Mathematics Mock Test - 1

Concept:

1. Direction angles: If α, β, and γ are the angles made by the line segment with the coordinate axis then these angles are termed to be the direction angles.
2. Direction cosines: The cosines of direction angles are the direction cosines of the line. Hence, cos α, cos β and cos γ are called as the direction cosines

It is denoted by l, m and n. ⇔ l = cos α, m = cos β and n = cos γ

3. The sum of squares of the direction cosines of a line is equal to unity.
4. l² + m² + n² = 1 or cos² α + cos² β + cos² γ = 1
4. Direction ratios: Any numbers which are proportional to the direction cosines of a line are called as the direction ratios. It is denoted by ‘a’, ‘b’ and ‘c’.
5. a ∝ l, b ∝ m and c ∝ n ⇔ a = kl, b = km and c = kn Where k is a constant.

Calculation:

We have to find the value of sin² α + sin² β + sin² γ

We know that sum of squares of the direction cosines of a line is equal to unity.

⇒ cos² α + cos² β + cos² γ = 1

⇒ 1 - sin² α + 1 - sin² β + 1 - sin² γ = 1 (∵ sin² θ + cos² θ = 1)

⇒ 3 – (sin² α + sin² β + sin² γ) = 1

⇒ 3 – 1 = sin² α + sin² β + sin² γ

Concept:

The highest order derivative present in the differential equation is the order of the differential equation.

Degree is the highest power of the highest order derivative in the differential equation, after the equation has been cleared from fractions and the radicals as for as the derivatives are concerned.

Calculations:

We know that, 

The highest order derivative present in the differential equation is the order of the differential equation.

Degree is the highest power of the highest order derivative in the differential equation, after the equation has been cleared from fractions and the radicals as for as the derivatives are concerned.

Given, the differential equation is, 

\(\rm {\left[ {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^3}} \right]^{\frac{7}{3}}} = 7\frac{{{d^2}y}}{{d{x^2}}}\)

The order and degree of differential equation are the integers not in fraction form.

To find the order and degree of differential equation, take cube on both side.

\(\rm \left({\left[ {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^3}} \right]^{\frac{7}{3}}}\right )^3= [7\frac{{{d^2}y}}{{d{x^2}}}]^3\)

\(\rm {\left[ {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^3}} \right]^{7}} = [7\frac{{{d^2}y}}{{d{x^2}}}]^3\)

The highest order derivative present in the differential equation \(\rm \dfrac {d^2y}{dx^2}\) . Its order is 2.

Hence, the order of the given differential equation is 2

The highest degree of the derivative present in the differential equation \(\rm [\dfrac {d^2y}{dx^2}]^3\) . Its degree is 3.

Hence, Order and degree of the differential equation

\(\rm {\left[ {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^3}} \right]^{\frac{7}{3}}} = 7\frac{{{d^2}y}}{{d{x^2}}}\) are 2 and 3 respectively

Concept:

\(\rm \int x^n dx = \frac{x^{n+1}}{n+1}+c\)

Calculation: 

I = \(\rm \int_{1}^{\infty} \frac{4}{x^4}dx\)

= \(\rm \int_{1}^{\infty}4{x^{-4}}dx\)

= \(\rm \left[\frac{4x^{-3}}{-3} \right ]_1^{\infty}\)

= \(\rm \frac{-4}{3}\left[\frac{1}{x^3} \right ]_1^{\infty}\)

= \(\rm \frac{-4}{3}\left[\frac{1}{\infty} - \frac{1}{1}\right ]\)

= \(\rm \frac{-4}{3}[0-1]\)

= \(\frac 4 3\)

Concept:

• For two events A and B, we have: P(A ∪ B) = P(A) + P(B) - P(A ∩ B).
• If A and B are independent events, then P(A ∩ B) = P(A) × P(B).

Calculation:

Using the concept above, because A and B are independent events, we can write:

P(A ∪ B) = P(A) + P(B) - P(A) × P(B)

⇒ 0.7 = 0.4 + P - 0.4 × P

⇒ 0.6P =0.3

⇒ P = 0.5.

Concept:

Let R be a binary relation on a set A.

1. Reflexive: Each element is related to itself.

• R is reflexive if for all x ∈ A, xRx.

2. Symmetric: If any one element is related to any other element, then the second element is related to the first.

• R is symmetric if for all x, y ∈ A, if xRy, then yRx.

3. Transitive: If any one element is related to a second and that second element is related to a third, then the first element is related to the third.

• R is transitive if for all x, y, z ∈ A, if xRy and yRz, then xRz.

4. R is an equivalence relation if A is nonempty and R is reflexive, symmetric and transitive.

Calculation:

Given: x < y if y at least 5 years older than x

⇒ y ≥ x + 5

For Reflexive: (x, x) should ∈R for all x ∈ X

Now, x cannot be 5 years older than himself. So the relation is not reflexive.

For Symmetric: If (x, y) ∈ R ⇒(y, x) ∈ R

(x, y) ∈ R ⇒ y is at least 5 years older than x.

(y, x) ∈ R ⇒ x is at least 5 years older than y. This contradicts the above statement. Hence the relation is not symmetric

For Transitive: If (x, y) ∈ R and (y, z) ∈ R ⇒ (x, z) ∈ R.

(x, y) ∈ R ⇒ y is at least 5 years older than x.

(y, z) ∈ R ⇒ z is at least 5 years older than y.

Then, (x, z) ∈ R ⇒ z is at least 5 years older than x.

Since, z is at least 10 years older than x. The relation is transitive.

Concept:

One–One Function / Injective Function:
A function f: A → B is said to be a one–one function, if different elements in A have different images or associated with different elements in B i.e if

f (x1) = f (x2) ⇒  x1 = x2, ∀ x1, x2 ∈ A.

Into Function:

Any function f: A → B is said to be into function if there exist at least one element in B which does not has a pre-image in A, then the function f is said to be into function.

i.e If Range of function f ⊂ Co-domain of function f, then f is into

Many-one Function:

Any function f: A → B is said to be many-one, if two (or more than two) distinct elements in A have same images in B.

