Ncert Solutions Maths class 12th
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New answer posted
4 months agoContributor-Level 10
We have, f (x) = log sin x
So, f (x) =
When
f (x) = cot x> 0 (Ist quadrat )
So, f (x) is strictly increasing on
When x ∈
f (x) = cot x< 0. (IIndquadrant ).
So, f (x) is strictly decreasing on
New answer posted
4 months agoContributor-Level 10
We have, f (x) =
So, f (x) =
So, for every x∈ I, where I is disjoint from [-1,1]
f (x) =
and f (x) = = ( + ve) > 0 when x< -1.
∴f (x) is strictly increasing on I .
New answer posted
4 months agoContributor-Level 10
We have, f (x) = x2 + x + 1
So, f (x) = 2x + a
If f (x) is strictly increasing onx
So, 1
The minimum value of f (x) is 2 + a and that of men value is 4 + a.
∴ 2 +a> 0and 4 + a> 0
a> -2 and a> -4.
a> -2.
∴The best value of a is -2.
New answer posted
4 months agoContributor-Level 10
We have, f (x) = x 100 + sin x - 1.
So, f (x) = 100x99 + cosx.
(A) When x (0,1). We get.
x>0
x99> 0
100 x99> 0.
Now, 0 radian = 0 degree
and 1 radian =
So, cosx> 0 for x∈ (0,1) radian = (0,57)
∴f (x) > 0 for x∈ (0,1).
(B) When x ∈ we get,
So, x> 1
x99> 1
100x99> 100.
And cosx is negative between -1 and 0.
So, f (x) = 100x99 + cosx> 100 - 1 = 99 > 0.
∴f(x) is strictly increasing on
(c) When x ∈ we get,
x> 0
x99> 0
100x99> 0
and cosx> 0. (firstquadrant).
I e, f(x) > 0.
∴f(x) is strictly increasing on
Hence, option (D) is correct.
New answer posted
4 months agoContributor-Level 10
(A) We have,
f(x) = cosx
So, f(x) = -sin x
When x we know that sin x> 0.
-sinx< 0. f(x) < 0
∴f(x) is strictly decreasing on
(B) We have, f(x) = cos 2x
So, f(x) = -2sin 2x
When we know that sin x> 0.
i e, 0
=> 0<2x<
So, sin 2x> 0 (sinØ is ( +ve) in 1st and 2ndquadrant).
-2sin 2x< 0.
f (x) < 0.
∴f (x) is strictly decreasing on
(c) We have, f(x) = cos 3x
f(x) = -3 sin 3x.
As
We can divide the interval into two
Case I At, 0 < 3x
sin 3x> 0.
-3 sin 3x< 0
f(x) < 0.
∴f(x) is strictly decreasing on
case II. At
sin 3x< 0.
-3 sin 3x> 0.
f(x) > 0.
∴f(x) is strictly increasing on
Hence, f(x) is with a increasing or
New answer posted
4 months agoContributor-Level 10
We have, f (x) = x2-x + 1.
So, f (x) =
Atf (x) = 0.
2x - 1 = 0
I e, x = divides the real line into two disjoint interval the interval ( -1, 1) into
Two disjoint interval
And f (x) is strictly increasing in and strictly decreasing in
Hence, f (x) is with a increasing or decreasing on ( -1,1).
New answer posted
4 months agoContributor-Level 10
We have, f (x) = log x.
So, f (x) =
i e, f (x) > 0. When
Hence, the logarithmic fx is strictly increasing on
New answer posted
4 months agoContributor-Level 10
We have, y =
Differentiating w rt.Ø we get,
When we know that,
So,
And also, (2 + cos)2> 0.
Hence, y is an increasing fxn of in
New answer posted
4 months agoContributor-Level 10
We have, y = [x (x- 2)]2.
Differentiating the above w rt. x we get,
= 2 [x (x- 2)] (x + x- 2)
= 2x (x - 2) (2x - 2)
= 4x (x - 2) (x - 1).
Now,
4x (x - 2) (x - 1) = 0.
i e, x = 0, x = 2, x = 1 divides the real line into
four disjoint interval. [0,1],[1,2] and
when x
∴f (x) is decreasing in
When
∴f (x) is increasing in [0,1].
When x
∴f (x) is decreasing.
When
∴f (x) is in creasing
Hence, f (x) is increasing for x
New answer posted
4 months agoContributor-Level 10
We have, y = log
Differentiating the above wrt.x.we get,
The given domain of the given function isx> -1.
(x + 1) > 0.
Also,
(2 + x)2> 0.
Hence,
∴ y is an increasing function of x throughout its domain.
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