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Question : 67 of 160
Marks:
+1,
-0
Solution:
Given, equation
y=x3−ax2+48x+7 Differentiating the above equation with respect to the variable
x, we have
‌=3x2−2ax+48 Thus, above function is a quadratic function.
So,
D<0.In
y′=3x2−2ax+48,A=3,B=−2a and
c=48∴‌‌D<0‌⇒‌‌(−2a)2−4⋅3⋅48<0⇒4a2−4⋅3⋅48<0
‌⇒‌‌4(a2−144)<0⇒a2−(12)2<0‌⇒‌‌(a−12)(a+12)<0‌⇒‌‌a∈(−12,12)
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