If f(x)=mx2−6x−3 and f′(1)=12, find the value of the constant m.
9
3
-3
-4
Correct answer is A
f(x)=mx2−6x−3
f′(x)=dydx=2mx−6
f′(1)=2m−6=12
2m=18⟹m=9
4
3
2
0
Correct answer is A
\lim \limits_{x \to 3} \frac{x^{2} - 2x - 3}{x - 3}
\frac{x^{2} - 2x - 3}{x - 3} = \frac{x^{2} - 3x + x - 3}{x - 3}
\frac{(x - 3)(x + 1)}{x - 3} = x + 1
\lim \limits_{x \to 3} \frac{x^{2} - 2x - 3}{x - 3} \equiv \lim \limits_{x \to 3} (x + 1) (L'Hopital rule)
\lim \limits_{x \to 3} (x + 1) = 3 + 1 = 4
Calculate in surd form, the value of \tan 15°.
2 + \sqrt{3}
1 + \sqrt{3}
\sqrt{3} - 1
2 - \sqrt{3}
Correct answer is D
\tan 15 = \tan (60 - 45)
\tan (x - y) = \frac{\tan x - \tan y}{1 + \tan x \tan y}
\tan (60 - 45) = \frac{\tan 60 - \tan 45}{1 + \tan 60 \tan 45}
= \frac{\sqrt{3} - 1}{1 + (\sqrt{3} \times 1)}
= \frac{\sqrt{3} - 1}{1 + \sqrt{3}}
Rationalizing by multiplying denominator and numerator by 1 - \sqrt{3},
\tan 15 = 2 - \sqrt{3}