{"id":4493,"date":"2016-05-16T11:49:00","date_gmt":"2016-05-16T06:19:00","guid":{"rendered":"http:\/\/mycbseguide.com\/blog\/ncert-solutions-class-12-maths-exercise-5-5\/"},"modified":"2018-05-04T16:36:43","modified_gmt":"2018-05-04T11:06:43","slug":"ncert-solutions-class-12-maths-exercise-5-5","status":"publish","type":"post","link":"https:\/\/mycbseguide.com\/blog\/ncert-solutions-class-12-maths-exercise-5-5\/","title":{"rendered":"NCERT Solutions class 12 Maths Exercise 5.5"},"content":{"rendered":"

NCERT Solutions class 12 Maths Exercise 5.5 Class 12 Maths book solutions are available in PDF format for free download. These ncert book chapter wise questions and answers are very helpful for CBSE board exam. CBSE recommends NCERT books and most of the questions in CBSE exam are asked from NCERT text books. Class 12 Maths chapter wise NCERT solution for Maths part 1 and Maths part 2 for all the chapters can be downloaded from our website and myCBSEguide mobile app for free.<\/p>\n

Download\u00a0NCERT solutions for\u00a0Continuity and Differentiability\u00a0<\/a>as PDF.<\/strong><\/p>\n

\"NCERT<\/p>\n

NCERT Solutions class 12\u00a0Continuity & Differentiability<\/h2>\n

Differentiate the functions with respect to <\/strong><\/strong>\u00a0in Exercise 1 to 5.<\/strong><\/p>\n

1. <\/strong><\/strong><\/p>\n

Ans. <\/strong>Let \u00a0\u00a0\u2026\u2026\u2026.(i)<\/p>\n

Taking logs on both sides, we have<\/p>\n

\u00a0= <\/p>\n

\u00a0<\/p>\n

\u00a0\u00a0<\/p>\n

\u00a0\u00a0<\/p>\n

\u00a0\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0\u00a0[From eq. (i)]\n

\n

2. <\/strong><\/strong><\/p>\n

Ans. <\/strong>Let \u00a0= \u00a0\u00a0\u2026\u2026\u2026.(i)<\/p>\n

Taking logs on both sides, we have<\/p>\n

<\/p>\n

\u00a0\u00a0 <\/p>\n

\u00a0<\/p>\n

\u00a0\u00a0[From eq. (i)]\n

\n

3. <\/strong><\/strong>\u00a0<\/strong><\/p>\n

Ans. <\/strong>Let \u00a0\u2026\u2026\u2026.(i)<\/p>\n

Taking logs on both sides, we have<\/p>\n

<\/p>\n

\u00a0<\/p>\n

\u00a0\u00a0 [By Product rule]\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0\u00a0= <\/p>\n

\n

4. <\/strong><\/strong><\/p>\n

Ans. <\/strong>Let <\/p>\n

Putting \u00a0and <\/p>\n

\u00a0<\/p>\n

\u00a0\u00a0\u2026\u2026\u2026.(i)<\/p>\n

Now, <\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0\u00a0= \u00a0\u2026\u2026\u2026.(ii)<\/p>\n

Again,\u00a0 <\/p>\n

\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 <\/p>\n

\u00a0\u00a0<\/p>\n

\u00a0\u00a0\u2026\u2026\u2026.(iii)<\/p>\n

Putting the values from eq. (ii) and (iii) in eq. (i),<\/p>\n

<\/p>\n

\n

5. <\/strong><\/strong>\u00a0<\/strong><\/p>\n

Ans. <\/strong>Let \u00a0\u2026\u2026.(i)<\/p>\n

Taking logs on both sides, we have<\/p>\n

<\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

[From eq. (i)<\/p>\n

\n

Differentiate the functions with respect to <\/strong><\/strong>\u00a0in Exercise 6 to 11.<\/strong><\/p>\n

6. <\/strong><\/strong><\/p>\n

Ans. <\/strong>Let <\/p>\n

Putting \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0and <\/p>\n

<\/p>\n

\u00a0\u00a0\u2026\u2026\u2026.(i)<\/p>\n

Now <\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

= \u00a0\u2026\u2026\u2026.(ii)<\/p>\n

Again <\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0\u00a0\u2026\u2026\u2026.(iii)<\/p>\n

