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

Question 1

If $I_{n} = \int t a n^{n} x d x, t h e n I_{0} + I_{1} + 2 (I_{2} + . . . I_{8}) + I_{9} + I_{10}$ , is equal to

\sum_{n = 1}^{9} \frac{t a n^{n} x}{n}

1 + \sum_{n = 1}^{8} \frac{t a n^{n} x}{n}

\sum_{n = 1}^{9} \frac{t a n^{n} x}{n + 1}

\sum_{n = 2}^{10} \frac{t a n^{n} x}{n + 1}

SOLUTION

Solution : A

We have $I_{n} = \int t a n^{n} x d x = \int t a n^{n - 2} x (s e c^{2} x - 1) d x = \frac{t a n^{n - 1} x}{n - 1} - I_{n - 2} i . e . I_{n - 2} + I_{n} = \frac{t a n^{n - 1} x}{n - 1} (n \geq 2)$
Thus, we have
$I_{0} + I_{1} + 2 (I_{2} + . . . . + I_{8}) + I_{9} + I_{10} = (I_{0} + I_{2}) + (I_{1} + I_{3}) + (I_{2} + I_{4}) + (I_{3} + I_{5}) + (I_{4} + I_{6}) + (I_{5} + I_{7}) + (I_{6} + I_{8}) + (I_{7} + I_{9}) + (I_{8} + I_{10}) . = \sum_{n = 2}^{10} \frac{t a n^{n - 1} x}{n - 1} = \sum_{n = 1}^{9} \frac{t a n^{n} x}{n}$

Question 2

Primitive of $f (x) = x {.2}^{I n (x^{2} + 1)}$ with respect to x is

\frac{2^{I n (x^{2} + 1)}}{(x^{2} + 1)} + C

\frac{(x^{2} + 1) 2^{I n (x^{2} + 1)}}{I n 2 + 1} + C

\frac{(x^{2} + 1)^{I n 2 + 1}}{2 (I n 2 + 1)} + C

\frac{(x^{2} + 1)^{I n 2}}{2 (I n 2 + 1)} + C

SOLUTION

Solution : C

$I = \int x 2^{I n (x^{2} + 1)} d x$ let $x^{2} + 1 = t; x d x = \frac{d t}{2}$
Hence $I = \frac{1}{2} \int 2^{I n t} d t = \frac{1}{2} \int t^{I n 2} d t = \frac{1}{2} . \frac{t^{I n 2 + 1}}{I n 2 + 1} + C = \frac{1}{2} . \frac{(x^{2} + 1)^{I n 2 + 1}}{I n 2 + 1} + C$

Question 3

$\int \frac{c o s^{3} x + c o s^{5} x}{s i n^{2} x + s i n^{4} x} d x$ equals

s i n x - 6 t a n^{- 1} (s i n x) + c

s i n x - 2 s i n^{- 1} x + c

s i n x - 2 (s i n x)^{- 1} - 6 t a n^{- 1} (s i n x) + c

s i n x - 2 (s i n x)^{- 1} + 5 t a n^{- 1} (s i n x) + c

SOLUTION

Solution : C

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

Question 4

$\int \frac{d x}{(x - 3)^{(4 / 5)} (x + 1)^{6 / 5}} =$

((x - 3) (x + 1)^{(1 / 5)}) + c

\frac{5}{4} {(\frac{x - 3}{x + 1})}^{1 / 5} + c

{(\frac{x - 3}{x + 1})}^{1 / 5} + c

(x - 3)^{6 / 5} (x + 1)^{4 / 5} + c

SOLUTION

Solution : B

$I = \int \frac{d x}{(x - 3)^{4 / 5} (x + 1)^{6 / 5}} P u t t = \frac{x - 3}{x + 1} d t = \frac{4}{(x + 1)^{2}} d x . I = \frac{1}{4} \int \frac{d t}{t^{4 / 5}} = \frac{1}{4} (\frac{t^{- 4 / 5} + 1}{- 4 / 5} + 1) = \frac{5}{4} t^{1 / 5} = \frac{5}{4} {(\frac{x - 3}{x + 1})}^{1 / 5}$

Question 5

Let $x^{2} \neq n π - 1, n ϵ N$ . Then, the value of $\int x \sqrt{\frac{2 s i n (x^{2} + 1) - s i n 2 (x^{2} + 1)}{2 s i n (x^{2} + 1) + s i n 2 (x^{2} + 1)}} d x$ is equal to

l o g | \frac{1}{2} s e c (x^{2} + 1) | + C

l o g | s e c (\frac{x^{2} + 1}{2}) | + C

\frac{1}{2} l o g | s e c (x^{2} + 1) | + C

D. None of these

SOLUTION

Solution : B

We have, $\int x \sqrt{\frac{2 s i n (x^{2} + 1) - s i n 2 (x^{2} + 1)}{2 s i n (x^{2} + 1) + s i n 2 (x^{2} + 1)}} d x = \int x \sqrt{\frac{2 s i n (x^{2} + 1) - 2 s i n (x^{2} + 1) c o s (x^{2} + 1)}{2 s i n (x^{2} + 1) + 2 s i n (x^{2} + 1) c o s (x^{2} + 1)}} d x = \int x \sqrt{\frac{1 - c o s (x^{2} + 1)}{1 + c o s (x^{2} + 1)}} d x = \int x t a n (\frac{x^{2} + 1}{2}) d x = \int t a n (\frac{x^{2} + 1}{2}) d (\frac{x^{2} + 1}{2}) = l o g | s e c (\frac{x^{2} + 1}{2}) | + C$

Question 6

$\int \frac{d x}{c o s (2 x) c o s (4 x)}$ is equal to

\frac{1}{2 \sqrt{2}} l o g | \frac{1 + \sqrt{2} s i n 2 x}{1 - \sqrt{2} s i n 2 x} | - \frac{1}{2} (l o g | s e c 2 x + t a n 2 x |) + C

