Q-1) Show that the function given below is strictly increasing on R.
g(y) = 4y + 19
Ans.)
Assuming y1 and y2 are the two real numbers.
Thus,
y1< y2à 4y1< 4y2à 4y1 + 19 < 4y2 +19 à g(y1) < g(y2)
Thus, the given function g is strictly increasing on R.
This problem can be solved by Alternate method as follows,
g’(y) = 4 > 0, for every interval of R.
Thus, the given function g is strictly increasing on R.
Q-2) Show that the function given below is strictly increasing on R.
g(y) = \(e^{3y}\)
Ans.)
Assuming y1 and y2 are the two real numbers.
Thus,
y1< y2à 3y1< 3y2à \(e^{3y_{1}}\)< \(e^{3y_{2}}\)à g(y1) < g(y2)
Thus, the given function g is strictly increasing on R.
Q-3) Show that the function given below is strictly increasing on R.
g(y) = \(\sin y\)
(i) g(y) is strictly decreasing in \((\frac{\pi }{2},\pi )\)
(ii) g(y) is strictly increasing in \((0, \frac{\pi }{2} )\)
(iii) g(y) is neither decreasing nor increasing in \((0,\pi )\)
Ans.)
Here, g(y) = \(\sin y\)
Now, g’(y) = \(\cos y\)
(i) As we know that for each \(y\epsilon (\frac{\pi }{2},\pi ),\; \cos y < 0\), we get g’(y) < 0.
Thus, g(y) is strictly decreasing in \((\frac{\pi }{2},\pi )\).
(ii) As we know that for each \(y\epsilon (0,\frac{\pi }{2} ),\; \cos y > 0\), we get g’(y) > 0.
Thus, g(y) is strictly increasing in \((\frac{\pi }{2},\pi )\).
(iii) From the (i) and (ii) it is clear that that g(y) is neither decreasing nor increasing in \((0,\pi )\).
Q-4) Find out in which intervals the given function is
(i) Strictly decreasing
(ii) Strictly increasing
g(y) = 3y2 – 5y
Ans.)
Here, g(y) = 3y2 – 5y
g’(y) = 6y – 5
Now, g’(y) = 0 à y = 5/6
Here, the point 5/6 splits the real line into 2 disjoint interval and they are\((-\infty ,\frac{5}{6})\; and\; (\frac{5}{6},\infty )\).
For interval \((-\infty ,\frac{5}{6})\),
g’(y) = 6y – 5 < 0
Thus, g(y) is strictly decreasing in \((-\infty ,\frac{5}{6})\).
For interval \((\frac{5}{6},\infty )\),
g’(y) = 6y – 5 > 0
Thus, g(y) is strictly increasing in \((-\infty ,\frac{5}{6})\).
Q-5) Find out in which intervals the given function is
(i) Strictly decreasing
(ii) Strictly increasing
g(y) = 2y3 – 3y2 – 36y + 7
Ans.)
Here, g(y) = 2y3 – 3y2 – 36y + 7
g’(y) = 6y2 – 6y – 36 = 6(y2 – y – 6) = 6(y + 3)(y + 2)
Thus, g’(y) = 0,
y = -2,3
Thus, these two points -2 and 3 divides the real line in 3 disjoint intervals and they are \((-\infty ,- 2), (- 2,3)\; and\;(3,\infty )\).
For interval \((-\infty ,- 2)\)
g’(y) = 6(y + 3)(y + 2) > 0
Thus, g(y) is strictly increasing in \((-\infty ,- 2)\)
For interval (- 2,3)
g’(y) = 6(y + 3)(y + 2) < 0
Thus, g(y) is strictly decreasing in (- 2,3)
For interval \((3,\infty )\)
g’(y) = 6(y + 3)(y + 2) > 0
Thus, g(y) is strictly increasing in \((3,\infty )\).
Q-6)Find out the intervals in which the functions given below are strictly increasing on R.
