Test Index

TS EAMCET 19-July-2022 Shift 2 Solved Paper

Section: Mathematics

© examsnet.com

Question : 10 of 160

Marks: +1, -0

Let

f(x)= A x^2+ B x, g(x)= L x^2+ M x+ N

f(2)-g(2)=1

f(3)-g(3)=4, f(4)-g(4)=9

. Then a root of

f(x)-g(x)=0

is

1
$-1$
0
$-2$

Solution:

f(x)=A x^2+B x, g(x)=L x^2+M x+N

f(x)-g(x)=A x^2+B x-L x^2-M x-N

=(A-L) x^2+(B-M) x-N

Let

A-L=P, B-M=Q

f(x)-g(x)=P x^2+Q x-N

\;f(2)-g(2)=1

\;4 P+2 Q-N=1

.....(i)

Also,

\;f(3)-g(3)=4

\;9 P+3 Q-N=4

......(ii)

Also,

f(4)-g(4)=9

16 P+4 Q-N=9

......(iii)

Now, Eq. (i)

\times 4-

Eq. (iii), we get

4 Q-3 N=-5

.......(iv)

Eq. (i)

\times 9-

Eq. (ii)

\times 4

, we get

6 Q-5 N=-7

.......(v)

Eq. (iv)

\times 3-

Eq. (v)

\times 2

, we get

-9 N+10 N=-15+14 \Rightarrow N=-1

From Eq. (iv),

\;4 Q-3(-1)=-5

\;4 Q+3=-5 \Rightarrow 4 Q=-8

\;Q=-2

From Eq. (i),

\;4 P+2(-2)-(-1)=1

\;4 P-4+1=1

\;P=1

\therefore \;f(x)-g(x)=x^2-2 x+1=(x-1)^2

f(x)-g(x)=0 \Rightarrow (x-1)^2=0

\;x=1

\therefore

The root of

f(x)-g(x)=0

is 1 .

© examsnet.com

Go to Question:

Prev Question

Next Question