1979-47: If y={ x }^{ 2 }-2x-3. Find the least value of y and the corresponding value of x
Solution
y={ x }^{ 2 }-2x-3\\ \frac { dy }{ dx } =2x-2\\ x\quad is\quad least\quad when\frac { dy }{ dx } =0\\ 2x-2=0\\ 2x=2\\ x=1\\ substituting\quad x=1\quad into\quad y={ x }^{ 2 }-2x-3\\ y={ 1 }^{ 2 }-2(1)-3\\ y=1-2-3\\ y=-4\\ \therefore x=1,y=-4
1987-28: The minimum value of y in the equation y={ x }^{ 2 }-6x+8 is?
Solution
y={ x }^{ 2 }-6x+8\\ \frac { dy }{ dx } =2x-6\\ when\frac { dy }{ dx } =0,\\ 2x-6=0\\ 2x=6\\ x=3\\ substituting\quad x=3\quad into\quad y={ x }^{ 2 }-6x+8\\ y=3^{ 2 }-6(3)+8\\ y=9-18+8\\ y=-1
1970-30: At what value of x is the function { x }^{ 2 }+x+1 minimum?
Solution
let\quad y={ x }^{ 2 }+x+1\\ \frac { dy }{ dx } =2x+1\\ when\quad \frac { dy }{ dx } =0,\quad x\quad is\quad minimum\\ 2x+1=0\\ 2x=-1\\ x=\frac { -1 }{ 2 }
1991-27: At what value of x is the function y={ x }^{ 2 }-2x-3 minimum?
Solution
y={ x }^{ 2 }-2x-3\\ \frac { dy }{ dx } =2x-2\\ at\quad minimum,\\ 2x-2=0\\ 2x=2\\ x=1
1992-39: Obtain a maximum value of the function f\left( x \right) ={ x }^{ 3 }-12x+11
Solution
f\left( x \right) ={ x }^{ 3 }-12x+11\\ \frac { df\left( x \right) }{ dx } =3{ x }^{ 2 }-12\\ when\quad \frac { df\left( x \right) }{ dx } =0\\ 3{ x }^{ 2 }-12=o\\ 3{ x }^{ 2 }=12\\ { x }^{ 2 }=4\\ x=\pm 2\\ when\quad x=-2\\ f\left( x \right) ={ (-2) }^{ 3 }-12(-2)+11\\ f\left( x \right) =-8+24+11\\ f\left( x \right) =27\\ however,\quad when\quad x=+2\\ f\left( x \right) ={ (2) }^{ 3 }-12(2)+11\\ f\left( x \right) =8-24+11\\ f\left( x \right) =-5\\ \therefore \quad the\quad maximum\quad value\quad of\quad the\quad function\quad is\quad 27
1994-40: Find the point(x,y) 0n the euclidean plane where the curve y=2{ x }^{ 2 }-2x+3 has 2 as a gradient
Solution
y=2{ x }^{ 2 }-2x+3\\ \frac { dy }{ dx } =4x-2\\ when\quad gradient(\frac { dy }{ dx } )=2,\\ 4x-2=2\\ 4x=4\\ x=1\\ substituting\quad x=1\quad into\quad y=2{ x }^{ 2 }-2x+3\\ y=2{ (1) }^{ 2 }-2(1)+3\\ y=2-2+3\\ y=3\\ \therefore \quad x=1,y=3\\ (x,y)=(1,3)
1999-35: If the maximum value of y=1+hx-3{ x }^{ 2 } is 13, find h
Solution
y=1+hx-3{ x }^{ 2 }\\ \frac { dy }{ dx } =h-6x\\ when\quad \frac { dy }{ dx } =0\\ h-6x=0\\ h=6x\\ substituting\quad h=6x\quad into\quad y=1+hx-3{ x }^{ 2 }\\ y=1+6x(x)-3{ x }^{ 2 }\\ maximum\quad value\quad of\quad y=13\\ \therefore \quad 1+6{ x }^{ 2 }-3{ x }^{ 2 }=13\\ 3{ x }^{ 2 }=12\\ { x }^{ 2 }=4\\ x=\pm 2\\ substitutiong\quad x=\pm 2\quad into\quad h=6x\\ h=6(\pm 2)\\ h=\pm 12
