Question

Kshitij invested a certain amount in scheme A which offers simple interest at 16
p.c.p.a. and invested a certain amount in scheme B which offers compound interest
compounded annually) at 20 p.c.p.a He invested in both the schemes for 2 years.If the amount invested in scheme B is 1/8th of the amount invested in scheme A and total interest accrued by him from both the schemes is Rs. 5.250/-. what is
the total amount invested by him in schemes A and B together ?​

Answer 1

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