Question

Let f(t) = `|(cos"t","t", 1),(2sin"t", "t", 2"t"),(sin"t", "t", "t")|`, then `lim_("t" - 0) ("f"("t"))/"t"^2` is equal to ______.

पर्याय

• 0

• – 1

• 2

• 3

Answer 1

Let f(t) = `|(cos"t","t", 1),(2sin"t", "t", 2"t"),(sin"t", "t", "t")|`, then `lim_("t" - 0) ("f"("t"))/"t"^2` is equal to 0.

Explanation:

We have f(t) = `|(cos"t","t", 1),(2sin"t", "t", 2"t"),(sin"t", "t", "t")|`

Expanding along R₁

= `cos "t"|("t", 2"t"),("t", "t")| - "t"|(2 sin "t", 2"t"),(sin"t", "t")| + 1|(2 sin "t", "t"),(sin"t", "t")|`

= cos t(t² – 2t) – t(2t sint – 2t sin t) + (2t sin t – t sin t)

= –t² cos t + t sin t

∴ `("f"("t"))/"t"^2 = ("t"^2 cos"t" + "t" sin"t")/"t"^2`

⇒ `("f"("t"))/"t"^2 = - cos "t" + (sin "t")/"t"`

⇒ `lim_("t" -> 0) ("f"("t"))/"t"^2 =  lim_("t" -> 0) (- cos "t")  + lim_("t" -> 0)  (sin "t")/"t"`

= – 1 + 1

= 0