Question

If for f(x)= k x2 + ux+12, f'(4)=15 and f'(2)=11, then find k and u.​

Answer 1

Answer:

k = 2 and  u = - 1

Step-by-step explanation:

Given :

f ( x ) = k x² + u x + 12

Diff. w.r.t. x

f' ( x ) = 2 k x + u + 0

f' ( x ) = 2 k x + u

Given :

f' ( 4 ) = 15

= > f' ( 4 ) = 8 k + u

= > 15 = 8 k + u

= > u = 15 - 8 k ... ( i )

Also given f' ( 2 ) = 11

= > f' ( 2 ) = 6 k + u

= > u = 11 - 6 k .. ( ii )

From ( i )  and  ( ii )  we get :

15 - 8 k = 11 - 6 k

= > 8 k - 6 k = 15 - 11

= > 2 k = 4

= > k = 2

Putting k = 2 in ( ii )

u = 11 - 6 k .. ( ii )

u = 11 - 12   [ k = 2 ]

u = - 1

Therefore , the value of k and u are 2 and - 1 respectively.

Answer 2

$\huge\bold\green{Question}$

If for f(x)= k x² + ux+12,

f'(4)=15 and f'(2)=11, then find k and u. ?

$\huge\bold\green{Answer}$

According to the question we have :-

→ f ( x ) = k x² + u x + 12

Simply by differentiate with respect to “ x ”

→ f' ( x ) = 2 k x + u + 0

→ f' ( x ) = 2 k x + u

We have second function is :-

→ f' ( 4 ) = 15

→ f' ( 4 ) = 8 k + u

→ 15 = 8 k + u

→ u = 15 - 8 k ______________(1)

And also given that f' ( 2 ) = 11

→ f' ( 2 ) = 6 k + u

u = 11 - 6 k _______________ ( 2 )

From the above eqn ( 1 ) and ( 2 ) by equating , we get

→ 15 - 8 k = 11 - 6 k

→ 8 k - 6 k = 15 - 11

→ 2 k = 4

→ k = 2

By substituting the value of k = 2 in eqn (2) we get

→ u = 11 - 12 {k = 2 }

→ u = - 1