The function f x -sin 2 x -cos 2 x - x -0- -2- is strictly decreasing in the intervalA- -8- -2-B- -8- -4-C- -8- 4-3-D- 0- -4-

The correct option is A (π/8 , π/2]f(x) = sin 2x + cos 2x ⇒ f'(x) = 2cos 2x 2 sin 2x For decreasing function, f'(x) 0 ⇒cos 2x sin 2x ∴ 2x ∈(π/4, π] x ∈ n ...

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Standard XII

Mathematics

Using Monotonicity to Find the Range of a Function

The function ...

Question

The function $f (x) = sin 2 x + cos 2 x \forall x \in [0, \frac{π}{2}]$ is strictly decreasing in the interval

(0, \frac{π}{4}]

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(\frac{π}{8}, \frac{π}{2}]

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(\frac{π}{8}, \frac{π}{4}]

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(\frac{π}{8}, \frac{π}{3}]

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Solution

The correct option is B $(\frac{π}{8}, \frac{π}{2}]$
$f (x) = sin 2 x + cos 2 x \Rightarrow f^{'} (x) = 2 cos 2 x - 2 sin 2 x$
For decreasing function,
$f^{'} (x) < 0 \Rightarrow cos 2 x < sin 2 x$
$∴ 2 x \in (\frac{π}{4}, π] x \in (\frac{π}{8}, \frac{π}{2}]$

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Similar questions

Q. Consider the function

f (x) = {sin}^{5} x + {cos}^{5} x - 1,

x \in [0, \frac{π}{2}] .

Which of the following is (are) CORRECT?

Q. Show that the function given by

f (x) = sin x

is (a) strictly increasing in

(0, \frac{π}{2})

(b) strictly decreasing in

(\frac{π}{2}, π)

(0, π)

Q. The function

f (x) = ln sin x \forall x \in (0, \frac{π}{2}]

is decreasing in the interval

Q. Verify Rolle's theorem for each of the following functions on the indicated intervals
(i) f(x) = cos 2 (x − π/4) on [0, π/2]

(ii) f(x) = sin 2x on [0, π/2]

(iii) f(x) = cos 2x on [−π/4, π/4]

(iv) f(x) = e^x sin x on [0, π]

(v) f(x) = e^xcos x on [−π/2, π/2]

(vi) f(x) = cos 2x on [0, π]

(vii) f(x) =

\frac{\sin x}{e^{x}}

on 0 ≤ x ≤ π

(viii) f(x) = sin 3x on [0, π]

(ix) f(x) =

{e^{1 - x}}^{2}

on [−1, 1]

(x) f(x) = log (x² + 2) − log 3 on [−1, 1]

(xi) f(x) = sin x + cos x on [0, π/2]

(xii) f(x) = 2 sin x + sin 2x on [0, π]

(xiii)

f (x) = \frac{x}{2} - \sin \frac{π x}{6} on [- 1, 0]

(xiv)

f (x) = \frac{6 x}{π} - 4 \sin^{2} x on [0, π / 6]

(xv) f(x) = 4^sin^x on [0, π]

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(xvii) f(x) = sin⁴ x + cos⁴ x on

[0, \frac{π}{2}]

(xviii) f(x) = sin x − sin 2x on [0, π]

Q. Show that the function given by

f (x) = sin x

is
(a) increasing in

(0, \frac{π}{2})

(b) strictly decreasing

(\frac{π}{2}, π)

(c)neither increasing nor decreasing in

(0, π)

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The function f(x)=sin2x+cos2x ∀x∈[0,π2] is strictly decreasing in the interval

The correct option is B (π8,π2]f(x)=sin2x+cos2x⇒f′(x)=2cos2x−2sin2x For decreasing function, f′(x)<0⇒cos2x<sin2x ∴2x∈(π4,π]x∈(π8,π2]

The function $f (x) = sin 2 x + cos 2 x \forall x \in [0, \frac{π}{2}]$ is strictly decreasing in the interval

The correct option is B $(\frac{π}{8}, \frac{π}{2}]$
$f (x) = sin 2 x + cos 2 x \Rightarrow f^{'} (x) = 2 cos 2 x - 2 sin 2 x$
For decreasing function,
$f^{'} (x) < 0 \Rightarrow cos 2 x < sin 2 x$
$∴ 2 x \in (\frac{π}{4}, π] x \in (\frac{π}{8}, \frac{π}{2}]$