Free Objective Test 01 Practice Test - 11th and 12th

Question 1

If y = $t a n^{- 1} \sqrt{(\frac{1 + s i n x}{1 - s i n x})}, \frac{π}{2} < x < π$ , then $\frac{d y}{d x}$ equals

\frac{- 1}{2}

- 1

\frac{1}{2}

1

SOLUTION

Solution : A

$y = t a n^{- 1} \sqrt{(\frac{1 - c o s (\frac{π}{2} + x)}{1 + c o s (\frac{π}{2} + x)})} = t a n^{- 1} ∣ ∣ t a n (\frac{π}{4} + \frac{x}{2}) ∣ ∣ \dots) (i)$
$N o w, \frac{π}{2} < x < π$
$∴ \frac{π}{4} < \frac{x}{2} < \frac{π}{2}$
$o r \frac{π}{2} < \frac{π}{4} + \frac{x}{2} < \frac{3 π}{4}$
$∴ ∣ ∣ t a n (\frac{π}{4} + \frac{x}{2}) ∣ ∣ = - t a n (\frac{π}{4} + \frac{x}{2}) (∴ i n s e c o n d q u a d r a n t)$
$= t a n {π - (\frac{π}{4} + \frac{x}{2})}$
$F r o m E q . (i),$
$y = t a n^{- 1} t a n {π - (\frac{π}{4} + \frac{x}{2})}$
$= π - (\frac{π}{4} + \frac{x}{2})$
$= \frac{3 π}{4} - \frac{x}{2}$
$(∵ p r i n c i p a l v a l u e o f t a n^{- 1} x i n - \frac{π}{2} t o \frac{π}{2})$
$∴ \frac{d y}{d x} = - \frac{1}{2}$

Question 2

If f(x) = |x|, then f’(x), where x $\neq$ 0 is equal to

- 1

0

1

\frac{| x |}{x}

SOLUTION

Solution : D

$∵ f (x) = {\begin{matrix} x, x \geq 0 - x, x < 0 \end{matrix} ∵ f^{'} (x) = ⎧ ⎨ ⎩ \begin{matrix} 1, x > 0, i . e, \frac{| x |}{x}, x > 0 - 1, x > 0, i . e, \frac{| x |}{x}, x > 0 \end{matrix} = \frac{| x |}{x}, x \neq 0$

Question 3

If $\sqrt{(1 - x^{2 n})} + \sqrt{(1 - y^{2 n})} = a (x^{n} - y^{n})$ , then $\sqrt{(\frac{1 - x^{2 n}}{1 - y^{2 n}})} \frac{d y}{d x}$ is equal to

\frac{x^{n - 1}}{y^{n - 1}}

\frac{y^{n - 1}}{x^{n - 1}}

\frac{x}{y}

1

SOLUTION

Solution : A

$P u t x^{n} = s i n θ a n d y^{n} = s i n ϕ t h e n, (c o s θ + c o s ϕ) = a (s i n θ - s i n ϕ) \Rightarrow 2 c o s (\frac{θ + ϕ}{2}) c o s (\frac{θ - ϕ}{2}) = 2 a c o s (\frac{θ + ϕ}{2}) s i n (\frac{θ - ϕ}{2}) \Rightarrow c o t (\frac{θ - ϕ}{2}) = a \Rightarrow (\frac{θ - ϕ}{2}) = c o t^{- 1} a \Rightarrow θ - ϕ = 2 c o t^{- 1} a o r s i n^{- 1} x^{n} - s i n^{- 1} y^{n} = 2 c o t^{- 1} a D i f f e r e n t i a t i n g b o t h s i d e s, w e h a v e \frac{n x^{n - 1}}{\sqrt{(1 - x^{2 n})}} - \frac{n y^{n - 1}}{\sqrt{(1 - y^{2 n})}} \frac{d y}{d x} = 0 ∴ \sqrt{(\frac{1 - x^{2 n}}{1 - y^{2 n}})} \frac{d y}{d x} = \frac{x^{n - 1}}{y^{n - 1}}$

Question 4

If $\frac{d^{2} x}{d y^{2}} {(\frac{d y}{d x})}^{3} + \frac{d^{2} y}{d x^{2}} = k$ , then k is equal to

