{"id":542321,"date":"2021-09-28T04:44:24","date_gmt":"2021-09-28T04:44:24","guid":{"rendered":"https:\/\/www.indcareer.com\/schools\/?p=542321"},"modified":"2022-12-26T09:38:27","modified_gmt":"2022-12-26T09:38:27","slug":"rd-sharma-solutions-for-class-12-maths-chapter-17-increasing-and-decreasing-functions","status":"publish","type":"post","link":"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-12-maths-chapter-17-increasing-and-decreasing-functions\/","title":{"rendered":"RD Sharma Solutions for Class 12 Maths Chapter 17\u2013Increasing and Decreasing Functions"},"content":{"rendered":"\n

Class 12: Maths Chapter 17 solutions. Complete Class 12 Maths Chapter 17 Notes.<\/p>\n\n\n\n

RD Sharma Solutions for Class 12 Maths Chapter 17\u2013Increasing and Decreasing Functions<\/h2>\n\n\n\n

RD Sharma 12th Maths Chapter 17, Class 12 Maths Chapter 17 solutions<\/p>\n\n\n\n

Exercise 17.1 Page No: 17.10<\/p>\n\n\n\n

1. Prove that the function f(x) = loge<\/sub> x is increasing on (0, \u221e).<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Let x1<\/sub>, x2<\/sub> \u2208 (0, \u221e)<\/p>\n\n\n\n

We have, x1 <\/sub>< x2<\/sub><\/p>\n\n\n\n

\u21d2 loge<\/sub> x1<\/sub> < loge<\/sub> x2<\/sub><\/p>\n\n\n\n

\u21d2 f (x1<\/sub>) < f (x2<\/sub>)<\/p>\n\n\n\n

So, f(x) is increasing in (0, \u221e)<\/p>\n\n\n\n

2. Prove that the function f(x) = loga<\/sub> x is increasing on (0, \u221e) if a > 1 and decreasing on (0, \u221e), if 0 < a < 1.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n
\"\"<\/figure>\n\n\n\n

3. Prove that f(x) = ax + b, where a, b are constants and a > 0 is an increasing function on R.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given,<\/p>\n\n\n\n

f (x) = ax + b, a > 0<\/p>\n\n\n\n

Let x1<\/sub>, x2<\/sub> \u2208 R and x1<\/sub> > x2<\/sub><\/p>\n\n\n\n

\u21d2 ax1<\/sub> > ax2<\/sub> for some a > 0<\/p>\n\n\n\n

\u21d2 ax1<\/sub> + b> ax2<\/sub> + b for some b<\/p>\n\n\n\n

\u21d2 f (x1<\/sub>) > f(x2<\/sub>)<\/p>\n\n\n\n

Hence, x1<\/sub> > x2 <\/sub>\u21d2 f(x1<\/sub>) > f(x2<\/sub>)<\/p>\n\n\n\n

So, f(x) is increasing function of R<\/p>\n\n\n\n

4. Prove that f(x) = ax + b, where a, b are constants and a < 0 is a decreasing function on R.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given,<\/p>\n\n\n\n

f (x) = ax + b, a < 0<\/p>\n\n\n\n

Let x1<\/sub>, x2<\/sub> \u2208 R and x1<\/sub> > x2<\/sub><\/p>\n\n\n\n

\u21d2 ax1<\/sub> < ax2<\/sub> for some a > 0<\/p>\n\n\n\n

\u21d2 ax1<\/sub> + b < ax2<\/sub> + b for some b<\/p>\n\n\n\n

\u21d2 f (x1<\/sub>) < f(x2<\/sub>)<\/p>\n\n\n\n

Hence, x1<\/sub> > x2<\/sub>\u21d2 f(x1<\/sub>) < f(x2<\/sub>)<\/p>\n\n\n\n

So, f(x) is decreasing function of R<\/p>\n\n\n\n

Exercise 17.2 Page No: 17.33<\/p>\n\n\n\n

1. Find the intervals in which the following functions are increasing or decreasing.<\/strong><\/p>\n\n\n\n

(i) f (x) = 10 \u2013 6x \u2013 2x2<\/sup><\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n
\"\"<\/figure>\n\n\n\n
\"\"<\/figure>\n\n\n\n