(i) 9 – 5y – y2
(ii) y2 + 4y – 7
(iii) (y + 5)3(y – 7)3
(iv) -2y3 – 9y2 – 12y + 1
(v) 7 – 7y – y2
Ans.)
(i) Here, g(y) = 9 – 5y –y2
g’(y) = – 5 – 2y
Now, g’(y) = 0
y = -5/2
Here, the point -5/2 splits the real line into 2 disjoint interval and they are \((-\infty ,\frac{-5}{2})\; and\; (\frac{-5}{2},\infty )\)
For interval \((-\infty ,\frac{-5}{2})\),
g’(y) = – 5 – 2y > 0
Thus, g(y) is strictly increasing in \((-\infty ,\frac{-5}{2})\).
For interval \((\frac{-5}{2},\infty )\),
g’(y) = – 5 – 2y< 0
Thus, g(y) is strictly decreasing in \((-\infty ,\frac{-5}{2})\).
(ii) Here, g(y) = y2 + 4y – 7
g’(y) = 2y + 4
Now, g’(y) = 0
y = – 2
Thus, this point divides the real line in 2 disjoint intervals and they are \((-\infty ,-2),\;and\;(-2,\infty )\).
For interval \((-\infty ,-2)\)
g’(y) = 2y + 4 < 0
Thus, g(y) is strictly decreasing in \((-\infty ,- 2)\)
For interval \((-2,\infty ) \)
g’(y) = 2y + 4 > 0
Thus, g(y) is strictly increasing in \((-2,\infty )\)
(iii) Here, g(y) =(y + 5)3(y – 7)3
g’(y) = 3(y + 5)2(y – 7)3 + 3(y + 5)3(y – 7)2
= 3(y + 5)2(y – 7)2(y + 5 + y – 7)
= 3(y + 5)2(y – 7)2(2y – 2)
= 6(y + 5)2(y – 7)2(y – 1)
Now, g’(y) = 0
y = -5,1 and 7
Thus, these three points -5, 1 and 7 divides the real line in 4 disjoint intervals and they are \((-\infty ,- 5), (- 5,1), (1,7)\; and\;(7,\infty )\).
For interval \((-\infty ,- 5)\)
g’(y) = 6(y + 5)2(y – 7)2(y – 1) < 0
Thus, g(y) is strictly decreasing in \((-\infty ,- 5)\)
For interval (-5,1)
g’(y) = 6(y + 5)2(y – 7)2(y – 1) < 0
Thus, g(y) is strictly decreasing in (-5,1).
For interval (1,7)
g’(y) = 6(y + 5)2(y – 7)2(y – 1) > 0
Thus, g(y) is strictly increasing in (1,7).
For interval \((7,\infty )\)
g’(y) = 6(y + 5)2(y – 7)2(y – 1) > 0
Thus, g(y) is strictly increasing in \((7,\infty )\)
(iv) Here, g(y) = -2y3 – 9y2 – 12y + 1
g’(y) = -6y2 – 18y – 12
Now, g’(y) = 0
à- 6y2 – 18y – 12 = 0
à-6(y2 + 3y + 2) = 0
à-6(y + 2)(y + 1) = 0
à y = -2,-1
Thus, these three points -5, and -1 divides the real line in 3 disjoint intervals and they are \((-\infty ,- 2), (- 2,-1)\; and\;(-1,\infty )\).
For, intervals \((-\infty ,- 2)\; and\;(-1,\infty )\)
g’(y) = -6(y + 2)(y + 1) < 0
Thus, g(y) is strictly decreasing in intervals \((-\infty ,- 2)\; and\;(-1,\infty )\)
For interval (- 2, – 1)
g’(y) = -6(y + 2)(y + 1) > 0
Thus, g(y) is strictly increasing in intervals (- 2, – 1)
(v) Here, g(y) = 7 – 7y – y2
g’(y) = – 7 – 2y
Now, g’(y) = 0
y = -7/2
Here, the point -7/2 splits the real line into 2 disjoint interval and they are \((-\infty ,\frac{-7}{2})\; and\; (\frac{-7}{2},\infty )\)
For interval \((-\infty ,\frac{-7}{2})\),
g’(y) = – 7 – 2y > 0
Thus, g(y) is strictly increasing in \((-\infty ,\frac{-7}{2})\).
