TS EAMCET · Maths · Differentiation

If $y=\left(\sin ^{-1} 2 x\right)^2+\left(\cos ^{-1} 2 x\right)^2$, then $\left(1-4 x^2\right) y_2-4 x y_1=$

A 0
B 4
C 16
D 12

Verified Solution

Answer & Solution

Correct Answer

(C) 16

Step-by-step Solution

Detailed explanation

We have, \[ y=\left(\sin ^{-1} 2 x\right)^2+\left(\cos ^{-1} 2 x\right)^2 \] On differentiating both sides w.r.t. $x$, we get \[ \frac{d y}{d x}=2 \sin ^{-1} 2 x \cdot \frac{1}{\sqrt{1-(2 x)^2}} \cdot 2+2 \cos ^{-1} 2 x \] \[ \left(\frac{-1}{\sqrt{1-(2 x)^2}}\right) \cdot 2 \]…

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

x_{1}, x_{2}, \dots, x_{n}

are

n

observations with mean

\bar{x}

and standard deviation

σ

. Match the items of List-I with those of List-II

	List - I		List - II
(a)	$\sum_{i = 1}^{n} (x_{i} - \bar{x})$	(i)	Median
(b)	Variance $(σ^{2})$	(ii)	Coefficient of variation
(c)	Mean deviation	(iiii)	Zero
(d)	Measure used to find the homogeneity of given two series	(iv)	Mean of the absolute deviations from any measure of central tendency
		(v)	Mean of the squares of the deviations from mean

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If \(y=\left(\sin ^{-1} 2 x\right)^2+\left(\cos ^{-1} 2 x\right)^2\), then \(\left(1-4 x^2\right) y_2-4 x y_1=\)

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