Question
The number of integral values of a for which (5a−1)<(a+1)2<7a−3 ___.
Answer:
:
First (5a−1)<(a+1)2⇒a2+1+2a>5a−1⇒(a−2)(a−1)>0⇒a<1;a>2
Second (a+1)2<7a−3⇒(a−4)(a−1)<0⇒1<a<4
So in this range its only a = 3 which satisfies the inequality. Therefore, no.of values of a satisfying the inequality is 1.
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:
First (5a−1)<(a+1)2⇒a2+1+2a>5a−1⇒(a−2)(a−1)>0⇒a<1;a>2
Second (a+1)2<7a−3⇒(a−4)(a−1)<0⇒1<a<4
So in this range its only a = 3 which satisfies the inequality. Therefore, no.of values of a satisfying the inequality is 1.
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