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 3 of 6 pages

y = l n {(\frac{x}{a + b x})}^{x}

,then

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

is equal to

(dydx+x)2
(dydx−y)2
(xdydx+y)2
(xdydx−y)2

Discuss Question

Answer: Option D. -> (xdydx−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 22.

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

- 1
0
1
None of these

Discuss Question

Answer: Option C. -> 1
:
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 23. If f(x) = x + tan x and f is inverse of g, then g’(x) is equal to

11+(g(x)−x)2
11−(g(x)−x)2
12+(g(x)−x)2
12−(g(x)−x)2

Discuss Question

Answer: Option C. -> 12+(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 24. 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

1
sin x
0
None of these

Discuss Question

Answer: Option C. -> 0
:
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 25. If

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

1x
1−xx
x1+x
None of these

Discuss Question

Answer: Option B. -> 1−xx
:
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 26.

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

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

Discuss Question

Answer: Option A. ->

\frac{- 1}{2}

:
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 27.

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

$- 1$
$0$
$1$
$\frac{| x |}{x}$

Discuss Question

Answer: Option D. ->

\frac{| x |}{x}

:
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 28.

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

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

Discuss Question

Answer: Option A. ->

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

:
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 29.

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

0
1
2
None of these

Discuss Question

Answer: Option A. -> 0
:
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 30.

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

0
-1
1
None of these

Discuss Question

Answer: Option C. -> 1
:
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

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