TS EAMCET · Maths · Application of Derivatives

$y=2 x^3-8 x^2+10 x-4$ is a function defined on $[1,2]$. If the tangent drawn at a point $(a, b)$ on the graph of this function is parallel to $X$-axis and $a \in(1,2)$, then $a=$

A $0$
B $5$
C $1$
D $\frac{5}{3}$

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Answer & Solution

Correct Answer

(D) $\frac{5}{3}$

Step-by-step Solution

Detailed explanation

$\frac{d y}{d x}=6 x^2-16 x+10$ Tangent is parallel to $X$ axis $\Rightarrow \frac{d y}{d x}=0$ $\begin{aligned} & 6 x^2+6 x+10=0 ; 3 x^2-8 x+5=0 \\ & (x-1)\left(x-\frac{5}{3}\right)=0 \quad \therefore a=\frac{5}{3} \in(1,2) . \end{aligned}$

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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From TS EAMCET

Match the items given in List-

I

with those of the items of List-

II

	List - $I$		List - $II$
$(a)$	If $y = \|x\| + \|x - 2\|$ then at $x = 2, \frac{d y}{d x} =$	$(i)$	$1$
$(b)$	If $f (x) = \|\cos 2 x\|$ , then $f^{'} (\frac{π}{4} +) =$	$(ii)$	$0$
$(c)$	If $f (x) = \sin π [x]$ where $[\cdot]$ denotes the greatest integer function, then $f^{'} (1 -) =$	$(iii)$	$- 2$
$(d)$	If $f (x) = \log \|x - 1\|, x \neq 1$ then $f^{'} (\frac{1}{2}) =$	$(iv)$	does not exits
		$(v)$	$\frac{1}{2}$

TS EAMCET 2018 Hard

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\(y=2 x^3-8 x^2+10 x-4\) is a function defined on \([1,2]\). If the tangent drawn at a point \((a, b)\) on the graph of this function is parallel to \(X\)-axis and \(a \in(1,2)\), then \(a=\)

Answer & Solution

Step-by-step Solution

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