IIFT 18th Dec 2022 Slot 1

$$64\times\ \frac{-11}{16}+4\times\ -\frac{11}{16}\times\ 4^2+32\times\ -\frac{11}{16}\times\ 4$$It is given that b=-14.

Then the eq $$5x^2 + bx + 8 = 0$$ becomes $$5x^2 -14x + 8 = 0$$.

$$x=\frac{-b\pm\ \sqrt{\ b^2-4ac}}{2a}=\frac{14\pm\ \sqrt{\ 196-160}}{10}$$

$$=\frac{14\pm\ 6}{10}$$

$$x=2\ or\ 0.8$$

Hence $$\beta=2$$.

If $$\beta=2$$, Then $$\alpha = \beta^2$$ = $$4$$.

As $$\alpha$$ is a root of $$ax^2 + 2x + 3 = 0$$

$$4^2a+8+3=0$$

$$a=-\frac{11}{16}$$.

By verifying options, Option D satisfies for the a & b values.

= $$64\times\ \frac{-11}{16}+4\times\ -\frac{11}{16}\times\ \left(-14\right)^2+32\times\ -\frac{11}{16}\times\ -14$$

$$=\ -275$$

Hence Option (D) is the answer.

Let $$\sqrt{\ 1-x}=a\ \&\ \sqrt{\ 1+x}=b$$

R.H.S = $$(2 + 2\sqrt{1 - x^2})^{\frac{3}{2}}$$

=$$(1+x+1-x + 2\sqrt{(1 - x)(1+x)})^{\frac{3}{2}}$$

=$$(a^2+b^2+ 2ab)^{\frac{3}{2}}$$

=$$\left(a+b\right)^{2\times\ \frac{3}{2}}$$ 

=$$\left(a+b\right)^3$$

L.H.S = $$(1 - x)^{\frac{3}{2}} + (1 + x)^{\frac{3}{2}} + 2(\sqrt{1 - x^2})$$

= $$a^3+b^3+2ab$$

=$$\left(a+b\right)^3-3ab(a+b)+2ab$$

Equating L.H.S & R.H.S, we get

$$\left(a+b\right)^3-3ab(a+b)+2ab = \left(a+b\right)^3$$

$$a+b=\frac{2}{3}$$

substituting the values of 'a' & 'b' we get

$$\sqrt{\ 1-x}+\sqrt{\ 1+x}=\frac{2}{3}$$

squaring on both sides, we get

$$1-x+1+x+2\sqrt{\ 1-x^2}=\frac{4}{9}$$

$$2\sqrt{\ 1-x^2}=\frac{4}{9}-2$$

$$\sqrt{\ 1-x^2}=\frac{-7}{9}$$

Squaring on both sides

$$1-x^2=\frac{49}{81}$$

$$x^2=\frac{32}{81}$$

$$x=\frac{4\sqrt{\ 2}}{9}$$

Hence, option (D) is the answer.

The given equation is $$(x - 12)^2 + (y - 10)^2 = 64$$

$$(x - 12)^2 + (y - 10)^2 = 8^2$$

It is an equation of a circle with a radius of 8 units & its centre is at (12,10).

It implies that the extreme coordinate of 'x' is = 12+8 = 20.

Similarly, the extreme coordinate of 'y' is = 10+8 = 18.

The square numbers which are less than 20 & 18 are 4, 9 & 16.

For a point (x,y) to lie inside the circle, $$(x - 12)^2 + (y - 10)^2 = 64$$, the value of $$(x - 12)^2 + (y - 10)^2$$ should be less than 64.

Case (i):- (4,4)

$$(4-12)^2+(4-10)^2=8^2+6^2=100 > 64$$. 

Hence, the point lies outside the circle.

As $$(4-12)^2=64$$, all the square points with '4' as x coordinate i.e (4,9), (4,16),  lie outside the circle.

Case (ii):- (9,9)

$$(9-12)^2+(9-10)^2=3^2+1^2=10 < 64$$.

Hence, the point lies inside the circle.

Case (iii):- (16,16)

$$(16-12)^2+(16-10)^2=4^2+6^2=52 < 64$$.