A. 0

B. 1

C. 2

D. None of these

SOLUTION

Solution : A

$\frac{d y}{d x} = {(\frac{d x}{d y})}^{- 1} \Rightarrow \frac{d^{2} y}{d x^{2}} = - {(\frac{d x}{d y})}^{- 2} {\frac{d}{d x} (\frac{d x}{d y})} \Rightarrow \frac{d^{2} y}{d x^{2}} = - {(\frac{d x}{d y})}^{- 2} {\frac{d}{d y} (\frac{d x}{d y}) \frac{d y}{d x}} \Rightarrow \frac{d^{2} y}{d x^{2}} = - {(\frac{d y}{d x})}^{2} {\frac{d^{2} x}{d y^{2}} . \frac{d y}{d x}} \Rightarrow \frac{d^{2} y}{d x^{2}} = - {(\frac{d y}{d x})}^{3} \frac{d^{2} x}{d y^{2}} \Rightarrow \frac{d^{2} y}{d x^{2}} + {(\frac{d y}{d x})}^{3} \frac{d^{2} x}{d y^{2}} = 0$

Question 5

If $y = (1 + x) (1 + x^{2}) (1 + x^{4}) \dots (1 + x^{2^{n}})$ then $\frac{d y}{d x}$ at x = 0 is

A. 0

B. -1

C. 1

D. None of these

SOLUTION

Solution : C

$S i n c e, y = (1 + x) (1 + x^{2}) (1 + x^{4}) \dots (1 + x^{2^{n}}), (1 - x) y = (1 - x^{2}) (1 + x^{2}) (1 + x^{4}) \dots (1 + x^{2^{n}}) = (1 - x^{4}) (1 + x^{4}) \dots (1 + x^{2^{n + 1}}) ∴ y = \frac{1 - x^{2^{n + 1}}}{(1 - x)} ∵ \frac{d y}{d x} = \frac{(1 - x) (- 2^{n + 1} . x^{2^{n}}) - (1 - x^{2^{n + 1}}) (- 1)}{(1 - x)^{2}} ∴ {\frac{d y}{d x} ∣ ∣}_{x = 0} = \frac{(1) (0) - (1) (- 1)}{(1)^{2}} = 1$

Question 6

The derivative of $s i n^{- 1} (\frac{2 x}{1 + x^{2}})$ with respect to $t a n^{- 1} (\frac{2 x}{1 - x^{2}})$ is

0

1

\frac{1}{1 - x^{2}}

\frac{1}{1 + x^{2}}

SOLUTION

Solution : B

$L e t u = s i n^{- 1} (\frac{2 x}{1 + x^{2}}) = 2 t a n^{- 1} x$
$∴ \frac{d u}{d x} = \frac{2}{1 + x^{2}}$
$a n d v = t a n^{- 1} (\frac{2 x}{1 + x^{2}}) = 2 t a n^{- 1} x$
$∴ \frac{d v}{d x} = \frac{2}{1 + x^{2}}$
$∴ \frac{d u}{d v} = \frac{(\frac{d u}{d x})}{(\frac{d v}{d x})} = 1$

Question 7

Let $f (x) = l o g_{e} {\frac{u (x)}{v (x)}}, u^{'} (2) = 4, v^{'} (2) = 2, u (2) = 2, v (2) = 1$ , then $f^{'} (2)$ is equal to

A. 0

B. 1

C. - 1

D. None of these

SOLUTION

Solution : A

$f (x) = l o g_{e} {\frac{u (x)}{v (x)}}$
$= l o g_{e} u (x) - l o g_{e} v (x)$
$∴ f^{'} (x) = \frac{u^{'} (x)}{u (x)} - \frac{v^{'} (x)}{v (x)}$
$f^{'} (2) = \frac{u^{'} (2)}{u (2)} - \frac{v^{'} (2)}{v (2)}$
$= \frac{4}{2} - \frac{2}{1}$
$= 2 - 2 = 0$

Question 8

If sin y = x sin (a + y) and $\frac{d y}{d x} = \frac{A}{1 + x^{2} - 2 x c o s a^{'}},$ then the value of A is

A. 2

B. cos a

C. sin a

D. None of these

SOLUTION

Solution : C

$x = \frac{s i n y}{s i n (a + y)} . . . . . (i)$
$\Rightarrow \frac{d x}{d y} = \frac{s i n (a)}{s i n^{2} (a + y)}$
$∴ \frac{d y}{d x} = \frac{s i n^{2} (a + y)}{s i n (a)} = \frac{A}{1 + x^{2} - 2 x c o s a}$
$P u t x = 0, y = 0,$
$t h e n A = s i n a$

