Question
The function f(x)=x3 β62x2+ax+9
has a local maximum at x=1. Then the value of a is:Solution
Given the function f(x)=x3 β62x2+ax+9. If f(x) has a local maximum at x=1, then the first derivative fβ²(x) must be zero at x=1, and the second derivative fβ²β²(x) must be negative at x=1. First, find the first derivative of f(x): Since there is a local maximum at x=1, we must have fβ²(1)=0: fβ²(1) = 3(1)2 β124(1) + a = 0 3 β 124 + a = 0 β121 + a = 0 a = 121 Now, we need to check the second derivative to ensure it's a local maximum. Find the second derivative of f(x):
Evaluate the second derivative at x=1: fββ(1) = 6(1)β124 fββ(1) = 6β124 fββ(1) = β118 Since fββ(1) = β118 < 0, there is indeed a local maximum at x=1. Therefore the final answer is option (B), a = 121.
What will come in the place of question mark (?) in the given expression?
{(3600 Γ· 20 of 18) Γ 30 + 240} = ? % of 1200
1.25 Γ 36 + 2.75 Γ 40 = ? Γ 3.1
72% of 486 β 64% of 261 = ?
(49 x ?) / 3.5 + 389 = 627
(25.111 % of 200) × 26 ÷ 12.99 – 18.88 × 15.82 + 150.33% of 3√ 4917 = ? – 200
...72.5% of 400 – 23.25% of 1020 = 105% of ?
(3984 ÷ 24) x (5862 ÷ 40) = ?
154 × 7 + 480 × 5 =?% of 6956
40% of (362 Γ· 0.05) = ?
(392 + 427 + 226 β 325) Γ· (441 + 128 β 425) = ?Β