Hence, the point lies inside the circle.

Case (iv):- (9,4)

$$(9-12)^2+(4-10)^2=3^2+6^2=45 < 64$$.

Hence, the point lies inside the circle.

Case (v):- (9,16)

$$(9-12)^2+(16-10)^2=3^2+6^2=45 < 64$$.

Hence, the point lies inside the circle.

Case (vi):- (16, 4)

$$(16-12)^2+(4-10)^2=4^2+6^2=52 < 64$$.

Hence, the point lies inside the circle.

Case (vii):- (16, 16)

$$(16-12)^2+(16-10)^2=4^2+6^2=52 < 64$$.

Hence, the point lies inside the circle.

A total of 6 perfect square points lie inside the circle.

Option 'C' is the answer.

Let's consider an n-digit number.

$$_{-------------n\ digits}$$

The 900 distinct numbers have to be formed using digits 2, 3, 4, 5 and 7.

The first and last digit numbers are the same & the first and third are prime numbers

$$abc_{-----------}a\ total\ 'n'\ digits.$$

the values of a & c can be anything from 2,3,5, and 7. i.e 4 values.

The second digit should be greater than or equal to 4. Hence, b can take any value from 4, 5 & 7. i.e. 3 values.

There are no conditions from the 4th digit to the 'n-1' digit.

Hence those digits can take any value from 2, 3, 4, 5 and 7, i.e. 5 values.

For n = 4, the number of digits that can be formed $$=4\times\ 4\times\ 3=48 < 900$$

For n = 5, the number of digits that can be formed $$=4\times\ 4\times\ 3\times\ 5 = 240 < 900$$

For n = 6, the number of digits that can be formed $$=4\times\ 4\times\ 3\times\ 5\times\ 5=1200 > 900$$

Hence a minimum of 6-digit number is required to form 900 distinct numbers with given conditions.

Option (B) is correct.

Let the probability of an even outcome of dice be $$'x'$$.

Then the probability of the odd outcome of the dice is $$'2x'$$.

The sum of all probabilities of all events is 1.

It implies $$x+x+x+2x+2x+2x=1$$

$$x=\frac{1}{9}$$

Probability of drawing a green ball from pot 'A' = Probability of selecting pot 'A' x probability of drawing a green ball.

= $$\left(\frac{2}{9}+\frac{1}{9}+\frac{2}{9}\right)\times\ \frac{10}{15}$$ =$$=\frac{5}{9}\times\ \frac{10}{15}=\frac{10}{27}$$

Similarly, the Probability of drawing a green ball from pot 'B' = $$\left(\frac{1}{9}+\frac{2}{9}+\frac{1}{9}\right)\times\ \frac{2}{9}$$

= $$=\frac{4}{9}\times\ \frac{2}{9}=\frac{8}{81}$$

Probability of drawing a green ball = Probability of drawing a green ball from pot 'A' +Probability of drawing a green ball from pot 'B'.

= $$=\frac{10}{27}+\ \frac{8}{81}=\frac{38}{81}$$

Option (A) is the answer.

The total number of members of Sahitya Shakti is 500.

Total members after removing people who read Shakespeare = 500-200=300

Total members after removing people who read Tolstoy = 300-100=200

Society added back people who read Shakespeare but not Tolstoy, and the number increased to 350.

Hence society added back 150 people. But initially, society removed 200 members.

Therefore, the other 50 members who were not added back also read Tolstoy. Those 50 members read all three books.

Option (A) False. Only 150 original members read both Premchand and Tolstoy.

