2.
The number of tangents to the curve $${x^{\frac{3}{2}}} + {y^{\frac{3}{2}}} = 2{a^{\frac{3}{2}}},\,a > 0,$$ which are equally inclined to the axes, is :
Given the curve is $${x^{\frac{3}{2}}} + {y^{\frac{3}{2}}} = 2{a^{\frac{3}{2}}}.....(1)$$
$$\therefore \,\frac{3}{2}\sqrt x + \frac{3}{2}\sqrt y \frac{{dy}}{{dx}} = 0\,\,{\text{or, }}\frac{{dy}}{{dx}} = - \frac{{\sqrt x }}{{\sqrt y }}$$
Since the tangent is equally inclined to the axes,
$$\eqalign{
& \frac{{dy}}{{dx}} = \pm 1 \cr
& \therefore \, - \frac{{\sqrt x }}{{\sqrt y }} = \pm 1{\text{ or }} - \frac{{\sqrt x }}{{\sqrt y }} = - 1 \cr
& \therefore \,\sqrt x = \sqrt y \,\,\,\,\,\left[ {\because \,\sqrt x > 0,\,\sqrt y > 0} \right] \cr
& {\text{Putting }}\sqrt y = \sqrt x {\text{ in equation}}\left( 1 \right){\text{,we get}} \cr
& 2{x^{\frac{3}{2}}} = 2{a^{\frac{3}{2}}}{\text{ or }}{x^3} = {a^3} \cr
& {\text{Therefore, }}x = a{\text{ and, so }}y = a \cr} $$
3.
Let $$f\left( x \right) = {\tan ^{ - 1}}\left\{ {\phi \left( x \right)} \right\},$$ where $$\phi \left( x \right)$$ is $$m.i.$$ for $$0 < x < \frac{\pi }{2}.$$ Then $$f\left( x \right)$$ is :
A.
increasing in $$\left( {0,\,\frac{\pi }{2}} \right)$$
B.
decreasing in $$\left( {0,\,\frac{\pi }{2}} \right)$$
C.
increasing in $$\left( {0,\,\frac{\pi }{4}} \right)$$ and decreasing in $$\left( {\frac{\pi }{4},\,\frac{\pi }{2}} \right)$$
D.
none of these
Answer :
increasing in $$\left( {0,\,\frac{\pi }{2}} \right)$$
$$f'\left( x \right) = \frac{{\phi '\left( x \right)}}{{1 + {{\left\{ {\phi \left( x \right)} \right\}}^2}}} > 0{\text{ for }}0 < x < \frac{\pi }{2}$$ because $$\phi '\left( x \right) > 0,\,\phi \left( x \right)$$ being m.i.
4.
The function $$f\left( x \right) = 1 + x\left( {\sin \,x} \right)\left[ {\cos \,x} \right],\,0 < x \leqslant \frac{\pi }{2}$$ (where [.] is G.I.F.)
A.
is continuous on $$\left( {0,\,\frac{\pi }{2}} \right)$$
B.
is strictly increasing in $$\left( {0,\,\frac{\pi }{2}} \right)$$
C.
is strictly decreasing in $$\left( {0,\,\frac{\pi }{2}} \right)$$
D.
has global maximum value $$2$$
Answer :
is continuous on $$\left( {0,\,\frac{\pi }{2}} \right)$$
For $$0 < x \leqslant \frac{\pi }{2}\,;\,\left[ {\cos \,x} \right] = 0$$
Hence, $$f\left( x \right) = 1$$ for all $$\left( {0,\,\frac{\pi }{2}} \right]$$
Trivially $$f\left( x \right)$$ is continuous on $$\left( {0,\,\frac{\pi }{2}} \right)$$
This function is neither strictly increasing nor strictly decreasing and its global maximum is $$1.$$
5.
A lamp of negligible height is placed on the ground $${l_1}\,m$$ away from a wall. A man $${l_2}\,m$$ tall is walking at a speed of $$\frac{{{l_1}}}{{10}}\,m/s$$ from the lamp to the nearest point on the wall. When he is midway between the lamp and the wall, the rate of change in the length of this shadow on the wall is :
Two roots of the first equation are 0 and $$\alpha .$$ The second equation is the derived equation of the first. So, it has root in $$\left( {0,\,\alpha } \right).$$ .
7.
The function $$f\left( x \right) = {\sin ^4}x + {\cos ^4}x$$ increases if :
A.
$$0 < x < \frac{\pi }{8}$$
B.
$$\frac{{\pi }}{4} < x < \frac{{3\pi }}{8}$$
C.
$$\frac{{3\pi }}{4} < x < \frac{{5\pi }}{8}$$
D.
$$\frac{{5\pi }}{8} < x < \frac{{3\pi }}{4}$$
Answer :
$$\frac{{\pi }}{4} < x < \frac{{3\pi }}{8}$$
$$y = {\log _e}x\,\,\,\, \Rightarrow \frac{{dy}}{{dx}} = \frac{1}{x}\,\,\,\,\,\,\therefore {\left. {\frac{{dy}}{{dx}}} \right)_{1,\,0}} = 1$$
$$\therefore $$ the equation of the normal at $$\left( {1,\,0} \right)$$ is
$$\eqalign{
& y - 0 = - \frac{1}{1}\left( {x - 1} \right){\text{ or }}x + y = 1 \cr
& \therefore {\text{ area}} = \frac{1}{2}.1.1 = \frac{1}{2} \cr} $$
9.
A function $$y = f\left( x \right)$$ has a second order derivative $$f''\left( x \right) = 6\left( {x - 1} \right).$$ If its graph passes through the point (2,1) and at that point the tangent to the graph is $$y = 3x - 5,$$ then the function is