For interval \((\frac{-7}{2},\infty )\),
g’(y) = – 7 – 2y< 0
Thus, g(y) is strictly decreasing in \((-\infty ,\frac{-7}{2})\).
Q-7) Show that the given function is increasing function throughout its domain
x = \(\log (1 + y) – \frac{2y}{2 + y}\), y > – 1.
Ans.)
Here, x = \(\log (1 + y) – \frac{2y}{2 + y}\)
\( \frac{\mathrm{d} x}{\mathrm{d} y} = \frac{1}{1 + y} – \frac{(2 + y)(2) – 2y(1)}{(2 + y)^{2}} = \frac{1}{1 + y} – \frac{4}{(2 + y)^{2}} = \frac{y^{2}}{(2 + y)^{2}}\)Now, \(\frac{\mathrm{d} x}{\mathrm{d} y} = 0\)
So, \(\frac{y^{2}}{(2 + y)^{2}} = 0\)
\( y^{2} = 0\) (as (2 + y) is not 0; y > -1)
y = 0
As y >- 1, the point y = 0 splits the domain \((- 1,\infty )\) in 2 disjoint intervals i.e. (-1,0) and \((0,\infty )\).
For interval (-1,0)
\(\frac{\mathrm{d} x}{\mathrm{d} y}\)> 0…………………………..(a)
Also,
For interval \((0,\infty )\)
\(\frac{\mathrm{d} x}{\mathrm{d} y}\)> 0……………………………(b)
Thus, the given function is increasing throughout its domain.
Q-8) b = [a(a – 2)]2 is an increasing function , then find out the values of b.
Ans.)
b = [a(a – 2)]2 = [a2 – 2a]
b’ = 2(a2 – 2a)(2a – 2) = 4a(a – 1)(a – 2)
Now, b’ = 0,
à b = 0,1 and 2
Thus, these three points 0, 1 and 2 divides the real line in 4 disjoint intervals and they are \((-\infty ,0), (0,1), (1,2)\; and\;(2,\infty )\).
For intervals \((-\infty ,0)\; and\;(1,2)\)
b’ < 0
Thus, b(a) is strictly decreasing in intervals \((-\infty ,0)\; and\;(1,2)\)..
For intervals \((0,1)\; and\;(2,\infty )\)
b’ > 0
Thus, b(a) is strictly increasing in intervals \((0,1)\; and\;(2,\infty )\).
Thus, b(a) is strictly increasing for 0 < a < 1 and a > 2.
Q-9)Show that the given function is increasing function in the domain \([0,\frac{\pi }{2}]\).
f(x) = \(\frac{4\sin \alpha }{(2 + \cos \alpha )} – \alpha\)
Ans.)
