Given,
4x2−16x+=0 ⇒
16x2−64x+λ=0...(i)
It is given that α and β are the roots of Eq. (i)
∴α=−(−64)+√(−64)2−4×16×λ |
2×16 |
and
β=−(−64)−√(−64)2−4×16×λ |
2×16 |
⇒
α= and
β=⇒
α= and
β=⇒
α= and
β=Since, 1< α < 2 and 2 < β < 3
∴
1<<2 and
2<<3⇒
4<8+√64−λ<8 and
8<8−√64−λ<12⇒
4−8<√64−λ<8−8 and
8−8<√64−λ<12−8⇒
−4<√64−λ<0 and
0<√64−λ<4On squaring each term, we get
16 < 64 − λ < 0 and 0 < 64 − λ < 16
⇒ 16 − 64 < − λ < 0 − 64 and 0 − 64 < − λ < 16 − 64
⇒ − 48 < − λ < − 64 and − 64 < − λ < − 48
⇒ 48 > λ > 64 and 64 > λ > 48 [∵48 > λ > 64 not possible]
Hence, λ can take 15 values.