Free Objective Test 02 Practice Test - 11th and 12th

Question 1

If f(x) = x + tan x and f is inverse of g, then g’(x) is equal to

\frac{1}{1 + (g (x) - x)^{2}}

\frac{1}{1 - (g (x) - x)^{2}}

\frac{1}{2 + (g (x) - x)^{2}}

\frac{1}{2 - (g (x) - x)^{2}}

SOLUTION

Solution : C

$L e t y = f (x) \Rightarrow x = f^{- 1} (y)$
$t h e n f (x) = x + t a n x$
$\Rightarrow x = f^{- 1} (y) + t a n (f^{- 1} (y))$
$\Rightarrow x = g (y) + t a n (g (y)) o r x = g (x) + t a n (g (x)) \dots (i)$
Differentiating both sides, then we get
$1 = g^{'} (x) + s e c 2 g (x) . g^{'} (x)$
$∴ g^{'} (x) = \frac{1}{1 + s e c^{2} (g (x))} = \frac{1}{1 + 1 + t a n^{2} (g (x))}$
$= \frac{1}{2 + (x - g (x))^{2}} [f r o m E q . (i)]$
$\frac{1}{2 + (g (x) - x)^{2}}$

Question 2

If $x = s e c θ - c o s θ, y = s e c^{10} θ - c o s^{10} θ$ and $(x^{2} + 4) {(\frac{d y}{d x})}^{2} = k (y^{2} + 4)$ , then k is equal to

\frac{1}{100}

1

10

100

SOLUTION

Solution : D

$∵ x^{2} + 4 = (s e c θ - c o s θ)^{2} + 4 = (s e c θ + c o s θ)^{2} \dots (i) S i m i l a r l y, y^{2} + 4 = (s e c^{10} θ + c o s^{10} θ)^{2} \dots (i i) N o w, \frac{d x}{d θ} = s e c θ t a n θ + s i n θ = t a n θ (s e c θ + c o s θ) a n d \frac{d y}{d θ} = 10 s e c^{9} θ s e c θ t a n θ - 10 c o s^{9} θ (- s i n θ) = 10 t a n θ (s e c^{10} θ - c o s^{10} θ) \Rightarrow \frac{d y}{d x} = \frac{(\frac{d y}{d θ})}{(\frac{d x}{d θ})} = \frac{10 t a n θ (s e c^{10} θ + c o s^{10} θ)}{t a n θ (s e c θ + c o s θ)} ∴ {(\frac{d y}{d x})}^{2} = 100 \frac{(s e c^{10} θ + c o s^{10} θ)}{(s e c θ + c o s θ)^{2}} = \frac{100 (y^{2} + 4)}{(x^{2} + 4)} o r (x^{2} + 4) {(\frac{d y}{d x})}^{2} = 100 (y^{2} + 4)$
On comparing with the expression given we get k = 100

Question 3

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

{(\frac{d y}{d x} + x)}^{2}

{(\frac{d y}{d x} - y)}^{2}

{(x \frac{d y}{d x} + y)}^{2}

{(x \frac{d y}{d x} - y)}^{2}

SOLUTION

Solution : D

$∵ y = l n {(\frac{x}{a + b x})}^{x} = x (l n x - l n (a + b x))$
$o r (\frac{y}{x}) = l n x - l n (a + b x)$
Differentiating both sides w.r.t.x,then
$\frac{x \frac{d y}{d x} - y .1}{x^{2}} = \frac{1}{x} - \frac{b}{a + b x} = \frac{a}{x (a + b x)} \dots (i)$
$o r (x \frac{d y}{d x} - y) = (\frac{a x}{a + b x})$
Again taking logarithm on both sides, then
$l n (x \frac{d y}{d x} - y) = l n (a x) - l n (a + b x)$
Differetiating both sides w.r.t.x, then
$\frac{x \frac{d^{2} y}{d x^{2}} + \frac{d y}{d x} - \frac{d y}{d x}}{(x \frac{d y}{d x} - y)} = \frac{1}{x} - \frac{b}{a + b x} = \frac{a}{x (a + b x)} = \frac{(x \frac{d y}{d x} - y)}{x^{2}} [F r o m E q . (i)] o r x^{3} \frac{d^{2} y}{d x^{2}} = {(x \frac{d y}{d x} - y)}^{2}$