Here, f(x) = \(\frac{4\sin \alpha }{(2 + \cos \alpha )} – \alpha\)
f’(x) = \(\frac{(2 + \cos\alpha )(4\cos \alpha ) – 4\sin \alpha (-\sin \alpha )}{(2 + \cos \alpha )^{2}} – 1\)
= \(\frac{8\cos\alpha + 4\cos^{2} \alpha + 4\sin^{2} \alpha}{(2 + \cos \alpha )^{2}} – 1\)
= \(\frac{8\cos \alpha + 4}{(2 + \cos\alpha )^{2}} – 1\)
Now, f’(x) = 0
à\(\frac{8\cos \alpha + 4}{(2 + \cos\alpha )^{2}} – 1\) = 0
à\(\frac{8\cos \alpha + 4}{(2 + \cos\alpha )^{2}} = 1\)
à\(8\cos \alpha + 4 = 4 + \cos ^{2}\alpha + 4\cos \alpha\)
à\(\cos ^{2}\alpha – 4\cos \alpha = 0\)
à\(\cos \alpha(\cos \alpha – 4) = 0\)
à\(\cos \alpha(\cos \alpha – 4) = 0\)
As, \(\cos \alpha \neq 4 , \cos\alpha = 0\)
\(\cos\alpha = 0 \Rightarrow \cos\alpha = \frac{\pi }{2}\)Now,
\(\frac{8\cos\alpha + -(4 +\cos^{2}\alpha + 4\cos\alpha)}{(2 + \cos\alpha )^{2}} = \frac{4\cos\alpha – \cos^{2}\alpha }{(2 + \cos\alpha )^{2}} = \frac{\cos\alpha (4 – \cos\alpha )}{(2 + \cos\alpha )^{2}}\)In \([0,\frac{\pi }{2}]\)\(\cos\alpha > 0\).
Also, \(4 >\cos\alpha\)
à\(4 – \cos\alpha> 0\)
\( \cos\alpha(4 – \cos\alpha) > 0\) and \((2 + \cos\alpha )^{2} > 0\)
à\(\frac{\cos\alpha(4 – \cos\alpha)}{(2 + \cos\alpha )^{2}} > 0\)
àf’(x) > 0
Thus, f is strictly increasing in \((0,\frac{\pi }{2})\)
Ans as it is also, continuous at x = 0, and x = \(\frac{\pi }{2}\)
Thus, the given function is increasing function f(x) in the domain \([0,\frac{\pi }{2}]\).
Q-10) Show that \(\log y\) is strictly increasing on \((0,\infty )\).
Ans.)
Here, let g(y) = \(\log y\).
\( \) g’(y) = 1/y
This implies that y > 0,
So, g’(y) > 0.
Thus, \(\log y\) is strictly increasing on \((0,\infty )\).
Q-11) Show that the given function is neither strictly increasing nor strictly decreasing on (- 1, 1).
g(y) = 2y2 – 2y + 1
Ans.)
Here, g(y) = 2y2 – 2y + 1
g’(y) = 4y – 2
Now, g’(y) = 0
à 4y – 2 = 0
ày = 1/2
Thus, the point ½ splits the interval (- 1, 1) in 2 disjoint intervals and they are (- 1, ½) and (1/2, 1).
For interval (- 1, 1/2)
g’(y) = 4y – 2 < 0.
Thus, g(y) is strictly decreasing in (- 1, 1/2).
For interval (1/2, 1)
g’(y) = 4y – 2 > 0.
Thus, g(y) is strictly increasing in (1/2, 1).
Hence proved.
Q-12) Find out from the following functions that which strictly functions decreasing on \((0, \frac{\pi }{2})\)?
(i) \(\tan y\)
(ii) \(\cos4y\)
(iii) \(\cos2y\)
(iv) \(\sin y\)
Ans.)
(i) Here, g(y) = \(\tan y\)
g’(y) = \(\sec ^{2}y\)
For interval \((0, \frac{\pi }{2})\),
g’(y) = \(\sec ^{2}y\)> 0
Thus, g(y) is strictly increasing in \((0, \frac{\pi }{2})\).
(ii) Here, g(y) = \(\cos4y\)
g’(y) = \(-4\sin 4y\)
Now, g’(y) = 0,
à\(\sin 4y\) = 0
à4y = \(\pi\); \(y\;\epsilon\; (0,\frac{\pi }{2})\)
ày = \(\frac{\pi }{4}\)
Thus, this point y = \(\frac{\pi }{4}\) splits the interval \((0, \frac{\pi }{2})\) into 2 disjoint intervals and they are \((0, \frac{\pi }{4})\;and\;(\frac{\pi }{4},\frac{\pi }{2})\).
