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**SECTION A**

**Question 1: Let a, b∈R. If the mirror image of the point P(a, 6, 9) with respect to the line (x - 3)/7 = (y - 2)/5 = (z - 1)/(-9) is (20, b, -a, -9), then |a + b| is equal to:**

- a. 86
- b. 88
- c. 84
- d. 90

Solution:

Answer: (b)

P(a, 6, 9), Q (20, b, -a-9)

Mid point of PQ=((a+20)/2, (b+6)/2, -a/2) lie on the line.

\(\left ( \frac{\frac{a+20}{2}-3}{7} \right )= \left ( \frac{\frac{b+6}{2}-2}{5} \right )=\left ( \frac{\frac{-a}{2}-1}{-9} \right )\)=> (a + 20 - 6)/14 = (b + 6 - 4)/10 = (-a - 2)/(-18)

=> (a + 14)/14 = (a + 2)/18

=> 18a + 252 = 14a + 28

=> 4a = -224

a = -56

(b + 2)/10 = (a + 2)/18

=> (b + 2)/10 = (-54)/18

=>(b + 2)/10 = -3

=> b = -32

|a + b| = |-56 - 32|

= 88

**Question 2: Let f be a twice differentiable function defined on R such that f(0) = 1, f'(0) = 2 and f'(x) ≠ 0 for all x∈R. If |f(x) f'(x) f'(x) f''(x)| = 0, for all x∈R, then the value of f(1) lies in the interval:**

- a. (9, 12)
- b. (6, 9)
- c. (3, 6)
- d. (0, 3)

Solution:

Answer: (b)

Given f(x) f''(x) - f'(x)

^{2 }=0Let h(x) = f(x)/f'(x)

Then h'(x) = 0

⇒ h(x) = k

⇒ f(x)/f'(x) = k

⇒ f(x) = k f’(x)

⇒ f(0) = k f'(0)

⇒ k = 1/2

Now, f(x) = ½ f'(x)

⇒∫ 2dx = ∫ f'(x)/f(x) dx

⇒ 2x = ln |f(x)| + C

As f(0) = 1 ⇒ C = 0

⇒ 2x = ln |f(x)|

⇒ f(x) = ±e

^{2x}As f(0) = 1 ⇒ f(x) = e

^{2x}∴ f(1) = e

^{2 }≈ 7.38

**Question 3: A possible value of tan (¼ sin ^{-1} √63/8) is:**

- a. 1/(2√2)
- b. 1/√7
- c. √7 - 1
- d. 2√2 - 1

Solution:

Answer: (b)

tan (¼ sin

^{-1}√63/8)Let sin

^{-1}(√63/8) = θsin θ = √63/8

cos θ = 1/8

2 cos

^{2}(θ/2) - 1 = 1/8⇒ cos

^{2}θ/2 = 9/16cos θ/2 = 3/4

⇒(1- tan

^{2}θ/4 )/(1 + tan^{2}θ/4) = 3/4tan θ/4 = 1/√7

**Question 4: The probability that two randomly selected subsets of the set {1, 2, 3, 4, 5} have exactly two elements in their intersection, is:**

- a. 65/2
^{7} - b. 135/2
^{9} - c. 65/2
^{8} - d. 35/2
^{7}

Solution:

Answer: (b)

Let A and B be two subsets.

For each x∈{1,2,3,4,5} , there are four possibilities:

x ∈ A∩B, x∈ A'∩B, x ∈ A∩B', x ∈ A'∩B'

So, the number of elements in sample space =4

^{5}Required probability

= (

^{5}C_{2 }× 3^{3})/4^{5}= (10×27)/2

^{10}= 135/2

^{9}

**Question 5: The vector equation of the plane passing through the intersection of the planes **

\(\vec{r}.(\hat{i}-2\hat{j}) = -2\)

and the point (1, 0, 2) is:- a. \(\vec{r}.(\hat{i}-7\hat{j}+3\hat{k}) = \frac{7}{3}\)
- b. \(\vec{r}.(\hat{i}+7\hat{j}+3\hat{k}) = 7\)
- c. \(\vec{r}.(3\hat{i}+7\hat{j}+3\hat{k}) = 7\)
- d. \(\vec{r}.(\hat{i}+7\hat{j}+3\hat{k}) = \frac{7}{3}\)

Solution:

Answer: (b)