Putting the values from eq. (ii) and (iii) in eq. (i),<\/p>\n

<\/p>\n

\n

7. <\/strong><\/strong><\/p>\n

Ans. <\/strong>Let \u00a0= \u00a0 where \u00a0and <\/p>\n

\u00a0 \u00a0\u00a0\u2026\u2026\u2026.(i)<\/p>\n

Now <\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0\u00a0\u2026\u2026\u2026.(ii)<\/p>\n

Again <\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0 \u00a0\u2026\u2026\u2026.(iii)<\/p>\n

Putting the values from eq. (ii) and (iii) in eq. (i),<\/p>\n

<\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\n

8. <\/strong><\/strong>\u00a0<\/strong><\/p>\n

Ans. <\/strong>Let \u00a0= \u00a0where \u00a0and <\/p>\n

\u00a0\u00a0\u2026\u2026\u2026.(i)<\/p>\n

Now <\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0\u00a0\u00a0\u2026\u2026\u2026.(ii)<\/p>\n

Again\u00a0 <\/p>\n

\u00a0<\/p>\n

\u00a0\u00a0<\/p>\n

\u00a0<\/p>\n

= <\/p>\n

= \u00a0\u2026\u2026\u2026.(iii)<\/p>\n

Putting the values from eq. (ii) and (iii) in eq. (i),<\/p>\n

<\/p>\n

\n

9. <\/strong><\/strong><\/p>\n

Ans. <\/strong>Let <\/p>\n

Putting \u00a0and , we get\u00a0 <\/p>\n

\u00a0\u00a0 \u2026\u2026\u2026.(i)<\/p>\n

Now <\/p>\n

\u00a0\u00a0= <\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0\u00a0\u2026..(ii)<\/p>\n

Again <\/p>\n

\u00a0\u00a0= <\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0\u00a0\u2026\u2026\u2026.(iii)<\/p>\n

Putting values from eq. (ii) and (iii) in eq. (i),<\/p>\n

<\/p>\n

\n

10. <\/strong><\/strong><\/p>\n

Ans. <\/strong>Let <\/p>\n

Putting \u00a0and , we have <\/p>\n

\u00a0\u00a0 \u2026\u2026\u2026.(i)<\/p>\n

Now\u00a0\u00a0\u00a0 <\/p>\n

\u00a0\u00a0= <\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0\u00a0\u2026\u2026\u2026.(ii)<\/p>\n

Again\u00a0 <\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0\u00a0\u2026\u2026\u2026.(iii)<\/p>\n

Putting the values from eq. (ii) and (iii) in eq. (i),<\/p>\n

<\/p>\n

\n

11. <\/strong><\/strong>\u00a0<\/strong><\/p>\n

Ans. <\/strong>Let <\/p>\n

Putting \u00a0and , we have <\/p>\n

\u00a0\u00a0\u2026\u2026\u2026.(i)<\/p>\n

Now <\/p>\n

\u00a0\u00a0= <\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0\u00a0\u2026\u2026\u2026.(ii)<\/p>\n

Again <\/p>\n

\u00a0\u00a0= <\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0\u00a0\u2026\u2026\u2026.(iii)<\/p>\n

Putting the values from eq. (ii) and (iii) in eq. (i)<\/p>\n

,<\/p>\n

\n

Find <\/strong><\/strong>\u00a0in the following Exercise 12 to 15<\/strong><\/p>\n

12. <\/strong><\/strong><\/p>\n

Ans. <\/strong>Given: <\/p>\n

\u00a0\u00a0where \u00a0and <\/p>\n

\u00a0<\/p>\n

\u00a0\u00a0\u2026\u2026\u2026.(i)<\/p>\n

Now <\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0\u00a0\u2026\u2026\u2026.(ii)<\/p>\n

Again <\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0\u2026\u2026\u2026.(iii)<\/p>\n

Putting values from eq. (ii) and (iii) in eq. (i),<\/p>\n

<\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\n

13. <\/strong><\/strong><\/p>\n

Ans. <\/strong>Given: <\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\n

14. <\/strong><\/strong>\u00a0<\/strong><\/p>\n

Ans. <\/strong>Given: <\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\n

15. <\/strong><\/strong>\u00a0<\/strong><\/p>\n

Ans. <\/strong>Given: <\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0\u00a0<\/p>\n

<\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\n

16. Find the derivative of the function given by <\/strong><\/strong>\u00a0and hence <\/strong><\/strong>\u00a0<\/strong><\/p>\n