\frac{1}{2 \sqrt{2}} l o g | \frac{1 + \sqrt{2} s i n 2 x}{1 + \sqrt{2} s i n 2 x} | - \frac{1}{2} (l o g | s e c 2 x + t a n 2 x |) + C

\frac{1}{\sqrt{2}} l o g | \frac{1 + \sqrt{2} s i n 2 x}{1 + \sqrt{2} s i n 2 x} | - \frac{1}{2} (l o g | s e c 2 x - t a n 2 x |) + C

D. None of these

SOLUTION

Solution : A

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

Question 7

$\int \frac{1 - 7 c o s^{2} x}{s i n^{7} x c o s^{2} x} d x = \frac{f (x)}{(s i n x)^{7}} + C$ , then f(x) is equal to

A. sin x

B. cos x

C. tan x

D. cot x

SOLUTION

Solution : C

$\int \frac{1 - 7 c o s^{2} x}{s i n^{7} x c o s^{2} x} d x = \int (\frac{s e c^{2} x}{s i n^{7} x} - \frac{7}{s i n^{7} x}) d x = \int \frac{s e c^{2} x}{s i n^{7} x} d x - \int \frac{7}{s i n^{7}} d x = I_{1} + I_{2} N o w, I_{1} = \int \frac{s e c^{2} x}{s i n^{7}} d x = \frac{t a n x}{s i n^{7} x} + 7 \int \frac{t a n x c o s x}{s i n^{8} x} = \frac{t a n x}{s i n^{7} x} - I_{2} ∴ I_{1} + I_{2} = \frac{t a n x}{s i n^{7} x} + C \Rightarrow f (x) = t a n x$

Question 8

$\int e^{x} [\frac{x^{3} + x + 1}{(1 + x^{2})^{3 / 2}}] d x$ is equal to

$\frac{x^{2} e^{x}}{(1 + x^{2})^{1 / 2}} + c$

$\frac{e^{x}}{x (1 + x^{2})^{1 / 2}} + c$

$\frac{e^{x}}{(1 + x^{2})^{1 / 2}} + c$

$\frac{x e^{x}}{(1 + x^{2})^{1 / 2}} + c$

SOLUTION

Solution : D

$I = \int e^{x} [\frac{x}{\sqrt{1 + x^{2}}} + \frac{1}{(1 + x^{2})^{3 / 2}}] d x$

Let $f (x) = \frac{x}{\sqrt{1 + x^{2}}} \Rightarrow f^{^{'}} (x) = \frac{\sqrt{1 + x^{2}} - \frac{x^{2}}{\sqrt{1 + x^{2}}}}{(1 + x^{2})}$

$= \frac{1}{(1 + x^{2})^{3 / 2}}$

$∴ I = e^{x} f (x) + c$

$= \frac{e^{x} x}{\sqrt{1 + x^{2}}} + c$

Question 9

$\int \frac{dx}{{sin}^{4} x + {cos}^{4} x}$ is equal to

\frac{1}{\sqrt{2}} {tan}^{- 1} (\frac{1}{\sqrt{2}} tan 2 x) + C

\sqrt{2} {tan}^{- 1} (\frac{1}{\sqrt{2}} tan 2 x) + C

\frac{1}{\sqrt{2}} {tan}^{- 1} (\frac{1}{\sqrt{2}} cot 2 x) + C

D. None of these

SOLUTION

Solution : A

$\int \frac{dx}{{sin}^{4} x + {cos}^{4} x} = \int \frac{dx}{({sin}^{2} x + {cos}^{2} x)^{2} - 2 {sin}^{2} x {cos}^{2} x} = \int \frac{2 dx}{2 - {sin}^{2} 2 x} = \int \frac{2 {sec}^{2} 2 x}{2 {sec}^{2} 2 x - {tan}^{2} 2 x} dx = 2 \int \frac{{sec}^{2} 2 x}{2 + {tan}^{2} 2 x} dx = \int \frac{dt}{(\sqrt{2})^{2} + t^{2}} [p u t t i n g tan 2 x = t \Rightarrow 2 {sec}^{2} 2 x dx = dt] = \frac{1}{\sqrt{2}} {tan}^{- 1} (\frac{t}{\sqrt{2}}) + C = \frac{1}{\sqrt{2}} {tan}^{- 1} (\frac{1}{\sqrt{2}} tan 2 x) + C$

Question 10

If $\int \frac{d x}{x \sqrt{1 - x^{3}}} = a l o g ∣ ∣ ∣ \frac{\sqrt{1 - x^{3}} - 1}{\sqrt{1 - x^{3}} + 1} ∣ ∣ ∣ + C$ then a =

$\frac{1}{3}$

$\frac{2}{3}$

$- \frac{1}{3}$

$- \frac{2}{3}$

SOLUTION

Solution : A

Put<br> $1 - x^{3} = t^{2} \Rightarrow - 3 x^{2} d x = 2 t d t ∴ \int \frac{d x}{x \sqrt{1 - x^{3}}} = \int \frac{x^{2}}{x^{3} \sqrt{1 - x^{3}}} d x = \frac{2}{3} \int \frac{d t}{t^{2} - 1} = \frac{1}{3} l o g ∣ ∣ \frac{t - 1}{t + 1} ∣ ∣ + C = \frac{1}{3} l o g ∣ ∣ ∣ \frac{\sqrt{1 - x^{3}} - 1}{\sqrt{1 - x^{3}} + 1} ∣ ∣ ∣ + C ∴ a = \frac{1}{3}$

Question 11

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 12

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 13

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 14

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

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 15

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

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 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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