Quantitative Aptitude PYQs

Concept:

• Equation of XY plane  is z = o
• Equation of YZ plane  is x = o
• Equation of XZ plane  is y = o

 

Explanation:

Given that,

z = 7

We know that

z = 0 represents the XY-plane.

∴ z  = 7 represents a plane parallel to XY-plane.

Concept:

Direction cosine:

Direction cosines are the cosines of the angles made by a directed line segment with the coordinate axes. 

The sum of the square of direction cosine is always 1.

Direction Ratio:

• If a, b, c are three numbers proportional to the direction cosine l, m, n of a straight line, then a, b, c are called its direction ratios.
• They are also called direction numbers or direction components.
• Direction cosines of a line are unique but direction ratios of a line in no way unique but can be infinite.

Explanation:

Statement: 1

The direction cosines for any line must be unique. However, there are infinitely many sets of direction ratios since direction ratios are just a set of any three numbers proportional to the direction cosines.

Statement: 2

We know that the sum of the squares of the direction cosines is equal to one which can be equal to the sum of the direction cosines (l + m + n) for certain lines like \(\frac{x}{1} =\frac{z}{0} = \frac{z}{0} \)

as, l = 1, m = 0, n = 0 and  (l + m + n = 1) so statement 2 is also correct.

Concept:

Let XZ-plane divide line AB.at P(x,y,z) in the ratio k :

1. Co- ordinate of point P(x,y,z) that divide line segment joining

point A(x₁​, y₁​, z₁​) and B(x₂​, y₂​, z₂​) in the ratio m : n is

\(\rm \left ( \frac{mx_2 + mx_1}{m + n}, \frac{my_2 + my_1}{m + n}, \frac{mz_2 + mz_1}{m + n} \right )\)

In XY-plane, z coordinate is zero.

Calculation:

Let XY-plane divides the line joining the points A(−1, 3, 4) and

B(2, −5, 6) in the ratio k :1 at point P.

Therefore, by using section formula, co-ordinates of P are

P(x, y, z) =  (\(\rm \frac{2k-1}{k+1}\), \(\rm \frac{-5k+3}{k+1}\), \(\rm \frac{6k+4}{k+1}\))

Since in the XY-plane,  z = 0.

⇒ \(\rm \frac{6k+4}{k+1}\) = 0

⇒ 6k + 4 = 0

⇒ k = \(-​​\rm \frac{2}{3}\)

∴ The ratio is 2 : 3 externally.

Concept:

Octant:  An octant in geometry is one of the eight divisions of a

Euclidean three-dimensional coordinate system defined by the signs

of the coordinates.

Explanation:

Let the radius of the sphere be r,  then the distance of its center from

coordinate planes which it is touching should be equal to radius a.

Hence, its center is (r, r). 

Also,

There are 8 quadrant in 3D, so for a given radius there will be 8

possible sphere touching all the 3 coordinate planes.

Given:

ABCD is a base of cuboid ABCDEFGH and 

 A(0, 0, 0), B(12, 0, 0), C(12, 6, 0) and G(12, 6, 4) be the vertices.

Concept:

Consider two vectors a and b separated by some angle θ. Then according to the formula of the dot product is:

\(\rm cosθ = \frac{\vec{a}.\vec{b}}{|\vec{a}|.|\vec{b}|}\)

If \(\vec{p}=\rm a:\widehat{i} + b: \widehat{j} + c : \widehat{k} \) is a vector then its magnitude is given by

 \(|\vec{p}| = \sqrt{a^2 + b^2 + c^2}\)

Calculation:

The position vector of AB is 

\(\rm \overrightarrow{AB}\) = \((\rm 12:\widehat{i} + 0: \widehat{j} + 0 : \widehat{k} )-(\rm 0:\widehat{i} + 0: \widehat{j} + 0 : \widehat{k}​​​​)\)

⇒ \(\rm \overrightarrow{AB}\) = \(\rm 12: \widehat{i}\)