1999-36: Find the value of x for which the function y={ x }^{ 3 }-x has a minimum value
Solution
y={ x }^{ 3 }-x\\ \frac { dy }{ dx } =3{ x }^{ 2 }-1\\ at\quad minimum,\frac { dy }{ dx } =0\\ 3{ x }^{ 2 }-1=0\\ 3{ x }^{ 2 }=1\\ { x }^{ 2 }=\frac { 1 }{ 3 } \\ x=\pm \sqrt { \frac { 1 }{ 3 } } \\ \therefore \quad x=\pm \frac { \sqrt { 3 } }{ 3 } \\ substituting\quad x=+\frac { \sqrt { 3 } }{ 3 } into\quad y={ x }^{ 3 }-x\\ y={ (\frac { \sqrt { 3 } }{ 3 } ) }^{ 3 }-\frac { \sqrt { 3 } }{ 3 } \\ y=0.385\\ substituting\quad x=-\frac { \sqrt { 3 } }{ 3 } into\quad y={ x }^{ 3 }-x\\ y={ (\frac { -\sqrt { 3 } }{ 3 } ) }^{ 3 }-(\frac { -\sqrt { 3 } }{ 3 } )\\ y={ (\frac { -\sqrt { 3 } }{ 3 } ) }^{ 3 }+\frac { \sqrt { 3 } }{ 3 } \\ y=-0.385\\ \therefore \quad at\quad x=\frac { -\sqrt { 3 } }{ 3 } ,\quad the\quad function\quad is\quad minimum
2000-26: Find the minimum value of the function f\left( \theta \right) =\frac { 2 }{ 3-\cos { \theta } } for 0\le \theta \le 2\pi
Solution
f\left( \theta \right) =\frac { 2 }{ 3-\cos { \theta } } \\ let\quad u=2,\frac { dy }{ d\theta } =0\\ let\quad v=3-\cos { \theta ,\quad \frac { dv }{ d\theta } } =\sin { \theta } \\ using\quad quotient\quad rule\\ \frac { dy }{ d\theta } =\frac { v\frac { du }{ d\theta } -u\frac { dv }{ d\theta } }{ { v }^{ 2 } } \\ \frac { dy }{ d\theta } =\frac { (3-cos\theta )0-2(sin\theta ) }{ { (3-cos\theta ) }^{ 2 } } \\ \frac { dy }{ d\theta } =\frac { -2\sin { \theta } }{ 9-3\cos { \theta -3\cos { \theta +{ \cos ^{ 2 }{ \theta } } } } } \\ \frac { dy }{ d\theta } =\frac { -2\sin { \theta } }{ 9-6\cos { \theta -\cos ^{ 2 }{ \theta } } } \\ at\quad minimum,\quad \frac { dy }{ d\theta } =0\\ \frac { -2\sin { \theta } }{ 9-6\cos { \theta -\cos ^{ 2 }{ \theta } } } =0\\ -2\sin { \theta } =0\\ \sin { \theta =0 } \\ \theta =0\\ putting\quad \theta =0\quad into\quad f\left( \theta \right) =\frac { 2 }{ 3-\cos { \theta } } \\ f\left( \theta \right) =\frac { 2 }{ 3-\cos { 0 } } \\ f\left( \theta \right) =\frac { 2 }{ 3-1 } \\ f\left( \theta \right) =\frac { 2 }{ 2 } \\ f\left( \theta \right) =1\\
2000-34: If the volume of a hemisphere is increasing at a steady rate of 18\pi { m }^{ 3 }{ s }^{ -1 }, at what rate is its radius changing when it is 6m?
