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

Question 1

$\int_{0}^{\frac{π}{2}} \frac{c o s x}{1 + s i n x} d x =$

log 2

log e

$\frac{1}{2} l o g 3$

SOLUTION

Solution : A

put 1+ sin x =t
$T h e n \int_{0}^{\frac{π}{2}} \frac{c o s x}{1 + s i n x} d x = {[l o g | 1 + s i n x |]}_{0}^{\frac{π}{2}} = l o g 2$

Question 2

If area bounded by the curves $y^{2} = 4 a x a n d y = m x i s \frac{a^{2}}{3}$ then the value of m is

A. 2

B. -2

$\frac{1}{2}$

D. None of these

SOLUTION

Solution : A

The two curves $y^{2}$ = 4ax and y = mx intersect at $(\frac{4 a}{m^{2}}, \frac{4 a}{m})$ and the area enclosed by the two curves is given by $\int_{0}^{\frac{4 a}{m^{2}}} (\sqrt{4 a x} - m x) d x$
$∴ \int_{0}^{\frac{4 a}{m^{2}}} (\sqrt{4 a x} - m x) d x = \frac{a^{2}}{3}$
$\Rightarrow \frac{8}{3} \frac{a^{2}}{m^{3}} = \frac{a^{2}}{3} \Rightarrow m^{3} = 8 \Rightarrow m = 2$

Question 3

$\int_{- \frac{π}{2}}^{\frac{π}{2}} s i n^{2} x d x =$

π

$\frac{π}{2}$

\frac{π}{2} - \frac{1}{2}

π - 1

SOLUTION

Solution : B

$\int_{- \frac{π}{2}}^{\frac{π}{2}} s i n^{2} x d x = 2 \int_{0}^{\frac{π}{2}} s i n^{2} x d x = 2 \frac{r (\frac{3}{2}) r (\frac{1}{2})}{2 r (\frac{2 + 2}{2})} = \frac{π}{2}$

Question 4

$\int_{0}^{\frac{π}{2}} s i n^{2} x c o s^{3} x d x =$ [RPET 1984, 2003]

\frac{2}{15}

\frac{4}{15}

None of these

SOLUTION

Solution : B

Using gamma function,
$\int_{0}^{\frac{π}{2}} s i n^{2} x c o s^{3} x d x = \frac{r (\frac{3}{2}) r 2}{2 r (\frac{7}{2})} = \frac{2}{15}$

Question 5

$\int_{0}^{\infty} \frac{l o g (1 + x^{2})}{1 + x^{2}} d x =$

$π l o g \frac{1}{2}$

π l o g 2

2 π l o g \frac{1}{2}

2 π l o g 2

SOLUTION

Solution : B

$L e t I = \int_{0}^{\infty} \frac{l o g (1 + x^{2})}{1 + x^{2}} d x$
$P u t x = t a n θ \Rightarrow d x = s e c^{2} θ d θ,$
$∴ I = \int_{0}^{\frac{n}{2}} l o g (s e c θ)^{2} d θ = 2 \int_{0}^{\frac{n}{2}} l o g s e c θ d θ$
$= - 2 \int_{0}^{\frac{n}{2}} l o g c o s θ d θ$ $= - 2. \frac{π}{2} l o g \frac{1}{2} = - π l o g \frac{1}{2} = π l o g 2$

Question 6

${lim}_{π \to \infty} \frac{1^{99} + 2^{99} + 3^{99} + \dots \dots n^{99}}{n^{100}} =$ [EAMCET 1994]

$\frac{9}{100}$

$\frac{1}{100}$

$\frac{1}{99}$

$\frac{1}{101}$

SOLUTION

Solution : B

${lim}_{π \to \infty} \frac{1^{99} + 2^{99} + 3^{99} + \dots \dots n^{99}}{n^{100}} = {lim}_{π \to \infty} \sum_{r = 1}^{n} (\frac{r^{99}}{n^{100}})$
$= {lim}_{π \to \infty} \frac{1}{n} \sum_{r = 1}^{n} {(\frac{r}{n})}^{99} = \int_{0}^{1} x^{99} d x = {[\frac{x^{100}}{100}]}_{0}^{1} = \frac{1}{100}$

