VITEEE 2018 :: Mathematics :: General questions

• Mathematics - General questions

The correct evaluation of $\int_{0}^{\pi/2} |\sin^{4}x|dx$ is

Answer:

Explanation:

$\int_{0}^{\pi/2} |\sin^{4}x|dx$ = $\int_{0}^{\pi/2} \sin^{4}xdx$

Applying gamma function,

$2\int_{0}^{\pi/2} \sin^{4}xdx=2\frac{\Gamma(5/2).\Gamma(1/2)}{2.\Gamma(6/2)}=\frac{3\pi}{8}$

The value of $\int_{-1}^{1} \left(x - \left[x\right]\right)dx$ (where [ . ] denotes  greatest integer function) is

Answer:

Explanation:

I = $\int_{-1}^{1} \left(x - \left[x\right]\right)dx$

= $\int_{-1}^{1} xdx-\int_{-1}^{1}[x]dx $ 

= $\left[\frac{x^{2}}{2}\right]_{-1}^{1}$ - $\left[\int_{-1}^{0}[x]dx+\int_{0}^{1}[x]dx\right]$

= $\frac{1}{2}\left[1-1\right] - \left[\int_{-1}^{0}[-1]dx+\int_{0}^{1}0dx\right]$

                                                                                        (if -1≤x<0,[x]=-1)

                                                                                        (if 0≤x<1, [x] =0)

= $0 - [-x]_{-1}^{1} - 0 = 0 -.[-0 - (-1)]= 1$

 If at x = 1 the function x⁴ - 62x² + ax + 9 attains its maximum value on the interval [0, 2], then the value of a is

Answer:

Explanation:

Let f(x) = x⁴ - 62x² + ax + 9

$\Rightarrow f'(x) = 4x^{3} - 124x +a$

It is given that function f attains its maximum value on the interval [0, 2] at x = 1.

.'. f'(1) = 0 → 4 x 1³ - 124 x 1+ 1 = 0 

→ 4 - 124 + a = 0 → a = 120

The two curves x³ - 3xy² + 2 = 0 and 3x²y - y³ - 2 = 0 intersect at an angle of

Answer:

Explanation:

$x^{3}-3xy^{2} + 2 = 0$

differentiating w.r.t. x :

$3x^{2} -3x(2y)\frac{dy}{dx}-3y^{2} = 0$

$\Rightarrow\frac{dy}{dx}=\frac{3x^{2}-3y^{2}}{6xy}$ and $3x^{2}y-y^{3}-2 = 0$

differentiating w.r.t. x:

$3x^{2}\frac{dy}{dx}+6xy-3y^{2}\frac{dy}{dx}= 0$

$\Rightarrow\frac{dy}{dx} = -\left(\frac{6xy}{3x^{2}-3y^{2}}\right)$

Now, Product of slope

$\frac{3x^{2}-3y^{2}}{6xy}$× $-\left(\frac{6xy}{3x^{2}-3y^{2}}\right)$

.'. they are perpendicular. Hence, angle = π/2

The combined equation of the asymptotes olthe hyperbola 2x² + 5xy + 2y² + 4x + 5y = 0 is -

Answer:

Explanation:

Let the equation of asymptotes be

2x² + 5xy + 2y² + 4x + 5y + λ = 0 ....(1)

This equation represents a pair of straight lines, 

 .'. abc + 2fgh - af² - bg² - ch² = 0

$\therefore 4\lambda+25-\frac{25}{2}-8-\lambda\times\frac{25}{4}=0$

$\Rightarrow -\frac{9\lambda}{4}+\frac{9}{2}=0$ $\Rightarrow \lambda=2$

Putting the value of λ, in eq. (1), we get

$2x^{2} + 5xy + 2y^{2} + 4x + 5y+2 = 0$

this is the equation of the asymptotes.

• Mathematics - General questions