Question 7 - NCERT Exemplar - MCQs - Chapter 6 Class 12 Application of Derivatives

Last updated at April 16, 2024 by Teachoo

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

(A)Increasing in (π, 3π/2) (B) decreasing in (π/2,π)

(C) decreasing in ((-π)/2 ", " π/2) ` (D) decreasing in ("0, " π/2)

Transcript

Question 7
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)

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.

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