(ii) f (x) = x2<\/sup> + 2x \u2013 5<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n
\"\"<\/figure>\n\n\n\n

(iii) f (x) = 6 \u2013 9x \u2013 x2<\/sup><\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n
\"\"<\/figure>\n\n\n\n

(iv) f(x) = 2x3<\/sup> \u2013 12x2<\/sup> + 18x + 15<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

(v) f (x) = 5 + 36x + 3x2<\/sup> \u2013 2x3<\/sup><\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given f (x) = 5 + 36x + 3x2<\/sup> \u2013 2x3<\/sup><\/p>\n\n\n\n

\u21d2
<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

\u21d2 f\u2019(x) = 36 + 6x \u2013 6x2<\/sup><\/p>\n\n\n\n

For f(x) now we have to find critical point, we must have<\/p>\n\n\n\n

\u21d2 f\u2019(x) = 0<\/p>\n\n\n\n

\u21d2 36 + 6x \u2013 6x2<\/sup> = 0<\/p>\n\n\n\n

\u21d2 6(\u2013x2<\/sup> + x + 6) = 0<\/p>\n\n\n\n

\u21d2 6(\u2013x2<\/sup> + 3x \u2013 2x + 6) = 0<\/p>\n\n\n\n

\u21d2 \u2013x2<\/sup> + 3x \u2013 2x + 6 = 0<\/p>\n\n\n\n

\u21d2 x2<\/sup> \u2013 3x + 2x \u2013 6 = 0<\/p>\n\n\n\n

\u21d2 (x \u2013 3) (x + 2) = 0<\/p>\n\n\n\n

\u21d2 x = 3, \u2013 2<\/p>\n\n\n\n

Clearly, f\u2019(x) > 0 if \u20132< x < 3 and f\u2019(x) < 0 if x < \u20132 and x > 3<\/p>\n\n\n\n

Thus, f(x) increases on x \u2208 (\u20132, 3) and f(x) is decreasing on interval (\u2013\u221e, \u20132) \u222a (3, \u221e)<\/p>\n\n\n\n

(vi) f (x) = 8 + 36x + 3x2<\/sup> \u2013 2x3<\/sup><\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given f (x) = 8 + 36x + 3x2<\/sup> \u2013 2x3<\/sup><\/p>\n\n\n\n

Now differentiating with respect to x<\/p>\n\n\n\n

\u21d2
<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

\u21d2 f\u2019(x) = 36 + 6x \u2013 6x2<\/sup><\/p>\n\n\n\n

For f(x) we have to find critical point, we must have<\/p>\n\n\n\n

\u21d2 f\u2019(x) = 0<\/p>\n\n\n\n

\u21d2 36 + 6x \u2013 6x2<\/sup> = 0<\/p>\n\n\n\n

\u21d2 6(\u2013x2<\/sup> + x + 6) = 0<\/p>\n\n\n\n

\u21d2 6(\u2013x2<\/sup> + 3x \u2013 2x + 6) = 0<\/p>\n\n\n\n

\u21d2 \u2013x2<\/sup> + 3x \u2013 2x + 6 = 0<\/p>\n\n\n\n

\u21d2 x2<\/sup> \u2013 3x + 2x \u2013 6 = 0<\/p>\n\n\n\n

\u21d2 (x \u2013 3) (x + 2) = 0<\/p>\n\n\n\n

\u21d2 x = 3, \u2013 2<\/p>\n\n\n\n

Clearly, f\u2019(x) > 0 if \u20132 < x < 3 and f\u2019(x) < 0 if x < \u20132 and x > 3<\/p>\n\n\n\n

Thus, f(x) increases on x \u2208 (\u20132, 3) and f(x) is decreasing on interval (\u2013\u221e, 2) \u222a (3, \u221e)<\/p>\n\n\n\n

(vii) f(x) = 5x3<\/sup> \u2013 15x2<\/sup> \u2013 120x + 3<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given f(x) = 5x3<\/sup> \u2013 15x2<\/sup> \u2013 120x + 3<\/p>\n\n\n\n

