PGDBA 2019 Question Paper

$$f'\left(x\right)=2x-1\ \ ,\ x<1$$

$$f'\left(x\right)=\frac{2x}{3}\ \ ,\ x\ge\ 1$$

$$f'\left(1^+\right)=\frac{2}{3}\ \ \ and\ f'\left(1^-\right)=1$$ => Right hand Derivative is not equal to LHD. Hence is is not differentiable at x=1

Left hand limit = 0 and RHL = 0. Hence it is continuos at x=1

Let a(x)=f(x)-g(x).

Then a'(x) = f'(x)-g'(x)

Now a'(x) is an increasing function. This would mean that a(x) is an increasing function, and hence one-one function. Thus, there can be atmost 1 point of intersection between the two graphs.

The number of intersections can be 0 or 1 depending upon the function.

$$\int_{-\frac{1}{\sqrt{3}}}^{\frac{1}{\sqrt{3}}} (\frac{x^2 - \tan x}{1 + x^2})dx$$

=$$\int_{-1/\sqrt{3}}^{1/\sqrt{3}} \frac{x^2+1-1 - \tan x}{1 + x^2}dx$$

=$$\int_{-1/\sqrt{3}}^{1/\sqrt{3}} \frac{x^2+1-1 - \tan x}{1 + x^2}dx$$

=$$\int_{-1/\sqrt{3}}^{1/\sqrt{3}}1-\int_{-1/\sqrt{3}}^{1/\sqrt{3}}\frac{1}{1+x^{2}}+\int_{-1/\sqrt{3}}^{1/\sqrt{3}}\frac{\tan x}{1+x^{2}}$$

=x-$$\tan^{-1}x$$+0     [$$\because$$ 

$$\frac{\tan x}{1+x^2}\ is\ an\ odd\ function,\ and\ will\ become\ zero$$

=$$\frac{1}{\sqrt{3}}-\frac{\pi}{6}$$ - $$\frac{-1}{\sqrt{3}}-\frac{\pi}{6}$$

=$$2(\frac{1}{\sqrt{3}} - \frac{\pi}{6})$$

Hence A is the correct answer.

$$y^2 = 16(1 + x )$$ and $$y^2 = 16(1 - x)$$

Equating the parabolas ,we will get the point of intersection of both 

16(1+x) = 16(1-x)

x=0

Substitute the value of x in the parabola $$y^2 = 16(1 + x )$$

y = $$\pm$$ 4

Point of intersection = (0,4),(0,-4)

=$$2(\int_{-4}^{4} (\frac{y^{2}-16}{16})-(\frac{16-y^{2}}{16}))$$ dy

= $$2(\int_{0}^{4} (\frac{y^{2}-16}{16})-(\frac{16-y^{2}}{16}))$$ dy                      [$$\because$$ $$\int_{0}^{2a} F(x)dx$$=2$$(\int_{0}^{a} F(x)dx)$$]

=$$2(\int_{0}^{4} (\frac{y^{2}}{16})-1-1+(\frac{y^{2}}{16}))$$

=$$2(\int_{0}^{4} (\frac{y^{2}}{8})-2))$$

=2$$(\frac{y^{3}}{3*8}-2y))$$ y varies from 0-4 

=2(8/3 - 8)

=2(16/3)   ($$\because$$ Area cant be negative)

Hence C is the correct answer

=$$\int_{0}^{1} 0 e^{x}dx+ \int_{1}^{\sqrt{2}} 1 e^{x} dx$$

=0 +  $$e^{\sqrt{\ 2}}-e^1$$

=$$e^{\sqrt{2}} - e^{1}$$

Hence C is the correct answer.

$$f(x) = ax^2 + bx + c$$ < x

$$ax^2+x\left(b-1\right)+c\ <0$$

Hence discriminant < 0 and $$a\le\ 0$$

Let the point P be $$\left(\alpha\ ,B\right)$$

Let OS =a, OR=b and RS = c and $$\angle\ OSR=\theta\ $$ , PS = z

=> a=ccos$$\theta\ $$ and b = c sin$$\theta\ $$

$$\alpha\ =a\ -\ z\ \cos\theta\ $$ and $$\beta\ \ =z\ \sin\theta\ $$

$$\alpha\ \ =\ c\ \cos\theta\ \ -z\ \cos\theta\ $$

=> $$\cos\ \theta\ \ =\ \frac{\alpha}{c-z}$$ and $$\sin\theta\ \ \ =\ \frac{\beta\ }{z}$$

$$\ \frac{\alpha\ ^2}{\left(c-z\right)^2}+\ \frac{\beta\ ^2}{z^2}=1$$ which is an ellipse.

Since the expression $$\lim_{x\rightarrow0}\frac{ae^x - be^{-x}}{x + \sin x}$$ is in 0/0 form, as RHS is a finite number, and denominator is zero, hence, the numerator must also be zero to get a finite number. Hence, a-b=0 when x tends to 0.

L'Hospitals rule can be applied 

$$\lim_{x \rightarrow 0}\frac{ae^x + be^{-x}}{1 + \cos x}$$ = 1

$$\frac{a+b}{2}$$ = 1 which is valid only when a = b = 1

$$\therefore$$ ab=1

Hence D is the correct answer.

$$\mid \frac{x + 1}{x - 1} \mid will be greater than  \frac{x + 1}{x - 1}$$ only if  $$\frac{x + 1}{x - 1}$$ < 0

$$\therefore$$ x $$\in$$ (-1,1)

Hence B is the correct answer.

If the row sums are all odd, then the total number of 1s is the sum of these 6 odd numbers, hence even. If the column sums are all odd, then the total number of 1s is the sum of these 7 odd numbers, hence odd. A contradiction; hence the number of such matrices is 0