Free Continuity and Differentiability 03 Practice Test - 12th Grade - Commerce

Question 1

$I f f (x) = (l o g_{c o t x} t a n x) (l o g_{t a n x} c o t x)^{- 1} + t a n^{- 1} (\frac{x}{\sqrt{(4 - x^{2})}})$ 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 2

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

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 3

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

- 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 4

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

\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 5

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

\frac{2}{a} e^{- π / 2}

a e^{π / 2}

- \frac{2}{a} e^{- π / 2}

D. Does not exist

SOLUTION

Solution : C

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

Question 6

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 7

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 8

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 9

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 10

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

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

Free Continuity and Differentiability 03 Practice Test - 12th Grade - Commerce

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

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