## The value of $I = \int_{0}^{3}([x] + [x + \frac{1}{3}] + [x+\frac{2}{3}])dx$

Q: The value of $\large I = \int_{0}^{3}([x] + [x + \frac{1}{3}] + [x+\frac{2}{3}])dx$

where [.] denotes the greatest integer function, is equal to;

(A) 10

(B) 11

(C) 12

(D) none of these

Ans: (C)

## Let g (x) = sin x + cos x and f (x) = ….

Q: Let g (x) = sin x + cos x and $\large f(x) = \left\{\begin{array}{ll} \frac{|x|}{x} \; , x \ne 0 \\ 0 \; , x = 0 \end{array} \right.$ then value of $\large \int_{-\pi/4}^{2\pi}f(g(x))dx$ is equal to

(A) π/2

(B) π/4

(C) π

(D) none of these

Sol: $\large f(g(x)) = \left\{\begin{array}{lll} 1 \; , sinx + cosx > 0 \\ -1 \; , sinx + cosx < 0 \\ 0 \; , sinx + cosx = 0\end{array} \right.$

Now sin x + cos x > 0

$\large \sqrt{2}sin(x+\frac{\pi}{4}) > 0$

$\large \int_{-\pi/4}^{2\pi}f(g(x))dx$

$\large = \int_{-\pi/4}^{3\pi/4}1 dx + \int_{3\pi/4}^{7\pi/4}-1 dx + \int_{7\pi/4}^{2\pi}1 dx$

$\large = \pi – \pi + \frac{\pi}{4} = \frac{\pi}{4}$

Hence (B) is the correct answer.

## If for non-zero x , $3 f(x) + 4 f(\frac{1}{x}) = \frac{1}{x} – 10$ then $\int_{2}^{3}f(x) dx$ is equal to

Q: If for non-zero x , $\large 3 f(x) + 4 f(\frac{1}{x}) = \frac{1}{x} – 10$ then $\large \int_{2}^{3}f(x) dx$ is equal to

(A) $\frac{4}{7} ln\frac{2}{3}$

(B) $\frac{3}{7} ln\frac{3}{2}$

(C) $\frac{3}{7} ln\frac{2}{3}$

(D) None of these

Sol: $\large 3 f(x) + 4 f(\frac{1}{x}) = \frac{1}{x} – 10$

replace x by 1/x

$\large 3 f(\frac{1}{x}) + 4 f(x) = x – 10$

solving (1) and (2)

9f(x) – 16f (x) = 3/x – 30 – 4x + 40

– 7f (x) = 10 + 3/x – 4x

$\large f(x) = -\frac{10}{7}-\frac{3}{7x} + \frac{4x}{7}$

$\large \int_{2}^{3}f(x) dx = \int_{2}^{3} (-\frac{10}{7}-\frac{3}{7x} + \frac{4x}{7}) dx$

$\large [-\frac{10}{7}x -\frac{3}{7}ln|x| + \frac{4x^2}{14}]_{2}^{3}$

$\large = \frac{3}{7}ln\frac{2}{3}$

Hence (C) is the correct answer.

## Let T > 0 be a fixed real number. Suppose f is a continuous function such that for all x ∈ R f(x+T)= f(x)…

Q: Let T > 0 be a fixed real number. Suppose f is a continuous function such that for all x ∈ R f(x+T)= f(x). If $I = \int_{0}^{T} f(x) dx$ , then the value of $\int_{0}^{3+3T} f(2x) dx$  is

(A) $\frac{3}{2}I$

(B) 2I

(C) 3I

(D) 6I

Sol: Let $\large L = \int_{0}^{3 + 3T} f(2x) dx$

Put 2x = t so that

$\large L = \frac{1}{2}\int_{0}^{6 + 6T} f(t) dt$

$\large L = \frac{6}{2}\int_{0}^{T} f(t) dt = 3 I$

Hence (C) is the correct answer.

## The value of $\int_{0}^{sin^2 x} sin^{-1}(\sqrt{t}) dt + \int_{0}^{cos^2 x} cos^{-1}(\sqrt{t}) dt$

Q: The value of $\int_{0}^{sin^2 x} sin^{-1}(\sqrt{t}) dt + \int_{0}^{cos^2 x} cos^{-1}(\sqrt{t}) dt$ is

(A) π/4

(B) 0

(C) π/2

(D) none of these

Sol: Let $\large f(x) = \int_{0}^{sin^2 x} sin^{-1}(\sqrt{t}) dt + \int_{0}^{cos^2 x} cos^{-1}(\sqrt{t}) dt$

Here f'(x) = 0

⇒  f(x) = c (a constant).

But $\large f(\pi/4) = \int_{0}^{1/2} (sin^{-1}(\sqrt{t} + cos^{-1}(\sqrt{t}) dt$

$\large = \int_{0}^{1/2} \frac{\pi}{2} dt = \frac{\pi}{4}$

Hence (A) is the correct answer.

## $\lim_{x \rightarrow \infty} \frac{(\int_{0}^{x}e^{t^2}dt)^2}{\int_{0}^{x}e^{2t^2}dt}$ is equal to

Q: $\large \lim_{x \rightarrow \infty} \frac{(\int_{0}^{x}e^{t^2}dt)^2}{\int_{0}^{x}e^{2t^2}dt}$ is equal to

(A) 1

(B) 0

(C) –1

(D) none of these

Sol: Given limit $\large = \lim_{x \rightarrow \infty} \frac{(\int_{0}^{x}e^{t^2}dt)^2}{\int_{0}^{x}e^{2t^2}dt}$

$\large = \lim_{x \rightarrow \infty} \frac{2 \int_{0}^{x}e^{t^2}dt .e^{x^2}}{e^{2x^2}}$

$\large = \lim_{x \rightarrow \infty} \frac{2 \int_{0}^{x}e^{t^2}dt }{e^{x^2}}$

$\large = \lim_{x \rightarrow \infty} \frac{2 e^{x^2}}{e^{x^2}. 2x } = 0$

Hence (B) is the correct answer.

## Let a, b, c be non zero numbers such that…

Q: Let a, b, c be non zero numbers such that
$\int_{0}^{3} \sqrt{x^2 + x + 1} (ax^2 + bx + c) dx = \int_{0}^{5} \sqrt{x^2 + x + 1} (ax^2 + bx + c) dx$ . Then the quadratic equation ax2 + bx + c = 0 has

(A) no root in (0, 3)

(B) at least one root in (3, 5)

(C) a double root in (0, 3)

(D) two imaginary roots

Sol. The given equation implies

$\large \int_{3}^{5} \sqrt{x^2 + x + 1} (ax^2 + bx + c) dx = 0$

This is possible only if f (x) cross the x-axis atleast once because algebraic sum of the area from 3 to 5 is zero.

⇒ Function ax2 + bx + c has atleast one zero in (3, 5).

Hence (B) is the correct answer.