Question 4

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

{(\frac{d^{2} y}{d x^{2}})}^{2}

3 \frac{d^{2} y}{d x^{2}}

3 {(\frac{d^{2} y}{d x^{2}})}^{2}

3 \frac{d^{2} x}{d y^{2}}

SOLUTION

Solution : C

$∵ y = (\frac{a x + b}{c x + d}) o r c x y + d y = a x + b$
Differentiating both sides w.r.t.x, then
$c {x \frac{d y}{d x} + y .1} + d \frac{d y}{d x} = a o r x \frac{d y}{d x} + y + (\frac{d}{c}) \frac{d y}{d x} = (\frac{a}{c})$
Again differentiating both sides w.r.t.x, then
$o r x \frac{d^{2} y}{d x^{2}} + \frac{d y}{d x} + \frac{d y}{d x} + (\frac{d}{c}) \frac{d^{2} y}{d x^{2}} = 0 o r x + \frac{2 \frac{d y}{d x}}{(\frac{d^{2} y}{d x^{2}})} + \frac{d}{c} = 0$
Again differentiating both sides w.r.t.x, then
$1 + 2 \frac{(\frac{d^{2} y}{d x^{2}} . \frac{d^{2} y}{d x^{2}} - \frac{d y}{d x} . \frac{d^{3} y}{d x^{3}})}{{(\frac{d^{2} y}{d x^{2}})}^{2}} + 0 = 0 ∴ 2 \frac{d y}{d x} . \frac{d^{3} y}{d x^{3}} = 3 {(\frac{d^{2} y}{d x^{2}})}^{2}$

Question 5

If $x^{y} = e^{x - y}$ then $\frac{d y}{d x} =$

(1 + ln x)^{- 1}

(1 + ln x)^{- 2}

ln x (1 + ln x)^{- 2}

D. None of these

SOLUTION

Solution : C

$S i n c e, x^{y} = e^{x - y} \Rightarrow y ln x = x - y ∴ y = \frac{x}{1 + ln x} ∴ \frac{d y}{d x} = \frac{ln x}{(1 + ln x)^{2}}$

Question 6

If $x = e^{y + e^{y + e^{y + e^{y + \dots \infty}}}}$

\frac{1}{x}

\frac{1 - x}{x}

\frac{x}{1 + x}

N o n e o f t h e s e

SOLUTION

Solution : B

$x = e^{y + x} \Rightarrow l o g_{e} x = y + x$
$∴ \frac{1}{x} = \frac{d y}{d x} + 1$
$\Rightarrow \frac{d y}{d x} = \frac{1 - x}{x}$

Question 7

If $x^{y} . y^{x} = 16, t h e n \frac{d y}{d x} a t (2, 2)$ is

A. - 1

B. 0

C. 1

D. None of these

SOLUTION

Solution : A

$x^{y} . y^{x} = 16$
$∴ l o g_{e} x^{y} + l o g_{e} y^{x} = l o g_{e} 16$
$\Rightarrow y l o g_{e} x + x l o g_{e} y = 4 l o g_{e} 2$
Now, differentiating both sides w.r.t.x
$\frac{y}{x} + l o g_{e} x \frac{d y}{d x} + \frac{x}{y} \frac{d y}{d x} + l o g_{e} y .1 = 0$
$∴ \frac{d y}{d x} = - \frac{(l o g_{e} y + \frac{y}{x})}{(l o g_{e} x + \frac{x}{y})}$
$∴ \frac{d y}{d x} |_{(2, 2)} = - \frac{(l o g_{e} 2 + 1)}{(l o g_{e} 2 + 1)} = - 1$

Question 8

If 5f(x) + 3f $(\frac{1}{x})$ = x + 2 and y = x f(x), then ${(\frac{d y}{d x})}_{x = 1}$ is equal to

14

\frac{7}{8}

1

D. None of these

SOLUTION

Solution : B

$5 f (x) + 3 f (\frac{1}{x}) = x + 2 . . . . . . . . (i)$
$R e p l a c i n g x b y \frac{1}{x}$
$∴ 5 f (\frac{1}{x}) + 3 f (x) = \frac{1}{x} + 2 . . . (i i)$
$F r o m E q . (i) .$
$25 f (x) + 15 f (\frac{1}{x}) = 5 x + 10 . . . . (i i i)$
$a n d f r o m E q . (i i)$
$9 f (x) + 15 f (\frac{1}{x}) = \frac{3}{x} + 6 . . . . . (i v)$
$S u b t r a c t i n g E q . (i v) f r o m (i i i), w e g e t$
$∴ 16 f (x) = 5 x - \frac{3}{x} + 4$
$∴ x f (x) = \frac{5 x^{2} - 3 + 4 x}{16} = y$
$∴ \frac{d y}{d x} = \frac{10 x + 4}{16}$
$∴ \frac{d y}{d x} |_{x = 1} = \frac{10 + 4}{16}$
$= \frac{7}{8}$

Question 9

If $2^{x} + 2^{y} = 2^{x + y}$ then the value of $\frac{d y}{d x} a t x = y = 1 i s$

A. 0

B. - 1

C. 1

D. 2

SOLUTION

Solution : B

$2^{x} + 2^{y} = 2^{x + y}$
$∴ 2^{x} . l o g_{e} 2 + 2^{y} . l o g_{e} 2 \frac{d y}{d x} = 2^{x + y} . l o g_{e} 2 (1 + \frac{d y}{d x})$
$∴ \frac{d y}{d x} = \frac{2^{x + y} - 2^{x}}{2^{y} - 2^{x + y}}$
$\frac{d y}{d x} |_{x = y = 1} = - 1$

Question 10

If f(x) = $s i n^{- 1} (s i n x) + c o s^{- 1} (s i n x) a n d ϕ (x) = f (f (f (x))), t h e n ϕ^{'} (x)$ is equal to

