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B
(−∞,0]∪[25,∞)
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C
(0,25]
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D
[0,25]
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Solution
The correct option is C(0,25] f(x)=12x2−6x+7 2x2−6x+7>0,∀x∈R(∵D=36−56=−20<0)
Now, let y=f(x) ⇒y=12x2−6x+7⇒(2y)x2−(6y)x+7y−1=0
for quadratic polynomial y≠0
Now, for x∈R D≥0⇒36y2−4(2y)(7y−1)≥0⇒5y2−2y≤0⇒y∈[0,25]
But y≠0 ∴ Range is (0,25]