Now by differentiating above equation with respect x, we get<\/p>\n\n\n\n

\u21d2
<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

\u21d2 f\u2019(x) = 15x2<\/sup> \u2013 30x \u2013 120<\/p>\n\n\n\n

For f(x) we have to find critical point, we must have<\/p>\n\n\n\n

\u21d2 f\u2019(x) = 0<\/p>\n\n\n\n

\u21d2 15x2<\/sup> \u2013 30x \u2013 120 = 0<\/p>\n\n\n\n

\u21d2 15(x2<\/sup> \u2013 2x \u2013 8) = 0<\/p>\n\n\n\n

\u21d2 15(x2<\/sup> \u2013 4x + 2x \u2013 8) = 0<\/p>\n\n\n\n

\u21d2 x2<\/sup> \u2013 4x + 2x \u2013 8 = 0<\/p>\n\n\n\n

\u21d2 (x \u2013 4) (x + 2) = 0<\/p>\n\n\n\n

\u21d2 x = 4, \u2013 2<\/p>\n\n\n\n

Clearly, f\u2019(x) > 0 if x < \u20132 and x > 4 and f\u2019(x) < 0 if \u20132 < x < 4<\/p>\n\n\n\n

Thus, f(x) increases on (\u2013\u221e,\u20132) \u222a (4, \u221e) and f(x) is decreasing on interval x \u2208 (\u20132, 4)<\/p>\n\n\n\n

(viii) f(x) = x3<\/sup> \u2013 6x2<\/sup> \u2013 36x + 2<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given f (x) = x3<\/sup> \u2013 6x2<\/sup> \u2013 36x + 2<\/p>\n\n\n\n

\u21d2
<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

\u21d2 f\u2019(x) = 3x2<\/sup> \u2013 12x \u2013 36<\/p>\n\n\n\n

For f(x) we have to find critical point, we must have<\/p>\n\n\n\n

\u21d2 f\u2019(x) = 0<\/p>\n\n\n\n

\u21d2 3x2<\/sup> \u2013 12x \u2013 36 = 0<\/p>\n\n\n\n

\u21d2 3(x2<\/sup> \u2013 4x \u2013 12) = 0<\/p>\n\n\n\n

\u21d2 3(x2<\/sup> \u2013 6x + 2x \u2013 12) = 0<\/p>\n\n\n\n

\u21d2 x2<\/sup> \u2013 6x + 2x \u2013 12 = 0<\/p>\n\n\n\n

\u21d2 (x \u2013 6) (x + 2) = 0<\/p>\n\n\n\n

\u21d2 x = 6, \u2013 2<\/p>\n\n\n\n

Clearly, f\u2019(x) > 0 if x < \u20132 and x > 6 and f\u2019(x) < 0 if \u20132< x < 6<\/p>\n\n\n\n

Thus, f(x) increases on (\u2013\u221e,\u20132) \u222a (6, \u221e) and f(x) is decreasing on interval x \u2208 (\u20132, 6)<\/p>\n\n\n\n

(ix) f(x) = 2x3<\/sup> \u2013 15x2<\/sup> + 36x + 1<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given f (x) = 2x3<\/sup> \u2013 15x2<\/sup> + 36x + 1<\/p>\n\n\n\n

Now by differentiating above equation with respect x, we get<\/p>\n\n\n\n

\u21d2
<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

\u21d2 f\u2019(x) = 6x2<\/sup> \u2013 30x + 36<\/p>\n\n\n\n

For f(x) we have to find critical point, we must have<\/p>\n\n\n\n

\u21d2 f\u2019(x) = 0<\/p>\n\n\n\n

\u21d2 6x2<\/sup> \u2013 30x + 36 = 0<\/p>\n\n\n\n

\u21d2 6 (x2<\/sup> \u2013 5x + 6) = 0<\/p>\n\n\n\n

\u21d2 6(x2<\/sup> \u2013 3x \u2013 2x + 6) = 0<\/p>\n\n\n\n

\u21d2 x2<\/sup> \u2013 3x \u2013 2x + 6 = 0<\/p>\n\n\n\n

\u21d2 (x \u2013 3) (x \u2013 2) = 0<\/p>\n\n\n\n

\u21d2 x = 3, 2<\/p>\n\n\n\n

Clearly, f\u2019(x) > 0 if x < 2 and x > 3 and f\u2019(x) < 0 if 2 < x < 3<\/p>\n\n\n\n

