Question and Answers Forum

All Questions      Topic List

Differentiation Questions

Previous in All Question      Next in All Question      

Previous in Differentiation      Next in Differentiation      

Question Number 39921 by Raj Singh last updated on 13/Jul/18

Answered by tanmay.chaudhury50@gmail.com last updated on 13/Jul/18

iii)max value of sin2x=1  min value of sin2x=−1  s0 max value ofh(x)=1+5=6  min value −1+5=4  i)f(x)=∣x+2∣−1  critical value of x=−2  when x=−2   f(x)=−1  when x>−2   then ∣x+2∣ is +ve and its value  increases as we increase x...so it has no max  value  when x<−2  ∣x+2∣  is −ve  if we decrease  x  keeping x<−2...then value of ∣x+2∣ continue  to decrease...so ∣x+2∣ has no min value

$$\left.{iii}\right){max}\:{value}\:{of}\:{sin}\mathrm{2}{x}=\mathrm{1} \\ $$$${min}\:{value}\:{of}\:{sin}\mathrm{2}{x}=−\mathrm{1} \\ $$$${s}\mathrm{0}\:{max}\:{value}\:{ofh}\left({x}\right)=\mathrm{1}+\mathrm{5}=\mathrm{6} \\ $$$${min}\:{value}\:−\mathrm{1}+\mathrm{5}=\mathrm{4} \\ $$$$\left.{i}\right){f}\left({x}\right)=\mid{x}+\mathrm{2}\mid−\mathrm{1} \\ $$$${critical}\:{value}\:{of}\:{x}=−\mathrm{2} \\ $$$${when}\:{x}=−\mathrm{2}\:\:\:{f}\left({x}\right)=−\mathrm{1} \\ $$$${when}\:{x}>−\mathrm{2}\:\:\:{then}\:\mid{x}+\mathrm{2}\mid\:{is}\:+{ve}\:{and}\:{its}\:{value} \\ $$$${increases}\:{as}\:{we}\:{increase}\:{x}...{so}\:{it}\:{has}\:{no}\:{max} \\ $$$${value} \\ $$$${when}\:{x}<−\mathrm{2}\:\:\mid{x}+\mathrm{2}\mid\:\:{is}\:−{ve}\:\:{if}\:{we}\:{decrease} \\ $$$${x}\:\:{keeping}\:{x}<−\mathrm{2}...{then}\:{value}\:{of}\:\mid{x}+\mathrm{2}\mid\:{continue} \\ $$$${to}\:{decrease}...{so}\:\mid{x}+\mathrm{2}\mid\:{has}\:{no}\:{min}\:{value} \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 13/Jul/18

ii)g(x)=−∣x+1∣+3  critical value of x=−1  at x=−1  g(x)=3  now when x>−1 and we increase x keeping  condition x>−1 value of g(x) decreases  so it has no min value...  when x<−1  we continue decrease the  value of x...then the value of x+1<0  so −∣x+1∣ become +ve...thus the value  of g(x) increases ...thus g(x) has no max value

$$\left.{ii}\right){g}\left({x}\right)=−\mid{x}+\mathrm{1}\mid+\mathrm{3} \\ $$$${critical}\:{value}\:{of}\:{x}=−\mathrm{1} \\ $$$${at}\:{x}=−\mathrm{1}\:\:{g}\left({x}\right)=\mathrm{3} \\ $$$${now}\:{when}\:{x}>−\mathrm{1}\:{and}\:{we}\:{increase}\:{x}\:{keeping} \\ $$$${condition}\:{x}>−\mathrm{1}\:{value}\:{of}\:{g}\left({x}\right)\:{decreases} \\ $$$${so}\:{it}\:{has}\:{no}\:{min}\:{value}... \\ $$$${when}\:{x}<−\mathrm{1}\:\:{we}\:{continue}\:{decrease}\:{the} \\ $$$${value}\:{of}\:{x}...{then}\:{the}\:{value}\:{of}\:{x}+\mathrm{1}<\mathrm{0} \\ $$$${so}\:−\mid{x}+\mathrm{1}\mid\:{become}\:+{ve}...{thus}\:{the}\:{value} \\ $$$${of}\:{g}\left({x}\right)\:{increases}\:...{thus}\:{g}\left({x}\right)\:{has}\:{no}\:{max}\:{value} \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 13/Jul/18

iv)for any value of x,min value of sin4x=−1  max value of sin4x=+1  so min value of f(x)=∣−1+3∣=2  max value f(x)=∣1+3∣=4  if the problem is f(x)=∣sin(4x+3)∣  then min value of sin(4x+3)=−1  thdn max value of sin(4x+3)=1  but value of f(x)=∣−1∣=1  f(x)=∣1∣=1  so f(x) has no min no max value...  so

$$\left.{iv}\right){for}\:{any}\:{value}\:{of}\:{x},{min}\:{value}\:{of}\:{sin}\mathrm{4}{x}=−\mathrm{1} \\ $$$${max}\:{value}\:{of}\:{sin}\mathrm{4}{x}=+\mathrm{1} \\ $$$${so}\:{min}\:{value}\:{of}\:{f}\left({x}\right)=\mid−\mathrm{1}+\mathrm{3}\mid=\mathrm{2} \\ $$$${max}\:{value}\:{f}\left({x}\right)=\mid\mathrm{1}+\mathrm{3}\mid=\mathrm{4} \\ $$$${if}\:{the}\:{problem}\:{is}\:{f}\left({x}\right)=\mid{sin}\left(\mathrm{4}{x}+\mathrm{3}\right)\mid \\ $$$${then}\:{min}\:{value}\:{of}\:{sin}\left(\mathrm{4}{x}+\mathrm{3}\right)=−\mathrm{1} \\ $$$${thdn}\:{max}\:{value}\:{of}\:{sin}\left(\mathrm{4}{x}+\mathrm{3}\right)=\mathrm{1} \\ $$$${but}\:{value}\:{of}\:{f}\left({x}\right)=\mid−\mathrm{1}\mid=\mathrm{1} \\ $$$${f}\left({x}\right)=\mid\mathrm{1}\mid=\mathrm{1} \\ $$$${so}\:{f}\left({x}\right)\:{has}\:{no}\:{min}\:{no}\:{max}\:{value}... \\ $$$${so}\: \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 13/Jul/18

v)h(x)=x+1  as x→1  h(x)→2  x→−1  h(x)→0  so value of h(x)  lies between 0 and 2  h(x)  (0,2)...but the range of h(x) does not  include lowes value 0 highest value 2  thus h(x) has no max or min value...

$$\left.{v}\right){h}\left({x}\right)={x}+\mathrm{1}\:\:{as}\:{x}\rightarrow\mathrm{1}\:\:{h}\left({x}\right)\rightarrow\mathrm{2} \\ $$$${x}\rightarrow−\mathrm{1}\:\:{h}\left({x}\right)\rightarrow\mathrm{0} \\ $$$${so}\:{value}\:{of}\:{h}\left({x}\right)\:\:{lies}\:{between}\:\mathrm{0}\:{and}\:\mathrm{2} \\ $$$${h}\left({x}\right)\:\:\left(\mathrm{0},\mathrm{2}\right)...{but}\:{the}\:{range}\:{of}\:{h}\left({x}\right)\:{does}\:{not} \\ $$$${include}\:{lowes}\:{value}\:\mathrm{0}\:{highest}\:{value}\:\mathrm{2} \\ $$$${thus}\:{h}\left({x}\right)\:{has}\:{no}\:{max}\:{or}\:{min}\:{value}... \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com