Option(B) True. 50 original members read together Premchand and Shakespeare and Tolstoy

Option(C) False. 350 original members read both Premchand and Shakespeare but not Tolstoy

Option (D) False. 50 original members read both Shakespeare and Tolstoy

It is given that, $$P(A \cup B) < \frac{3}{4}, P(A) > \frac{1}{8}, P\left(\frac{A}{B}\right) < \frac{1}{2}$$

$$P\left(\frac{A}{B}\right)=\frac{P(A \cap B)}{P\left(B\right)}< \frac{1}{2}$$

$$2\ P(A \cap B) < P(B)$$

 $$P(A \cap B) < P(B)-P(A \cap B)$$.........(1)

$$ P(A \cup B) = P(A) + P(B) - P(A \cap B) < \frac{3}{4}$$

$$ P(A) + P(B) - P(A \cap B) < \frac{3}{4}$$

$$ P(A) + (P(A \cap B) <) < \frac{3}{4}$$      From Eq (1),

$$ P(A) + P(A \cap B) < \frac{3}{4}$$

Hence Options (A), (B) & (D) are false.

$$ P(A) + P(A \cap B) < \frac{3}{4}$$

$$ (> \frac{1}{8}) + P(A \cap B) < \frac{3}{4}$$       From  $$(P(A) > \frac{1}{8})$$

$$P(A \cap B) < \frac{3}{4}-\frac{1}{8}$$

$$P(A \cap B) < \frac{5}{8}$$

Option (C) is correct.

Given that,

$$\tan\left(\theta\ -\gamma\ \right)=\frac{1}{\sqrt{\ 2}}$$

$$\frac{\tan\theta-\tan\gamma}{1+\tan\theta\ \tan\gamma\ }\ =\frac{1}{\sqrt{\ 2}}$$

$$\frac{\tan\theta-\tan\gamma}{1+\tan^2\alpha\ }\ =\frac{1}{\sqrt{\ 2}}$$

$$\frac{\tan\theta-\tan\gamma}{\sec^2\alpha\ }\ =\frac{1}{\sqrt{\ 2}}$$

$$\ \tan\theta\ -\tan\gamma\ =\frac{1}{\sqrt{\ 2}\cos^2\alpha\ }$$

$$\ \frac{\tan^2\alpha\ }{\tan\gamma\ }-\tan\gamma\ =\frac{1}{\sqrt{\ 2}\cos^2\alpha\ }$$

$$\ \tan^2\alpha\ -\tan^2\gamma\ =\frac{\tan\gamma\ }{\sqrt{\ 2}\cos^2\alpha\ }$$

Squaring on both sides,

$$\tan^4\alpha\ +\tan^4\gamma-2\ \tan^2\alpha\ \tan^2\gamma\ =\frac{\tan^2\gamma\ }{2\cos^4\alpha\ }$$

$$\tan^4\alpha\ +\tan^4\gamma\ =2\ \tan^2\alpha\ \tan^2\gamma+\frac{\tan^2\gamma\ }{2\cos^4\alpha\ }$$

$$\tan^4\alpha\ +\tan^4\gamma\ =\tan^2\gamma\ \left(2\ \tan^2\alpha\ +\frac{1}{2\cos^4\alpha\ }\right)$$

$$\tan^4\alpha\ +\tan^4\gamma\ =\tan^2\gamma\ \left(\ \frac{4\sin^2\alpha\ \cos^2\alpha}{2\cos^4\alpha}\ +\frac{1}{2\cos^4\alpha\ }\right)$$

$$\tan^4\alpha\ +\tan^4\gamma\ =\frac{\tan^2\gamma}{2}\ \left(\ \frac{4\sin^2\alpha\ \cos^2\alpha+1}{\cos^4\alpha}\ \right)$$

$$=\frac{\tan^2\gamma}{2}\ \left(\ \frac{6\sin^2\alpha\ \cos^2\alpha+1-2\sin^2\alpha\ \cos^2\alpha}{\cos^4\alpha}\ \right)$$

$$=\frac{\tan^2\gamma}{2}\ \left(\ \frac{6\sin^2\alpha\ \cos^2\alpha+\left(\sin^2\alpha\ +\cos^2\alpha\right)^2-2\sin^2\alpha\ \cos^2\alpha}{\cos^4\alpha}\ \right)$$

$$=\frac{\tan^2\gamma}{2}\ \left(\ \frac{6\sin^2\alpha\ \cos^2\alpha+\sin^4\alpha\ +\cos^4\alpha+2\sin^2\alpha\ \cos^2\alpha-2\sin^2\alpha\ \cos^2\alpha}{\cos^4\alpha}\ \right)$$

$$=\frac{\tan^2\gamma}{2}\ \left(\ \frac{6\sin^2\alpha\ \cos^2\alpha+\sin^4\alpha\ +\cos^4\alpha}{\cos^4\alpha}\ \right)$$

$$=\frac{\tan^2\gamma}{2}\ \left(\ 6\tan^2\alpha+\tan^4\alpha\ +1\ \right)$$

$$=\frac{\tan^2\gamma}{2}\ \left(\ \left(\tan^2\alpha+3\right)^2\ -8\ \right)$$

Option (A) is the answer.