Onto Function / Surjective Function :

Any function f: A → B is said to be onto if every element in B has atleast one pre-image in A.

i.e If Range of function f = Co-domain of function f, then f is onto

Calculation:

Given: f: R → R is a function such that, \(f\left( x \right) = \;\left\{ {\begin{array}{*{20}{c}} {1,\;if\;x > 0}\\ {0,\;if\;x = 0}\\ { - \;1,\;if\;x < 0} \end{array}} \right.\)

Let x₁ = 1 and x₂ = 2

Now according to the definition of the function we have

⇒ f(x₁) = 1 and f(x₂) = 1

So, we can see that, f(x1) = f(x₂) but x₁ ≠ x₂

So, the given function is many-one function.

As we can see that the range of the given function is {- 1, 0, 1} ⊂ R 

So, the given function is a into function.

Hence, the given function is many-one and into function.

Concept:

Composition of function:

(fog) (x) = f[g(x)]

Calculations:

Given, f, g : R → R be two functions defined as f(x) = |x| + x and g(x) = |x| - x ∀ x ∈ R

Given, x < 0

If(x) = |x| + x and g(x) = |x| - x 

Now, (fog) (x) = f[g(x)]

= |g(x)| + g(x)

= ||x| - x | + |x| - x 

= |x + x| + x + x          (∵ x < 0)

= |2x| + 2x                 

= 2x + 2x

= 4x

Concept:

\(\rm \displaystyle\int_{a}^{b} f(x) dx = \displaystyle\int_{a}^{b} f(a+b-x) dx\)

Calculation:

Consider I = \(\rm \displaystyle\int_0^{\pi/2} \dfrac{\sqrt{\sin^8 x}}{\sqrt{\sin^8 x}+ \sqrt{\cos^8 x}}dx\)      ----(i)

⇒ I =  \(\rm \displaystyle\int_0^{\pi/2} \dfrac{\sqrt{\sin^8 (π/2 -x)}}{\sqrt{\sin^8 (π/2 -x)}+ \sqrt{\cos^8 (π/2 -x)}}dx\)

⇒ I = \(\rm \displaystyle\int_0^{\pi/2} \dfrac{\sqrt{\cos^8 x}}{\sqrt{\sin^8 x}+ \sqrt{\cos^8 x}}dx\)      ----(ii)

Add (i) and (ii), we get

⇒ 2I = 

⇒ 2I = \(\rm \displaystyle\int_0^{\pi/2}dx\)

⇒ 2I = \(\rm[x]^\frac{\pi}{2}_0\)

⇒ I = \(\dfrac{\pi}{4}\)

Concept:

An operation * on a non-empty set S, is said to be a binary operation if it satisfies the closure property.

Closure Property:

Let S be a non-empty set and a, b ∈ S, if a * b ∈ S for all a, b ∈ S then S is said to be closed with respect to operation *.

Calculation:

Given: * is a binary operation on A = {1, 2, 3, 4, 5} given by the table shown below:

Here we have to find the value of (3 * 3) * (4 * 4)

⇒ (3 * 3) = 3 and (4 * 4) = 4-----------(from the table given above)

⇒ (3 * 3) * (4 * 4) = 3 * 4 = 1-----------(From the table given above)

Concept:

The range of $f\left(x\right)$ is all the $y$-values where there is a number $x$ with $y=f\left(x\right)$.

Calcuations: 

To find the range of the function \(f(x) = \frac {|x|}{x}, x \neq 0\), first split the function.
\(\rm f(x) = \frac {x}{x} , x>0\)

\(\rm f(x) = \frac {-x}{x} , x<0\)

We know that, the range of $f\left(x\right)$ is all the $y$-values where there is a number $x$ with $y=f\left(x\right)$.

Now to find the range, take the limit of the function.

\(\lim_{x \to\infty } \rm \frac{x}{x} = \lim_{x \to\infty }1 = 1\)

Now, 

\(\lim_{x \to\infty }\rm \frac{-x}{x} = \lim_{x \to\infty }-1 = -1\).

The range of the function \(\rm f(x)=\dfrac{|x|}{x}, \ x \neq 0 \ \) is (1, -1).

Concept:

gof(x) = g(f(x))

Chain rule : [f(g(h(x))]' = f'(g(h(x))g'(h(x)h'(x)   where f'(x) means derivative of f(x) wrt x

Formula:

d/dx(e^x) = e^x

d/dx(sinx) = cos x

Calculation:

Given:

f : R → R, f(x) = sin(sinx) and g : R → R, g(x) = ex

∵ gof(x) = g(f(x))

⇒ [gof(x)]' = g'(f(x))f'(x)

⇒ [gof(x)]' = e^sin(sinx)[sin(sinx)]'

⇒ [gof(x)]' = esin(sinx)cos(sinx)[(sinx)]'

⇒ [gof(x)]' = esin(sinx)cos(sinx)cosx

So, the correct answer is option 1.

Concept :

For any fixed integer n > 1 and for any arbitrary integers a, b, c and d.

