NCERT Solutions for Class 12 Maths Chapter 6 – Application of Derivatives Ex 6.2
Page No 205:
Question 1:
Show that the function given by f(x) = 3x + 17 is strictly increasing on R.
Answer:
Let
be any two numbers in R.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7148/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_6c7add03.gif)
Then, we have:
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7148/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_m3e960391.gif)
Hence, f is strictly increasing on R.
Question 2:
Show that the function given by f(x) = e2x is strictly increasing on R.
Answer:
Let
be any two numbers in R.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7150/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_6c7add03.gif)
Then, we have:
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7150/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_m43f7255a.gif)
Hence, f is strictly increasing on R.
Question 3:
Show that the function given by f(x) = sin x is
(a) strictly increasing in
(b) strictly decreasing in ![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7153/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_8a338de.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7153/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_m23698395.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7153/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_8a338de.gif)
(c) neither increasing nor decreasing in (0, π)
Answer:
The given function is f(x) = sin x.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7153/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_m6d8f8ff7.gif)
(a) Since for each
we have
.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7153/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_15116a35.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7153/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_41687cb0.gif)
Hence, f is strictly increasing in
.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7153/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_m23698395.gif)
(b) Since for each
, we have
.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7153/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_m24b92dc1.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7153/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_3f4a8bc3.gif)
Hence, f is strictly decreasing in
.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7153/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_8a338de.gif)
(c) From the results obtained in (a) and (b), it is clear that f is neither increasing nor decreasing in (0, π).
Question 4:
Find the intervals in which the function f given by f(x) = 2x2 − 3x is
(a) strictly increasing (b) strictly decreasing
Answer:
The given function is f(x) = 2x2 − 3x.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7157/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_m7ecc3881.gif)
Now, the point
divides the real line into two disjoint intervals i.e.,
and![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7157/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_2867fc9d.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7157/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_1f30e4cd.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7157/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_m5b0ff7a5.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7157/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_2867fc9d.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7157/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_14145f1.jpg)
In interval![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7157/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_m3c0eeb7d.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7157/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_m3c0eeb7d.gif)
Hence, the given function (f) is strictly decreasing in interval
.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7157/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_m5b0ff7a5.gif)
In interval![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7157/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_42ec1354.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7157/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_42ec1354.gif)
Hence, the given function (f) is strictly increasing in interval
.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7157/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_m78e10100.gif)
Question 5:
Find the intervals in which the function f given by f(x) = 2x3 − 3x2 − 36x + 7 is
(a) strictly increasing (b) strictly decreasing
Answer:
The given function is f(x) = 2x3 − 3x2 − 36x + 7.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7162/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_20862ae8.gif)
∴![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7162/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_m604d8e20.gif)
x = − 2, 3
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7162/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_m604d8e20.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7162/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_74849e29.gif)
The points x = −2 and x = 3 divide the real line into three disjoint intervals i.e.,
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7162/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_150de9f1.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7162/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_m4c204261.jpg)
In intervals
is positive while in interval
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7162/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_2324160c.gif)
(−2, 3),
is negative.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7162/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_2240d1fb.gif)
Hence, the given function (f) is strictly increasing in intervals
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7162/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_371ba27.gif)
(−2, 3).
Question 6:
Find the intervals in which the following functions are strictly increasing or decreasing:
(a) x2 + 2x − 5 (b) 10 − 6x − 2x2
(c) −2x3 − 9x2 − 12x + 1 (d) 6 − 9x − x2
(e) (x + 1)3 (x − 3)3
Answer:
(a) We have,
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7168/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_m3aa12125.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7168/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_34a23d00.gif)
Now,
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7168/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_m604d8e20.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7168/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_74849e29.gif)
Point x = −1 divides the real line into two disjoint intervals i.e., ![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7168/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_m75d77264.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7168/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_m75d77264.gif)
In interval![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7168/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_7a84014.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7168/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_7a84014.gif)
∴f is strictly decreasing in interval![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7168/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_m45bbb4ac.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7168/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_m45bbb4ac.gif)
Thus, f is strictly decreasing for x < −1.
In interval![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7168/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_m497d6178.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7168/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_m497d6178.gif)
∴ f is strictly increasing in interval![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7168/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_m736c3a8c.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7168/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_m736c3a8c.gif)
Thus, f is strictly increasing for x > −1.
