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i. The number of selections one ¹⁰C₄ way.

ii. The number of selections are ¹⁰C₄ way.

(10 - 1)! = 9! ways

“UNIQUE” has 6 letter; U = 2 times 

∴ Total words = \(\frac{6!}{2!}\)ways 

(i) The words end with “QUE”=3! 

(ii) The work begin with V and end with E = 4!

(iii) The words begin with a consonant (2 consonants are there N and Q) = \(\frac{4!}{2!}\)

x² = -16y Comparing with 

x² = -4ay we get 4a = 16 ⇒ a = 4 

∴ Focus = S = (0,-4) 

Directrix is y = -4. 

Length of L.R = 4a = 16.

Express the money received by B and C as Ratio 

i.e. B receives = \(\frac{120}{100}\)A, if A = 100 then B = 120 

∴ \(\frac{A}{B} = \frac{100}{120} =\frac{5}{6}\)

C receives \(\frac{B}{C} = \frac{100}{80} = \frac{5}{4}\) B. If B = 100, then C = 80

\(\frac{B}{C} = \frac{100}{80} = \frac{5}{4}\)

A : B = 5 : 6; B : C = 5 : 4 

A : B : C = 25 : 30 : 24

∴ 25x + 30x + 24x = 632

79x = 632 ⇒ x = \(\frac{632}{79}\) = 8

∴ A receives 25 × 8 = Rs. 200

B receives 30 × 8 = Rs. 240

C receives 24 × 8 = Rs. 192

(10 – 1)! = 9!

Let [~(p→~q)] = x and (~p↔q) = y

∴ [~(p→~q)] v (~p↔q) is neither tautology nor contradictia

PVQ = x is an integer or 5 is odd number.

5³ : 4³ or 125 : 64

4 is not an even integer and 7 is not a prime number.

Rs. 2 as 1 unit 

∴ No. of letters = 6 – 2 + 1 = 5

∴ No. of ways 5! = 120

3³ : 5³ = 27:125

No. of ways 

= ⁸C₅ × ⁷C₆ + ⁸C₆ × ⁷C₅ + ⁸C₇ × ⁷C₄ 

= 56 × 7 + 28 × 21 + 8 × 35 = 1260

Total-11, N – 3, G – 2, 1 – 2, E = 3 

Total permutations = \(\frac{11!}{3! \times 3! \times 2! \times2!}\) 

I. Number of permutations which begin with E & end with E is = \(\frac{9!}{3! \times 2! \times 2!}\)

II. Number of Permutations in Which all 3E’s are together (Total 11 – 3 + 1 = 9′) = \(\frac{9!}{3! \times 2! \times 2!}\)

\(^{25}C_2\) = \(\frac{25 \times 24}{2 \times 1}\) = 25 x 12 = 300

Let x be the 3rd proportional of 4 and 6, 

∴ 4 : 6 :: 6 : x 

4x = 36 ⇒ x = 9

P → (q → r) 

F → (T → F) 

F → F = T

~ q → p is true. 

∴ ~(T) → F 

F → F = T

Concept:

• If f′(x) > 0 then the function is said to be strictly increasing.
• If f′(x) < 0 then the function is said to be decreasing.

Calculation:

Given: f(x) = 1 - x - x3

Differentiating with respect to x, we get

⇒ f'(x) = 0 - 1 - 3x2

⇒ f'(x) =  - 1 - 3x2

For decreasing function, f'(x) < 0

⇒ -1 - 3x2 < 0

⇒ -(1 + 3x2) < 0

As we know, Multiplying/Dividing each side of an inequality by a negative number reverses the direction of the inequality symbol.

⇒ (1 + 3x2)  > 0

As we know, x2 ≥ 0,  x ∈ R

So, 1 + 3x2 > 0, x ∈ R

Hence, the function is decreasing for all values of x

Concept:

If x = f(θ), y =  f(θ) then, 

By chain Rule, we have 
\(\rm \frac {dy}{dx} = \dfrac {\frac {dy}{d\theta}}{\frac {dx}{d\theta}}\)

Calculations:

Given, x = k(θ + sin θ)

⇒\(\rm \dfrac {dx}{dθ} = k (1+cos \;θ)\)

y = k (1 + cos θ) 

⇒\(\rm \dfrac {dy}{dθ} = k (-sin \;θ)\)

By chain rule, we have

\(\rm \dfrac {dy}{dx} = \dfrac {\dfrac {dy}{d\theta}}{\dfrac {dx}{d\theta}}\)

⇒ \(\rm \dfrac{dy}{dx} = \dfrac {-ksin \;\theta}{k(1+ cos\;\theta)}\)

⇒ \(\rm \dfrac{dy}{dx} = \dfrac {-sin \;\theta}{(1+ cos\;\theta)}\)

Put θ = π/2

\(\rm \dfrac{dy}{dx} = \dfrac {-sin \;\dfrac {\pi}{2}}{(1+ cos\;\dfrac {\pi}{2})}\)

⇒ \(\rm \dfrac{dy}{dx} = -1\)

Hence, if x = k(θ + sin θ) and y = k (1 + cos θ), then the derivative of y with respect to x at θ = π/2 is -1.