If a ≡ b (mod n) and c ≡ d (mod n) then

• a + c ≡ b + d (mod n) and 
• ac ≡ bd (mod n)

Proof:

Firstly,

Given , a ≡ b (mod n) and c ≡ d (mod n) 

⇒ n | (a - b) and n | (c - d)

⇒ a - b = hn and c - d = kn ; for some integers h and k    ---(1)

Now consider, a - b + c - d = nh + nk

⇒ (a + c) - (b + d) = n (h + k)         ( ∵ h + k is an integer )

⇒ n | [ (a + c) - (b + d) ]

⇒ a + c ≡ b + d (mod n)

Next,

From (1) we have,

a = b + hn and c = d + kn

Consider, ac = (b + hn) × (d + kn)

⇒ ac = bd + bkn + hnd + hknn

⇒ ac - bd = n (bk + hd + hkn) 

⇒ n | ac - bd                                  ( ∵ (bk + hd + hkn) is an integer )

⇒ ac ≡ bd (mod n)

Hence, the correct answer is option 3)

Concept:

If area of triangle formed by the vertices is zero, then three points are collinear

Formula:

Area of triangle = \(|\frac{1}{2}\begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1\\x_3 & y_3 & 1 \end{vmatrix}|\)

Calculation:

Given:

The points A(a , b), B(b , a) and C(2a - 3b , 3b - a) 

Area = \(|\frac{1}{2}\begin{vmatrix} a & b & 1 \\ b & a & 1\\2a-3b & 3b-a & 1 \end{vmatrix}|\)

= 1/2[a(a × 1 - (3b - a) × 1) -b(b × 1 - (2a - 3b) × 1) +1(b × (3b - a) - (2a - 3b) × a)]

⇒ Area = 0 

Hence points are collinear.

Concept:

The associative property of matrix is given by:

X (YZ) = (XY) Z      ----(1)

Given:

AB = B and BA = A      ----(2)

Calculation:

A2 + B2

⇒ AA + BB

⇒ A (BA) + B (AB)      [using (2)]

⇒ (AB) A + (BA) B      [using (1)]

⇒ BA + AB

⇒ A + B

Hence, A2 + B2 = A + B.

Concept:

Formula:

Area of triangle = \(\frac{1}{2}\begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1\\x_3 & y_3 & 1 \end{vmatrix}\)

Calculation:

Given: The area of the triangle formed by vertices (-k , k), (1 , 0) and (5 , 0) is 8 square units.

Area = \(\frac{1}{2}\begin{vmatrix} -k & k& 1 \\ 1 & 0 & 1\\5 & 0 & 1 \end{vmatrix}\)

⇒ |- k(0 × 1 - 0 × 1) - k(1 × 1 - 5 × 1) + 1(1 × 0 - 5 × 0)| = 8 × 2

⇒ |4k| = 16

⇒ k = ± 4

So, the correct answer is option 3

Concept

Let the system of equations be,

a₁x + b₁y + c₁z = d₁

a₂x + b₂y + c₂z = d₂

a₃x + b₃y + c₃z = d₃

\(\; \Rightarrow \left[ {\begin{array}{*{20}{c}} {{a_1}}&{{b_1}}&{{c_1}}\\ {{a_2}}&{{b_2}}&{{c_2}}\\ {{a_3}}&{{b_3}}&{{c_3}} \end{array}} \right]\;\left[ {\begin{array}{*{20}{c}} x\\ y\\ z \end{array}} \right] = \;\left[ {\begin{array}{*{20}{c}} {{d_1}}\\ {{d_2}}\\ {{d_3}} \end{array}} \right]\)

⇒ AX = B

⇒ X = A^-1 B = \(\frac{{{\rm{adj\;}}\left( {\rm{A}} \right)}}{{\det {\rm{\;}}({\rm{A}})}}\;B\)

⇒ If det (A) ≠ 0, the system is consistent having unique solution.

⇒ If det (A) = 0 and (adj A). B = 0, system is consistent, with infinitely many solutions.

⇒ If det (A) = 0 and (adj A). B ≠ 0, system is inconsistent (no solution)

Calculation:

Given: 

kx + y + z = 1, x + ky + z = k and x + y + kz = k²

\( \Rightarrow {\rm{A}} = {\rm{}}\left[ {\begin{array}{*{20}{c}} {\rm{k}}&1&1\\ 1&{\rm{k}}&1\\ 1&1&{\rm{k}} \end{array}} \right],{\rm{B}} = {\rm{}}\left[ {\begin{array}{*{20}{c}} {\rm{x}}\\ {\rm{y}}\\ {\rm{z}} \end{array}} \right]{\rm{and\;C}} = \left[ {\begin{array}{*{20}{c}} 1\\ {\rm{k}}\\ {{{\rm{k}}^2}} \end{array}} \right]\)

⇒ For the given equations to have no solution, |A| = 0

\(\Rightarrow \left| {\begin{array}{*{20}{c}} {\rm{k}}&1&1\\ 1&{\rm{k}}&1\\ 1&1&{\rm{k}} \end{array}} \right| = 0\)

⇒ k(k² – 1) - 1(k – 1) + 1(1 – k) = 0

⇒ k³ – k – k + 1 + 1 – k = 0

⇒ k³ -3k +2 = 0

⇒ (k – 1) (k – 1) (k + 2) = 0

⇒ k = 1, -2

If we put k = 1 in the above given equations, then all the equations will become the same.

Hence, the given equations have no solution if k = - 2. 

EXPLANATION:

• Index numbering is classified into two categories -

1.   Weighted method - Weighted Aggregative Index and Weighted Average of Relatives
2.  Simple or unweighted method - Simple aggregate method and Simple average of price relatives

• Hence statement (i) is wrong.
• The index number of the base year is always chosen to be 100 hence statement (ii) is correct.
• The price index of a year tells how much percent price in the current year increased or decreased concerning the base year. For example, if the price index is 110, prices increased by 10% regarding the base year hence statement (iii) is wrong.

Concept:

• Marginal revenue is the instantaneous rate of change of revenue function with respect to the production output x.
• Marginal revenue = R'(x) = \( \frac{dR(x)}{dx}\)
• Revenue function R(x) = x p(x) 

Calculation:

Given: Price per unit p(x) = 400 - αx2, Marginal revenue when 10 units were in demand = 0

• Revenue function, R(x) = x p(x)

⇒ R(x) = x(400 - αx2)

⇒ R(x) = 400x - αx³

• Marginal revenue,

⇒ R'(x) = \( \frac{dR(x)}{dx}\) 

\(⇒ R'(x) = \frac{d( 400x - αx^3)}{dx}\)

⇒ R'(x) = 400 - 3αx2

So, Marginal revenue at (x = 10),

⇒ R'(x = 10) = 400 - 3α(10)²

⇒ R'(x = 10) = 400 - 300α

⇒ 400 - 300α = 0 (given)

⇒ α = 4/3

Concept:

• The formula for a simple average of price relative, 

 \(P_{01} = \frac{Σ (\frac{P_{1} }{P_0} \times 100)}{N}\)

where, 

P₁ = The price of the current year, P₀  = The price of the base year., N is the number of items.