Thus, f(x) increases on (\u2013\u221e, 2) \u222a (3, \u221e) and f(x) is decreasing on interval x \u2208 (2, 3)<\/p>\n\n\n\n

(x) f (x) = 2x3<\/sup> + 9x2<\/sup> + 12x + 20<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given f (x) = 2x3<\/sup> + 9x2<\/sup> + 12x + 20<\/p>\n\n\n\n

Differentiating above equation we get<\/p>\n\n\n\n

\u21d2
<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

\u21d2 f\u2019(x) = 6x2<\/sup> + 18x + 12<\/p>\n\n\n\n

For f(x) we have to find critical point, we must have<\/p>\n\n\n\n

\u21d2 f\u2019(x) = 0<\/p>\n\n\n\n

\u21d2 6x2<\/sup> + 18x + 12 = 0<\/p>\n\n\n\n

\u21d2 6(x2<\/sup> + 3x + 2) = 0<\/p>\n\n\n\n

\u21d2 6(x2<\/sup> + 2x + x + 2) = 0<\/p>\n\n\n\n

\u21d2 x2<\/sup> + 2x + x + 2 = 0<\/p>\n\n\n\n

\u21d2 (x + 2) (x + 1) = 0<\/p>\n\n\n\n

\u21d2 x = \u20131, \u20132<\/p>\n\n\n\n

Clearly, f\u2019(x) > 0 if \u20132 < x < \u20131 and f\u2019(x) < 0 if x < \u20131 and x > \u20132<\/p>\n\n\n\n

Thus, f(x) increases on x \u2208 (\u20132,\u20131) and f(x) is decreasing on interval (\u2013\u221e, \u20132) \u222a (\u20132, \u221e)<\/p>\n\n\n\n

2. Determine the values of x for which the function f(x) = x2<\/sup> \u2013 6x + 9 is increasing or decreasing. Also, find the coordinates of the point on the curve y = x2<\/sup> \u2013 6x + 9 where the normal is parallel to the line y = x + 5.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given f(x) = x2<\/sup> \u2013 6x + 9<\/p>\n\n\n\n

\u21d2
<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

\u21d2 f\u2019(x) = 2x \u2013 6<\/p>\n\n\n\n

\u21d2 f\u2019(x) = 2(x \u2013 3)<\/p>\n\n\n\n

For f(x) let us find critical point, we must have<\/p>\n\n\n\n

\u21d2 f\u2019(x) = 0<\/p>\n\n\n\n

\u21d2 2(x \u2013 3) = 0<\/p>\n\n\n\n

\u21d2 (x \u2013 3) = 0<\/p>\n\n\n\n

\u21d2 x = 3<\/p>\n\n\n\n

Clearly, f\u2019(x) > 0 if x > 3 and f\u2019(x) < 0 if x < 3<\/p>\n\n\n\n

Thus, f(x) increases on (3, \u221e) and f(x) is decreasing on interval x \u2208 (\u2013\u221e, 3)<\/p>\n\n\n\n

Now, let us find coordinates of point<\/p>\n\n\n\n

Equation of curve is f(x) = x2<\/sup> \u2013 6x + 9<\/p>\n\n\n\n

Slope of this curve is given by<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n
\"\"<\/figure>\n\n\n\n

3. Find the intervals in which f(x) = sin x \u2013 cos x, where 0 < x < 2\u03c0 is increasing or decreasing.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n
\"\"<\/figure>\n\n\n\n

4. Show that f(x) = e2x<\/sup> is increasing on R.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given f (x) = e2x<\/sup><\/p>\n\n\n\n

\u21d2
<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

\u21d2 f\u2019(x) = 2e2x<\/sup><\/p>\n\n\n\n

For f(x) to be increasing, we must have<\/p>\n\n\n\n

\u21d2 f\u2019(x) > 0<\/p>\n\n\n\n

\u21d2 2e2x<\/sup> > 0<\/p>\n\n\n\n

\u21d2 e2x<\/sup> > 0<\/p>\n\n\n\n

Since, the value of e lies between 2 and 3<\/p>\n\n\n\n

So, whatever be the power of e (that is x in domain R) will be greater than zero.<\/p>\n\n\n\n

