Free Integrals 03 Practice Test - 12th Grade - Commerce

Question 1

If $\int \frac{d x}{x \sqrt{1 - x^{3}}} = a l n (\frac{\sqrt{1 - x^{3}} + b}{- \sqrt{1 - x^{3}} + 1}) + k$ , then

b = 1, a = 1

b = 1, a = - \frac{1}{3}

b = 1, a = - \frac{2}{3}

D. None of these

SOLUTION

Solution : B

$I = \int \frac{d x}{x \sqrt{1 - x^{3}}} P u t 1 - x^{3} = t^{2} 3 x^{2} d x = - 2 t d t I = \frac{- 2}{3} \int \frac{d t}{(1 - t^{2})} = \frac{- 1}{3} \int (\frac{1}{1 - t} + \frac{1}{1 + t}) d t . = \frac{- 1}{3} l o g \frac{1 + t}{1 - t} + c = \frac{- 1}{3} l o g \frac{1 + \sqrt{1 - x^{3}}}{1 - \sqrt{1 - x^{3}}} + c .$

Question 2

$\frac{s i n^{3} x}{(c o s^{4} x + 3 c o s^{2} x + 1) t a n^{- 1} (s e c x + c o s x)} d x =$

t a n^{- 1} (s e c x + c o s x) + c

l o g | t a n^{- 1} (s e c x + c o s x) | + c

\frac{1}{(s e c x + c o s^{2} x)^{2}} + c

l o g | s e c x + c o s x | + c

SOLUTION

Solution : B

Put $t a n^{- 1} (s e c x + c o s x) = f (x) f^{1} (x) = \frac{s i n^{3} x}{c o s^{4} x + 3 c o s^{2} x + 1} ∴ \int \frac{f^{1} (x)}{f (x)} = l o g | f (x) | + c$

Question 3

If $\int \frac{2 c o s x - s i n x + λ}{c o s x + s i n x - 2} d x = A ℓ n | c o s x + s i n x - 2 | + B x + C$ . Then the ordered triplet $A, B, λ$ is

(\frac{1}{2}, \frac{3}{2}, - 1)

(\frac{3}{2}, \frac{1}{2}, - 1)

(\frac{1}{2}, - 1, - \frac{3}{2})

(\frac{3}{2}, - 1, - \frac{1}{2})

SOLUTION

Solution : B

$\frac{d}{d x} (A | c o s x + s i n x - 2 | + B x + C) = A \frac{c o s x - s i n x}{c o s x + s i n x - 2} + B = \frac{A c o s x + B c o s x - A s i n x + B s i n x - 2 B}{c o s x + s i n x - 2} ∴ A + B = 2, A + B = - 1, - 2 B = λ ∴ A = \frac{3}{2}, B = \frac{1}{2}, λ = - 1$

Question 4

$\int \frac{c o s^{4} x d x}{s i n^{3} x (s i n^{5} x + c o s^{5} x)^{\frac{3}{5}}} = - \frac{1}{2} (1 + c o t^{A} x)^{B} + C$ then AB=

A. 1

B. 2

\frac{1}{2}

D. None of these

SOLUTION

Solution : B

$I = \int \frac{c o s x^{4} d x}{s i n^{3} x (s i n^{5} x + c o s^{5} x)^{\frac{3}{5}}} = \int \frac{c o s e c^{2} x c o t^{4} x d x}{(1 + c o t^{5} x)^{3 / 5}} 1 + c o t^{5} x = t . 5 c o t^{4} x (- c o s e c^{2} x) d x = d t = \frac{- 1}{5} \int \frac{d t}{t^{3 / 5}} = \frac{- t^{2 / 5}}{2} = \frac{- 1}{2} (1 + c o t^{5} x)^{2 / 5} + C .$

Question 5

If $f (x) = \sqrt{x}, g (x) = e^{x - 1}, a n d \int f o g (x) d x = A f o g (x) + B t a n^{- 1} (f o g (x)) + C$ , then A + B is equal to

A. 1

B. 2

C. 3

D. 0

SOLUTION

Solution : D

$f o g (x) = \sqrt{e^{x} - 1} ∴ I = \int \sqrt{e^{x} - 1} d x = \int \frac{2 t^{2}}{t^{2} + 1} d t {w h e r e t = \sqrt{e^{x} - 1}} = 2 t - 2 t a n^{- 1} t + C = 2 \sqrt{e^{x} - 1} - 2 t a n^{- 1} (\sqrt{e^{x} - 1}) + C = 2 f o g (x) - 2 t a n^{- 1} (f o g (x)) + C ∴ A + B = 2 + (- 2) = 0$

