AP ICET 2013

Solution

Hence

Option A Answer

None of the options

Solution

For maximum no. of elements, they have to be disjoint

n(AUB)= n(A) + n(B)

= 5+10 =15 Answer

Solution

$$f(x) = x^{2} + 1$$, then $$f^{-1}(26) =$$

Let f(x) = y

$$y = x^{2} + 1$$, then $$f^{-1}(26) =$$

$$x\ =\ \sqrt{\ \left(y-1\right)}$$

putting y =26

$$f^{-1}(26) =$$ = $$\sqrt{\ \left(26-1\right)}=\sqrt{\ 25}\ $$

Range = (-5,+5) Answer

Solution

$$y\ -y1\ =\ \frac{\left(y2-y1\right)}{\left(x2-x1\right)}\left(x-x1\right)$$

Putting the values, 

y2 = 7, x2 = -2

y1=-5, x1= 3

$$y\ -\left(-5\right)\ =\ \frac{\left(7-\left(-5\right)\right)}{\left(-2-\left(3\right)\right)}\left(x-3\right)$$

$$y\ +5\ =\ \frac{\left(12\right)}{\left(-5\right)}\left(x-3\right)$$

$$-5\left(y\ +5\right)\ =\ \left(12\right)\left(x-3\right)=>-5y-25=12x-36$$

on reaarranging,

12x + 5y =11

For ,x intercept

putting y=0

$$x=\frac{11}{12}$$ Answer

Solution

A/q

Here A = (2,5) and B = (-4,8)

intercept lenght = CD = AE.

Also , AEB is a right angle triangle right angled at E.

So ACC. to pythagoras theorem

$$\sqrt{\ BA^{2\ }-\ BE^2}\ =\ AE$$ = CD

= 

Solution 

Given $$5 \tan \theta = 4$$

So $$ tan \theta $$ = (4/5)

P = 4

B = 5

H=  $$\sqrt{\ \left(4^2\right)\ +\left(5^2\right)}$$ = $$\sqrt{41\ }$$

$$ sin \theta $$ = (4/$$\sqrt{41\ }$$)

$$ cos \theta $$ = (5/$$\sqrt{41\ }$$)

So 

= {5 (4/$$\sqrt{41\ }$$) - 3 (5/$$\sqrt{41\ }$$) } / {(4/$$\sqrt{41\ }$$) + 2 (5/$$\sqrt{41\ }$$) }

={20 $$\sqrt{41\ }$$ - 15 $$\sqrt{41\ }$$} / {4 $$\sqrt{41\ }$$ + 10 $$\sqrt{41\ }$$}

= $$\frac{5}{14}$$ Answer

Solution

If $$\theta +\phi = \frac {\pi}{4}$$, then $$(1 + \tan \theta) (1 + \tan \phi) =$$

We can assume, $$\theta\ $$ = 0

So $$\phi$$ = $$\frac{\pi}{4}$$

then putting values in,

$$(1 + \tan \theta) (1 + \tan \phi) =$$

= $$(1 + 0) (1 + \tan \frac{\pi}{4}) =$$ $$\left(1+0\right)\left(1+1\right)=1\times\ 2\ =\ 2$$

= 2

Solution:

$$(\cos^{2}15^\circ - \cos^{2} 75\circ) (\sin 65^\circ . \cos 35^\circ — \cos 65^\circ . \sin 35^\circ) =$$

$$\left(\cos\ 15\ +\cos\ 75\right)\left(\cos\ 15\ -\cos\ 75\right)\left(\sin65\cos35-\cos65\sin35\right)$$

$$\left(2\cos\ \left(\frac{90}{2}\right)\cos\left(\frac{60}{2}\right)\right)\left(2\sin\ \left(\frac{90}{2}\right)\sin\ \left(\frac{60}{2}\right)\right)\left(\sin65\cos35-\cos65\sin35\right)$$

$$\left(2\cos\ \left(45\right)\cos\left(30\right)\right)\left(2\sin\ \left(45\right)\sin\ \left(30\right)\right)\left(\sin65\ -35\right)$$

$$\left(2\ \left(\frac{1}{\sqrt{\ 2}}\right)\left(\ \frac{\sqrt{\ 3}}{2}\right)\right)\left(2\left(\frac{1}{\sqrt{\ 2}}\right)\left(\frac{1}{2}\right)\right)\left(\sin30\right)$$

$$\left(2\ \left(\frac{1}{\sqrt{\ 2}}\right)\left(\ \frac{\sqrt{\ 3}}{2}\right)\right)\left(2\left(\frac{1}{\sqrt{\ 2}}\right)\left(\frac{1}{2}\right)\right)\left(\frac{1}{2}\right)$$

On solving

$$\frac{\sqrt{\ 3}}{4}$$ Answer