Derivatives : part 2: IITJEE Maths : Tutorial problems for practice

Problem 1:

If f(a)=2, f^{'}(a)=1, g(a)=-1, g^{'}(a)=2, then the value of \lim_{x \rightarrow a}\frac{g(x)f(a)-g(a)f(x)}{x-a} is

(a) -5 (b) \frac{1}{5} (c) 5 (d) 0

Problem 2:

Let y = \arcsin{(\frac{2x}{1+x^{2}})}, 0 < x <1 and 0 < y < \frac{\pi}{2}, then \frac{dy}{dx} is equal to :

(a) \frac{2}{1+x^{2}} (b) \frac{2x}{1+x^{2}} (c) \frac{-2}{1+x^{2}} (d) none

Problem 3:

Let f(x) = ax^{2}+1 for x \leq 1

and f(x)= x+a for x \leq 1 then f is derivable at x=1, if

(a) a=0 (b) a = \frac{1}{2} (c) a=1 (d) a=2

Problem 4:

If f(x) = ax^{2}+b for x \leq 1

if f(x)=b x^{2}+ax+c for x>1, where b \neq 0, then f(x) is continuous and differentiable at x=1, if

(a) c=0, a=2b (b) a=2b, c \in \Re (c) a=b, c=0 (d) a=2b, c \neq 0

Problem 5:

\lim_{h \rightarrow 0} \frac{\cos^{2}(x+h)- \cos^{2}(x)}{h} is equal to

(a) \cos^{2}(x) (b) -\sin{2x} (c) \sin{x} \cos{x} (d) 2\sin{x}

Problem 6:

\lim_{h \rightarrow 0} \frac{\sin{\sqrt{x+h}-\sin{\sqrt{x}}}}{h} is equal to

(a) \cos {\sqrt{x}} (b) \frac{1}{2\sin{\sqrt{x}}} (c) \frac{\cos{\sqrt{x}}}{2\sqrt{x}} (d) \sin{\sqrt{x}}

Problem 7:

(\arccos{x})^{'}= \frac{-1}{\sqrt{1-x^{2}}} where

(a) -1 < x <1 (b) -1 \leq x \leq 1 (c) -1 \leq x < 1 (d) -1 < x \leq 1

Problem 8:

\frac{d}{dx}(\arctan{(\frac{3x-x^{2}}{1-3x^{2}})}) is equal to

(a) \frac{3}{1+x^{2}} (b) \frac{3}{1+9x^{2}} (c) \sec^{2}{x} (d) \frac{1}{9+x^{2}}

Problem 9:

If x=a\cos^{3}(t) and y=a\sin^{3}(t), then \frac{dy}{dx} is equal to

(a) \cos{t} (b) \cot{t} (c) cosec{(t)} (d) -\tan{t}

Problem 10:

If y = arcsin{\cos{x}}, then \frac{dy}{dx} is equal to

(a) -1 (b) \cos{t} (c) cosec{(t)} (d) -\tan{t}

Regards,

Nalin Pithwa

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