Family of planes passing through the intersection of planes is

\((\vec{r}.(\hat{i}+\hat{j}+\hat{k})-1)+\lambda (\vec{r}.(\hat{i}-2\hat{j})+2) =0\)The above curve passes through

\(\hat{i}+2\hat{k}\),(3 − 1) + λ(1 + 2) = 0

⇒ λ = -2/3

Hence, equation of plane is

\(3(\vec{r}.(\hat{i}+\hat{j}+\hat{k})-1)-2(\vec{r}.(\hat{i}-2\hat{j})+2)= 0\)⇒

\(\vec{r}.(\hat{i}+7\hat{j}+3\hat{k}) = 7\)**TRICK:**Only option (2) satisfies the point (1, 0, 2)

**Question 6: If P is a point on the parabola y = x ^{2} + 4 which is closest to the straight line y = 4x − 1, then the co-ordinates of P are :**

- a. (–2, 8)
- b. (1, 5)
- c. (3, 13)
- d. (2, 8)

Solution:

Answer: (d)

Tangent at P is parallel to the given line.

dy/dx|

_{P}= 4⇒2x

_{1 }= 4⇒x

_{1}= 2Required point is (2, 8)

**Question 7: Let a, b, c be in arithmetic progression. Let the centroid of the triangle with vertices (a, c), (2, b) and (a, b) be (10/3, 7/3). If α, β are the roots of the equation ax ^{2 }+ bx + 1 = 0, then the value of α^{2} + β^{2 }- αβ is:**

- a. 71/256
- b. -69/256
- c. 69/256
- d. -71/256

Solution:

Answer: (d)

2b = a + c

(2a + 2)/3 = 10/3 and (2b + c)/3 = 7/3

⇒ a = 4

2b + c = 7, 2b - c = 4 },solving

b = 11/4 and c = 3/2

∴ Quadratic equation is 4x

^{2}+ (11/4)x + 1 = 0∴ The value of (α + β)

^{2 }- 3αβ = (121/256) - (¾)= -71/256

**Question 8: The value of the integral, **

- a. –4
- b. –5
- c. -√2 - √3 - 1
- d. -√2 - √3 + 1

Solution:

Answer: (c)

\(I = \int_{1}^{3}-3 dx + \int_{1}^{3}\left [ (x-1)^{2} \right ]dx\)Put x - 1 = t ; dx = dt

\(I = (-6) + \int_{0}^{2}\; \; \; \; \; \;[t^{2}]dt\)\(I = -6 + \int_{0}^{2}\: \: \:0dt + \int_{1}^{\sqrt{2}}\: \: \:1dt + \int_{\sqrt{2}}^{\sqrt{3}}\: \: \:2dt + \int_{\sqrt{3}}^{2}\: \: \: 3dt\)I = -6 + (√2 - 1) + 2√3 - 2√2 + 6 - 3√3

I = -1 - √2 - √3

**Question 9: Let f: R → R be defined as**

**Let A = {x ∈ R ∶ f is increasing}. Then A is equal to :**

- a. (-5, -4) ∪ (4,∞)
- b. (-5, ∞)
- c. (-∞, -5) ∪ (4, ∞)
- d. (-∞, -5) ∪ (-4, ∞)

Solution:

Answer: (a)

f’(x) = {-55 ; x< -5 6(x

^{2 }- x - 20) ; -5 < x < 4 6(x^{2}- x - 6) ; x > 4⇒ f’(x) = {-55 ; x< -5 6(x - 5)(x + 4) ; –5 < x < 4 6(x - 3)(x + 2) ; x > 4

Hence, f(x) is monotonically increasing in (-5, -4) ∪ (4, ∞)

**Question 10: If the curve y = ax ^{2 }+ bx + c,x ∈ R passes through the point (1, 2) and the tangent line to this curve at origin is y = x, then the possible values of a, b, c are:**

- a. a = 1, b = 1, c = 0
- b. a = -1, b = 1, c = 1
- c. a = 1, b = 0, c = 1
- d. a = 1/2, b = 1/2 ,c = 1

Solution:

Answer: (a)

2 = a + b + c

dy/dx = 2ax + b, (dy/dx)

_{(0,0) }= 1⇒ b = 1 and a + c = 1

Since (0, 0) lies on curve,

∴ c = 0, a = 1

TRICK: (0, 0) lies on the curve. Only option (1) has c = 0

**Question 11: The negation of the statement ~p∧(p∨q) is∶**

- a. ~p∧q
- b. p∧~q
- c. ~p∨q
- d. p∨∼q

Solution:

Answer: (d)

Negation of ~p∧(p∨q) is

∼[~p∧(p∨q)]

≡ p∨ ∼(p∨q)

≡ p∨(∼p∧∼q)

≡ (p∨∼p)∧(p∨∼q)

≡ T∧(p∨∼q), where T is tautology.