(b) We have,
f(x) = 10 − 6x − 2x2
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7168/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_401cbdf9.gif)
The point
divides the real line into two disjoint intervals i.e.,![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7168/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_m4b119f7d.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7168/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_m2b1d04f3.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7168/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_m4b119f7d.gif)
In interval
i.e., when
,
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7168/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_3514a6fa.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7168/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_m23b1420d.gif)
f'(x)=-6-4x>0.
∴ f is strictly increasing for
.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7168/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_m23b1420d.gif)
In interval
i.e., when
,![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7168/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_428c90d6.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7168/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_m687695c6.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7168/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_6a632142.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7168/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_428c90d6.gif)
∴ f is strictly decreasing for
.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7168/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_6a632142.gif)
(c) We have,
f(x) = −2x3 − 9x2 − 12x + 1
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7168/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_m41662c6a.gif)
Points x = −1 and x = −2 divide the real line into three disjoint intervals i.e.,![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7168/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_m1c9aa1d9.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7168/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_m1c9aa1d9.gif)
In intervals
i.e., when x < −2 and x > −1,
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7168/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_m4c080802.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7168/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_m6c3b1af4.gif)
∴ f is strictly decreasing for x < −2 and x > −1.
Now, in interval (−2, −1) i.e., when −2 < x < −1,
.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7168/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_35667e23.gif)
∴ f is strictly increasing for
.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7168/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_m1cc5da32.gif)
(d) We have,
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7168/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_49835a4d.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7168/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_209c89be.gif)
The point
divides the real line into two disjoint intervals i.e.,
.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7168/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_2d02a6ce.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7168/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_400e3ae0.gif)
In interval
i.e., for
,
.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7168/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_m2c8c6cf0.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7168/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_72bfe241.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7168/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_m72d3b443.gif)
∴ f is strictly increasing for
.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7168/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_72bfe241.gif)
In interval
i.e., for
,![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7168/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_67e14f85.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7168/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_63556c31.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7168/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_m6c7c837f.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7168/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_67e14f85.gif)
∴ f is strictly decreasing for
.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7168/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_m6c7c837f.gif)
(e) We have,
f(x) = (x + 1)3 (x − 3)3
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7168/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_m793ceb7c.gif)
The points x = −1, x = 1, and x = 3 divide the real line into four disjoint intervals i.e.,
, (−1, 1), (1, 3), and
.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7168/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_398d3249.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7168/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_m3ad6980e.gif)
In intervals
and (−1, 1),
.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7168/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_398d3249.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7168/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_m1435f0ad.gif)
∴ f is strictly decreasing in intervals
and (−1, 1).
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7168/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_398d3249.gif)
In intervals (1, 3) and
,
.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7168/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_m3ad6980e.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7168/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_76998282.gif)
∴ f is strictly increasing in intervals (1, 3) and
.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7168/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_m3ad6980e.gif)
Question 7:
Show that
, is an increasing function of x throughout its domain.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7171/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_4c10f807.gif)
Answer:
We have,
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7171/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_m1de80fa8.gif)
∴
dydx=11+x-(2+x)(2)-2x(1)(2+x)2=11+x-4(2+x)2=x2(1+x)(2+x)2Now,
dydx=0
⇒x2(1+x)(2+x)2=0⇒x2=0 [(2+x)≠0 as x>-1]⇒x=0Since x > −1, point x = 0 divides the domain (−1, ∞) in two disjoint intervals i.e., −1 < x < 0 and x > 0.
When −1 < x < 0, we have:
x<0 x2="">0x>-1⇒(2+x)>0⇒(2+x2)>0∴0>
y’=x2(1+x)(2+x)2>0Also, when x > 0:
x>0⇒x2>0, (2+x)2>0∴
y’=x2(1+x)(2+x)2>0Hence, function f is increasing throughout this domain.
Question 8:
Find the values of x for which
is an increasing function.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7174/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_2ec2548e.gif)
Answer:
We have,
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7174/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_67d6550c.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7174/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_e9322c6.gif)
The points x = 0, x = 1, and x = 2 divide the real line into four disjoint intervals i.e., ![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7174/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_m7f8d995d.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7174/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_m7f8d995d.gif)
In intervals
,
.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7174/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_d9af8c5.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7174/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_600a283a.gif)
∴ y is strictly decreasing in intervals
.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7174/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_d9af8c5.gif)
However, in intervals (0, 1) and (2, ∞), ![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7174/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_339d45bf.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7174/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_339d45bf.gif)
∴ y is strictly increasing in intervals (0, 1) and (2, ∞).