Question 7

$l i m_{π \to \infty} \sum_{r = 1}^{n} \frac{1}{n} e^{\frac{r}{n}} i s$ [AIEEE 2004]

A. e + 1

B. e - 1

C. 1 - e

D. e

SOLUTION

Solution : B

$l i m_{π \to \infty} \sum_{r = 1}^{n} \frac{1}{n} e^{\frac{r}{n}} = \int_{0}^{1} e^{x} d x = [e^{x}]_{0}^{1} = e - 1$

Question 8

The value of the definite integral $\int_{0}^{1} \frac{x d x}{x^{3} + 16}$ lies in the interval [a, b]. The smallest such interval is.

[0, \frac{1}{17}]

$[0, 1]$

[0, \frac{1}{27}]

$N o n e o f t h e s e$

SOLUTION

Solution : A

$f (x) = \frac{x}{x^{3} + 16} \Rightarrow f^{'} (x) = \frac{16 - 2 x^{3}}{(x^{3} + 16)^{2}} ∴ f^{'} (x) = 0 \Rightarrow x = 2 Also f^{''} (2) < 0 \Rightarrow$ The function $f (x) = \frac{x}{x^{3} + 16}$ is an increasing function in [0,1], so Min f(x) = f(0) = 0 and Max f(x) = f(1) = $\frac{1}{17}$
Therefore by the property $m (b - a) \leq \int_{a}^{b} f (x) d x \leq M (b - a)$
(where m and M are the smallest and greatest values of function)
$\Rightarrow 0 \leq \int_{0}^{1} \frac{x}{x^{3} + 16} d x \leq \frac{1}{17}$

Question 9

If $\int_{0}^{x} f (t) d t = x + \int_{x}^{1} t f (t) d t$ , then the value of f(1) is [IIT 1998; AMU 2005]

\frac{1}{2}

C. 1

$- \frac{1}{2}$

SOLUTION

Solution : A

$\int_{0}^{x} d t = x + \int_{x}^{1} t f (t) d t \Rightarrow \int_{x}^{1} t f (t) d t = x - \int_{1}^{x} t f (t) d t$
Differentiating w.r.t x, we get $f (x) = 1 + {0 - x f (x)}$
$\Rightarrow f (x) = 1 - x f (x) \Rightarrow (1 + x) f (x) = 1 \Rightarrow f (x) = \frac{1}{1 + x}$
$∴ f (1) = \frac{1}{1 + 1} = \frac{1}{2}$

Question 10

Calculate $\int_{- 2}^{3} f (x) d x$ where
$f (x) = {\begin{matrix} 6 & i f x > 1 3 x^{2} & i f x \leq 1 \end{matrix}$

A. 19

B. 21

C. 17

D. 15

SOLUTION

Solution : B

Here, we can see that the integrand f(x) in not continuous in the interval (-2, 3) as it has a discontinuity at x = 1. So we can’t just integrate f(x) and put limits.We’ll have to break this integral into integrals which have limits such that integrands is continuous in those limits.
$\int_{- 2}^{3} f (x) d x = \int_{- 2}^{1} 3 x^{2} d x + \int_{1}^{3} 6 d x$
Now we can integrate these integrands and put limits.
$\int_{- 2}^{1} 3 x^{2} d x + \int_{1}^{3} 6 d x = (x^{3}) |_{- 2}^{1} + (6 x) |_{1}^{3} = 1 - (- 8) + 18 - 6$
= 21

Question 11

Find the integral $\infty \int 0 e^{- x} d x$ .