Solution
v=\frac { 2 }{ 3 } \pi { r }^{ 3 }\\ \frac { dv }{ dr } =2\pi { r }^{ 2 }\\ where\quad dv\quad is\quad change\quad in\quad area,\quad and\quad dr\quad is\quad change\quad in\quad radius\\ dv=18\pi { m }^{ 3 }{ s }^{ -1 },\quad r=6m,\quad dr=?\\ \frac { 18\pi }{ dr } =2\pi { (6) }^{ 2 }\\ \frac { 18\pi }{ dr } =72\pi \\ dr=\frac { 18\pi }{ 72\pi } \\ dr=0.25m/s
2000-37: The expression a{ x }^{ 2 }+bx+c equals 5 at x=1. If its derivative is 2x+1, what are the values of a,b,c respectively?
Solution
let\quad y=a{ x }^{ 2 }+bx+c\\ \frac { dy }{ dx } =2a{ x }+b\\ similarly,\quad \frac { dy }{ dx } =2x+1\\ 2a{ x }=2x\\ \therefore \quad a=1\\ b=1\\ at\quad x=1,y=5\\ 5=a{ (1) }^{ 2 }+b(1)+c\\ a+b+c=5\\ substituting\quad a=1\quad and\quad b=1\quad into\quad a+b+c=5\\ 1+1+c=5\\ c=3\\ \therefore \quad a=1,b=1,c=3
2001-35: Find the ratio of change of the volume of v of a sphere with respect to its radius r, with r=1
Solution
v=\frac { 4 }{ 3 } \pi { r }^{ 3 }\\ \frac { dv }{ dr } =4\pi { r }^{ 2 }\\ when\quad r=1\\ \frac { dv }{ dr } =4\pi { (1) }^{ 2 }\\ \frac { dv }{ dr } =4\pi
2002-17: The slope to the tangent to the curve y=3{ x }^{ 2 }-2x+5 at the point (1,6) is?
Solution
y=3{ x }^{ 2 }-2x+5\\ \frac { dy }{ dx } =6x-2\\ at\quad point\quad (1,6)\\ \frac { dy }{ dx } =6(1)-2\\ \frac { dy }{ dx } =6-2\\ \frac { dy }{ dx } =4
2002-21: A circle with a radius 5cm has its radius increasing at the rate of 0.2cm. What will be the corresponding increase in the area?
Solution
A=\pi { r }^{ 2 }\\ \frac { dA }{ dr } =2\pi r\\ when\quad r=5cm\quad and\quad dr=0.2cm\\ \frac { dA }{ 0.2 } =2\pi (5)\\ dA=0.2\times 10\pi \\ dA=2\pi \\
2003-22: Find the maximum value of y in the equation y=1-2x-3{ x }^{ 2 }
Solution
y=1-2x-3{ x }^{ 2 }\\ \frac { dy }{ dx } =-2-6x\\ when\quad \frac { dy }{ dx } =0\\ -2-6x=0\\ 6x=-2\\ x=\frac { -1 }{ 3 } \\ substituting\quad x=\frac { -1 }{ 3 } \quad into\quad y\\ y=1-2(\frac { -1 }{ 3 } )-3{ (\frac { -1 }{ 3 } ) }^{ 2 }\\ y=1\frac { 1 }{ 3 } \\
2004-33: Determine the maximum value of y=3{ x }^{ 2 }-{ x }^{ 3 }
Solution
y=3{ x }^{ 2 }-{ x }^{ 3 }\\ \frac { dy }{ dx } =6x-3{ x }^{ 2 }\\ when\quad \frac { dy }{ dx } =0\\ 6x-3{ x }^{ 2 }=0\\ 3{ x }^{ 2 }=6x\\ x=2\\ substitute\quad x=2\quad into\quad y\\ y=3{ (2) }^{ 2 }-{ (2) }^{ 3 }\\ y=12-8\\ y=4
2004-38: What is the ratio of change of volume v of a hemisphere with respect to its radius r when r=2?