Calculation;

Given: Data for the current and the base year are in the table.

 \(⇒P_{01} = \frac{Σ (\frac{P_{1} }{P_0} \times 100)}{N} =690/5\) 

⇒ P_{01} = 138

So, the correct answer is option 3.

Concept:

sin (π - θ) = sin θ 

\({\sin ^{ - 1}}(sinx)= x\), x ∈ [\(\rm \frac {-\pi}{2},\rm \frac {\pi}{2}\)]

Calculation:

\(\rm {\sin ^{ - 1}}(sin \frac{{3\pi }}{5})\)

\(\rm = {\sin ^{ - 1}}[sin \rm (\pi - \frac{{2\pi }}{5})]\)

\(= {\sin ^{ - 1}}(sin \frac{{2\pi }}{5})\)                       (∵ sin (π - θ) = sin θ)

\(= \frac{{2\pi }}{5}\)

Hence, option (2) is correct.

Concept:

• Total cost function C(x) = Variable cost function + Other expenses
• Marginal cost is the instantaneous rate of change of cost function with respect to the production output x.

Marginal cost = C'(x) = \( \frac{dC(x)}{dx}\) 

Calculation:

Given:

• Variable cost function C(x) = α2x(β√x + αβ) 
• The cost of storage and other expenses are Rs. 4500
• The marginal cost is 40 when 144 toys have been produced.
• αβ = 1

Total cost function C(x) =  α2x(β√x + αβ) + 4500

⇒ C'(x) = \( \frac{d}{dx} \ [α^2x(β√x + αβ) + 4500]\)

= \(⇒ C'(x) = \frac{d}{dx} [ α^2βx^{3/2} + αβx + 4500]\)

⇒ C'(x) = (3/2)α2βx^(1/2) + α³β

• At x = 144, Marginal cost = C'(x) = (3/2)α2β(144)(1/2) + α³β

⇒ C'(x) = 18α2β + α3β

⇒ C'(x) = α2β(18 + α)

⇒ C'(x) = α(αβ)(18 + α)

⇒ C'(x) = α² + 18α     (∵ αβ = 1)

⇒ α2 + 18α = 40

⇒ (α + 20)(α - 2) = 0

⇒ α = - 20 or 2

CONCEPT:

• For the function to be monotonically decreasing function,

f'(x) < 0

CALCULATION

Given:

f(x) = tan-1x - x 

⇒ f'(x) < 0

\(f'(x) = \frac{1}{{1 + {x^2}}} - 1 = \frac{{ - {x^2}}}{{1 + {x^2}}} < 0\)

• So, the expression is true for all the values of x ϵ R except 0, as the square of any real number will always be positive.

⇒ x ϵ R - {0}

• So, the correct answer is option 3.

Concept:

The rate of change of the value of a function f(x) with respect to a variable t, is given by: \(\rm \frac {df(x)}{dt}\)

Calculation:

Given the side of the cube L = 10cm and \(\rm dL\over dt\) = 4 cm/s

Now volume of the cube V = L3 

\(\rm dV\over dt\) = \(\rm dV\over dL\) × \(\rm dL\over dt\)

\(\rm dV\over dt\) = \(\rm dL^3\over dL\) × 4

\(\rm dV\over dt\) = 3L2 × 4

\(\rm dV\over dt\) = 12 × 102 

\(\rm dV\over dt\) = 1200 cm3/s

Calculation:

y + sin-1 (1 - x2) = ex 

y = e^x - sin^-1 (1 - x²)

Differentiating w.r.t x, we get

\(\rm {dy\over dx} = e^x - {1\over\sqrt{1-(1-x^2)^2}}(-2x)\)

\(\rm {dy\over dx} = e^x + {2x\over\sqrt{1-(1-2x^2+x^4)}}\)

\(\rm {dy\over dx} = e^x + {2x\over\sqrt{2x^2-x^4}}\)

\(\boldsymbol{\rm {dy\over dx} = e^x + {2\over\sqrt{2-x^2}}}\)

CONCEPT:

•  if the entity inside the modulus function is positive then the modulus function will open as positive otherwise it will open by taking a negative sign to make it overall positive as the modulus function can not be negative. 

CALCULATION

Given:

f(x) = | x2 - 3x -4 | 

• To check whether function inside the modulus function (x2- 3x - 4 ) is positive or negative for -1 ≤ x ≤ 4,

∵ x2- 3x - 4 = (x - 4)(x + 1)

• Here the points of inversion for the given function are -1 and 4 for which this function will change its sign.

⇒ For -1 ≤ x ≤ 4, x2 - 3x - 4 ≤  0.

• So, the modulus function will open by taken the negative sign.

∴ f(x) = - (x2 - 3x -4) = (4 - x)(x + 1)      ...(1)

• For maximum and minimum values, using equation 1,
• As  f(x) is continuous in -1 ≤ x ≤ 4 so it will be differentiable for these values of x,

∴ f(x) = - (x2 - 3x -4)

⇒  f'(x) = 3 - 2x = 0

⇒ x = \( \frac{3}{2}\) ϵ [-1, 4]

⇒  f''(x) = - 2 < 0 for x = 3/2 =1.5

• So at x = 1.5 the function will be having maxima and its maximum value will be f(1.5). 

⇒ f(3/2) = | 1.52 - 3 × 1.5  - 4 |

 ⇒ f(3/2) = 6.25

• So, the correct answer will be option 1.

