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Differentiation

Section: Mathematics

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Question : 12 of 34

Marks: +1, -0

If

y = (ax + b) \cos x

y_2 + y_1 \sin 2x + y (1 + \sin^2 x) =

[AP EAPCET 21 May 2025 S2]

$y_2 \cos^2 x$
$y_2 \sin^2 x$
$y_1 \sin^2 x$
$y \sin^2 x$

Solution:

We have,

y=(a x+b) \cos x

\;y \sec x=a x+b

\;y \sec x \tan x+\sec x y_1=a

Again, differentiating w.r.t.

x

, we get

\;\Rightarrow y \sec x \sec^2 x+\sec x \tan x \cdot y_1+y \tan x \sec x \tan x+\sec x y_2+y_1 \sec x \tan x=0

\;\Rightarrow y \sec^3 x+y_1 \sec x \tan x+y \sec x \tan^2 x+y_1 \sec x \tan x+\sec x \cdot y_2=0

\;\Rightarrow y_2 \sec x+2 y_1 \sec x \tan x+y (\sec^2 x+\tan^2 x) \sec x=0

On dividing both sides by

\sec^3 x

, we get

y_2 \cos^2 x+2 y_1 \sin x \cdot \cos x+y (1+\sin^2 x)=0

\;y_2 - y_1 \sin^2 x + y_1 \sin 2x + y (1 + \sin^2 x) = 0

\therefore y_2 + y_1 \sin 2x + y(1+\sin^2 x) = y_2 \sin^2 x

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