English, asked by amrithaadithya44691, 1 month ago

In the aqueous sol containing at+ cu2+ and ti4+ which will respond to magnetic field and why

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Answered by ᎷᎪᎠᎪᎡᎪ
16

Answer:

f′(x)f′(x) gives you the slope of ff in x

f′(x)f′(x) gives you the slope of ff in xQuite easily, if f′(x)f′(x) is positive, f(x)f(x) increases. If f′(x)f′(x) is negative, f(x)f(x) decreases.

f′(x)f′(x) gives you the slope of ff in xQuite easily, if f′(x)f′(x) is positive, f(x)f(x) increases. If f′(x)f′(x) is negative, f(x)f(x) decreases.We know that, for y∈R∗+y∈R∗+

f′(x)f′(x) gives you the slope of ff in xQuite easily, if f′(x)f′(x) is positive, f(x)f(x) increases. If f′(x)f′(x) is negative, f(x)f(x) decreases.We know that, for y∈R∗+y∈R∗+0<y<1⇔ln(y)<00<y<1⇔ln(y)<0

f′(x)f′(x) gives you the slope of ff in xQuite easily, if f′(x)f′(x) is positive, f(x)f(x) increases. If f′(x)f′(x) is negative, f(x)f(x) decreases.We know that, for y∈R∗+y∈R∗+0<y<1⇔ln(y)<00<y<1⇔ln(y)<0ln(1)=0ln(1)=0

f′(x)f′(x) gives you the slope of ff in xQuite easily, if f′(x)f′(x) is positive, f(x)f(x) increases. If f′(x)f′(x) is negative, f(x)f(x) decreases.We know that, for y∈R∗+y∈R∗+0<y<1⇔ln(y)<00<y<1⇔ln(y)<0ln(1)=0ln(1)=01<y⇔ln(y)>01<y⇔ln(y)>0

f′(x)f′(x) gives you the slope of ff in xQuite easily, if f′(x)f′(x) is positive, f(x)f(x) increases. If f′(x)f′(x) is negative, f(x)f(x) decreases.We know that, for y∈R∗+y∈R∗+0<y<1⇔ln(y)<00<y<1⇔ln(y)<0ln(1)=0ln(1)=01<y⇔ln(y)>01<y⇔ln(y)>0So we can write that

f′(x)f′(x) gives you the slope of ff in xQuite easily, if f′(x)f′(x) is positive, f(x)f(x) increases. If f′(x)f′(x) is negative, f(x)f(x) decreases.We know that, for y∈R∗+y∈R∗+0<y<1⇔ln(y)<00<y<1⇔ln(y)<0ln(1)=0ln(1)=01<y⇔ln(y)>01<y⇔ln(y)>0So we can write thatf′(x)>0⇔ln(x2x+1)>0⇔x2x+1>1f′(x)>0⇔ln(x2x+1)>0⇔x2x+1>1

f′(x)f′(x) gives you the slope of ff in xQuite easily, if f′(x)f′(x) is positive, f(x)f(x) increases. If f′(x)f′(x) is negative, f(x)f(x) decreases.We know that, for y∈R∗+y∈R∗+0<y<1⇔ln(y)<00<y<1⇔ln(y)<0ln(1)=0ln(1)=01<y⇔ln(y)>01<y⇔ln(y)>0So we can write thatf′(x)>0⇔ln(x2x+1)>0⇔x2x+1>1f′(x)>0⇔ln(x2x+1)>0⇔x2x+1>1f′(x)<0⇔ln(x2x+1)<0⇔x2x+1<1f′(x)<0⇔ln(x2x+1)<0⇔x2x+1<1

f′(x)f′(x) gives you the slope of ff in xQuite easily, if f′(x)f′(x) is positive, f(x)f(x) increases. If f′(x)f′(x) is negative, f(x)f(x) decreases.We know that, for y∈R∗+y∈R∗+0<y<1⇔ln(y)<00<y<1⇔ln(y)<0ln(1)=0ln(1)=01<y⇔ln(y)>01<y⇔ln(y)>0So we can write thatf′(x)>0⇔ln(x2x+1)>0⇔x2x+1>1f′(x)>0⇔ln(x2x+1)>0⇔x2x+1>1f′(x)<0⇔ln(x2x+1)<0⇔x2x+1<1f′(x)<0⇔ln(x2x+1)<0⇔x2x+1<1If x<−1

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