For interval \((0, \frac{\pi }{4})\),
g’(y) = \(-4 \sin 4y\)< 0 ……………\((because 0 < y < \frac{\pi }{4}\Rightarrow 0 < 4y <\pi)\)
Thus, g(y) is strictly decreasing in \((0, \frac{\pi }{2})\).
For interval \((\frac{\pi }{4},\frac{\pi }{2})\),
g’(y) = \(-4 \sin 4y\)> 0 ……….\((because \frac{\pi }{4} < y < \frac{\pi }{2}\Rightarrow \pi < 4y < 2\pi)\)
Thus, g(y) is strictly increasing in \((\frac{\pi }{4}, \frac{\pi }{2})\).
(iii) g(y) = \(\cos2y\)
g’(y) = \(-2\sin 2y\)
Now, \(0 < y < \frac{\pi }{2}\Rightarrow 0 < 2y < \pi\Rightarrow \sin 2y > 0\Rightarrow -2\sin 2y < 0\)
Thus, g(y) is strictly decreasing in \((0, \frac{\pi }{2})\).
(iv) g(y) = \(\sin y\)
g’(y) = \(\cos y\)
For, interval \((0, \frac{\pi }{2})\),
g’(y) = \(\cos y\) > 0
Thus, g(y) is strictly increasing in \((0, \frac{\pi }{2})\).
So, the correct answer is (iii)
Q-13) From the intervals given below, find out in which interval the given function is strictly decreasing?
g(y) = \(y^{100} + \sin y – 1\)
(i) \((0, \frac{\pi }{2})\)
(ii) (0, 1)
(iii) \((\frac{\pi }{2}, \pi)\)
(iv) None of above
Ans.)
(iv) None of above
Explanation:
Here, g(y) = \(y^{100} + \sin y – 1\)
\( \;g'(y)\; 100y^{99} + \cos y\)àFor interval \((0, \frac{\pi }{2})\)
\(\cos y\) > 0, \(y^{100}\)> 0
\( \; 100y^{99} + \cos y\)> 0
Thus, g(y) is strictly increasing in \((0, \frac{\pi }{2})\).
à For interval (0, 1)
\(\cos y\)> 0, \(y^{100}\)> 0
\( \; 100y^{99} + \cos y\)> 0
Thus, g(y) is strictly increasing in (0, 1).
àFor interval \((\frac{\pi }{2}, \pi)\)
\(\cos y\)> 0, \(y^{100}\)> 0
\( \; 100y^{99} + \cos y\)> 0
Thus, g(y) is strictly increasing in \((\frac{\pi }{2}, \pi)\).
Thus, the answer is (iv) None of above
Q-14) Find out the least value of m such that given function is strictly increasing on (1, 2).
g(y) = y2 +my + 1
Ans.)
Here, g(y) = y2 +my + 1
àg(y) = 2y + m
Now, the g(y) will be increasing on (1, 2) only if,
g’(y) > 0 in (1, 2)
So, g’(y) > 0
à2y + m > 0
à2y > -m
ày = -m/2
Thus, for least value of m,
y > -m/2; \(y\;\epsilon \;(1,2)\)
à y > -m/2; ( 1 < y < 2)
Thus, the least requiredvalue of m is
-m/2 = 1
-m/2 = 1
à m = – 2
Thus, the required least value of m is – 2.
Q-15) Assuming D to be any interval disjoint from (-1, 1). Show that the given function is strictly increasing on D.
g(y) = \(y + \frac{1}{y}\)
Ans.)
Here, g(y) = \(y + \frac{1}{y}\)
à g’(y) = \(1 – \frac{1}{y^{2}}\)
Now, g’(y) = 0
à\(\Rightarrow \frac{1}{y^{2}} = 1 \Rightarrow y = \pm 1\)
Thus, y = -1 and 1.
