Question 13
Are the following statements ‘True’ or ‘False’? Justify your answer.
(i) If the zeroes of a quadratic polynomial ax2+bx+c are both positive, then a,b and c all have the same sign.
(ii) If the graph of a polynomial intersects the X-axis at only one point, it cannot be a quadratic polynomial.
(iii) If the graph of a polynomial intersects the X-axis at exactly two points, it need not ve a quadratic polynomial.
(iv) If two of the zeroes of a cubic polynomial are zero, then it does not have linear and constant terms.
(v) If all the zeroes of a cubic polynomial are negative, then all the coefficients and the constant term of the polynomial have the same sign.
(vi) If all three zeroes of a cubic polynomial x3+ax2−bx+c are positive, then atleast one of a,b and c is non-negative.
(vii) The only value of k for which the quadratic polynomial kx2+x+k has equal zeroes is 1/2.
Solution:
(i) False, if the zeroes of a quadratic polynomial ax2+bx+c are both positive, then,
α+β=−b/a and α.β=c/a
where α and β are the zeroes of quadratic polynomial.
∴c<0 a<0 and b>0
or c>0 a>0 and b<0
(ii) True, if the graph of a polynomial intersects the X-axis at only one point, then it cannot be a quadratic polynomial because a quadratic polynomial may touch the X-axis at exactly one point or intersects X-axis at exactly two points or do not touch then X-axis.
(iii) True, if the graph of a polynomial intersects the X-axis at exactly two points, then it may or may not be a quadratic polynomial. As, a polynomial of degree more than z is possible which intersects the X-axis at exactly two points when it has two real roots and other imaginary roots.
(iv) True, let α,β and γ be the zeroes of the cubic polynomial and given that two of the zeroes have value 0.
Let β=γ=0
and f(x)=(x−α)(x−β)(x−γ)
=(x−α)(x−0)(x−0)
=x3−ax2
which does not have linear and constant terms.
(v) True, if f(x)=ax3+bx2+cx+d. Then, for all negative roots, a,b,c and d must have same sign.
(vi) False, let α,β and γ be the three zeroes of cubic polynomial x3+ax2−bx+c.
Then, product of zeroes =(−1)3 [constant term/Coefficient of x2]
⇒αβγ=−(+c)/1
⇒αβγ=−c ….(i)
Given that, all three zeroes are positive. So, the product of all three zeroes is also positive
i.e., αβγ>0
⇒−c>0 [from Eq.(i)]
⇒c<0
Now, sum of the zeroes =α+β+γ=(−1)[Coefficient of x2/Coefficient of x3]
⇒α+β+γ=−a/1=−a
But α,β and γ are all positive.
Thus, its sum is also positive.
So, α+β+γ>0
⇒−a>0
⇒a<0
and sum of the product of two zeroes at a time =(−1)2.[Coefficient of x/Coefficient of x3]=−b/1
⇒αβ+βγ+γα=−b
∵αβ+βγ+αγ>0⇒−b>0
⇒b<0
So, the cubic polynomial x3+ax2−bx+c has all three zeroes which are positive only when all constants a,b and c are negative.
(vii) False, let f(x)=kx2+x+k
For equal roots. Its discriminant should be zero i.e., D=b2−4ac=0
⇒k=±1/2
So, for two values of k, given quadratic polynomial has equal zeroes.