Question

Consider the quadratic equation $(c-5){{x}^{2}}-2cx+(c-4)=0,c\ne 5$. Let S be the set of all integral values of c for which one root of the equation lies in the interval ( 0, 2 ) and it’s another root lies in the interval ( 2, 3 ). Then the number of elements in S is,
( a ) 11
( b ) 18
( c ) 10
( d ) 12

Answer 1

Hint: To solve this question we will use the concept, which is stated as, if we have an open interval ( a, b ) and if f ( a )f ( b ) < 0, for real-valued function f(x), then there exist only one root in the interval ( a, b ). Using this concept, we will solve the inequalities we got. And, hence we will get the cardinality of set S.

Complete step-by-step solution:
Now, it is given that there is a quadratic equation $(c-5){{x}^{2}}-2cx+(c-4)=0,c\ne 5$and also there be any set, say S which is set of all integral values for which one root of the equation lies in the interval ( 0, 2 ) and it’s another root lies in the interval ( 2, 3 ), then we are asked to find the number of integral values in set S.
Now, it is given that one root of the equation lies in the interval ( 0, 2 ).
Remember, if we have open interval ( a, b ) and if f ( a )f ( b ) < 0, for real valued function f(x), then there exist only one root in interval ( a, b ).
So, for ( 0, 2 ), there is one root of equation $(c-5){{x}^{2}}-2cx+(c-4)=0,c\ne 5$, then
f ( 0 )f ( 2 ) < 0
now, at x = 0,
$(c-5){{(0)}^{2}}-2c(0)+(c-4)=0$
( c – 4 ) = 0……( i )
At x = 2,
$(c-5){{(2)}^{2}}-2c(2)+(c-4)=0$
( c – 24 ) = 0……( ii )
From ( i ) and ( ii ), we get
( c – 4 )( c – 24 ) < 0
Similarly for, ( 2, 3 ), there is one root of equation $(c-5){{x}^{2}}-2cx+(c-4)=0,c\ne 5$, then
f ( 2 )f ( 3 ) < 0
now, at x = 2,
$(c-5){{(2)}^{2}}-2c(2)+(c-4)=0$
( c – 4 ) = 0……( iii )
At x = 2,
$(c-5){{(3)}^{2}}-2c(3)+(c-4)=0$
( 4c – 49 ) = 0……( iv )
Form, ( iii ) and ( iv ), we get
( c – 4 )( 4c – 49 ) < 0
So, number of integral values of c will be lie between ( c – 4 )( c – 24 ) < 0 and ( c – 4 )( 4c – 49 ) < 0.
For, ( 4, 24 ), we get, f ( 0 )f ( 2 ) < 0
And for, $\left( \dfrac{49}{4},24 \right)$ , we get, f ( 2 )f ( 3 ) < 0
As, $(4,24)\subset \left( \dfrac{49}{4},24 \right)$
So, $\dfrac{49}{4}< c <24$
Thus, set S containing values of c will be,
S = { 13, 14, 15, 16, 17, 18, 18, 19, 20, 21, 22, 23 }
So, the cardinality of Set S is 11.
Hence, option ( a ) true.

Note: To solve questions based on the root, the concept of the single root in the open interval which is for real-valued function f(x), if we have an open interval ( a, b ) and if f ( a )f ( b ) < 0, then there exist only one root in the interval ( a, b ). Try not to make any calculation error as it will make the solution go wrong.