Solution
v=\frac { 2 }{ 3 } \pi { r }^{ 3 }\\ \frac { dv }{ dr } =2\pi { r }^{ 2 }\\ when\quad r=2\\ \frac { dv }{ dr } =2\pi { (2) }^{ 2 }\\ \frac { dv }{ dr } =8\pi
2005-38: The radius of a circular disc is increasing at the rate of 0.5cm/sec. At what rate is the area of the disc increasing when its radius is 6cm?
Solution
A=\pi { r }^{ 2 }\\ \frac { dA }{ dr } =2\pi r\\ when\quad dr=0.5\quad and\quad r=6cm\\ \frac { dA }{ 0.5 } =2\pi (6)\\ dA=0.5\times 2\pi (6)\\ dA=6\pi { cm }^{ 2 }/sec
2005-39: The maximum value of the function f\left( x \right) =2+x-{ x }^{ 2 } is?
Solution
f\left( x \right) =2+x-{ x }^{ 2 }\\ \frac { df\left( x \right) }{ dx } =1-2x\\ when\quad \frac { df\left( x \right) }{ dx } =0\\ 1-2x=0\\ 2x=1\\ x=\frac { 1 }{ 2 } \\ substitute\quad x=\frac { 1 }{ 2 } \quad into\quad f\left( x \right) \\ f\left( x \right) =2+\frac { 1 }{ 2 } -{ (\frac { 1 }{ 2 } ) }^{ 2 }\\ f\left( x \right) =\frac { 5 }{ 2 } -\frac { 1 }{ 4 } \\ f\left( x \right) =2\frac { 1 }{ 4 }
2007-5: Find the value of x for which the function f\left( x \right) =2{ x }^{ 3 }-{ x }^{ 2 }-4x+4 has a minimum value
Solution
f\left( x \right) =2{ x }^{ 3 }-{ x }^{ 2 }-4x+4\\ \frac { df\left( x \right) }{ dx } =6{ x }^{ 2 }-2x-4\\ when\quad \frac { df\left( x \right) }{ dx } =0\\ 6{ x }^{ 2 }-2x-4=0\\ 6{ x }^{ 2 }-6x+4x-4=0\\ (6{ x }^{ 2 }-6x)+(4x-4)=0\\ 6x(x-1)+4(x-1)=0\\ (x-1)(6x+4)=0\\ x-1=0\quad or\quad 6x+4=0\\ x=1\quad or\quad x=\frac { -2 }{ 3 } \\ when\quad x=1\\ f\left( x \right) =2{ (1) }^{ 3 }-{ 1 }^{ 2 }-4(1)+4\\ f\left( x \right) =1\\ when\quad x=\frac { -2 }{ 3 } \\ f\left( x \right) =2{ (\frac { -2 }{ 3 } ) }^{ 3 }-{ (\frac { -2 }{ 3 } ) }^{ 2 }-4(\frac { -2 }{ 3 } )+4\\ f\left( x \right) =\frac { 152 }{ 27 } =5.62\\ 5.62>1\\ \therefore \quad at\quad x=\frac { -2 }{ 3 } ,\quad the\quad function\quad is\quad maximum\\ \\
2008-38: Find the minimum value of the function y=x(1+x)
Solution
y=x(1+x)\\ y=x+{ x }^{ 2 }\\ \frac { dy }{ dx } =1+2x\\ when\quad \frac { dy }{ dx } =0\\ 1+2x=0\\ 2x=-1\\ x=\frac { -1 }{ 2 } \\ substituting\quad x=\frac { -1 }{ 2 } \quad into\quad y\\ y=\frac { -1 }{ 2 } +{ (\frac { -1 }{ 2 } ) }^{ 2 }\\ y=\frac { -1 }{ 2 } +\frac { 1 }{ 4 } \\ y=\frac { -1 }{ 4 } \\
2009-38: What value of x will make the function x(4-x) a maximum?
