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Note that V = x³ 

or dV = (dV/dx) Δx = (3x²) Δx

= (3x²) (0.02x) = 0.06 x³m³ 

Thus, the approximate change in volume is 0.06 x³m³

(A)  Since ΔPRQ = 90° , in general, the locus represented by R is a circle with P and Q as ends of the diameter. Because area of ΔPQR is 2 sq. unit, there will be four positions for R (two each in the two semicircles for which PQ is a diameter).

 Answer: (A) → (r)

(B)  It is known that a = 1 + 5 - 4 = 2 and b = 2 + 7 - 6 = 3 Therefore a + b = 5

 Answer: (B) → (t)

 (C) Centroid (10/3,15/3) 

P + r/q + s - 1 = 25/5 = 5

 Answer: (C) → (t)

 (D) We have p = x Δm is even and cos Δm π = 1)

Similarly, q = x. Since p = x is rational and q = x is irrational, we have p = q = 0. Therefore, (p,q) =  (0, 0). Hence the area of the triangle is 

1/2|2(1) - (-2)(1)| = 2

  Answer: (D) → (P)

(A) Slope of BC is

4 + 3/5 - 2 = 7/2

Therefore, the equation of the altitude AD is

y - 1 = -3/7(x + 2)

3x + 7y - 1 = 0

Answer: (A) → (t)

(B) Slope of CA is

1 + 3/-2 - 2  = -1

Therefore, the equation of the altitude BE is 

y - 4 = (x - 5)

x - y - 1 = 0

Answer: (B) → (p)

(C) The midpoint of BC is

(7/2,1/2)

and the slope of the median AM is -1/ 11 so that the equation of the median AM is

y - 1 = -1/11(x + 2)

x + 11y - 9 = 0

Answer: (C) → (q)

(D) Lastly, the slope of AB is 

4 - 1/5 + 2 = 3/7

and hence the equation of the altitude CF is

y + 3 = -7/3(x - 2)

7x + 3y - 5 = 0

Answer: (D) → (r)

w.r.t volume of a cube = v = x³.

⇒ dv/dx = 3x² dx

Δ_x = 0.02 x

∴ dv = (dv/dx) Δ_x = (3x²) Δ_x = 3x² x 0.02x

= 0.06 x³m³

y = sin √x

⇒ dy/dx = cos√x/2√x

(b) 50

If f(x) = 1 − x + x² − x³ + ⋯ − x⁹⁹ + x¹⁰⁰, then f′(1) is 50.

Take logarithm on Both sides, we get 

logy = logx^x ⇒ logy = xlogx 

Differentiate wr.t. x

1/y.dy/dx = x^x 1/x + logx

dy/dx = y[1 + logx]

dy/dx = x^x[1 + logx]

Correct answer is

(b) 2 × n

(c) 3584

In the series 7, 14, 28, 3584, the 10th term is.

Correct answer is x = 4.

Level of output is x = 4.

x/a + y/b + z/c = 1

Correct answer is (100111)₂.

The correct answer is Rs. 55,000.

A comer point of a feasible region is a point in the region which is the intersection of the two boundary lines.

Correct answer is 122.32.

The conditional probability is \(P\left(\frac{E}{F}\right) = \frac{2}{5}\).

P(S/F) = P(SnF)/P(F)

Let tan^-1 x = y 

x = tan y = cot (π/2 - y) 

cot^-1 x = π/2 - y 

cot^-1 x + y = π/2

cot^-1 x – tan^-1 x = π/2

Correct answer is 153.33.

The correct answer is Rs. 11,365.96.

f: N → N by f(1) = f(2) = 1 and f(x) = x - 1. 

f is not one-one because 

f (1) = 1 and f(2) = 1 

∴ f(1) = f(2) 

but 1 ≠ 2 

∴ f is not one-one 

for every y ∈ N then f(x) = y - x - 1 then y = x - 1 

⇒ x ∈ N 

∴ y ∈ N ∋ x ∈ N 

∴ f is onto.