Ans. <\/strong>Given:\u00a0\u00a0\u00a0\u00a0 \u00a0\u2026\u2026\u2026.(i)<\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

Putting the value of \u00a0from eq. (i),<\/p>\n

<\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0\u00a0= 8 x 15 = 120<\/p>\n

\n

17. Differentiate <\/strong><\/strong>\u00a0in three ways mentioned below:<\/strong><\/p>\n

(i)\u00a0\u00a0\u00a0 by using product rule.<\/strong><\/p>\n

(ii)\u00a0\u00a0 by expanding the product to obtain a single polynomial<\/strong><\/p>\n

(iii)\u00a0 by logarithmic differentiation.<\/strong><\/p>\n

Do they all give the same answer?<\/strong><\/p>\n

Ans. <\/strong>Let \u00a0\u2026\u2026\u2026.(i)<\/p>\n

(i)<\/strong> <\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0\u00a0\u2026\u2026\u2026.(ii)<\/p>\n

(ii) <\/strong><\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0\u00a0\u2026\u2026\u2026.(iii)<\/p>\n

(iii) <\/strong><\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0\u00a0[From eq. (i)]\n

\u00a0\u00a0\u2026\u2026\u2026.(iv)<\/p>\n

From eq. (ii), (iii) and (iv), we can say that value of \u00a0is same obtained by three different methods.<\/p>\n

\n

18. If <\/strong><\/strong>\u00a0and <\/strong><\/strong>\u00a0are functions of <\/strong><\/strong>\u00a0then show that <\/strong><\/strong>\u00a0in two ways \u2013 first by repeated application of product rule, second by logarithmic differentiation.<\/strong><\/p>\n

Ans. <\/strong>Given: \u00a0and \u00a0are functions of <\/p>\n

To prove: <\/p>\n

(i)<\/strong> By repeated application of product rule:<\/p>\n

L.H.S.\u00a0\u00a0 = <\/p>\n

= <\/p>\n

= <\/p>\n

= <\/p>\n

= <\/p>\n

= <\/p>\n

= R.H.S\u00a0 Hence proved.<\/p>\n

(ii) <\/strong>By Logarithmic differentiation:<\/p>\n

Let <\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

Putting , we get<\/p>\n

<\/p>\n

\u00a0\u00a0Hence proved.<\/p>\n

NCERT Solutions class 12 Maths Exercise 5.5<\/h2>\n

NCERT Solutions Class 12 Maths PDF (Download) Free from myCBSEguide app and myCBSEguide website. Ncert solution class 12 Maths includes text book solutions from both part 1 and part 2. NCERT Solutions for CBSE Class 12 Maths have total 20 chapters. 12 Maths NCERT Solutions in PDF for free Download on our website. Ncert Maths class 12 solutions PDF and Maths ncert class 12 PDF solutions with latest modifications and as per the latest CBSE syllabus are only available in myCBSEguide<\/p>\n

CBSE App for Class 12<\/h2>\n

To download\u00a0NCERT Solutions for class 12 Physics, Chemistry, Biology, History, Political Science, Economics, Geography, Computer Science, Home Science, Accountancy, Business Studies and Home Science; do check myCBSEguide app or website. myCBSEguide provides sample papers with solution, test papers for chapter-wise practice, NCERT solutions, NCERT Exemplar solutions, quick revision notes for ready reference, CBSE guess papers and CBSE important question papers. Sample Paper all are made available through\u00a0the best app for CBSE students<\/strong><\/a>\u00a0and myCBSEguide website.<\/p>\n","protected":false},"excerpt":{"rendered":"

NCERT Solutions class 12 Maths Exercise 5.5 Class 12 Maths book solutions are available in PDF format for free download. These ncert book chapter wise questions and answers are very helpful for CBSE board exam. CBSE recommends NCERT books and most of the questions in CBSE exam are asked from NCERT text books. Class 12 … Read more<\/a><\/p>\n","protected":false},"author":1,"featured_media":-1,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1432,281],"tags":[1486,283,321,216],"class_list":["post-4493","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-mathematics-cbse-class-12","category-ncert-solutions","tag-cbse-12","tag-cbse-study-material","tag-mathematics","tag-ncert-solutions"],"yoast_head":"\nNCERT Solutions class 12 Maths Exercise 5.5 | myCBSEguide<\/title>\n<meta name=\"description\" content=\"NCERT Solutions class 12 Maths Exercise 5.5 in PDF format for free download. 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