CONCEPT:

• For the function to be monotonically increasing function,

f'(x) > 0

CALCULATION

Given:

 f(x) = sin4 x + cos4 x

⇒ 4 sin³ x.cos x - 4 cos³ x . sin x > 0

⇒  sinx .cosx(sin² x - cos² x) > 0

⇒  2sinx .cosx ( cos2 x - sin2 x)/2 < 0

• Multiply and divide by 2,

⇒ 2sin 2x. cos2x/2 < 0

⇒ sin 4x < 0

∴ π < 4x < 2π or 3π < 4x < 4π

∴ π/4 < x < π/2 or 3π/4 < x < π

• So, the correct answer is option 2.

Concept:

Rolle's theorem:

Rolle's theorem states that if a function f(x) is continuous in the closed interval [a, b] and differentiable on the open interval (a, b) such that,

f(a) = f(b), then, for some c ∈ [a, b]

f′(c) = 0 

Calculation:

The given function is f(x) = \(\rm \cos \dfrac{x}{2}\) on [π, 3π].

f(π) = \(\rm \cos \dfrac{\pi}{2}\) = 0 and f(3π) = \(\rm \cos \dfrac{3\pi}{2}\) = 0.

Since, f(π) = f(3π), there must exist a c ∈ [π, 3π] such that f'(c) = 0.

f'(x) = \(\rm \dfrac{d}{dx}\left (\cos \dfrac{x}{2} \right )=-\dfrac{1}{2}\left (\sin \dfrac {x}{2} \right )\)

⇒ f'(c) = \(\rm -\dfrac{1}{2}\left (\sin \dfrac {c}{2} \right )\) = 0

⇒ \(\rm \sin \dfrac {c}{2}\) = 0

⇒ \(\rm \dfrac {c}{2}\) = nπ

⇒ c = 2nπ, where n is an integer.

We want c ∈ [π, 3π], therefore c = 2π.

Concept:

For a function say f, \(\mathop {\lim }\limits_{x \to a} f(x)\) exists

\( ⇒ \mathop {\lim }\limits_{x \to {a^ - }} f\left( x \right) = \;\mathop {\lim }\limits_{x \to {a^ + }} f\left( x \right) = l = \;\mathop {\lim }\limits_{x \to a} f(x)\), where l is a finite value.

Any function say f is said to be continuous at point say a if and only if \(\mathop {\lim }\limits_{x \to a} f(x) = l = f\left( a \right)\), where l is a finite value.

Calculation:

Given: \(\rm f(x)=\left\{\begin{matrix} \frac{ksin(π -x)}{π -x} & if & x \neq π \\ 1 & if & x =π \end{matrix}\right.\) is continuous at x = π.

As we know that, if a function f is continuous at point say a then \(\mathop {\lim }\limits_{x \to a} f(x) = l = f\left( a \right)\)

\(\Rightarrow \rm \lim_{x\rightarrow π}\frac{ksin(π-x)}{π-x} = f(π) = 1\)

∵ \(\rm \lim_{x\rightarrow π}\frac{ksin(π-x)}{π-x} = \frac{0}{0}\), we can use L'Hospitals rule.

⇒ \(\rm \lim_{x\rightarrow π}\frac{-kcos(π-x)}{-1} = 1\)

⇒ \(\rm \frac{kcos(π-π)}{1} = 1\) 

⇒ k cos(0) = 1

⇒ k = 1.

Hence, option 2 is correct.

Calculation:

Given:

f(x3) = x5 for all x ∈ R, x ≠ 0. Then the value of \(\dfrac{df}{dx} (8)\) 

⇒ f (x³) = x⁵ for all x ∈ R, x ≠ 0

Differentiating w.r.to x 

⇒ (d/dx) f (x³) = (d/dx) x⁵

⇒ f' (x³) (d/dx) x³ = 5x⁴ 

⇒ f' (x3) 3x² = 5x4 

⇒ f'(x³) = (5/3) x² 

Putting x³ = 8 to find f' (8)

Such that f'(8) =( 5 × (2)²)/3 = 20/3

⇒ 20/3

Calculation:

Given: f(x) = xⁿ, n ≠ 0.

F’(x) = nx^{n – 1}

For f(x) to be differentiable for all values of x, n – 1 ≥ 0

So, n ≥ 1.

Concept:

Parametric Derivative

Let's define a function by the pair of parametric equations:

\( \begin{cases} x=x(t) \\ y=y(t) \end{cases} \), and a ≤ t ≤ b,

Where x(t), y(t) are differentiable functions and x'(t) ≠ 0. Then the derivative \(\frac{dy}{dx}\) is defined by the formula:

\(\frac{dy}{dx}={\frac{dy}{dt}\over\frac{dx}{dt}}, \ and \ a \leq t\leq b,\)

Where \(\frac{dy}{dt}\) - The derivative of the parametric equation y(t) by the parameter t and

\(\frac{dx}{dt}\) - The derivative of the parametric equation x(t), by the parameter t.

Calculation:

Given equation of curve is x = t2 + 3t - 8 and y = 2t2 - 2t - 5

⇒\(\frac{dx}{dt}\) = 2t + 3 and \(\frac{dy}{dt}\) = 4t - 2

⇒ \(\frac{dy}{dx}={\frac{dy}{dt}\over\frac{dx}{dt}}={{4t-2}\over{2t+3}} \)

Slope of tangent,\(\big(\frac{dy}{dx}\big)_{t=2}={4\times 2-2\over2\times2+3}=\frac{6}{7} \)

∴ The slope of the tangent to the curve is \(\frac{6}{7}\)

Concept:

Integration is the inverse process of differentiation and therefore it is called anti-differentiation.

i.e If g (x) = f’(x) then \(\smallint g\left( x \right)\;dx = \smallint f'\left( x \right)\;dx = f\left( x \right) + C\)

\(\smallint \left( {f\left( x \right) + g\left( x \right)} \right)\;dx = \smallint f\left( x \right)\;dx + \smallint g\left( x \right)\;dx\)

\(\smallint a{x^n}\;dx = a \times \frac{{{x^{n\; + \;1}}}}{{n + 1}} + C\)

Calculation:

Given: \(f'\left( x \right) = \frac{{{x^2}}}{2} - kx + 1\), such that f(0) = 0 and f(3) = 15.