Thus f(x) is increasing on interval R<\/p>\n\n\n\n

5. Show that f (x) = e1\/x<\/sup>, x \u2260 0 is a decreasing function for all x \u2260 0.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n
\"\"<\/figure>\n\n\n\n

6. Show that f(x) = loga<\/sub> x, 0 < a < 1 is a decreasing function for all x > 0.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

7. Show that f(x) = sin x is increasing on (0, \u03c0\/2) and decreasing on (\u03c0\/2, \u03c0) and neither increasing nor decreasing in (0, \u03c0).<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

8. Show that f(x) = log sin x is increasing on (0, \u03c0\/2) and decreasing on (\u03c0\/2, \u03c0).<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n
\"\"<\/figure>\n\n\n\n

9. Show that f(x) = x \u2013 sin x is increasing for all x \u03f5 R.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given f (x) = x \u2013 sin x<\/p>\n\n\n\n

\u21d2
<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

\u21d2 f\u2019(x) = 1 \u2013 cos x<\/p>\n\n\n\n

Now, as given x \u03f5 R<\/p>\n\n\n\n

\u21d2 \u20131 < cos x < 1<\/p>\n\n\n\n

\u21d2 \u20131 > cos x > 0<\/p>\n\n\n\n

\u21d2 f\u2019(x) > 0<\/p>\n\n\n\n

Hence, condition for f(x) to be increasing<\/p>\n\n\n\n

Thus f(x) is increasing on interval x \u2208 R<\/p>\n\n\n\n

10. Show that f(x) = x3<\/sup> \u2013 15x2<\/sup> + 75x \u2013 50 is an increasing function for all x \u03f5 R.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given f(x) = x3<\/sup> \u2013 15x2<\/sup> + 75x \u2013 50<\/p>\n\n\n\n

\u21d2
<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

\u21d2 f\u2019(x) = 3x2<\/sup> \u2013 30x + 75<\/p>\n\n\n\n

\u21d2 f\u2019(x) = 3(x2<\/sup> \u2013 10x + 25)<\/p>\n\n\n\n

\u21d2 f\u2019(x) = 3(x \u2013 5)2<\/sup><\/p>\n\n\n\n

Now, as given x \u03f5 R<\/p>\n\n\n\n

\u21d2 (x \u2013 5)2<\/sup> > 0<\/p>\n\n\n\n

\u21d2 3(x \u2013 5)2<\/sup> > 0<\/p>\n\n\n\n

\u21d2 f\u2019(x) > 0<\/p>\n\n\n\n

Hence, condition for f(x) to be increasing<\/p>\n\n\n\n

Thus f(x) is increasing on interval x \u2208 R<\/p>\n\n\n\n

11. Show that f(x) = cos2<\/sup> x is a decreasing function on (0, \u03c0\/2).<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given f (x) = cos2<\/sup> x<\/p>\n\n\n\n

\u21d2
<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

\u21d2 f\u2019(x) = 2 cos x (\u2013sin x)<\/p>\n\n\n\n

\u21d2 f\u2019(x) = \u20132 sin (x) cos (x)<\/p>\n\n\n\n

\u21d2 f\u2019(x) = \u2013sin2x<\/p>\n\n\n\n

Now, as given x belongs to (0, \u03c0\/2).<\/p>\n\n\n\n

\u21d2 2x \u2208 (0,
\"https:\/\/gradeup-question-images.grdp.co\/liveData\/PROJ23971\/1543903130748996.png\"\u03c0)<\/p>\n\n\n\n

\u21d2 Sin (2x)> 0<\/p>\n\n\n\n

\u21d2 \u2013Sin (2x) < 0<\/p>\n\n\n\n

\u21d2 f\u2019(x) < 0<\/p>\n\n\n\n

Hence, condition for f(x) to be decreasing<\/p>\n\n\n\n

Thus f(x) is decreasing on interval (0, \u03c0\/2).<\/p>\n\n\n\n

Hence proved<\/p>\n\n\n\n

12. Show that f(x) = sin x is an increasing function on (\u2013\u03c0\/2, \u03c0\/2).<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given f (x) = sin x<\/p>\n\n\n\n