Question 6

If $\int \frac{c o s 8 x + 1}{t a n 2 x - c o t 2 x} d x = a c o s 8 x + C$ , then

a = \frac{- 1}{16}

a = \frac{1}{8}

a = \frac{1}{16}

a = \frac{- 1}{8}

SOLUTION

Solution : C

$\int \frac{c o s 8 x + 1}{t a n 2 x - c o t 2 x} d x = \int \frac{2 c o s^{2} 4 x}{s i n^{2} 2 x - c o s^{2} 2 x} . s i n 2 x c o s 2 x d x = - \int s i n 4 x c o s 4 x d x = - \frac{1}{2} \int s i n 8 x d x = \frac{1}{16} c o s 8 x + C ∴ a = \frac{1}{16}$

Question 7

The anti – derivative of the function (3x + 4) |sin x|, when $0 < x < π$ , is given by

3 s i n x - (3 x + 4) c o s x + c

3 s i n x + (3 x + 4) c o s x + c

- 3 s i n x - (3 x + 4) c o s x + c

D. None of these

SOLUTION

Solution : A

In the interval $(0, π)$ , sin x is positive, therefore, $(3 x + 4) | s i n x | = (3 x + 4) s i n x$ .
$∴$ The antiderivative of $(3 x + 4) | s i n x |$ is
$= \int (3 x + 4) s i n x d x = - (3 x + 4) c o s x + \int 3 c o s x d x = - (3 x + 4) c o s x + 3 s i n x + c$

Question 8

If $Φ (x) = \int \frac{d x}{s i n^{\frac{1}{2}} x c o s^{\frac{7}{2}} x}$ , then $Φ (\frac{π}{4}) - Φ (0) =$

\frac{12}{5}

\frac{9}{5}

\frac{6}{5}

D. 0

SOLUTION

Solution : A

$t a n x = t \Rightarrow s e c^{2} x d x = d t ∴ f (x) = \int \frac{d x}{\frac{s i n^{\frac{1}{2}} x}{c o s^{\frac{1}{2}} x} . c o s^{4} x} = \int \frac{(1 + t a n^{2} x) s e c^{2} x}{\sqrt{t a n x}} d x = \int \frac{(1 + t^{2})}{\sqrt{t}} d t = \int (t^{- 1 / 2} + t^{3 / 2}) d t = 2 t^{1 / 2} + \frac{2}{5} t^{5 / 2} = 2 \sqrt{t a n x} + \frac{2}{5} (t a n x)^{5 / 2} ∴ ϕ (\frac{π}{4}) - ϕ (0) = 2 + \frac{2}{5} = \frac{12}{5}$

Question 9

$\int x {f (x^{2}) g^{''} (x^{2}) - f^{''} (x^{2}) g (x^{2})} d x$

f (x^{2}) g^{^{'}} (x^{2}) - g (x^{2}) f^{^{'}} (x^{2}) + c

\frac{1}{2} {f (x^{2}) g (x^{2}) f^{^{'}} (x^{2})} + c

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

D. None of these

SOLUTION

Solution : C

Put $x^{2} = t \Rightarrow I = \frac{1}{2} \int {f (t) g^{''} (t) - g (t) f^{''} (t)} d t = \frac{1}{2} {f (t) g^{^{'}} (t) - \int f^{^{'}} (t) g^{^{'}} (t) - g (t) f^{^{'}} (t) + \int g^{^{'}} (t) f^{^{'}} (t) d t} + c ∴ I = \frac{1}{2} {f (t) g^{^{'}} (t) - g (t) f^{^{'}} (t)} + c = \frac{1}{2} {f (x^{2}) g^{^{'}} (x^{2}) - g (x^{2}) f^{^{'}} (x^{2})} + c$

Question 10

$\int \frac{x^{4} + 1}{1 + x^{6}} d x =$

t a n^{- 1} (x) - t a n^{- 1} (x^{3}) + c

t a n^{- 1} (x) - \frac{1}{3} t a n^{- 1} (x^{3}) + c

t a n^{- 1} (x) + t a n^{- 1} (x^{3}) + c

t a n^{- 1} (x) + \frac{1}{3} t a n^{- 1} (x^{3}) + c

SOLUTION

Solution : D

$I = \int \frac{x^{4} + 1}{1 + x^{6}} d x = \int \frac{(x^{4} - x^{2} + 1) + x^{2}}{(1 + x^{6})} d x = \int \frac{x^{4} - x^{2} + 1}{1 + x^{6}} d x + \int \frac{x^{2}}{1 + x^{6}} d x = \int \frac{1}{1 + x^{2}} d x + \frac{1}{3} \int \frac{3 x^{2}}{1 + x^{6}} d x = t a n^{- 1} (x) + \frac{1}{3} t a n^{- 1} x^{3} + c$

Free Integrals 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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