And, its magnitude is

\(\rm |\overrightarrow{AB}| = \sqrt{12^2 + 0^2 + 0^2}\)

⇒ \(\rm |\overrightarrow{AB}| = 12\)

Similarly,

\(\rm \overrightarrow{AG}\) = \(\rm 12:\widehat{i} + 6: \widehat{j} + 4 : \widehat{k} \)

⇒\(\rm |\overrightarrow{AG}| = \sqrt{144 + 36 + 16}\)

⇒ \(\rm |\overrightarrow{AG}| = 14\)

\(\rm \overrightarrow{AC}\) = \(\rm 12:\widehat{i} + 6: \widehat{j}\)

⇒ \(\rm |\overrightarrow{AC}| = \sqrt{144 + 36}\)

⇒ \(\rm |\overrightarrow{AC}| = \sqrt{180}\)

Hence, angle b/w \(\rm \overrightarrow{AB}\) and \(\rm \overrightarrow{AG}\) is

\(\rm cos(α) = \frac{\overrightarrow{AB}.\overrightarrow{AG}}{|\overrightarrow{AB}|.|\overrightarrow{AG}|}\)

⇒ \(\rm cos(α) = \frac{144}{12 \times 14}\)

⇒ cos α = \(\rm \frac{6}{7}\)

⇒ cos β = \(\rm \frac{\sqrt{180}}{14}\)

⇒ cos 2α + cos 2β

⇒ 2cos²α - 1 + 2 cos²β - 1  (∵ cos2θ = 2cos2θ  - 1)

⇒  \(2\times (\frac{6}{7})^2-1+2\times (\frac{\sqrt{180}}{14})^2-1+\)

⇒ \(\rm \frac{72}{49} + \rm \frac{90}{49} - 2\) 

⇒ cos 2α + cos 2β = \(\frac{64}{49}\)

∴ The value of cos 2α + cos 2β is \(\rm \frac{64}{49}\)

Given:

|\(\rm⃗ {a}\)| = || = |\(\rm⃗{c}\)| = 1

\(\rm⃗{a} \times ⃗{b}\) ⊥ \(⃗{c}\)

Concept:

If \(\rm⃗ {a}\) and \(\rm⃗ {b}\) are two vectors, then

\(\rm⃗{a} \times ⃗{b} = |⃗{a}| |⃗{b}| sin \theta\ \hat{n}\)

Where, \(\hat{n}\) is a unit vector that represents the direction of \(\rm⃗{a} \times ⃗{b}\). 

Direction of \(\hat{n}\) or \(\rm⃗{a} \times ⃗{b}\) is perpendicular to both  \(\rm⃗ {a}\) and \(\rm⃗ {b}\)

Scalar triple product:

\([\overrightarrow a \overrightarrow b \overrightarrow c ] = a.(b \times c) \)

 if any two vectors among a,b and c are equal, the scalar triple product is zero

Explanation:

Statement: 1

We know that, the cross-product of vector  \(\rm⃗ {a}\) and \(\rm⃗ {b}\) is

\(\rm⃗{a} \times ⃗{b} = ⃗{a}~ ⃗{b} ~sin \theta~ \hat n \)

\(\hat n\) is unit vectors in the direction of \(\rm\vec{a} \times \vec{b}\) so the first statement cannot be correct as \(\vec {c}\) and \(\hat n\) are perpendicular to each other. 

Hence, statement 1 is incorrect.

Statement: 2

∵ \([ ⃗ a\ \overrightarrow b \overrightarrow c ] = [ ⃗ b\ \overrightarrow c \overrightarrow a] = [⃗ c\ \overrightarrow a \overrightarrow b ] \)

So,

\([\overrightarrow a \overrightarrow b \overrightarrow c ] = [⃗ c\ \overrightarrow a \overrightarrow b ] = \vec c.(⃗ a \times ⃗ b) \) 

Since \(\rm⃗{a} \times ⃗{b}\) and \(\vec{c}\) are perpendicular to each other so,

 \(\rm\vec {c} \cdot (\vec{a}\times \vec{b})=0\)

⇒ \(\rm\vec {a} \cdot (\vec{b}\times \vec{c})=0\)

So, statement 2 is correct.