* is defined on Q by a*b = ab/2 for associative we have to prove that a* (b* c) = (a* b)* c 

∴ a*(b*c) = a * ab/2 b*c = bc/2 

= abc/4  ……….. (1) 

(a*b)*c = ab/2 *c 

= abc/4 ……….. (2) 

∴ from (1) and (2) 

∴ * is Associative

∴ * Satisfies the associative property.

Given:

n(A) = 55%, n(B) = 45% and n(A∩B) = 20%

Formula used:

n(A/B) = n(A∩B)/n(B)

Calculation:

n(A) = 55%, n(B) = 45% and n(A∩B) = 20%

n(A/B) = n(A∩B)/n(B)

⇒ 20/45

⇒ 4/9

⇒ 0.44

∴ The value of n(A/B) is 44.4%

Correct option:

(B) 5π/6

(D) k = ±1/√3

The Common Region determined by all the constraints including the noh-negative constraints (x ≥ 0, y ≥ 0) of a linear programming problem is called the feasible region.

∴ The Required Equation of the plane is z = 4.

Correct option:

(C) 1/2 

Correct option:

(C) 2e√x + c 

correct option:

(C) 2. e^√x

Correct option:

(A) 13

∫(sinx + cosx).dx 

= ∫ sin x.dx + ∫ cos x.dx. 

= -cosx + sin x + c 

= sinx – cosx + c

(B) 1/x log x. log(log x)

Correct option:

(D) π/12

Correct option:

(C) z = 0 

Correct option :

(A) π/4

Given system of equations are

x + 2y - 5z = -9

3x -y + 2z = 5

2x + 3y - z = 3

It's matrix form is AX = B

Where A = \(\begin{vmatrix}1&2&-5\\3&-1&2\\2&3&-1\end{vmatrix} \times \begin{vmatrix}x\\y\\z\end{vmatrix} \&\, B = \begin{vmatrix}-9\\5\\3\end{vmatrix}\) 

We have to find inverse of matrix A

∵ A = IA

= \(\begin{vmatrix}1&2&-5\\3&-1&2\\2&3&-1\end{vmatrix} = \begin{vmatrix}1&0&0\\0&1&0\\0&0&1\end{vmatrix}A\) 

Applying R₂ → R₂ -3R₁ 

& R₃ → R₃ -2R₁

\(\begin{vmatrix}1&2&-5\\0&-7&17\\0&-1&9\end{vmatrix} = \begin{vmatrix}1&0&0\\-3&1&0\\-2&0&1\end{vmatrix}A\)

Applying R₃ → R₃ - \(\frac 17\) R₂

\(\begin{vmatrix}1&2&-5\\0&-7&17\\0&0&\frac {46}{7}\end{vmatrix} = \begin{vmatrix}1&0&0\\-3&1&0\\\frac {-11}{7}&\frac {-1}{7}&1\end{vmatrix}A\)

Applying R₃ → \(\frac {R_3}{\frac {46}{7}}\) 

\(\begin{vmatrix}1&2&-5\\0&-7&17\\0&0&1\end{vmatrix} = \begin{vmatrix}1&0&0\\-3&1&0\\\frac {-11}{46}&\frac {-1}{46}&\frac {7}{46}\end{vmatrix}A\)

Applying R₂ → R₂ - 17 R₃ 

\(\begin{vmatrix}1&2&-5\\0&-7&0\\0&0&1\end{vmatrix} = \begin{vmatrix}1&0&0\\\frac {43}{46}&\frac {63}{46}&\frac {-119}{46}\\\frac {-11}{46}&\frac {-1}{46}&\frac {7}{46}\end{vmatrix}A\)

Applying R₂ → \(\frac {R_2}{-7}\)

\(\begin{vmatrix}1&2&-5\\0&1&0\\0&0&1\end{vmatrix} = \begin{vmatrix}1&0&0\\\frac {-7}{46}&\frac {-9}{46}&\frac {17}{46}\\\frac {-11}{46}&\frac {-1}{46}&\frac {7}{46}\end{vmatrix}A\)