A. 1

B. sin x

C. 0

D. None of these

SOLUTION

Solution : C

$f (x) = \frac{π}{2} (a c o n s t a n t f u n c t i o n)$
$\Rightarrow ϕ^{'} (x) = 0$

Question 11

$I f f (x) = (l o g_{c o t x} t a n x) (l o g_{t a n x} c o t x)^{- 1} + t a n^{- 1} (\frac{x}{\sqrt{(4 - x^{2})}})$ then f'(0) is equal to

- 2

2

\frac{1}{2}

0

SOLUTION

Solution : C

$f (x) = \frac{(l o g_{c o t x} t a n x)}{(l o g_{t a n x} c o t x)} + t a n^{- 1} (\frac{x}{\sqrt{(4 - x^{2})}})$
$f (x) = 1 + t a n^{- 1} (\frac{x}{\sqrt{(4 - x^{2})}})$
Now Put $x = 2 s i n θ$ , we get
$f (x) = 1 + t a n^{- 1} (\frac{2 s i n θ}{\sqrt{4 - 4 s i n^{2} θ}}) \Rightarrow f (x) = 1 + t a n^{- 1} (\frac{2 s i n θ}{2 c o s θ}) \Rightarrow f (x) = 1 + t a n^{- 1} (t a n θ) \Rightarrow f (x) = 1 + θ \Rightarrow f (x) = 1 + s i n^{- 1} (\frac{x}{2}) \Rightarrow f^{'} (x) = 0 + \frac{1}{\sqrt{1 - (\frac{x}{2})^{2}}} \cdot \frac{1}{2} \Rightarrow f^{'} (0) = \frac{1}{2}$

Question 12

$I f x^{2} + y^{2} = t - \frac{1}{t} a n d x^{4} + y^{4} = t^{2} + \frac{1}{t^{2}}, t h e n x^{3} y \frac{d y}{d x} e q u a l s$

A. - 1

B. 0

C. 1

D. None of these

SOLUTION

Solution : C

$x^{2} + y^{2} = t - \frac{1}{t}$
On squaring both the sides
$= x^{4} + y^{4} + 2 x^{2} y^{2}$ = $t^{2} + \frac{1}{t^{2}} - 2$
Given that -
$x^{4} + y^{4} = t^{2} + \frac{1}{t^{2}}$
$∴ x^{2} y^{2} = - 1$
$\Rightarrow x^{2} .2 y \frac{d y}{d x} + y^{2} . 2 x = 0$
$\Rightarrow x^{3} y \frac{d y}{d x} = - x^{2} y^{2} = 1$

Question 13

$I f y = \sqrt{(\frac{1 + c o s 2 θ}{1 - c o s 2 θ})}, \frac{d y}{d θ} a t θ = \frac{3 π}{4} i s$

- 2

2

\pm 2

D. None of these

SOLUTION

Solution : B

$y = \sqrt{(\frac{1 + c o s 2 θ}{1 - c o s 2 θ})}$
$= | c o t θ | = - c o t θ (θ = \frac{3 π}{4})$
$∴ \frac{d y}{d θ} = c o s e c^{2} θ$
$\frac{d y}{d θ} |_{θ = 3 π / 4} = (\sqrt{2})^{2} = 2$

Question 14

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

\frac{2 y - 1}{c o s x}

\frac{c o s x}{2 y - 1}

\frac{2 x - 1}{c o s x}

\frac{c o s x}{2 x - 1}

SOLUTION

Solution : B

$y = \sqrt{s i n x + y} \Rightarrow y^{2} - y = s i n x$
$∴ (2 y - 1) \frac{d y}{d x} = c o s x$

Question 15

$I f \sqrt{(x^{2} + y^{2})} = a . e^{t a n^{- 1 (y / x)}} a > 0, t h e n y " (0) i s e q u a l t o$

\frac{2}{a} e^{- π / 2}

a e^{π / 2}

- \frac{2}{a} e^{- π / 2}

D. Does not exist

SOLUTION

Solution : C

$\sqrt{(x^{2} + y^{2})} = a . e^{t a n^{- 1 (y / x)}}$
$\Rightarrow \frac{1}{2 \sqrt{x^{2} + y^{2}}} (2 x + 2 y y^{'}) = a . e^{t a n^{- 1} (y / x)} \times \frac{1}{1 + \frac{y^{2}}{x^{2}}} \times \frac{x y^{'} - y}{x^{2}} . . . . . (i)$
$\Rightarrow \frac{x + y y^{'}}{\sqrt{x^{2} + y^{2}}} = \sqrt{(x^{2} + y^{2})} \times \frac{x y^{'} - y}{x^{2} + y^{2}}$
$[f r o m E q . (i)]$
$∴ x + y y^{'} = x y^{'} - y \Rightarrow y^{'} = \frac{x + y}{x - y}$
$∴ y "= \frac{2 (x y^{'} - y)}{(x - y)^{2}}$
$y " (0) = \frac{2 (0 - y (0))}{{0 - y (0)}^{2}} = \frac{- 2}{a e^{π / 2}} = \frac{- 2}{a} e^{- π / 2}$

Free Objective Test 02 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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