Question 20 - NCERT Exemplar - MCQs - Chapter 6 Class 12 Application of Derivatives (Term 1)

Last updated at Dec. 4, 2021 by Teachoo

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The function f (x) = 4 sin
^{
3
}
x β 6 sin
^{
2
}
x + 12 sin x + 100 is strictly

(A)Increasing in (Ο, 3Ο/2)Β Β Β Β Β Β Β Β Β Β Β Β Β (B) decreasing in (Ο/2,Ο)

(C) decreasing in ((-Ο)/2 ", "Β Ο/2)Β `Β Β Β Β Β (D) decreasingΒ in ("0, "Β Ο/2)

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Question 20
The function f (x) = 4 sin3 x β 6 sin2 x + 12 sin x + 100 is strictly
Increasing in (π, 3π/2) (B) decreasing in (π/2,π)
(C) decreasing in ((βπ)/2 ", " π/2) ` (D) decreasing in ("0, " π/2)
f (π₯) = 4 sin^3β‘γπ₯β6 sin^2β‘γπ₯+12 sinβ‘γπ₯+100γ γ γ
Differentiating w.r.t π
fβ (π) = 4 Γβ‘γ3 γβ‘γγπ ππγ^2 π₯γ Γ (sinβ‘γπ₯)β²γβ6 Γ 2 sinβ‘π₯Γ (sinβ‘π₯ )β²+12 cosβ‘π₯+0
= 12 sin^2 π₯ cos π₯ β 12 sin π₯ cos π₯ + 12 cos π₯
= 12 cos π₯ (γsin^2 π₯γβ‘γβsinβ‘γπ₯+1γ γ )
= 12 cos π (γγπππγ^π πγβ‘γ+ (πβπ¬π’π§β‘π )γ )
Now, we need to check sign of fβ (π)
Checking sign of γπππγ^πβ‘γπ+(γπβπππγβ‘π ) γ
Since, γπππγ^π π β₯ 0
And
β1 β€ sin π β€ 1
1 β₯ β sin π₯ β₯ β1
1 + 1 β₯ 1 β sin π₯ β₯ 1 β 1
2 β₯ 1 β sin π₯ β₯ 0
0 β€ 1 β sin π₯ β€ 2
So, (1 β sin π) β₯π
Therefore, γπππγ^πβ‘γπ+(γπβπππγβ‘π ) γβ₯ 0
Either cos πβ₯ 0
π₯ π(0, π/2)βͺ(3π/2, 2π)
i.e., 1st and 4th quadrant, which can be written as
π β ((βπ )/π,π /π)
Or cos π β€ 0
π₯ π(π/2,π)βͺ(π,3π/2)
i.e., 2nd and 3rd quadrant, which can be written as
π β(π /π,ππ /π)
Thus,
f is increasing in the interval ((βπ )/π,π /π)
f is decreasing in the interval (π /π,ππ /π)
Our options are
(A) Increasing in (π, 3π/2) (B) decreasing in (π/2,π)
(C) decreasing in ((βπ)/2 ", " π/2) ` (D) decreasing in ("0, " π/2)
As (π/2,π) lies in the interval (π/2,3π/2)
So, f is decreasing in this interval (π /π,π )
So, the correct answer is (B)

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