\u21d2
<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

\u21d2 f\u2019(x) = cos x<\/p>\n\n\n\n

Now, as given x \u2208 (\u2013\u03c0\/2, \u03c0\/2).<\/p>\n\n\n\n

That is 4th<\/sup> quadrant, where<\/p>\n\n\n\n

\u21d2 Cos x> 0<\/p>\n\n\n\n

\u21d2 f\u2019(x) > 0<\/p>\n\n\n\n

Hence, condition for f(x) to be increasing<\/p>\n\n\n\n

Thus f(x) is increasing on interval (\u2013\u03c0\/2, \u03c0\/2).<\/p>\n\n\n\n

13. Show that f(x) = cos x is a decreasing function on (0, \u03c0), increasing in (\u2013\u03c0, 0) and neither increasing nor decreasing in (\u2013\u03c0, \u03c0).<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given f(x) = cos x<\/p>\n\n\n\n

\u21d2
<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

\u21d2 f\u2019(x) = \u2013sin x<\/p>\n\n\n\n

Taking different region from 0 to 2\u03c0<\/p>\n\n\n\n

Let x \u2208 (0, \u03c0).<\/p>\n\n\n\n

\u21d2 Sin(x) > 0<\/p>\n\n\n\n

\u21d2 \u2013sin x < 0<\/p>\n\n\n\n

\u21d2 f\u2019(x) < 0<\/p>\n\n\n\n

Thus f(x) is decreasing in (0, \u03c0)<\/p>\n\n\n\n

Let x \u2208 (\u2013\u03c0, o).<\/p>\n\n\n\n

\u21d2 Sin (x) < 0<\/p>\n\n\n\n

\u21d2 \u2013sin x > 0<\/p>\n\n\n\n

\u21d2 f\u2019(x) > 0<\/p>\n\n\n\n

Thus f(x) is increasing in (\u2013\u03c0, 0).<\/p>\n\n\n\n

Therefore, from above condition we find that<\/p>\n\n\n\n

\u21d2 f (x) is decreasing in (0, \u03c0) and increasing in (\u2013\u03c0, 0).<\/p>\n\n\n\n

Hence, condition for f(x) neither increasing nor decreasing in (\u2013\u03c0, \u03c0)<\/p>\n\n\n\n

14. Show that f(x) = tan x is an increasing function on (\u2013\u03c0\/2, \u03c0\/2).<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given f (x) = tan x<\/p>\n\n\n\n

\u21d2
<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

\u21d2 f\u2019(x) = sec2<\/sup>x<\/p>\n\n\n\n

Now, as given<\/p>\n\n\n\n

x \u2208 (\u2013\u03c0\/2, \u03c0\/2).<\/p>\n\n\n\n

That is 4th<\/sup> quadrant, where<\/p>\n\n\n\n

\u21d2 sec2<\/sup>x > 0<\/p>\n\n\n\n

\u21d2 f\u2019(x) > 0<\/p>\n\n\n\n

Hence, Condition for f(x) to be increasing<\/p>\n\n\n\n

Thus f(x) is increasing on interval (\u2013\u03c0\/2, \u03c0\/2).<\/p>\n\n\n\n

15. Show that f(x) = tan\u20131<\/sup> (sin x + cos x) is a decreasing function on the interval (\u03c0\/4, \u03c0 \/2).<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

16. Show that the function f (x) = sin (2x + \u03c0\/4) is decreasing on (3\u03c0\/8, 5\u03c0\/8).<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Thus f (x) is decreasing on the interval (3\u03c0\/8, 5\u03c0\/8).<\/p>\n\n\n\n

17. Show that the function f(x) = cot\u20131<\/sup> (sin x + cos x) is decreasing on (0, \u03c0\/4) and increasing on (\u03c0\/4, \u03c0\/2).<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given f(x) = cot\u20131<\/sup> (sin x + cos x)<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

