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Question : 2 of 160
Marks:
+1,
-0
Solution:
‌f(x)=‌ ‌∵‌ Range ‌=[‌,3] For range let
f(x)=y y‌=‌ ⇒‌‌x2(y−1)−x(y+1)+ky−k=0 ∴‌D≥0 ⇒‌‌(y+1)2−4k(y−1)(y−1)≥0 ⇒‌‌(y+1)2−(2√k(y−1))2≥0 ⇒‌‌(y+1+2√k(y−1)) ‌‌‌(y+1−2√k(y−1)≥0 ⇒‌‌(y(2√k+1)+1−2√k)‌ ‌‌‌(y(2√k−1)−2√k−1)≤0‌‌‌⋅⋅⋅⋅⋅⋅⋅(i) ∵‌y∈[‌,3] ‌(3y−1)(y−3)≤0‌‌‌⋅⋅⋅⋅⋅⋅⋅(ii) Comparing Eqs. (i) and (ii),
2√k+1=3⇒2√k=2⇒k=1
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