { or } \quad y^1=\frac{y}{y^2-x} \end{aligned} \] $\because$ tangent is…

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.

Same subject

Match the items of List-

I

with the items of List-

II

and choose the correct option

List- $I$		List- $II$
$(A)$	If $A$ is a non-singular matrix of order $3$ and $\| A \| = a,$ then $\|{(adj A^{- 1})}^{- 1}\| =$	$(I)$	null matrix
$(B)$	A is a non-singular matrix of order $3$ and $B$ is any matrix of order $3$ such that $A B = O$ , then $B$ is	$(II)$	$a^{2}$
$(C)$	$\|\begin{matrix} 1 & x & x^{2} \\ \cos (a - b) y & \cos α y & \cos (a + b) y \\ \sin (a - b) y & \sin α y & \sin (a + b) y \end{matrix}\|$ does not depend on	$(III)$	$b$
$(D)$	$A$ is a square matrix of order $3$ and $B = A - A^{T}$ , then $\|B\|$ is	$(IV)$	$a$
		$(V)$	$0$

The correct answer is

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\(V\) is the set of points on the curve \(y^3-3 x y+2=0\) where the tangent is vertical then \(\mathrm{V}=\)

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