Question 2 - If f(x) = px² + qx + r, p ≠ 0 and p + r = q, then one of

Question 2 If 𝑓(𝑥)=𝑝𝑥^(2)+𝑞𝑥+𝑟,𝑝≠0 and 𝑝+𝑟=𝑞, then one of the zeroes of 𝑓(𝑥) is: (A) (q)/(p) (B) (r)/(p) (C) −(r)/(p) (D) −(q)/(p) Given 𝑓(𝑥)=𝑝𝑥^(2)+𝑞𝑥+𝑟 And, 𝑝+𝑟=𝑞 𝑝+𝑟−𝑞=0 𝒑−𝒒+𝒓=𝟎 Comparing it with 𝑓(𝑥) Putting x = –1 in 𝑓(𝑥) will give us the same equation Let’s try 𝑓(−1)=𝑝(−1)^(2) +𝑞(−1)+𝑟 𝒇(−𝟏)=𝒑−𝒒+𝒓 Putting 𝑝−𝑞+𝑟=0 𝒇(−𝟏)=𝟎 Thus, –1 is a zero of 𝒇(𝒙) We need to find the other zero Now, we know that Product of zeroes = (𝒄)/(𝒂) For equation 𝑓(𝑥)=𝑝𝑥^(2)+𝑞𝑥+𝑟 𝐜=𝒓, 𝒂=𝒑 And one zero is –1 Thus, –1 × Second zero = (𝑟)/(𝑝) Second zero = (−𝒓)/(𝒑) So, correct answer is (C) 𝒇(−𝟏)=𝒑−𝒒+𝒓 Putting 𝑝−𝑞+𝑟=0 𝒇(−𝟏)=𝟎 Thus, –1 is a zero of 𝒇(𝒙) We need to find the other zero Now, we know that Product of zeroes = (𝒄)/(𝒂) For equation 𝑓(𝑥)=𝑝𝑥^(2)+𝑞𝑥+𝑟 𝐜=𝒓, 𝒂=𝒑 And one zero is –1 Thus, –1 × Second zero = (𝑟)/(𝑝) Second zero = (−𝒓)/(𝒑) So, correct answer is (C)

Question 2 - CBSE Class 10 Sample Paper for 2027 Boards - Maths Standard - Solutions of Sample Papers for Class 10 Boards

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