Now, by integrating f’(x), we get

\(\Rightarrow f\left( x \right) = \smallint f'\left( x \right)\;dx = \smallint \left( {\frac{{{x^2}}}{2} - kx + 1} \right)\;dx\)

\(\Rightarrow f\left( x \right) = \smallint \frac{{{x^2}}}{2}\;dx - k\smallint x\;dx + \;\smallint dx\)

\(\Rightarrow f\left( x \right) = \frac{{{x^3}}}{6} - k \times \frac{{{x^2}}}{2} + x + C\)

As it is given that, f(0) = 0 and f(3) = 15.

\(\Rightarrow f\left( 0 \right) = C = 0\)

\(\Rightarrow f\left( x \right) = \frac{{{x^3}}}{6} - k \times \frac{{{x^2}}}{2} + x\)

\(\Rightarrow f\left( 3 \right) = \frac{9}{2} - \frac{9}{2}\;k + 3 = 15\)

Concept:

Lagrange’s Mean Value Theorem:
If a function f is defined on the closed interval [a, b] satisfying:

• The function f is continuous on the closed interval [a, b].
• The function f is differentiable on the open interval (a, b).

Then there exists a value x = c such that f'(c) = \(\rm \frac{f(b)-f(a)}{b-a}\).

Calculation:

The given function f(x) = \(\rm x+\frac1x\) is both differentiable and continuous in the interval [1, 3].

f'(x) = \(\rm 1-\frac1{x^2}\)

By the Mean Value Theorem, there exists a c ∈ [1, 3], such that:

f'(c) = \(\rm \frac{f(b)-f(a)}{b-a}\)

⇒ \(\rm 1-\frac1{c^2}=\frac{f(3)-f(1)}{3-1}\)

⇒ \(\rm 1-\frac1{c^2}=\frac{\left(3+\frac13\right)-\left(1+\frac11\right)}{2}\)

⇒ \(\rm 1-\frac1{c^2}=\frac{\frac43}{2}=\frac23\)

⇒ \(\rm \frac1{c^2}=1-\frac23=\frac13\)

⇒ c = √3.

Concept: 

\(\rm \int x^{n}\space dx = \frac{x^{n + 1}}{n + 1} + C\)

\(\rm \int f(x) \space dx^2\) = \(\rm \int (1 + x^{2} + x^{4}) \space d(x^2)\)      ....(i)

Calculation:

Let, x² = u

From equation (i)

​\(\rm \int f(x) \space dx^2\) = \(\rm \int (1 + u + u^{2}) \space du\)

⇒ u + \(\rm \frac{u^{2}}{2}\) + \(\rm \frac{u^{3}}{3}\)+ C

Now putting the value of u,

​⇒ \(\rm \int f(x)dx^2\) = x² +​ \(\rm \frac{x^{4}}{2}\) + \(\rm \frac{x^{6}}{3}\) + C

∴ The required integral is x2 +​ \(\rm \frac{x^{4}}{2}\) + \(\rm \frac{x^{6}}{3}\) + C.

Concept:

• Parabola: A parabola is a plane curve that is mirror-symmetrical and is approximately U-shaped.
• The general equation of a parabola is: y = a(x - h)² + k or x = a(y - k)² + h, where (h, k) denotes the vertex.
• The standard equation of a regular parabola is y² = 4ax
• The general equation of a straight line is. y = m x + c, where m is the slope and c is the intercept.

Calculations:

Given the parabola 4y = 3x2 is cut off by the straight line 2y = 3x + 12.

Let us find the intersection point of the parabola and the straight line by substituting one equation in the other 

i.e.,  2y = 3x + 12 in 4y = 3x², we get

2(3x + 12)= 3x²

⇒ x² - 2x - 8 = 0

⇒ x = 4 or -2 are the roots of the quadratic equation x2 - 2x - 8 = 0

Now putting these values in the equation of straight line 2y = 3x + 12 we get the value of y 

i.e., \(y=\frac{3x+12}{2}\)

put x = 4 in this,we get

\(y=\frac{3(4)+12}{2}\)

⇒ y = 12

Now, put x = - 2 in the equation of straight line 2y = 3x + 12  we get the value of y 

\(y=\frac{3(-2)+12}{2}\) 

⇒ y = 3

Therefore the intersection points of the parabola and the straight line are (-2, 3) and (4, 12)

Finally, we need to find the area of the shaded region in the above diagram i.e., the required area of the region is an area of OABO

And this region is bounded by the straight-line 2y = 3x + 12 and the parabola 4y = 3x².

y coordinates of a straight line is \(y=\frac{3x+12}{2}\) = f(x) and parabola is \(y=\frac{3x^2}{4}\) = g(x)

So on integrating f(x) - g(x) we get the area under the straight line and the curve.

Therefore, the required area is

= \(\int\limits_{-2}^4[{f(x)-g(x)]}dx\)

= \(\int\limits_{-2}^4\bigg({\frac{3x+12}{2}-\frac{3x^2}{4}}\bigg)dx\)

on simplifying this and putting limits we get,

= 27 sq.units

Hence, the correct answer is option 4).