Concept:

Suppose that we have two functions f(x) and g(x) and they are both differentiable.

​Product Rule: 

\(\rm \dfrac {d}{dx} [f(x)g(x)] = f(x) \dfrac {d}{dx} g(x) + g(x) \dfrac {d}{dx} f(x)\)

\(\rm \dfrac {d}{dx} (\log x) = \dfrac {1}{x}\)

Calculation:

Consider, y = x log x

Taking derivative w. r. to x on both side, we get

⇒ \(\rm \dfrac {dy}{dx} = \dfrac {d}{dx} (x \log x)\)

⇒\(\rm \dfrac {dy}{dx} = x \dfrac {d}{dx} (\log x) + (\log x) \dfrac {d}{dx} (x)\)

⇒\(\rm \dfrac {dy}{dx} = x \;(\dfrac {1}{x} )+ \log x \;(1)\)

 Hence,  the first order derivative of  x log x is \(\rm 1+ \log x \)

Concept:

• If f′(x) > 0 then the function is said to be strictly increasing.
• If f′(x) < 0 then the function is said to be decreasing.

Calculation:

Given: f(x) = 1 + x2 + x4

Differentiating with respect to x, we get

⇒ f'(x) = 0 + 2x + 4x³

⇒ f'(x) = 2x + 4x3

For strictly increasing function, f'(x) > 0

⇒ 2x + 4x³ > 0

⇒ 2x(1 + 2x²) > 0

As we know, x2 ≥ 0,  x ∈ R

So, 1 + 2x2 > 0, x ∈ R

Now, 2x > 0

⇒ x > 0

Hence function is strictly increasing for x > 0

Calculations:

Given, f(x) = e4log x

f(x) = \(\rm e^{\log {x^4}}\)

f(x) = x⁴               (∵ e^{log x} = x)

Taking derivative w. r. to x on both side, we get 

f'(x) = 4x³

Concept:

Arc length or curve length is the distance between two points along the section of the curve. Determining the length of an irregular section of the arc is termed as rectification of the curve.

The length of the curve y = f(x) from x = a to x = b is given as:

\(l= \mathop \smallint \limits_{{\rm{x}} = {\rm{a}}}^{{\rm{x}} = {\rm{b}}} \sqrt {1 + {{\left( {\frac{{{\rm{dy}}}}{{{\rm{dx}}}}} \right)}^2}} {\rm{dx}}\)

Calculation:

Given curve is y = log sec x

\(\frac{{dy}}{{dx}} = \frac{1}{{\sec x}} \cdot \sec x\tan x = \tan x\)

Hence the required length will be:

\(S = \mathop \smallint \nolimits_{x = 0}^{\pi /6} \sqrt {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^2}dx} \)

= \(\mathop \smallint \nolimits_0^{\pi /6} \sqrt {1 + {{\tan }^2}x} \;dx\)

= \(\mathop \smallint \nolimits_0^{\pi /6} \sec x\;dx = \left[ {\log \left( {\tan x + \sec x} \right)} \right]_0^{\pi /6}\)

= \(\log \left( {\frac{1}{{\sqrt 3 }} + \frac{2}{{\sqrt 3 }}} \right) - \log 1 = \log \sqrt 3 \)

= \(\frac{1}{2}\log 3\)

Important Note:

If the curve is parametrized in the form x = f(t) and y = g(t) with the parameter t going from a to b then

\(l = \mathop \smallint \limits_{{\rm{t}} = {\rm{a}}}^{{\rm{t}} = {\rm{b}}} \sqrt {{{\left( {\frac{{{\rm{dx}}}}{{{\rm{dt}}}}} \right)}^2} + {{\left( {\frac{{{\rm{dy}}}}{{{\rm{dt}}}}} \right)}^2}} {\rm{dt\;}}\)

When three coins are tossed, the sample space is 

S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT} 

‘X’ is the random variable denotes the number of heads. 