18. Show that f(x) = (x \u2013 1) ex<\/sup> + 1 is an increasing function for all x > 0.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given f (x) = (x \u2013 1) ex<\/sup> + 1<\/p>\n\n\n\n

Now differentiating the given equation with respect to x, we get<\/p>\n\n\n\n

\u21d2
<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

\u21d2 f\u2019(x) = ex<\/sup> + (x \u2013 1) ex<\/sup><\/p>\n\n\n\n

\u21d2 f\u2019(x) = ex<\/sup>(1+ x \u2013 1)<\/p>\n\n\n\n

\u21d2 f\u2019(x) = x ex<\/sup><\/p>\n\n\n\n

As given x > 0<\/p>\n\n\n\n

\u21d2 ex<\/sup> > 0<\/p>\n\n\n\n

\u21d2 x ex<\/sup> > 0<\/p>\n\n\n\n

\u21d2 f\u2019(x) > 0<\/p>\n\n\n\n

Hence, condition for f(x) to be increasing<\/p>\n\n\n\n

Thus f(x) is increasing on interval x > 0<\/p>\n\n\n\n

19. Show that the function x2<\/sup> \u2013 x + 1 is neither increasing nor decreasing on (0, 1).<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given f(x) = x2<\/sup> \u2013 x + 1<\/p>\n\n\n\n

Now by differentiating the given equation with respect to x, we get<\/p>\n\n\n\n

\u21d2
<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

\u21d2 f\u2019(x) = 2x \u2013 1<\/p>\n\n\n\n

Taking different region from (0, 1)<\/p>\n\n\n\n

Let x \u2208 (0, \u00bd)<\/p>\n\n\n\n

\u21d2 2x \u2013 1 < 0<\/p>\n\n\n\n

\u21d2 f\u2019(x) < 0<\/p>\n\n\n\n

Thus f(x) is decreasing in (0, \u00bd)<\/p>\n\n\n\n

Let x \u2208 (\u00bd, 1)<\/p>\n\n\n\n

\u21d2 2x \u2013 1 > 0<\/p>\n\n\n\n

\u21d2 f\u2019(x) > 0<\/p>\n\n\n\n

Thus f(x) is increasing in (\u00bd, 1)<\/p>\n\n\n\n

Therefore, from above condition we find that<\/p>\n\n\n\n

\u21d2 f (x) is decreasing in (0, \u00bd)  and increasing in (\u00bd, 1)<\/p>\n\n\n\n

Hence, condition for f(x) neither increasing nor decreasing in (0, 1)<\/p>\n\n\n\n

20. Show that f(x) = x9<\/sup> + 4x7<\/sup> + 11 is an increasing function for all x \u03f5 R.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given f (x) = x9<\/sup> + 4x7<\/sup> + 11<\/p>\n\n\n\n

Now by differentiating above equation with respect to x, we get<\/p>\n\n\n\n

\u21d2
<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

\u21d2 f\u2019(x) = 9x8<\/sup> + 28x6<\/sup><\/p>\n\n\n\n

\u21d2 f\u2019(x) = x6<\/sup>(9x2<\/sup> + 28)<\/p>\n\n\n\n

As given x \u03f5 R<\/p>\n\n\n\n

\u21d2 x6<\/sup> > 0 and 9x2<\/sup> + 28 > 0<\/p>\n\n\n\n

\u21d2 x6 <\/sup>(9x2<\/sup> + 28) > 0<\/p>\n\n\n\n

\u21d2 f\u2019(x) > 0<\/p>\n\n\n\n

Hence, condition for f(x) to be increasing<\/p>\n\n\n\n

Thus f(x) is increasing on interval x \u2208 R<\/p>\n\n\n\n

RD Sharma Solutions for Class 12 Maths Chapter 17: Download PDF<\/strong><\/h2>\n\n\n\n

RD Sharma Solutions for Class 12 Maths Chapter 17\u2013Increasing and Decreasing Functions<\/p>\n\n\n\n

Download PDF<\/strong>: RD Sharma Solutions for Class 12 Maths Chapter 17\u2013Increasing and Decreasing Functions PDF<\/a><\/p>\n\n\n\n

Chapterwise RD Sharma Solutions for Class 12 Maths :<\/strong><\/h2>\n\n\n\n