It is given that $$\angle\ OAM=60^{\circ\ },\ \angle\ OBM=45^{\circ},\ \angle\ OCM=30^{\circ\ },\ AB=y,\ BC=15-5\sqrt{\ 3}\ $$

Let the height of the tree be OM = $$'h'$$.

Then $$AM\ =\frac{OM}{\tan\ 60^{\circ\ }}=\frac{h}{\sqrt{\ 3}}$$

From $$\triangle\ OBM,\ \tan\ 45^{\circ\ }=\frac{h}{\frac{h}{\sqrt{\ 3}}+y}$$

$$\frac{h}{\sqrt{\ 3}}+y=h$$

$$y=h-\frac{h}{\sqrt{\ 3}}$$

From $$\triangle\ OCM,\ \tan30^{\circ\ }=\frac{h}{15-5\sqrt{\ 3}+y+\frac{h}{\sqrt{\ 3}}}$$

$$\frac{1}{\sqrt{\ 3}}=\frac{h}{15-5\sqrt{\ 3}+h-\frac{h}{\sqrt{\ 3}}+\frac{h}{\sqrt{\ 3}}}$$

$$\frac{1}{\sqrt{\ 3}}=\frac{h}{15-5\sqrt{\ 3}+h}$$

$$15-5\sqrt{\ 3}+h=h\sqrt{\ 3}$$

$$15-5\sqrt{\ 3}=h\left(\sqrt{\ 3}-1\right)$$

$$h=\frac{15-5\sqrt{\ 3}}{\left(\sqrt{\ 3}-1\right)}$$

$$h=\frac{5\sqrt{\ 3}\times\ \left(\sqrt{\ 3}-1\right)}{\left(\sqrt{\ 3}-1\right)}=5\sqrt{\ 3}$$

Option (C) is the answer.

Given that $$x = a(a - d) (a + d) (a + 2d) + d^4$$ 

$$x=a\left(a-d\right)\left(a+d\right)\left(a+2d\right)+d^4$$

$$=a\left(a+d\right)\left(a-d\right)\left(a+2d\right)+d^4$$

$$=\left(a^2+ad\right)\left(a^2+2ad-ad-2d^2\right)+d^4$$

Let $$t=a^2+ad$$

$$x=\left(t\right)\left(t-2d^2\right)+d^4$$

$$x=t^2-2td^2+d^4$$

$$x=\left(t-d^2\right)^2$$

Substituting $$t=a^2+ad$$

$$x=\left(a^2+ad-d^2\right)^2$$.........(1)

Given that,  $$y = a(a + d) (a + 2d) (a + 3d) + d^4$$

$$y=a\left(a+d\right)\ \left(a+2d\right)\ \left(a+3d\right)+d^4$$

$$=a\left(a+3d\right)\ \left(a+d\right)\ \left(a+2d\right)+d^4$$

$$=\left(a^2+3ad\right)\ \left(a^2+2ad+ad+2d^2\right)+d^4$$

$$=\left(a^2+3ad\right)\ \left(a^2+3ad+2d^2\right)+d^4$$

Let  $$t=a^2+3ad$$

$$=\left(t\right)\ \left(t+2d^2\right)+d^4$$

$$=t^2+2td^2+d^4$$

$$=\left(t+d^2\right)^2$$

Substituting  $$t=a^2+3ad$$

$$y=\left(a^2+3ad+d^2\right)^2$$.........(2)

From Eq (1) & (2)

We get, $$x+y=\left(a^2-d^2+ad\right)^2+\left(a^2+d^2+3ad\right)^2$$

Option (D) is correct.