[Download PDF] AP ICET 2016 Question Paper (16th May) Paper with Solutions

Instructions

For the following questions answer them individually

AP ICET 2016 (16th May) - Question 121

$$(1 + \tan 95^\circ)(1 + \tan 130^\circ) =$$

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AP ICET 2016 (16th May) - Question 122

If $$0 < \theta < \frac{\pi}{2}$$ and $$\tan \theta + \sec \theta = 3$$, then $$\sin \theta =$$

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Solution

We have $$\tan\ \theta\ +\sec\theta\ =3$$

$$\frac{\sin\theta}{\cos\theta\ }\ \ +\frac{1}{\cos\theta\ }=3\ \ or\ 1+\sin\theta\ =3\cos\theta\ $$

Squaring both sides

$$1+\sin^2\theta\ +2\sin\theta\ =9\cos^2\theta\ $$

$$1+\sin^2\theta\ +2\sin\theta\ =9-9\sin^2\theta\ \ $$

$$10\sin^2\theta\ +2\sin\theta\ -8=0\ or\ 5\ \sin^2\theta\ +\sin\theta\ -4=0$$

$$\ \ 5\ \sin^2\theta\ +5\sin\theta\ -4\sin\theta\ \ -4=0$$

$$5\sin\theta\ \left(\sin\theta\ +1\right)-4\left(\sin\theta\ +1\right)=0$$

$$\left(5\sin\theta\ -4\right)\left(\sin\theta\ +1\right)=0$$

$$0<\theta\ <\pi\ \ so\ \sin\theta\ \ne\ -1\ $$

Hence $$\sin\theta\ =\frac{4}{5}$$

AP ICET 2016 (16th May) - Question 123

The angles of elevation of the top of a building measured from two points A, B on the horizontal ground and on the sameside of the building are respectively $$30^\circ, 60^\circ$$. If A, B and the foot of the building are collinear and AB = 400 metres, then the height of that building, in metres, is

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AP ICET 2016 (16th May) - Question 124

When the polynomial $$7x^3 - 3x^2 + ax - 5$$ is divided by $$x + 3$$, the remainder is -13. Then a =

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Solution

$$Let\ f\left(x\right)=7x^3-3x^2+ax-5$$ Now according to remainder theorem, x+3=0 or x=-3 . Hence f(-3)=-13

-189-27-3a-5=-13 or -3a=308 or a=-308/3

AP ICET 2016 (16th May) - Question 125

The relation R is defined on the set of all real numbers by R = {(a, b) $$\in$$ R $$\times$$ R/a - b is an integer}. Then R is

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AP ICET 2016 (16th May) - Question 126

The sum of all the 3 digit numbers which leave the remainder 1 when divided by 4 is

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Solution

The First 3 digit number, which leaves a remainder of 1 when divided by 4 is 101 and the last is 997.

Total number of terms is 225. Hence applying the formula of AP and finding the sum

$$\frac{225}{2}\left(997+101\right)=\frac{1098}{2}\times\ 225=123525$$

AP ICET 2016 (16th May) - Question 127

If $$x + y + z = 0$$ and $$xyz = 5$$, then $$x^3 + y^3 + z^3 =$$

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Solution

Given, $$x + y + z = 0$$ and $$xyz = 5$$

When $$x+y+z=0$$, then $$x^3+y^3+z^3=3xyz$$

$$\Rightarrow$$ $$x^3+y^3+z^3=3\left(5\right)$$

$$\Rightarrow$$ $$x^3+y^3+z^3=15$$

Hence, the correct answer is Option B

AP ICET 2016 (16th May) - Question 128

In an arithmetic progression the ratio of $$15^{th}$$ term to $$7^{th}$$ term is 11 : 9. Then the ratio of $$12^{th}$$ term to $$8^{th}$$ term is

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AP ICET 2016 (16th May) - Question 129

The value of the term independent of x in the expansion of $$\left(3x - \frac{4}{x^2} \right)^9$$ is

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AP ICET 2016 (16th May) - Question 130

If $$C_r = ^{10}C_r$$, then the value of $$C_0 + 3.C_1 + 5.C_2 + ..... + 21.C_{10} =$$

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CAT Content

AP ICET 2016 Question Paper (16th May)

AP ICET 2016 Question Paper (16th May)

AP ICET 2016 (16th May) - Question 121

AP ICET 2016 (16th May) - Question 122

AP ICET 2016 (16th May) - Question 123

AP ICET 2016 (16th May) - Question 124

AP ICET 2016 (16th May) - Question 125

AP ICET 2016 (16th May) - Question 126

AP ICET 2016 (16th May) - Question 127

AP ICET 2016 (16th May) - Question 128

AP ICET 2016 (16th May) - Question 129

AP ICET 2016 (16th May) - Question 130

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