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7174/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_4dd19828.gif)
Question 9:
Prove that
is an increasing function of θ in
.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7176/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_m293ef05b.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7176/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_m7b0578ae.gif)
Answer:
We have,
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7176/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_m293ef05b.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7176/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_1420e4a6.gif)
Since cos θ ≠ 4, cos θ = 0.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7176/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_m7a465faa.gif)
Now,
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7176/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_230d5bf8.gif)
In interval
, we have cos θ > 0. Also, 4 > cos θ ⇒ 4 − cos θ > 0. ![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7176/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_m194b904f.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7176/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_600e5254.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7176/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_m194b904f.gif)
Therefore, y is strictly increasing in interval
.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7176/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_600e5254.gif)
Also, the given function is continuous at ![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7176/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_m7a1caa7f.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7176/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_m7a1caa7f.gif)
Hence, y is increasing in interval
.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7176/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_6a55f665.gif)
Page No 206:
Question 10:
Prove that the logarithmic function is strictly increasing on (0, ∞).
Answer:
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7180/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_m4be78744.gif)
It is clear that for x > 0, ![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7180/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_m332da90c.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7180/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_m332da90c.gif)
Hence, f(x) = log x is strictly increasing in interval (0, ∞).
Question 11:
Prove that the function f given by f(x) = x2 − x + 1 is neither strictly increasing nor strictly decreasing on (−1, 1).
Answer:
The given function is f(x) = x2 − x + 1.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7183/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_259de1bb.gif)
The point
divides the interval (−1, 1) into two disjoint intervals i.e.,![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7183/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_4d8ac8dc.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7183/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_eeecab0.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7183/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_4d8ac8dc.gif)
Now, in interval![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7183/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_1b7f2973.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7183/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_1b7f2973.gif)
Therefore, f is strictly decreasing in interval
.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7183/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_m24764fb1.gif)
However, in interval![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7183/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_4a9d0174.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7183/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_4a9d0174.gif)
Therefore, f is strictly increasing in interval
.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7183/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_702d8f76.gif)
Hence, f is neither strictly increasing nor decreasing in interval (−1, 1).
Question 12:
Which of the following functions are strictly decreasing on
?
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7184/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_m23698395.gif)
(A) cos x (B) cos 2x (C) cos 3x (D) tan x
Answer:
(A) Let![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7184/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_m62dad420.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7184/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_m62dad420.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7184/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_38390f59.gif)
In interval![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7184/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_mb299efe.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7184/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_mb299efe.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7184/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_me780183.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7184/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_600e5254.gif)
(B) Let![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7184/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_m22cbc460.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7184/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_m22cbc460.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7184/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_m54ecbf95.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7184/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_43ddfac0.jpg)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7184/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_m163bb329.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7184/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_7db3ce36.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7184/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_600e5254.gif)
(C) Let![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7184/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_m4cdd0a06.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7184/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_m4cdd0a06.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7184/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_m74d85fe4.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7184/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_m5ec50b02.jpg)
The point
divides the interval
into two disjoint intervals
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7184/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_21f205dc.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7184/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_600e5254.gif)
i.e., 0![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7184/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_5c4c12a1.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7184/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_5c4c12a1.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7184/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_202bb64e.jpg)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7184/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_6f6c805c.jpg)
∴ f3 is strictly increasing in interval
.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7184/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_m589f3e9f.gif)
Hence, f3 is neither increasing nor decreasing in interval
.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7184/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_600e5254.gif)
(D) Let![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7184/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_m75020c6d.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7184/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_m75020c6d.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7184/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_m4a595ba1.gif)
In interval![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7184/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_31eceb12.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7184/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_31eceb12.gif)
∴ f4 is strictly increasing in interval![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7184/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_46839c89.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7184/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_46839c89.gif)
Therefore, functions cos x and cos 2x are strictly decreasing in![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7184/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_46839c89.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7184/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_46839c89.gif)
Hence, the correct answers are A and B.
Question 13:
On which of the following intervals is the function f given by
strictly decreasing?
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7187/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_m672b9ec0.gif)
(A)
(B) ![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7187/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_42065bae.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7187/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_m6faf5861.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7187/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_42065bae.gif)
(C)
(D) None of these
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7187/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_m3379a2ff.gif)
Answer:
We have,
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7187/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_m1176112d.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7187/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_m17d7fcc3.gif)
In interval![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7187/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_7bfd720d.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7187/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_7bfd720d.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7187/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_fa63c60.gif)
Thus, function f is strictly increasing in interval (0, 1).