∴ Option 2 is correct.

Concept:

If θ is the angle between vectors \(\vec a\) and \(\vec b\) then 

\(\rm cosθ = \frac{⃗{a}.⃗{b}}{|⃗{a}|.|⃗{b}|}\)

Calculation:

Let the angle between vectors \(\vec a\) and \(\vec b\) be θ

\(\vec{a}+3\vec{b} = 3\hat{i}- \hat{j}\)       ....(i)

\(2\vec{a}+\vec{b} = \hat{i}- 2\hat{j}\)       ....(ii)

On doing (i) × 2 - (ii), we get 

\(⇒ 5\vec b = 5\hat i \)

\(⇒ \vec b = \hat i \)

On putting the value of the vector \( \vec b \) in equation (i), we get

⇒ \(\rm ⃗{a} = -\hat{j}\)

Now, \(\vec {a}. \vec{b} = \hat{i} .(- \hat{j}) = 0\)

According to the concept used

\(\rm cosθ = \frac{⃗{a}.⃗{b}}{|⃗{a}|.|⃗{b}|}\)

⇒ cosθ = 0

⇒ θ = cos^-10

⇒ \(\rm θ = \frac{\pi}{2}\)

∴ The angle between \(\rm⃗{a}\) and \(\rm⃗{b}\) is \(\rm \frac{\pi}{2}\).

Concept:

If \(\rm(\vec {a} + \vec{b})\) is perpendicular to \(\rm\vec {a}\), then 

\(\rm(\vec {a} + \vec{b})\).\(\vec {a}\) = 0

Calculation:

\(|\vec{a}|^{2}\) + \(\rm \vec {a}.\rm \vec {b}\) = 0

Given: \(\rm |\vec{b}| = 2\times |\vec{a}|\)

So, we can write

\(\rm \frac{|\vec{b}|^{2}}{4}\) +  \(\rm \vec {a}.\rm \vec {b}\) = 0

\(\rm \vec {a}.\rm \vec {b}\) = - \(\rm \frac{|\vec{b}|^{2}}{4}\)      -----(1)

To find: \(\rm (4\vec {a} + \vec{b})\cdot \vec{b}\)

\(\rm (4\vec {a} + \vec{b})\cdot \vec{b}\) = 4\(\rm \vec {a}.\rm \vec {b}\) + \(\rm |\vec{b}|^{2}\)      ----(2)

From equations (1) & (2) we can write,

​\(\rm (4\vec {a} + \vec{b})\cdot \vec{b}\) = -​\(\rm |\vec{b}|^{2}\) + \(\rm |\vec{b}|^{2}\) = 0​​

∴ The value of \(\rm(4\vec {a} + \vec{b})\cdot \vec{b}\) equal to 0.

Concept:

1. If \(\rm\vec {a}, \vec{b}\) and \( \rm \vec {c}\) be three vectors such that  \(\rm\vec {a}, \vec{b}\) and \( \rm \vec {c}\) are co-planar.

\(\rm \vec{a} . (\rm \vec{b} \times \rm \vec{c})\) = 0

2.Vector \(\rm \vec{a}\times\rm \vec{b}\) is perpendicular to both vector  \(\rm\vec {a}\) and \(\rm\vec {b}\).

Calculation:

We have,

\(\rm(\vec{a}\times \vec{b})\times \vec{c}\)  = \(\rm\vec {d}\) (Let)

⇒ \(\rm\vec {d}\) is ⊥ to both \(\rm \vec{a}\times\rm \vec{b}\) and \( \rm \vec {c}\).

