The curve y(x)=a x³+b x²+c x+5 touches the x-axis at the point P(-2,0) and cuts the y-axis at the point Q, where y′ is equal to 3 . Then the local maximum value of y(x) is:

Solution

Maths sir · 4 steps

Touching the x-axis at x=-2 gives two equations at once: y=0 and y'=0 there.

1. Read the conditions. Q is on the y-axis, so x=0 there. Since y'=3x²a+2bx+c, we get y'(0)=c=3. Touching the x-axis at P(-2,0) means y(-2)=0 and also y'(-2)=0.
2. Find a and b. y(-2)=-8a+4b-6+5=0 gives -8a+4b=1. y'(-2)=12a-4b+3=0 gives 12a-4b=-3. Adding them, 4a=-2, so a=-1/2 and then b=-3/4.

   y'(x)=-3/2x²-3/2x+3=-3/2(x+2)(x-1)

3. Test the critical points. y''(x)=-3x-3/2. At x=1, y''=-9/2<0, so x=1 is the local maximum. At x=-2, y''=9/2>0, so that is the local minimum, where y=0.
4. So. y(1)=-1/2-3/4+3+5=27/4. The local maximum value is 27/4, option (a).

The idea to keep. A curve that touches the x-axis at a point has a double root there, so both y and y' vanish at that point. One touch gives you two equations.

Answer from the NTA final answer key. Solution by PrepWiser.