In interval ![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7187/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_132a5fd3.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7187/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_132a5fd3.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7187/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_fcd9a51.gif)
Thus, function f is strictly increasing in interval
.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7187/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_42065bae.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7187/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_m6f6be90f.gif)
∴ f is strictly increasing in interval
.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7187/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_600e5254.gif)
Hence, function f is strictly decreasing in none of the intervals.
The correct answer is D.
Question 14:
Find the least value of a such that the function f given
is strictly increasing on [1, 2].
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7190/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_m4b9bc312.gif)
Answer:
We have,
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7190/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_m18535e96.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7190/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_m5c59af04.gif)
Now, function f is increasing on [1,2].
∴ f’x≥0 on 1,2
Now, we have 1⩽x⩽2
⇒2⩽2x⩽4
⇒2+a⩽2x+a⩽4+a
⇒2+a⩽f’x⩽4+a
Since f’x≥0
⇒2+a≥0
⇒a≥-2
So, least value of a is -2.
Question 15:
Let I be any interval disjoint from (−1, 1). Prove that the function f given by
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7193/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_9339e43.gif)
Answer:
We have,
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7193/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_m16d4b55f.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7193/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_33903ff4.gif)
The points x = 1 and x = −1 divide the real line in three disjoint intervals i.e.,
.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7193/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_64e86749.gif)
In interval (−1, 1), it is observed that:
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7193/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_m49673c51.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7193/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_eec6305.gif)
∴ f is strictly decreasing on
.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7193/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_15b3d33b.gif)
In intervals
, it is observed that:
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7193/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_64bafc2d.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7193/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_3852d11b.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7193/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_1b2acfa1.gif)
∴ f is strictly increasing on
.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7193/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_m458f929d.gif)
Hence, function f is strictly increasing in interval I disjoint from (−1, 1).
Hence, the given result is proved.
Question 16:
Prove that the function f given by f(x) = log sin x is strictly increasing on
and strictly decreasing on![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7195/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_m262f622d.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7195/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_7bda2029.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7195/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_m262f622d.gif)
Answer:
We have,
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7195/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_38ad72fd.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7195/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_565acb79.gif)
In interval![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7195/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_7aed1f36.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7195/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_7aed1f36.gif)
∴ f is strictly increasing in
.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7195/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_600e5254.gif)
In interval![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7195/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_33a4678e.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7195/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_33a4678e.gif)
∴f is strictly decreasing in![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7195/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_m262f622d.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7195/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_m262f622d.gif)
Question 17:
Prove that the function f given by f(x) = log cos x is strictly decreasing on
and strictly increasing on![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7199/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_m262f622d.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7199/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_m3379a2ff.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7199/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_m262f622d.gif)
Answer:
We have,
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7199/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_9db93de.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7199/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_7b9c15ef.gif)
In interval![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7199/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_m56807ba6.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7199/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_m56807ba6.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7199/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_m9f38ddb.gif)
∴f is strictly decreasing on
.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7199/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_m3379a2ff.gif)
In interval![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7199/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_m7199f30d.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7199/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_m7199f30d.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7199/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_5adfd3c.gif)
∴f is strictly increasing on
.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7199/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_42065bae.gif)
Question 18:
Prove that the function given by
is increasing in R.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7202/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_1d6f6528.gif)
Answer:
We have,
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7202/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_7be9a707.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7202/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_5ae43178.gif)
For any x∈R, (x − 1)2 > 0.
Thus,
is always positive in R.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7202/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_2240d1fb.gif)
Hence, the given function (f) is increasing in R.
Question 19:
The interval in which
is increasing is
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7204/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_710acaac.gif)
(A)
(B) (−2, 0) (C)
(D) (0, 2)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7204/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_4b85d1ba.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7204/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_578ba5d7.gif)
Answer:
We have,
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7204/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_59d3ec82.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7204/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_m60898d17.gif)
The points x = 0 and x = 2 divide the real line into three disjoint intervals i.e.,![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7204/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_7f094d9.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7204/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_7f094d9.gif)
In intervals
is always positive.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7204/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_m72fe8da6.gif)
∴f is decreasing on![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7204/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_17509f26.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7204/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_17509f26.gif)
In interval (0, 2),![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7204/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_m107bda09.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/1/235/7204/Grade12_CBSE_NCERTSolu_Chapter%206_D_html_m107bda09.gif)
∴ f is strictly increasing on (0, 2).
Hence, f is strictly increasing in interval (0, 2).
The correct answer is D.