But \(\rm \vec{a}\times\rm \vec{b}\) is ⊥ to  \(\rm\vec {a}\) and \(\rm\vec {b}\).

So, \(\rm\vec {d}\) or \(\rm(\vec{a}\times \vec{b})\times \vec{c}\) will be coplanar to vector \(\rm\vec {a}\) and \(\rm\vec {b}\).

**∴ Statement

(1) is correct.**

For statement 2:

Statement 2 is only possible when the vector \(\rm\vec {d}\) is parallel or anti-parallel to vector \(\rm \vec{a}\times\rm \vec{b}\), which is not possible as per the above explanation for statement

1. So, this statement 2 is incorrect.

So, the best answer will be option 1.

Concept:

If A and B are points with position vectors \(\rm \vec{a}\) and \(\rm \vec{b}\) respectively then \(\rm \vec{AB}\) = \(\rm \vec{b}\) - \(\rm \vec{a}\).

If \(\rm \vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}\) is a vector then the magnitude of the vector is given by \(\rm |\vec{a}| = \sqrt{a_{1}^{2} + a_{2}^{2} + a_{3}^{2}}\)

Calculation:

A = \(\rm (\sqrt{2}-1) \hat{i}-\hat{j}\) and B = \(\rm \hat{i} + (\sqrt{2}+1) \hat{j}\)

⇒  \(\rm \vec{AB}=\)  \(\rm \vec{b}\) - \(\rm \vec{a}\).

⇒ \(\rm \vec{AB}=\)  \(\rm \hat{i} + (\sqrt{2}+1) \hat{j}\) - \(\rm (\sqrt{2}-1) \hat{i}+\hat{j}\) =  \(\rm (-\sqrt{2}+2) \hat{i}+(\sqrt{2}+2)\hat{j}\)

⇒ \(\rm \vec{AB}=\) \(\rm (-\sqrt{2}+2) \hat{i}+(\sqrt{2}+2)\hat{j}\)

⇒   \(|\rm \vec {AB}|=\rm \sqrt{(-\sqrt{2} + 2)^{2} + (\sqrt{2} + 2)^{2}}\) 

⇒ ​\(\rm \vec{AB} = \sqrt{12}=2\sqrt 3\)​ 

∴ The magnitude of \(\rm \vec{AB}\) is  \(\rm 2\rm \sqrt{3}\).

Concept:

• \(\rm \frac{d}{dx}(x^{n}) = nx^{n - 1}\)
• \(\rm \frac{d}{dx}(ln \ x )= \frac{1}{x}\)
• log m + log n = log mn

 

Calculation:

y = (1 + x)(1 + x2)(1 + x4)(1 + x8)(1 + x16)  -----(1)

At x = 0

y(0) = (1 + 0)(1 + 0)(1 + 0)(1 + 0)(1 + 0) 

⇒ y(0) = 1    -----(2)

Taking log on both side in equation (1)

ln y = ln[(1 + x)(1 + x2)(1 + x4)(1 + x8)(1 + x16)]

We know that,

log m + log n = log mn

⇒ ln y = ln(1 + x) + ln(1 + x2) + ln(1 + x4) + ln(1 + x8) + ln(1 + x16)

Differentiating both side with respect to x,

⇒ \(\frac{1}{y}\frac{dy}{dx}=\frac{1}{1+x}+\frac{2x}{1+x^2}+\frac{4x^3}{1+x^4}+\frac{8x^7}{1+x^8}+\frac{16x^{15}}{1+x^{16}}\)

⇒ \(\frac{dy}{dx}=y[\frac{1}{1+x}+\frac{2x}{1+x^2}+\frac{4x^3}{1+x^4}+\frac{8x^7}{1+x^8}+\frac{16x^{15}}{1+x^{16}}]​​\)

When x = 0, then

\(\frac{dy}{dx}=(1)\Big[\frac{1}{1+0} + 0 + 0 +0 +0\Big]\)  (∵ y(0) = 1)