Concept:

projection of \(\rm \vec{a}\) in the direction of  \(\rm \vec{b}\)  = \(\rm\dfrac {1}{|\vec b|}(\vec a.\vec b)\)

Calculations:

Given, the angle between \(\rm \vec{a} \ and \ \vec{b} \ is \ \dfrac{2\pi}{3}\) and the projection of \(\vec{a}\) in the direction of \(\vec{b}\) is -2

We know that, 

projection of \(\vec{a}\) in the direction of \(\rm \vec{b}\)= \(\rm\dfrac {1}{|\vec b|}(\vec a.\vec b)\)

⇒ - 2 = \(\rm\dfrac {1}{|\vec b|}(|\vec a||\vec b|\;cos\;\dfrac {2\pi}{3})\)

⇒ - 2 = \(\rm |\vec a|\;cos\;\dfrac {2\pi}{3}\)

⇒ - 2 = \(\rm |\vec a|\;(\dfrac {-1}{2})\)

⇒  \(\rm |\vec a| = 4\)

Hence, if the angle between \(\rm \vec{a} \ and \ \vec{b} \ is \ \dfrac{2\pi}{3}\) and the projection of \(\vec{a}\) in the direction of \(\vec{b}\) is -2, then \(|\vec{a}|=\) 4

Concept:

• If two lines are parallel, then the distance between them is fixed.
• The distance between two parallel lines \(\rm \vec{r}=\vec{a}_1 +\lambda \vec{b}\) and \(\rm r=\vec{a}_2 + \mu\vec{b}\) is given by the formula: \(\rm d=\left|\dfrac{\vec{b}\times (\vec{a}_2-\vec{a}_1)}{|\vec{b}|}\right|\).

Calculation:

Using the formula for the distance between two parallel lines, we can say that the distance is \(\rm d=\left|\dfrac{\vec{b}\times (\vec{a}_2-\vec{a}_1)}{|\vec{b}|}\right|\).

Concept:

Dot product of two perpendicular lines is zero.

Distance between two points (x₁, y_1, z₁) and (x₂, y₂, z₂) is given by, \(\rm \sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}\)

Calculation:

Let M be the foot of perpendicular drawn from the the point P(2, 3, 4)

Let , \(\rm \dfrac{x-0}{1}=\dfrac{y-0}{0}=\dfrac{z-0}{0} =k\)

x = k, y  = 0, z = 0

So M = (k, 0, 0)

Now direction ratios of PM = (2 - k, 3 - 0, 4 - 0) = (2- k, 3, 4) and direction ratios of given line are 1, 0, 0

PM is perepedicular to the given line so,

(2 - k) (1) + 3(0) + 4 (0) = 0

∴ k = 2

M = (2, 0, 0)

Perpendicular distance PM =

 \(\rm \sqrt {(2-2)^2+(0-3)^2+(0-4)^2}\\ =\sqrt{9+16}\\ =5\)

Hence, option (2) is correct. 

Concept:

If θ is the angle between the line \(\vec r = \;\vec a + \lambda \;\vec b\) and the plane \(\vec r \cdot \;\vec n = q\) is given by: \(\sin \theta = \frac{{\vec b \cdot \;\vec n}}{{\left| {\vec b} \right| \times \left| {\vec n} \right|}}\)

Calculation:

Given: Equation of line \(\vec r = \left( {\hat i + 2\hat j - \;\hat k} \right) + \lambda \;\left( {\hat i - \;\hat j + \;\hat k} \right)\) and equation of plane \(\vec r \cdot \left( {2\hat i - \;\hat j + \;\hat k} \right) = 6\)

Here, we have to find the angle between the given line and plane.

As we know that, if θ is the angle between the line \(\vec r = \;\vec a + \lambda \;\vec b\) and the plane \(\vec r \cdot \;\vec n = q\) is given by: \(\sin \theta = \frac{{\vec b \cdot \;\vec n}}{{\left| {\vec b} \right| \times \left| {\vec n} \right|}}\)

Here, \(\vec b = \hat i - \;\hat j + \;\hat k\) and \(\vec n = 2\hat i - \;\hat j + \;\hat k\)

⇒ \(\vec b \cdot \;\vec n = 2 + 1 + 1 = 4\), \(\left| {\vec b} \right| = \sqrt {{1^2} + {{\left( { - \;1} \right)}^2} + {1^2}} = \sqrt 3 \;and\;\left| {\vec n} \right| = \sqrt {{2^2} + {{\left( { - \;1} \right)}^2} + {{\left( 1 \right)}^2}} = \sqrt 6 \)

By substituting the above given values in \(\sin \theta = \frac{{\vec b \cdot \;\vec n}}{{\left| {\vec b} \right| \times \left| {\vec n} \right|}}\), we get

\(\Rightarrow \sin \theta = \frac{4}{{\sqrt 3 \times \sqrt 6 }} = \frac{{2\sqrt 2 }}{3}\)

\(\Rightarrow \theta = {\sin ^{ - 1}}\left( {\frac{{2\sqrt 2 }}{3}} \right)\)

Explanation:

After equating each constraint to zero (0) we get the equations of lines on cartesian coordinate.

On comparing the inequality with (0,0) and shading the common area we get the feasible region as follows,

Hence we can easily conclude that point (60,0) and (0,100) are outside the feasible region.

Additional Information

Points outside the feasible region do not contribute to the objective function.

Concept:

Probability of an event happening = \(\rm \dfrac{\text{(Number of ways it can happen)}}{\text{ (Total number of outcomes)}}\)

If a die thrown, Number of sample space = 6, If two dice are thrown n(S) = 6² = 36

Calculation:

Here, four dice are thrown, 

n(S) = 6⁴

Now, sum of the numbers appearing on them 25 = { }       

⇒ n = 0               

(∵maximum sum = 6 + 6 + 6 + 6 = 24)

∴ Probability = 0/(6⁴) = 0

Hence, option (1) is correct.

Explanation:

Two dice are thrown and sum = 7

Now, the possible cases are

(1, 6) (6, 1) (2, 5) (5, 2) (3, 4) (4, 3)

∴ Total number of cases for sum (7) = 6

Now, favourable cases will be,

Favourable cases for 2 appears at least once = 2 {(2, 5), (5, 2)}

∴ The Probability that number 2 has appeared at least once \(= \frac{2}{6} = \frac{1}{3}\)

EXPLANATION:

Properties of Linear programming problem:

Proportionality → Relationship between two variables is assumed to be linear (or) proportional

Non-negativity → All decision variables & RHS values are non-negative in nature.