Question 9

$I f \sqrt{(x + y)} + \sqrt{(y - x)} = a, t h e n \frac{d^{2} y}{d x^{2}} e q u a l s$

\frac{2}{a}

\frac{- 2}{a^{2}}

\frac{2}{a^{2}}

D. None of these

SOLUTION

Solution : C

Given,
$\sqrt{(x + y)} + \sqrt{(y - x)} = a . . . . . . . . . . (i)$
$∴ \sqrt{x + y} - \sqrt{y - x} = \frac{2 x}{a} . . . . . . . (i i)$
$A d d i n g E q s . (i) a n d (i i), t h e n$
$2 \sqrt{x + y} = a + \frac{2 x}{a}$
$S q u a r i n g, 4 x + 4 y = a^{2} + \frac{4 x^{2}}{a^{2}} + 4 x$
$∴ 4 + 4 \frac{d y}{d x} = 0 + \frac{8 x}{a^{2}} + 4$
$∴ 0 + 4 \frac{d^{2} y}{d x^{2}} = \frac{8}{a^{2}}$
$∴ \frac{d^{2} y}{d x^{2}} = \frac{2}{a^{2}}$

Question 10

$I f s i n (x + y) = l o g_{e} (x + y), t h e n \frac{d y}{d x} i s e q u a l t o$

A. 2

B. - 2

C. 1

D. - 1

SOLUTION

Solution : D

$∵ s i n (x + y) = l o g_{e} (x + y),$
$c o s (x + y) (1 + \frac{d y}{d x}) = \frac{1}{(x + y)} (1 + \frac{d y}{d x})$
$\Rightarrow (1 + \frac{d y}{d x}) (c o s (x + y) - \frac{1}{x + y}) = 0$
$∵ c o s (x + y) - \frac{1}{(x + y)} \neq 0$
$∴ 1 + \frac{d y}{d x} = 0$
$∴ \frac{d y}{d x} = - 1$

Question 11

$I f x^{c o s y} + y^{c o s x} = 5.$ Then

A. at x = 0, y = 0, y’ = 0

B. at x = 0, y = 1, y’ = 0

C. at x = y = 1, y’ = – 1

D. at x = 1, y = 0, y’ = 1

SOLUTION

Solution : C

$x^{c o s y} + y^{c o s x} = 5$
$\Rightarrow e^{c o s y l o g_{e} x} + e^{c o s x l o g_{e} y} = 5$
$∴ e^{c o s y l o g_{e} x} {\frac{c o s y}{x} - l o g_{e} x (s i n y) \frac{d y}{d x}} + e^{c o s x l o g_{e} y} {\frac{c o s x}{y} \frac{d y}{d x} - s i n x l o g_{e} y} = 0$
$P u t x = y = 1, (c o s 1 - 0) + (c o s 1 \frac{d y}{d x} - 0) = 0$
$∴ \frac{d y}{d x} = - 1$
$o r y^{'} = - 1$

Question 12

$I f \sqrt{(1 - x^{6})} + \sqrt{(1 - y^{6})} = a (x^{3} - y^{3}) a n d \frac{d y}{d x} = f (x, y) \sqrt{(\frac{1 - y^{6}}{1 - x^{6}})}, t h e n$

f (x, y) = \frac{y}{x}

f (x, y) = \frac{y^{2}}{x^{2}}

f (x, y) = \frac{2 y^{2}}{x^{2}}

f (x, y) = \frac{x^{2}}{y^{2}}

SOLUTION

Solution : D

$P u t x^{3} = s i n θ, y^{3} = s i n ϕ,$
$t h e n c o s θ + c o s ϕ = a (s i n θ - s i n ϕ)$
$\Rightarrow 2 c o s (\frac{θ + ϕ}{2}) c o s (\frac{θ - ϕ}{2})$ $= 2 a c o s (\frac{θ + ϕ}{2}) s i n (\frac{θ - ϕ}{2})$
$\Rightarrow c o t (\frac{θ - ϕ}{2})$ = a
$\Rightarrow (\frac{θ - ϕ}{2}) = c o t^{- 1} a$
$\Rightarrow s i n^{- 1} x^{3} - s i n^{- 1} y^{3} = 2 c o t^{- 1} a$
$∴ \frac{3 x^{2}}{\sqrt{(1 - x^{6})}} - \frac{3 y^{2}}{\sqrt{(1 - y^{6})}} \frac{d y}{d x} = 0$
$\Rightarrow \frac{d y}{d x} = \frac{x^{2}}{y^{2}} \sqrt{(\frac{1 - y^{6}}{1 - x^{6}})}$
$∴ f (x, y) = \frac{x^{2}}{y^{2}}$