\(\frac{dy}{dx} = 1 \times 1\)

\(\therefore\ (\frac{dy}{dx})_{x=0}=1\)

Concept:

• log m + log n = log mn
• sin x/cos x = tan x
• \(\rm \frac{d}{dx} log\ x = \frac{1}{x}\)
• \(\rm \frac{d}{dx} cos x = -sin x\)

 

Calculation:

y = cos x ⋅ cos 4x ⋅ cos 8x  

Taking log on both side

⇒ log y = log (cos x ⋅ cos 4x ⋅ cos 8x) 

⇒ log y = log cos x + log cos 4x + log cos 8x 

Differentiating both side with respect to x

⇒ \(\frac{1}{y}\frac{dy}{dx} = \frac{-sin\ x}{cos\ x}+\frac{-4sin\ 4x}{cos\ 4 x}+\frac{-8sin\ 8x}{cos\ 8x}\)

⇒ \(\frac{1}{y}\frac{dy}{dx}\) = -1[tan x + tan 4x + tan 8x]

Now putting \(\rm x = \frac{π}{4}\)

⇒ \(\frac{1}{y}\frac{dy}{dx}\) = -1[\(tan\ \frac{π}{4}+tan \frac{4π}{4}+tan\ \frac{8π}{4}\)]

⇒  \(\frac{1}{y}\frac{dy}{dx}\) = -[1 + tan π + tan 2π]     [∵ tan(π/4) = 1]

⇒  \(\frac{1}{y}\frac{dy}{dx}\) = -1    (∵ tan nπ = 0)

 ∴​ \(\rm \frac{1}{y}\frac{dy}{dx}\) at \(x = \frac{π}{4}\) equal to is -1.

Concept:

\(\rm \frac{d}{dx}f(x) = f'(x)\)

Calculation:

fof(x) = x4

⇒ fof(x) = (x²)²

⇒ f(x) = x² 

f'(x) = 2x

Hence, at x = 1

f'(1) = 2 × 1

f'(1) = 2

∴ f'(1) equal to 2.

Given:

f(x) = \(\rm \displaystyle\lim_{n → ∞} \frac{a^n+b^n}{a^n-b^n}\) and

a > b > 1

Calculation:

We have,

a > b > 1

⇒ \(\frac{a}{b}>1\) or \(\frac{b}{a}<1\)

Given that,

f(x) = \(\rm \displaystyle\lim_{n → ∞} \frac{a^n+b^n}{a^n-b^n}\)    

⇒ f(x) = \(\rm \displaystyle\lim_{n → ∞} \frac{a^n[1+(\frac{b}{a}) ^{n}]}{a^n[1 - (\frac{b}{a}) ^{n}]}\) 

⇒ f(x) = \(\rm \displaystyle\lim_{n → ∞} \frac{[1+(\frac{b}{a}) ^{n}]}{[1 - (\frac{b}{a}) ^{n}]}\) 

Taking limit n→∞

⇒ f(x) = \(\rm \displaystyle\lim_{n → ∞} \frac{[1+(\frac{b^{\infty}}{a^{\infty}}) ]}{[1 - (\frac{b^{\infty}}{a^{\infty}}) ]}\)  

⇒ f(x) = \(\rm \frac{1 + 0}{1 - 0}\)       (∵ \(\frac{b}{a}<1\) )

∴  f(x) = 1

Concept:

Step: 1

Check f(a) is defined. If it is not defined then no need to go further.  The function is not continuous at a. If f(a) is defined then

Step: 2

Check Left-hand limit (LHL) and Right-hand limit (RHL) 

\(LHL = \rm \lim_{x\rightarrow a^{-}}f(x) \)

\(RHL= \lim_{x\rightarrow a^{+}} f(x) \)

If LHL = RHL then limit exist.