Additivity → Two difference variables can be added

Certainty → All the parameters are deterministic in nature & resources are limited

Hence, negativity is not a property whereas non-negativity is a property.

Let E₁, E₂ and E₃ denote the events of selecting box A, B, C respectively and A be the event that a screw selected at random is defective.

Then,

P(E₁) = P(E₂) = P(E₃) = 1/3,

\({\rm{P}}\left( {{\rm{A}}/{{\rm{E}}_1}} \right) = \frac{1}{5}\)

\({\rm{P}}\left( {\frac{{\rm{A}}}{{{{\rm{E}}_2}}}} \right) = \frac{1}{6} \Rightarrow {\rm{P}}\left( {{\rm{A}}/{{\rm{E}}_3}} \right) = \frac{1}{7}\)

Then, by Baye’s theorem, required probability

= P(E₁/A)

Concept:

\({\rm{Probability\;of\;exactly\;r\;success\;in\;n\;trials}} = {\rm{p}}\left( {{\rm{x}} = {\rm{r}}} \right) = {{\rm{\;}}^{\rm{n}}}{{\rm{C}}_{\rm{r}}}{\rm{\;}}{{\rm{p}}^{\rm{r}}}{{\rm{q}}^{{\rm{n}} - {\rm{r}}}}\)

Where n = number of trials; p = probability of success; q = (1 - p) = probability of failure

Calculation:

Given: 75% patient is cured by the medicine

Number of patients = 5

∴ n = 5

Let p be the probability that a patient is cured by the medicine

⇒ p = 75% = 3/4

Probability that a patient is not cured by medicine = q

⇒ q = 1 – p = 1 – 3/4 = 1/4

Probability that at least one patient is cured = P(X ≥ 1)

⇒ P(X ≥ 1) = 1 - Probability that no patient is cured 

= 1 − P(X = 0)

\(= 1{ - ^5}{{\rm{C}}_0}{\frac{3}{4}^0} \times {\left( {\frac{1}{4}} \right)^{5 - 0}}{\rm{\;}}\)

\(= 1 - {\left( {\frac{1}{4}} \right)^5} = \frac{{1023}}{{1024}}{\rm{\;}}\)

Concept:

Expected Mean:

E(x) = np  

Calculations:

Given that x ~ B ( n = 10, p)

E(x) = 8 

We know that E(x) = np

⇒ np = 8 

⇒(10)p = 8

⇒ p = \(\rm\dfrac {8}{10}\)

⇒ p = 0.8

Given that x ~ B ( n = 10, p) if E(x) = 8 then the value of P is 0.8

Calculation:

Probability Distribution of Z is

0.4 + k + 2k = 1

3k = 1 - 0.4

k = 0.2

P(you study atleast two hours) 

P(Z ≥ 2) = P(2) + P(3) + P(4) + .....

              = 2k

              = 2 × 0.2

P(Z ≥ 2) = 0.4

Concept:

For two events A and B:

• P(A ∪ B) = P(A) + P(B) - P(A ∩ B).
• The conditional probability of A given B is defined as: P(A|B) = \(\rm \dfrac{P(A\cap B)}{P(B)}\), when P(B) > 0.

Calculation:

Using the relation P(B|A) = \(\rm \dfrac{P(A\cap B)}{0.4}\), we get:

0.6 = \(\rm \dfrac{P(A\cap B)}{0.4}\)

⇒ P(A ∩ B) = 0.24

Now using the relation P(A ∪ B) = P(A) + P(B) - P(A ∩ B), we get:

P(A ∪ B) = 0.4 + 0.8 - 0.24 = 0.96.

CONCEPT:

• For a non-singular matrix A,

A^-1 = \(\frac{{adjA}}{{|A|}}\)      ....(1)

• The adj A can be calculated by keeping the diagonal elements as it is and changing the sign of anti diagonal elements (This trick is valid only for 2 × 2 matrix)

⇒ Adj A = \({\left( {\begin{array}{*{20}{c}} 4&{ - 2}\\ { - 9}&2 \end{array}} \right)}\)

CALCULATION

Given:

A = \(\left( {\begin{array}{*{20}{c}} 2&2\\ 9&4 \end{array}} \right)\)

⇒ | A | = 8 - 18 = -10 ≠ 0 

• So, matrix A is the non-singular matrix.

Using equation 1,

\(\Rightarrow A^{-1} = \frac{{\left( {\begin{array}{*{20}{c}} 4&{ - 2}\\ { - 9}&2 \end{array}} \right)}}{{ - 10}}\)

⇒ 10A^-1 = \({\left( {\begin{array}{*{20}{c}} -4&{ 2}\\ { 9}&-2 \end{array}} \right)}\)= A - 6I.

• So, the correct answer is option 2.

Concept:

\(\rm \frac{d\sin x}{dx} = \cos x\\ \rm \frac{d\cos x}{dx} = -\sin x\)

Calculation:

Given:

y = a sin (λx + α)               .... (1)

Now differentiating both sides, we get

\(⇒ \rm \frac{dy}{dx} = \frac{a d\sin{(\lambda x + \alpha)}}{d(\lambda x + \alpha)}\times\frac{{d(\lambda x + \alpha)}}{dx} \)

⇒ \(\rm \frac{dy}{dx} = a\lambda\cos\;(\lambda x+\alpha)\)

Again differentiating both sides, we get

\(\Rightarrow \frac{{{{\rm{d}}^2}{\rm{y}}}}{{{\rm{d}}{{\rm{x}}^2}}} = \rm -a\lambda^2\) sin (λx + α)

From equation 1^st, 

\(\Rightarrow \frac{{{{\rm{d}}^2}{\rm{y}}}}{{{\rm{d}}{{\rm{x}}^2}}} = \rm -\lambda^2 y\)

∴ \(\rm \frac {d^2y}{dx^2} + \lambda^2y = 0\)