Solution
let\quad y=x(4-x)\\ y=4x-{ x }^{ 2 }\\ \frac { dy }{ dx } =4-2x\\ when\quad \frac { dy }{ dx } =0\\ 4-2x=0\\ 2x=4\\ x=2
2009-39: The distance travelled by a particle from a fixed point is given as s=({ t }^{ 3 }-{ t }^{ 2 }-t+5)cm. Find the minimum distance that the particle can cover from the fixed point
Solution
s=({ t }^{ 3 }-{ t }^{ 2 }-t+5)cm\\ \frac { ds }{ dt } =3{ t }^{ 2 }-2t-1\\ when\quad \frac { ds }{ dt } =0\\ 3{ t }^{ 2 }-2t-1=0\\ 3{ t }^{ 2 }-3t+t-1=0\\ (3{ t }^{ 2 }-3t)+(t-1)=0\\ 3t(t-1)+1(t-1)=0\\ (t-1)+(3t+1)=0\\ t-1=0\quad or\quad 3t+1=0\\ t=1\quad or\quad t=\frac { -1 }{ 3 } \\ when\quad t=1\\ s={ 1 }^{ 3 }-{ 1 }^{ 2 }-1+5\\ s=4\\ when\quad t=\frac { -1 }{ 3 } \\ s={ (\frac { -1 }{ 3 } ) }^{ 3 }-{ (\frac { -1 }{ 3 } ) }^{ 2 }-(\frac { -1 }{ 3 } )+5\\ s=\frac { -1 }{ 27 } -\frac { 1 }{ 9 } +\frac { 1 }{ 3 } +5\\ s=5\frac { 5 }{ 27 } \\ since\quad 4<5\frac { 5 }{ 27 } ,\quad 4cm\quad is\quad the\quad minimum\quad distance
2010-42: At what value of x does the function y=-3-2x+{ x }^{ 2 } attain a minimum value?
Solution
y=-3-2x+{ x }^{ 2 }\\ \frac { dy }{ dx } =-2+2x\\ when\frac { dy }{ dx } =0\\ -2+2x=0\\ 2x=2\\ x=1
2011-36: Find the value of x at the minimum of the curve y={ x }^{ 3 }+{ x }^{ 2 }-x+1
Solution
y={ x }^{ 3 }+{ x }^{ 2 }-x+1\\ \frac { dy }{ dx } =3{ x }^{ 2 }+2x-1\\ when\quad \frac { dy }{ dx } =0\\ 3{ x }^{ 2 }+2x-1=0\\ 3{ x }^{ 2 }+3x-x-1=0\\ (3{ x }^{ 2 }+3x)-(x+1)=0\\ 3x(x+1)-1(x+1)=0\\ (x+1)(3x-1)=0\\ x+1=0\quad or\quad 3x-1=0\\ x=-1\quad or\quad x=\frac { 1 }{ 3 } \\ when\quad x=-1\\ y={ (-1) }^{ 3 }+{ (-1) }^{ 2 }-(-1)+1\\ y=-1+1+1+1\\ y=2\\ when\quad x=\frac { 1 }{ 3 } \\ y={ (\frac { 1 }{ 3 } ) }^{ 3 }+{ (\frac { 1 }{ 3 } ) }^{ 2 }-(\frac { 1 }{ 3 } )+1\\ y=\frac { 1 }{ 27 } +\frac { 1 }{ 9 } -\frac { 1 }{ 3 } +1\\ y=1\frac { 13 }{ 27 } \\ since\quad 1\frac { 13 }{ 27 } <2,\\ at\quad minimum\quad point\quad of\quad curve,\quad x=\frac { 1 }{ 3 }
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