**Calculation:**​

\(\rm \lim_{x\rightarrow 2^{-}} f(x) = \lim_{x\rightarrow 2}(1 + \frac{x}{2k} )\)

⇒ \(\rm \lim_{x\rightarrow 2^{-}}f(x)\) = \(\rm 1 + \frac{2}{2k} = 1 + \frac{1}{k}\)

⇒​ \(\rm \lim_{x\rightarrow 2^{-}}f(x)\) = \( 1 + \frac{1}{k}\)         ----(1)

Also,

\(\rm \lim_{x\rightarrow 2^{+}}f(x)\) = \(\rm \lim_{x\rightarrow 2^{+}}(kx)\)

⇒ \(\rm \lim_{x\rightarrow 2^{+}}f(x)\) =   2k    ----(2)

If \(\displaystyle\lim_{x\rightarrow 2}\) f(x) exists, then 

\(\rm 1 + \frac{1}{k} = 2k\)

⇒ 2k² - k - 1 = 0

⇒ 2k2 - 2k  + k - 1 = 0

⇒ 2k(k - 1) +1(k - 1) = 0

⇒ (k - 1)(2k + 1) = 0

⇒ k = 1 & -1/2

∴ The value of k is 1.

Additional Information Compare f(a), LHL & RHL . 

If \(\rm \lim_{x\rightarrow a^{-}}f(x) = \lim_{x\rightarrow a^{+}} f(x) = f(a)\) then,

The function is continuous at f(a).

Concept:

Differentiable Functions:

• If a graph has a sharp corner at a point, then the function is not differentiable at that point.
• If a graph has a break at a point, then the function is not differentiable at that point.
• If a graph has a vertical tangent line at a point, then the function is not differentiable at that point.

 

Calculation:

Givne that,

f(x) = |x| - 1     ----(1)

Step: 1

Step: 2 

Step: 3 

Clearly, we can see that,

f(x) is continuous at x = 1 and

f(x) is not differentiable at x = 0 because there is a corner at x = 0.

∴ Only statement 1 is correct.

Additional Information 

• Differentiable functions are those functions whose derivatives exist.
• If a function is differentiable, then it is continuous.
• If a function is continuous, then it is not necessarily differentiable.
• The graph of a differentiable function does not have breaks, corners, or cusps.

Concept:

Modullus funtion: f(x) = |x| 

f(x) = \(\rm \left{\begin{matrix} x & x > 0\ -x & x < 0 \end{matrix}\right.\)

Greatest Integer Function:

Greatest Integer Function [x] indicates an integral part of the real number x which is the nearest and smaller integer to x. It is also known as the floor of x

• In general, If n≤ x ≤ n+1 Then [x] = n     (n ∈ Integer)
• This means if x lies in [n, n+1) then the Greatest Integer Function of x will be n.

Continuity of function at any point:

\(\rm \lim_{x\rightarrow a^{-}}f(x) = \lim_{x\rightarrow a^{+}} f(x) = \lim_{x\rightarrow a}f(x) \)

A function f(x) is differentiable at x = a, if LHD = RHD

LHD = \(\rm \lim_{x\rightarrow a^{-}}f'(x) = \rm \lim_{h\rightarrow 0^{-}}\frac{f(a-h) - f(a)}{-h}\)

RHD = \(\rm \lim_{x\rightarrow a^{+}}f'(x) = \rm \lim_{h\rightarrow 0^{+}}\frac{f(a+h) - f(a)}{h}\)

$\lim_{x\rightarrow a^{-1}}x→a− \lim_{x\rightarrow a^{+}x→a+

Calculation:$

\(f(x) = \frac{[x]}{|x|}\)

f(x) = \(\rm \left{\begin{matrix} \frac{-1}{-x} = \frac{1}{x} & -1 < x < 0 & \ \frac{0}{x} = 0 & 0 < x < 1 & \ \frac{1}{x} = \frac{1}{x} & 1 < x < 2 & \end{matrix}\right.\)

At x = 1

RHL = \(\lim_{x \to 0^+} =\lim_{x \to 0^+} \frac{1}{x} = 1\)

∴ The right-hand limit of f(x) at x = 1 is 1.

