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Question : 12 of 160
Marks:
+1,
-0
Solution:
Let
y=‌| 9⋅32x+6⋅3x+4 |
| 9⋅32x−6⋅3x+4 |
Put,
3x=t ∴ y=‌ ⇒y(9t2−6t+4)=9t2+6t+4 ⇒9t2(y−1)−6t(y+1)+(4y−4)=0 ⇒‌‌(y−1)t2‌(y+1)t+‌(y−1)=0 ⇒(y−1)t2−‌(y+1)t+‌(y−1)=0 ∵t is real
∴D≥0 ⇒‌(y+1)2−4(y−1)‌(y−1)≥0 ⇒‌‌‌(y+1)2−‌(y−1)2≥0 ⇒‌‌(y+1)2−4(y−1)2≥0 ⇒‌‌y2+1+2y−4y2−4+8y≥0 ⇒‌‌−3y2+10y−3≥0 ⇒‌‌3y2−10y+3≤0 ⇒(y−3)(3y−1)≤0⇒‌≤y≤3 Hence, minimum value
=‌.
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