0

1

2

\infty

SOLUTION

Solution : B

We can see that the given integral is an improper integral as one of its limits is not finite.
In such cases where we have to deal with infinity as the limits of definite integral, we’ll change the limit which is not finite to a variable and then put the limits.
$\infty \int 0 e^{- x} d x = lim a \to \infty a \int 0 e^{- x} d x$
$= lim a \to \infty {(- e^{- x})}_{0}^{- x}$
$= lim a \to \infty (- e^{- a} - (- e^{0}))$
$= 0 - (- 1)$
$= 1$

Question 12

Area enclosed by curve $y^{3} - 9 y + x = 0$ and Y - axis is -

\frac{9}{2}

9

\frac{81}{2}

81

SOLUTION

Solution : C

The given equation of curve can be written as $x = f (y) = 9 y - y^{3}$ . Now to calculate the area we need to find the boundaries of this curve i.e ordinates or the point where this curve is meeting Y - axis.
$f (y) = y (9 - y^{2})$
$f (y) = y . (3 + y) (3 - y)$
So, the points where f(y) is meeting y - axis are y = -3, y = 0 & y = 3.
Important thing to note here is that the function is changing its signs.
I.e. from y = -3 to y = 0 f(y) is negative.
& from y = 0 to y = 3 f(y) is positive.
Since, the function in negative in the interval (-3, 0 ) we’ll take absolute value of it. Because we are interested in the area enclosed and not the algebraic sum of area.
Let the area enclosed be A.
$A = \int_{- 3}^{0} | f (y) | d y + \int_{0}^{3} f (y) d y A = \int_{- 3}^{0} | 9 y - y^{3} | d y + \int_{0}^{3} (9 y - y^{3}) d y A = - \int_{- 3}^{0} (9 y - y^{3}) d y + \int_{0}^{3} 9 y - y^{3} d y A = - [\frac{9 y^{2}}{2} - \frac{y^{4}}{4}]_{- 3}^{0} + [\frac{9 y^{2}}{2} - \frac{y^{4}}{4}]_{0}^{3} A = - [0 - 0 - \frac{81}{2} + \frac{81}{4}] + [\frac{81}{2} - \frac{81}{4}] A = 81 - \frac{81}{2} A = \frac{81}{2}$

Question 13

$\int_{0}^{10 π} | s i n x | d x i s$

A. 20

B. 8

C. 10

D. 18

SOLUTION

Solution : A

$∴ \int_{0}^{10 π} | s i n x | d x = 10 \int_{0}^{π} | s i n x | d x$
$∴ | s i n x | i s$ positive in I & II quadrant and has period $π$
$= 10 \int_{0}^{π} s i n x d x = 10 [- c o s x]_{0}^{x} = 20$

Question 14

$\int_{0}^{2} [x^{2}] d x$ is (where [.] is greastest integral function

2 - \sqrt{2}

2 + \sqrt{2}

\sqrt{2} - 1

- \sqrt{2} - \sqrt{3} + 5

SOLUTION

Solution : D

$\int_{0}^{2} [x^{2}] d x$
$= \int_{0}^{1} [x^{2}] d x + \int_{0}^{\sqrt{2}} [x^{2}] d x + \int_{\sqrt{2}}^{\sqrt{3}} [x^{2}] d x + \int_{\sqrt{3}}^{2} [x^{2}] d x$
$= \int_{0}^{1} 0 d x + \int_{0}^{\sqrt{2}} 1 d x + \int_{\sqrt{2}}^{\sqrt{3}} 2 d x + \int_{\sqrt{3}}^{2} 3 d x$
$= \sqrt{2} - 1 + 2 \sqrt{3} - 2 \sqrt{2} + 6 - 3 \sqrt{3}$
$= 5 - \sqrt{3} - \sqrt{2}$

Question 15

$\int_{0}^{\frac{π}{2}} l o g (t a n x + c o t x) d x =$

π l o g 2

- π l o g 2

- \frac{π}{2} l o g 2

\frac{π}{2} l o g 2

SOLUTION

Solution : A

$\int_{0}^{\frac{π}{2}} l o g (t a n x + c o t x) d x = \int_{0}^{\frac{π}{2}} l o g [\frac{2}{s i n 2 x}] d x$
$= \int_{0}^{\frac{π}{2}} (l o g 2 - l o g s i n 2 x) d x = l o g 2 (\frac{π}{2}) + (\frac{π}{2}) l o g 2 = π l o g 2$

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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