Concept:

Let y = f(x) be a function. Then for a function, we say,

The function is continuous if it satisfies the following conditions.

\(\mathop {\lim }\limits_{{\rm{x}} \to {{\rm{a}^{-}}{}}} {\rm{f}}\left( {\rm{x}} \right) = \mathop {\lim }\limits_{{\rm{x}} \to {{\rm{a^+}}{ }}} {\rm{f}}\left( {\rm{x}} \right) = {\rm{f}}\left( {\rm{a}} \right){\rm{}}\)

Calculation:

Given that,

\(\rm f(x) = sin \left(\frac{1}{x^2}\right)\)           ----(1)

Statement: 1

\(LHL=\rm \displaystyle\lim_{x→0^-}sin(\frac{1}{x^2})\)

⇒ \(LHL=\rm \displaystyle\lim_{h→ 0}sin[\frac{1}{(0-h)^2}]=sin\ ∞\)     ----(2)

We know that, 

sin θ ∈  [ -1 , 1] 

⇒ sin (∞) is a definite value.

Similarly,

\(RHL=\rm \displaystyle\lim_{h→ 0}sin[\frac{1}{(0+h)^2}]=sin\ ∞\)   ----(3)

According to the question,

f(0) = 0        ----(4)

From equations (2), (3), (4),

Statement 1 is incorrect.

Statement: 2

\(LHL=\rm \displaystyle\lim_{h→ 0}sin[\frac{1}{(0-\frac{2}{\sqrt \pi})^2}]=sin\ \frac{\pi}{4}\)

⇒ LHL = \(\frac{1}{\sqrt 2}\)    ----(5)

Similarly, 

RHL =  \(\frac{1}{\sqrt 2}\)       -----(6)

And,

f(\(\frac{2}{\sqrt{\pi}}\)) = sin \(\rm \left ( \frac{1}{(\frac{2}{\sqrt{\pi}})^{2}} \right ) \)

⇒ f(\(\frac{2}{\sqrt{\pi}}\)) = \(\rm \frac{1}{\sqrt{2}} \)       ----(7)

From equation (5), (6), (7)

We can say that f(x) is continuous at x = \(\frac{2}{\sqrt{\pi}}\)

∴  Statement 2 is incorrect.

Concept:

The range of a function is the set of all its outputs.

Example:

Let us consider the function f: A→ B,

where f(x) = 2x and A and B = {set of natural numbers}.

Here we say A is the domain and B is the co-domain.

The set of ƒ images of all the elements of A is known as the range of ƒ.  

Calculation:

f(x) = 1 - sinx 

Here the domain of f(x) is (-∞, ∞)

At this domain, the range of sinx is [-1, 1]

Range of f(x) = 1 - sin2x = 1 - [-1, 1] 

⇒ f(x) = [0 , 2]

∴ Range of f(x) is [0 , 2].

Concept:

If we let Δx and Δy be the distances (along the x and y axes, respectively) between two points on a curve, then the slope given by the above definition,

m = \(\rm \frac{\Delta y}{\Delta x}\)

Formula Used:

cos-1 (cos x) = -x , -π ≤ x ≤ 0

cos-1 (cos x) = x , 0 < x ≤ π 

Calculation:

y = cos-1 (cos x)

Since, x = -π/4 which belongs to -π to 0 

So, y = - x

⇒ dy/dx = -1

Since on reducing the equation from y = cos^-1 (cos x) it becomes y = -x.

As x belongs (-π to 0 )

So Slope is d/dx [y = -x]

So Slope "m" = -1.

∴ The slope of the tangent of y = cos-1 (cos x) at x = \(-\frac{\pi}{4}\) is -1.