∴ ‘X’ can take the values of 0, 1, 2 and 3 

Hence, the probabilities 

P(X = 0) = P (No heads) = 1/8; 

P(X = 1) = P (1 head) = 3/8; 

P(X = 2) = P (2 heads) = 3/8; 

P(X = 3) = P (3 heads)=1/8;

∴ The probability mass function is

f(x) = \(\begin{cases} 1/8&for&x=0,3\\3/8&for &x=1,2\end{cases}\)

The correct answer is : (b) 4

To prove sin2x = \(\frac{2tanx}{1+tan^2x}\)

Let us start from RHS and prove it equal to LHS.

RHS = \(\frac{2tanx}{1+tan^2x}\) 

⇒\(\frac{2tanx}{sec^2x}\), as 1 + tan²x  = sec² x,identity

⇒ \(\frac{2\Big(\frac{sinx}{cosx}\Big)}{\frac{1}{cos^2x}}\)

⇒2sinxcosx = sin2x as per sin2x=2sinxcosx identity

P = secx Q = tan x 

IF = e^{∫sec xdx} = e^{log(sec x + tan x)} = secx + tanx 

y x IF = ∫Q x IF 

y (sec + tan x) = ∫(sec x + tan x) tan x.dx 

= ∫sec x.tan x + ∫ tan² x.dx 

= secx + ∫(sec² x - 1) dx 

y(sec x + tan x) = sec x + tan x - x + c

Take cos(A+B) =cos Acos B - sin Asin B 

Put A = x, B = 2x 

∴ cos(x + 2x) = cos x.cos2x - sin x sin 2x 

cos(3x) = cos x(2cos²x - 1) - sin x(2 sin cos x) 

= 2cos³x - cosx - 2cosx(sin²x) = 2cos³x - cosx - 2cosx(1 - cos2x). 

= 2 cos³x - cosx - 2 cosx + 2cos³x [4cos³r - 3cosx.]

P(X ≤ 5) = P(0) + P(1) + P(2) + P(3) + P(4) + P(5) = [1 + 6 + 15 + 20 + 15 + 6] = 63/64

Given, secx = \(\sqrt{2}\),    \(\frac{3\pi}{2}<x<2\pi\) 

∴ cos x = \(\frac{1}{\sqrt{2}}\), \(\frac{3\pi}{2}\) < x <\(2\pi\) 

and sinx = \(-\sqrt{1-cos^2x} \) as  \(\frac{3\pi}{2}\) < x <\(2\pi\)  

= \(-\sqrt{1-(\frac{1}{\sqrt{2}})^2}\)

= \(-\sqrt{\frac{1}{2}}\) =\(-\frac{1}{\sqrt{2}}\)

∴ cosecx = \(-\sqrt{2}\) 

Now, \(tanx=\frac{sinx}{cosx}=\frac{\frac{-1}{\sqrt{2}}}{\frac{1}{\sqrt{2}}}=-1\)

∴ \(cotx=\frac{1}{tanx}=-1\) 

Now, 

\(\frac{1+tanx+cosecx}{1+cotx-cosecx}=\frac{1+(-1)+(-\sqrt{2})}{1+(-1)-(-\sqrt{2})}=\frac{-\sqrt{2}}{2}=-1\)

Correct Answer is: (b) tan^-1 (1/2) 

Correct Answer is: (a) 2√(gl sin θ)

Correct Answer is: (c) ω² = 8β/5 

Correct Answer is:  (d) 4r/π

(c) 3√x+2x+7 

Mean – Mode = 3 (Mean – Median)

Which of the following number is irrational √15

P: The number 0 is a rational number which is a true statement

Hence, truth value of statement P is T.

Q: The zero is a multiple of any integer.

which is false statement

Hence, truth value of statement Q is F.

Statement 12 : P which is True (T)

13 : P ʌ ~Q = T ʌ T = T

14 : P v Q = T v F = T

15: If P then T

for its truth value ~P ʌ Q = F ʌ F = F.

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Crossing provides protection and safeguard to the owner of the cheque as by securing payment through a banker it can be easily detected to whose use the money is received. Where the cheque is crossed, the paying banker shall not pay it except to a banker.

An applicant for a Demand Draft is required to fill in a DD Request Slip, mentioning the amount, payee's name, issuing branch, location the draft should be payable at, his name, signature and account number etc. In most cases, the purchaser of the draft is an account holder with the bank, hence he can authorise the bank to debit (that is, take out funds from) his account either through a Cheque or a debit mandate. The Cheque should be drawn in favor of “Yourself for the issue of DD favoring XYZ ", where XYZ refers to the payee of the DD. The banker will debit the account of the account holder and issue a DD in favour of the beneficiary. The purchaser of the Demand Draft will send it to the beneficiary who will deposit the DD in his Bank account. Beneficiary’s Banker will clear the cheque through the clearing process. On clearance, Beneficiary’s Bank will credit the Beneficiary account