Methods Of Differentiation(11th And 12th > Mathematics ) Questions and answers for exam Preparation

11th And 12th > Mathematics

METHODS OF DIFFERENTIATION MCQs

Methods Of Differentiation

Total Questions : 55 | Page 5 of 6 pages

Question 41.

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

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

Discuss Question

Answer: Option C. ->

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

:
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 42.

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

$\frac{1}{100}$
$1$
$10$
$100$

Discuss Question

Answer: Option D. ->

100

:
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 43.

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

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

Discuss Question

Answer: Option D. ->

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

:
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 44.

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

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

Discuss Question

Answer: Option C. ->

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

:
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 45.

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

$(1 + ln x)^{- 1}$
$(1 + ln x)^{- 2}$
$ln x (1 + ln x)^{- 2}$
None of these

Discuss Question

Answer: Option C. ->

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

:
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 46.

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

$\frac{1}{x}$
$\frac{1 - x}{x}$
$\frac{x}{1 + x}$
$N o n e o f t h e s e$

Discuss Question

Answer: Option B. ->

\frac{1 - x}{x}

:
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 47.

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

- 1
0
1
None of these

Discuss Question

Answer: Option A. -> - 1
:
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 48.

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

$14$
$\frac{7}{8}$
$1$
None of these

Discuss Question

Answer: Option B. ->

\frac{7}{8}

:
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 49.

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

Discuss Question

Answer: Option B. -> - 1
:
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 50.

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

$- 2$
$2$
$\frac{1}{2}$
$0$

Discuss Question

Answer: Option C. ->

\frac{1}{2}

:
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}

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