AP ICET 26th April 2019 Shift-2

As per the given question,

Let the height of the tower is h.

Now, in the $$\triangle$$ECD,

$$\tan 30=\dfrac{CD}{EC}---------------(i)$$

In the $$\triangle$$ABD

$$\Rightarrow \tan 60=\dfrac{BD}{AB}$$

$$\Rightarrow \tan 60=\dfrac{15+CD}{AB}--------------(ii)$$

As per the given equation,

$$x^3 + 6x^2 + 11x + 6=0$$
If the given options are satisfying the equation then it will give zero.
x-1=0, so x =1 putting in the equation,

$$\Rightarrow x^3 + 6x^2 + 11x + 6=0$$

$$\Rightarrow 1+6+11+6 \neq 0$$

Now, trying option b, x-2=0 , so x=2

$$\Rightarrow x^3 + 6x^2 + 11x + 6=0$$

$$\Rightarrow 8+24+22+6 \neq0$$

Now, putting option c, x+3=0, x=-3

$$\Rightarrow (-3)^3+6\times (-3)^2+11\times (-3)+6=0$$

$$\Rightarrow -27+54-33+6=0$$

Hence it is satisfying the condition.

Now, putting option D, x-3=0, x=3

$$\Rightarrow x^3 + 6x^2 + 11x + 6=0$$

$$\Rightarrow 27+54+33+6\neq 0$$

Hence, option C is the correct answer.

$$x^2 - x - 2$$ = (x+1)(x-2)

Let f(x) = $$ax^4 + bx^3 + 2x^2 + 4$$

f(-1) = 0 => a - b + 6 =0

f(2) = 0 => 16a+8b+12 = 0

Solving above two, we get a=-5/2 and b = 7/2

Let P have p mangos and Q have q mangoes

Condition 1:  p+30 = 2(q-30)

Condition 2: q+10 = 3(p-10)

Solving both these equations we get p=34 and q=62

Let the fraction be x/y

Given, $$\frac{\left(x+1\right)}{y+1}=\frac{4}{5}$$

$$\frac{\left(x-5\right)}{y-5}=\frac{1}{2}$$

Solving both the equation, we get x=7 and y=9

The first 25 terms of the given series is $$\frac{4}{2}$$, $$\frac{5}{2}$$ , $$\frac{6}{2}$$ , $$\frac{7}{2}$$, ......$$\frac{28}{2}$$

$$\therefore\ $$ Sum of first 25 terms of the given series = $$\frac{\left(4+5+6+7+.....28\right)}{2}$$

$$=\frac{\frac{25}{2}\left(4+28\right)}{2}$$

$$=\frac{\frac{25}{2}\left(32\right)}{2}$$

$$=\frac{400}{2}$$

$$=200$$

Hence, the correct answer is Option B

Let the 3 terms of the geometric progression be a/r , a , ar

Given, product of the terms = 27

$$\Rightarrow$$ $$\frac{a}{r}\times a\times ar=27$$

$$\Rightarrow$$  $$a^3=27$$

$$\Rightarrow$$  a = 3

Hence middle term = 3

kth term of series $$\left(x+a\right)^n\ =\ n_{c_{k-1}}\left(x\right)^{n-\left(k-1\right)}\left(a\right)^{k-1}$$

$$9^{th}$$ term in the binomial expansion of $$\left(x - \frac{1}{x}\right)^{17}$$ = $$\ 17_{c_8}x^9\left(-\frac{1}{x}\right)^8=17_{c_8}x$$