≡ p∨∼q

**Question 12: For the system of linear equations: **x − 2y = 1, x − y + kz = −2, ky + 4z = 6,k ∈ R

**Consider the following statements:**

**(A) The system has unique solution if k ≠ 2,k ≠ −2.**

**(B) The system has unique solution if k = -2.**

**(C) The system has unique solution if k = 2.**

**(D) The system has no-solution if k = 2.**

**(E) The system has infinite number of solutions if k ≠ -2.**

**Which of the following statements are correct?**

- a. (B) and (E) only
- b. (C) and (D) only
- c. (A) and (D) only
- d. (A) and (E) only

Solution:

Answer: (c)

x -2y + 0.z = 1

x - y + kz = -2

0.x + ky + 4z = 6

∆ =

\(\begin{vmatrix} 1 & -2 & 0\\ 1& -1 & k\\ 0& k & 4 \end{vmatrix}\)= 4 - k

^{2}For unique solution, 4 - k

^{2}≠ 0k ≠ ±2

For k = 2,

x - 2y + 0.z =1

x - y + 2z = -2

0.x + 2y + 4z = 6

Δ

_{x}=\(\begin{vmatrix} 1 & -2 & 0\\ -2& -1 & 2\\ 6& 2 & 4 \end{vmatrix}\)= -8 + 2 (-20)

⇒Δ

_{x}= -48 ≠ 0For k = 2,Δ

_{x}≠ 0So, for k = 2, the system has no solution.

**Question 13: For which of the following curves, the line x + √3y = 2√3 is the tangent at the point (3√3/2, 1/2)?**

- a. x
^{2 }+ 9y^{2 }= 9 - b. 2x
^{2 }- 18y^{2}= 9 - c. y
^{2 }= x/(6√3) - d. x
^{2 }+ y^{2 }= 7

Solution:

Answer: (a)

Tangent to x

^{2 }+ 9y^{2}= 9 at point (3√3/2, 1/2) is x(3√3)/2 + 9y(1/2) = 9⇒3√3 x + 9y = 18

⇒x +√3 y = 2√3

⇒ Option (a) is true.

**Question 14: The angle of elevation of a jet plane from a point A on the ground is 60 ^{0}. After a flight of 20 seconds at the speed of 432 km/hour, the angle of elevation changes to 30^{0}. If the jet plane is flying at a constant height, then its height is:**

- a. 1200√3 m
- b. 1800√3 m
- c. 3600√3 m
- d. 2400√3 m

Solution:

Answer: (a)

v = 432 × 1000/(60×60) m/sec

= 120 m/sec

Distance PQ = v × 20 = 2400 m

In ΔPAC

tan 60

^{0}= h/AC⇒ AC = h/(√3)

In ΔAQD

tan 30

^{0}= h/AD⇒AD = √3h

AD = AC + CD

⇒√3 h = h/√3 + 2400

⇒2h/√3 = 2400

⇒ h = 1200√3 m

**Question 15: For the statements p and q, consider the following compound statements:**

**(a) (~q∧(p→q)) → ~p**

**(b) ((p∨q))∧~p) → q**

**Then which of the following statements is correct?**

- a. (a) is a tautology but not (b)
- b. (a) and (b) both are not tautologies
- c. (a) and (b) both are tautologies
- d. (b) is a tautology but not (a)

Solution:

Answer: (c)

(a) is tautology.

(b) is tautology.

∴ (a) and (b) both are tautologies.

**Question 16: Let A and B be 3×3 real matrices such that A is symmetric matrix and B is skew-symmetric matrix. Then the system of linear equations (A ^{2}B^{2 }− B^{2}A^{2})X = O, where X is a 3×1 column matrix of unknown variables and O is a 3×1 null matrix, has:**

- a. a unique solution
- b. exactly two solutions
- c. infinitely many solutions
- d. no solution

Solution:

Answer: (c)

A

^{T }= A, B^{T}= –BLet A

^{2}B^{2 }− B^{2}A^{2}= PP

^{T}= (A^{2}B^{2}– B^{2}A^{2})^{T}= (A

^{2}B^{2})^{T}– (B^{2}A^{2})^{T}= (B

^{2})^{T}(A^{2})^{T}– (A^{2})^{T}(B^{2})^{T}= B

^{2}A^{2}– A^{2}B⇒ P is a skew-symmetric matrix.