 Applying R₂ → R₁ - 2R₂ + 5R₃

\(\begin{vmatrix}1&0&0\\0&1&0\\0&0&1\end{vmatrix} = \begin{vmatrix}\frac {5}{46}&\frac {13}{46}&\frac {1}{46}\\\frac {-7}{46}&\frac {-9}{46}&\frac {17}{46}\\\frac {-11}{46}&\frac {-1}{46}&\frac {7}{46}\end{vmatrix}A\)

∴ A₁ = \(\frac {1}{46}\begin{vmatrix}5&13&1\\-7&-9&17\\-11&-1&7\end{vmatrix} \)

Now, ∵ AX = B

∴ x = A' B

= \(\frac {1}{46}\begin{vmatrix}5&13&1\\-7&-9&17\\-11&-1&7\end{vmatrix} \begin{vmatrix}-9\\5\\3\end{vmatrix}\)

= \(\frac {1}{46}\begin{vmatrix}5\times-9 +13\times5+1\times3\\-7\times -9+(-9) \times 5+ 17\times 3\\-11\times-9 + (-1) \times 5 + 7\times 3\end{vmatrix} = \frac {1}{46}\begin{vmatrix}-45+65+3\\63-45+51\\99-5+21\end{vmatrix}\)

= \(\frac {1}{46}\begin{vmatrix}23\\69\\115\end{vmatrix} = \begin{vmatrix}23/46\\69/46\\115/46\end{vmatrix} = \begin{vmatrix}1/2\\3/2\\5/2\end{vmatrix} \)

Hence, x = \( \begin{vmatrix}x\\y\\z\end{vmatrix} = \begin{vmatrix}1/2\\3/2\\5/2\end{vmatrix} \)

∴ x = \(\frac 12\), y = \(\frac 32\) & z = \(\frac 52\) 

Hence, given system is consistent & its solution is 

  x = \(\frac 12\), y = \(\frac 32\) & z = \(\frac 52\) 

A = \(\begin{bmatrix}3&2\\1&1\end{bmatrix}\)

Then A² = \(\begin{bmatrix}3&2\\1&1\end{bmatrix}\)\(\begin{bmatrix}3&2\\1&1\end{bmatrix}\) = \(\begin{bmatrix}3\times3+2\times1&3\times2+2\times1\\1\times3+1\times1&1\times2+1\times1\end{bmatrix}\)

 = \(\begin{bmatrix}11&8\\4&3\end{bmatrix}\)

Given that A² + aA + bI = 0

\(\therefore\) \(\begin{bmatrix}11&8\\4&3\end{bmatrix}\) + a\(\begin{bmatrix}3&2\\1&1\end{bmatrix}\) + b\(\begin{bmatrix}1&0\\0&1\end{bmatrix}\) = \(\begin{bmatrix}0&0\\0&0\end{bmatrix}\)

⇒ \(\begin{bmatrix}11+3a+b&8+2a\\4+a&3+a+b\end{bmatrix}\) = \(\begin{bmatrix}0&0\\0&0\end{bmatrix}\)

\(\therefore\) 4 + a = 0 (By comparing a₂₁ element of both equal matrices)

⇒ a = -4

And 3 + a + b = 0

⇒ b = -(a + 3)

 = -(-4 + 3)

 = -(-1)

 = 1

\(\therefore\) a = -4 and b = 1

Correct option :

(D) kⁿ|A|

(AB)^T= BA

⇒ (AB)^T= BA = AB

The above expression is true if and only if AB = BA.

Therefore, it is shown that if A and B are symmetric matrices then AB is symmetric if and only if AB = BA.

(C) (1/√2, -1/√2, 0)

Let e^x = z

. ^.. e^xdx =dz

. ^.. I =  ∫e^x cos (e^x)dx 

=  ∫cos (e^x )e^x dx 

= ∫cos zdz = sin z = sin (e^x) + C

(B) (x – y) (y – z) (z – x)

(A) skew-symmetric matrix 

(B) a^x (loga)² 

Correct option:

(A) (0,1,0) 

Correct option :

(B) 2 

Correct option:

(B) y = kx

y = log(sin x) 

Differentiae w.r.t.x, we get

dy/dx = 1/sin x.cos x = cot x

dy/dx = cotx