Thus, these three points -1, and 1 divides the real line in 3 disjoint intervals and they are \((-\infty ,- 1), (- 1,1)\; and\;(1,\infty )\).
For interval (- 1, 1)
-1 < y < 1
à y2< 1
\(\Rightarrow 1 < \frac{1}{y^{2}} , y \neq 0\) \(\Rightarrow 1 – \frac{1}{y^{2}} < 0 \;on\;(- 1, 1)\sim (0)\)Thus, g(y) is strictly decreasing on \((-1, 1)\sim (0)\)
For intervals \((-\infty , -1) \;and\; (1, \infty )\)
y < -1
à1 < y
ày2<1
à1 > 1/y2
à 1 – 1/y2> 0
Thus, g’(y) > 0
So, g(y) is strictly increasing on \((-\infty , -1) \;and\; (1, \infty )\).
Thus, the given function g(y) is strictly increasing on D which is an interval disjoint of (-1, 1).
Q-16) Show that the given function is strictly increasing on \((0,\frac{\pi}{2})\) and strictly decreasing on \((\frac{\pi}{2}, \pi)\)
g(y) = \(\log \sin y\)
Ans.)
Here, g(y) = \(\log \sin y\)
g’(y) = \(\frac{1}{\sin y}\cos y = \cot y\)
For, interval \((0, \frac{\pi}{2})\)
g’(y) = cot y > 0
So, g(y) is strictly increasing on \((0, \frac{\pi}{2})\).
For interval \((\frac{\pi}{2}, \pi)\)
g’(y) = cot y < 0
So, g(y) is strictly decreasing on\((\frac{\pi}{2}, \pi)\).
Q-17) Show that the given function is strictly decreasing on \((0,\frac{\pi}{2})\) and strictly increasing on \((\frac{\pi}{2}, \pi)\)
g(y) = \(\log \cos y\)
Ans.)
Here, g(y) = \(\log \cos y\)
g’(y) = \(\frac{1}{\cos y}(-\sin y) = -\tan y\)
For, interval \((0, \frac{\pi}{2})\)
g’(y) = – tan y < 0
So, g(y) is strictly decreasing on \((0, \frac{\pi}{2})\).
For interval \((\frac{\pi}{2}, \pi)\)
g’(y) = – tan y > 0
So, g(y) is strictly increasing on \((\frac{\pi}{2}, \pi)\).
Q-18)Show that the given function is increasing in R.
g(y) = y3 – 3y2 + 3y -100
Ans.)
Here, g(y) = y3 – 3y2 + 3y -100
g’(y) = 3y2 – 6y + 3
= 3(y2 -2y + 1)
= 3(y – 1)2
Now, for any value of R, (x -1)2> 0.
So, g’(y) > 0 for R.
Thus, g(y) is increasing in R.
Q-19) In which of the following interval the function, x = y2e-y
(i) (-2, 0)
(ii) \((-\infty ,\infty )\)
(iii) (0, 2)
(iv) \((2,\infty )\)
Ans.)
(iii) (0, 2)
Explanation:
Here, x = y2e-y
So, \(\frac{\mathrm{d} x}{\mathrm{d} y} = 2ye^{-y} – y^{-2}e^{-y} = ye^{-y}(2 – y)\)
Now, \(\frac{\mathrm{d} x}{\mathrm{d} y} = 0\)
à y = 0 and y = 2
Thus, these two points 0 and 2 divides the real line in 3 disjoint intervals and they are \((-\infty ,0), (0,2)\; and\;(2,\infty )\).
For intervals \((-\infty ,0)\; and\;(2,\infty )\)
g’(y) < 0 (as e-y is always positive)
Thus, g(y) is decreasing on \((-\infty ,0)\; and\;(2,\infty )\).
For interval (0, 2)
g’(y) > 0
Thus, g(y) is increasing on (0, 2).
Thus, the correct answer is (iii).