Question 13

If x = a $c o s θ, y = b s i n θ, t h e n \frac{d^{3} y}{d x^{3}}$ is equal to

(\frac{- 3 b}{a^{3}}) c o s e c^{4} θ c o t^{4} θ

(\frac{3 b}{a^{3}}) c o s e c^{4} θ c o t θ

(\frac{- 3 b}{a^{3}}) c o s e c^{4} θ c o t θ

D. None of the above

SOLUTION

Solution : C

$∵ x = a c o s θ \Rightarrow \frac{d x}{d θ} = - a s i n θ a n d y = b s i n θ \Rightarrow \frac{d y}{d θ} = b c o s θ ∴ \frac{d y}{d x} = - \frac{b}{a} c o t θ \Rightarrow \frac{d^{2} y}{d x^{2}} = \frac{b}{a} c o s e c^{2} θ \frac{d θ}{d x} = - \frac{b}{a^{2}} c o s e c^{3} θ ∴ \frac{d^{3} y}{d x^{3}} = \frac{3 b}{a^{2}} c o s e c^{2} θ (- c o s e c θ c o t θ) \frac{d θ}{d x} = \frac{3 b}{a^{2}} c o s e c^{3} θ c o t θ (- \frac{1}{a s i n θ}) = - \frac{3 b}{a^{3}} c o s e c^{4} θ c o t θ$

Question 14

If $y = \sqrt{x + \sqrt{y + \sqrt{x + \sqrt{y + . . . . \infty,}}}}$ then $\frac{d y}{d x}$ is equal to

\frac{1}{2 y - 1}

\frac{y^{2} - x}{2 y^{3} - 2 x y - 1}

(2 y - 1)

D. None of these

SOLUTION

Solution : B

$∴ y = \sqrt{x + \sqrt{y + y}}$
$\Rightarrow (y^{2} - x) = \sqrt{2 y}$
$o r (y^{2} - x)^{2} = 2 y$
Differentiating both sides w.r.t. x, then
$2 (y^{2} - x) (2 y \frac{d y}{d x} - 1) = 2 \frac{d y}{d x}$
$∴ \frac{d y}{d x} = \frac{(y^{2} - x)}{2 y^{3} - 2 x y - 1}$

Question 15

The derivative of $c o s^{- 1} (\frac{x^{- 1} - x}{x^{- 1} + x}) a t x = - 1$ is

A. - 2

B. - 1

C. 0

D. 1

SOLUTION

Solution : D

We have, $y = {cos}^{- 1} (\frac{x^{- 1} - x}{x^{- 1} + x}) \Rightarrow y = {cos}^{- 1} (\frac{1 - x^{2}}{1 + x^{2}})$
Now, Put $x = tan θ$
We get $y = {cos}^{- 1} (\frac{1 - {tan}^{2} θ}{1 + {tan}^{2} θ}) \Rightarrow y = {cos}^{- 1} (cos 2 θ) \Rightarrow y = 2 θ \Rightarrow y = 2 {tan}^{- 1} x ∴ \frac{d y}{d x} = \frac{2}{1 + x^{2}} ∴ \frac{d y}{d x} |_{x = - 1} = \frac{2}{1 + (- 1)^{2}} = 2$

Free Objective Test 01 Practice Test - 11th and 12th

Question 1

SOLUTION

Question 2

SOLUTION

Question 3

SOLUTION

Question 4

SOLUTION

Question 5

SOLUTION

Question 6

SOLUTION

Question 7

SOLUTION

Question 8

SOLUTION

Question 9

SOLUTION

Question 10

SOLUTION

Question 11

SOLUTION

Question 12

SOLUTION

Question 13

SOLUTION

Question 14

SOLUTION

Question 15

SOLUTION

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