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Differential Calculus
A poly nomial of degree 3 as shown in figure and Gradient at Q is 3.
on the basis of above in formation...
1-the polynomial f(x) is givem by...
a)-1/2 x^3 -3/4 x^2 +3x +5
b)-1/2 x^3 +3x+5
c)-1/2 x^3- 3/4 x^2 + 5
d)-1/2 x^3+3/4 x^2 -3x+5
2-the value of f(2)...
a)2 b)4 c)6 d)8
3)the no. of solution of |f(|x|)|-3=0 is...
a)2 b)3 c)4 d)5
4)the equation of normal at R is....
a)6x-81y-15=0
b)8x-81y-20=0
c)2x-81y-5=0
d)4x-81y-10=0
5)the equation of tangent...........................?..........
please give solution or hint..
RATES assured.....
Comments (5)
since gradient at Q is 3
hence dy/dx at 5 = 3
also f(-2) = 0
also tangent at -2 is x axis
hence f'(-x) = 0
solving these we have ans of 1
once we have ploynomial tangent and normal can be found out using dy/dx
and for solution of | f(x) - 3 | draw the graph we have 4 solutions ( unable to post it due to slow speed )
point R is ( 5/2 , 0)
hence my answers are
1 a
2 b
3 c
4 b
5 equation of tangent at R ( 5/2,0) = 8y = -81x + 405/2..
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f(x)=ax^3 +bx^2 +cx +d
f(0)=5
=> 0+0+0+d=5
then, ax^3 +bx^2 +cx +5
f'(x)=3ax^2 + 2bx + c
f'(-2)=12a-4b+c =0................................................(1)
here, f(x) passws through (-2, 0)
then , 0=-8a+4b-2c+5............................................(2)
and f'(0)=3...c=3
from eq(1)....12a-4b+3=0.........................(3)
from eq(2).....-8a+4b-1=0...............................(4)
from(3)and (4)
we get..
a=-1/2 , b= -3/4
hence f(x)=-1/2 x3 + -3/4 x2 + 3x + 5