AP ICET 2015 Question Paper

Solution 

Sum of A.P. = $$\frac{n}{2}$$ (2a + (n-1)d)

Here, a = 2

         d = 8 - 5 = 5 - 2 = 3

and A/q n = 10.

Putting the values in the Formula

S = $$\frac{10}{2}$$ (2 x 2 + (10-1)3)

  =5 (4 + 27)

= 5 x 31 

= 165 Answer

Solution

Here a = 2 

and r = $$\sqrt{2}$$

nth term = 128

$$T\ \left(n\right)\ =\ a.r^{\left(n-1\right)}$$

Putting the values,

128 =$$\ 2.\sqrt{2}^{\left(n-1\right)}$$

$$64=2^{\frac{\left(n-1\right)}{2}}$$

$$2^6=2^{\frac{\left(n-1\right)}{2}}$$

Comparing the exponents

$$6={\frac{\left(n-1\right)}{2}}$$

On solving,n = 13 Answer

Solution:

The constant term in the bionomial expansion of $$\left(3x^2 - \frac{1}{4x}\right)^6$$ is

$$^{n}C_r$$  $$\left(3x^2\right)^{\left(6-r\right)}\left(-\frac{1}{4x}\right)^r$$

$$3^{\left(6-r\right)}x^{\left(12-2r\right)}\frac{\left(-1\right)^r}{4^rx^r}$$

= Constant $$x^{\left(12-2r\right)}x^(-r)$$

= Constant$$x^{\left(12-2r\ -r\right)}\ =Constant\ x^{\left(12-3r\right)}$$

For constant term , power of x should be 0

12-3r = 0

r =4

Constant = $$^6Cr\ 3^{\left(6-r\right)}\ \frac{\left(-1\right)^r}{\left(4\right)^r}\ =\ \frac{6!}{\left(6-2\right)!2!}.3^{\left(6-4\right)}\left(-\frac{1}{4}\right)^4$$

=$$=\ \frac{6!}{\left(4\right)!2!}.3^{\left(2\right)}\frac{1}{4^4}^{ }=\frac{5.6}{2}.9.\left(\frac{1}{256}\right)$$

= 135/256

Solution

$$2.^{10}C_0$$ +$$3.^{10}C_1$$ + $$4.^{10}C_2$$.....

=$$2.1\ +\ \frac{3.10!}{1!9!}\ +\frac{4.10!}{2!8!}\ +\frac{5.10!}{3!7!}$$

=  2 + 30 + 180 + 600...

Let A be an n × n matrix and c be a scalar then:

det (cA) = $$c^n\ \det\ \left(A\right)$$

det (2A) = $$2^4\ \det\ \left(A\right)$$ = 192

$$\det\ \left(A\right)=\ \frac{192}{2^4}\ =\ \frac{192}{16}\ =\ 12$$

12 Answer

Solution:

$$\left(\begin{array}{c}2 & 6\\ 1 & 3\end{array}\right)\left(\begin{array}{c}3 & -6\\ -1 & 2\end{array}\right) =$$

let the resultant matrix be, $$\left(\begin{array}{c}A11 & A12\\ A21 & A22\end{array}\right)$$

A11 = 2.3 + 6.(-1) = 0

A12 = 2.(-6) + 6.(2) = 0

A21 = 1.3 +3.(-1) = 0

A22 = 1.(-6) + 3.2 = 0

 = $$\left(\begin{array}{c}0 & 0\\ 0 & 0\end{array}\right)$$

Solution:

As on putting limit ,$$\lim_{x \rightarrow 3}$$ f(x) -> $$\frac{0}{0}\ format$$

So using L hospitals rule

Differentiating Nr and Dr. $$\lim_{x \rightarrow 3}$$  $$\frac{\left(3x^2\ -\ 6x\right)}{6x\ -\ 4}$$

Putting the limits, 

$$\frac{\left(3x^2\ -\ 6x\right)}{6x\ -\ 4}\ =\ \frac{\left(3.3^2\ -\ 6.3\right)}{6.3\ -\ 4}=\ \frac{\left(27-18\right)}{14}=\ \frac{9}{14}$$

$$\ \frac{9}{14}$$ Answer

Solution

$$\lim_{x \rightarrow 0} \frac{4^{3x} - 1}{x} =$$

As the fraction is in the form of $$of\ the\ \frac{0}{0}\ form$$

We can use L' Hospitals rule

Differentiating Nr and Dr.

$$\lim_{x \rightarrow 0} \frac{4^{3x} - 1}{x} =$$ $$\frac{d}{dx}\left(4^{3x}\ -\ 1\right)\ =\ \left(\ln4.\ 4^{3x}.3\right)\ -\ 0$$

Now again putting limits $$\left(\ln4.\ 4^{3x}.3\right)\ \ =\ \frac{\left(\ln\ 4.4^0.3\right)}{1}\ =\ 3\ln4$$ Answer

we will use chain rule, moving inwards

there are 3 functions in the form: cos K, 3M, and log Z

For cos K differentiation, -sin K = -sin (3log x)........where K = (3 logx)

For 3M , differentiation, =3 ..................

For log z, differentiation = 1/z = 1/x

Combining all the three

$$\frac{\text{d}}{\text{d}x}{(\cos(3\log_e x))} =$$ = {-sin (3log x)}{-3}{1/x} = 

Answer

Solution

Sum of interior angles = S = (n - 2)*180 

S = (5-2) X 180 = 540

Each interior angle = 540/5 = 108

in radians $$\frac{108}{180}\ \pi\ \ radian\ =\ \frac{3}{5}\pi\ radian$$ Answer