\(\begin{bmatrix} 0 & a & b\\ -a & 0 &c \\ -b &-c & 0 \end{bmatrix}\begin{bmatrix} x & y & z \end{bmatrix}=\begin{bmatrix} 0 & 0 & 0 \end{bmatrix}\)∴ ay + bz = 0 …(1)

–ax + cz = 0 …(2)

–bx – cy =0 ...(3)

From equation (1), (2), (3)

Δ = 0 and Δ

_{1 }= Δ_{2}= Δ_{3 }= 0∴ System of equations has infinite number of solutions.

**Question 17: If n ≥ 2 is a positive integer, then the sum of the series ^{n+1}C_{2} + 2(^{2}C_{2} + ^{3}C_{2} + ^{4}C_{2} + …. + ^{n}C_{2}) is**

- a. n(n + 1)
^{2}(n + 2)/12 - b. n(n - 1)(2n + 1)/6
- c. n(n + 1)(2n + 1)/6
- d. n(2n + 1)(3n + 1)/6

Solution:

Answer: (c)

^{2}C_{2}=^{3}C_{3}Let S =

^{3}C_{3}+^{3}C_{2}+ ……. +^{n}C_{2}=^{n+1}C_{3}(∵^{n}C_{r }+^{n}C_{r–1}=^{n+1}C_{r})∴

^{n+1}C_{2}+^{n+1}C_{3}+^{n+1}C_{3 }=

^{n+2}C_{3}+^{n+1}C_{3}=(n + 2)!/3!(n – 1)! + (n + 1)!/3!(n - 2)!

=(n + 2)(n + 1)n/6 + (n + 1)(n)(n - 1)/6 = (n(n + 1)(2 + 1))/6

TRICK : Put n = 2 and verify the options.

**Question 18: If a curve y = f(x) passes through the point (1, 2) and satisfies xdy/dx + y = bx ^{4}, then for what value of b, **

- a. 5
- b. 62/5
- c. 31/5
- d. 10

Solution:

Answer: (d)

dy/dx + y/x = bx

^{3}I.F. =

\(e^{\int \frac{dx}{x}}\)= x

∴ yx = ∫bx

^{4}dx = bx^{5}/5 + cThe above curve passes through (1, 2).

2 = b/5 + c

Also,

\(\int_{1}^{2}(\frac{bx^{4}}{5}+\frac{c}{x})dx=\frac{62}{5}\)⇒ (b/25) ×32 + c ln 2 - b/25 = 62/5

⇒ c = 0 and b = 10

**Question 19: The area of the region: R{(x, y): 5x ^{2} ≤ y ≤ 2x^{2} + 9} is:**

- a. 9√3 square units
- b. 12√3 square units
- c. 11√3 square units/
- d. 6√3 square units

Solution:

Answer: (b)

Required area

=

\(2\int_{0}^{\sqrt{3}}\: \: \:(2x^{2}+9-5x^{2})dx\)=

\(2\int_{0}^{\sqrt{3}}\: \: \:(9-3x^{2})dx\)=

\(2\left | 9x-x^{3} \right |_{0}^{\sqrt{3}}\)= 12√3

**Question 20: Let f(x) be a differentiable function defined on [0,2] such that f’(x) =f '(2 - x) for all x∈(0, 2), f(0) = 1 and f(2) = e ^{2}. Then the value of **

- a. 1 + e
^{2} - b. 1 - e
^{2} - c. 2(1 - e
^{2}) - d. 2(1+e
^{2})

Solution:

Answer: (a)

f'(x) = f'(2 - x)

On integrating both sides, we get

f(x) = -f(2 - x)+c

Put x = 0

f(0) + f(2) = c

⇒ c = 1+ e

^{2}⇒ f(x) + f(2 - x) = 1 + e

^{2}I =

\(\int_{0}^{2}f(x)dx= \int_{0}^{1}(f(x) + f(2-x))dx\)= 1+e

^{2}

**Section B**

**Question 1: **T**he number of the real roots of the equation (x + 1) ^{2} + |x - 5| = 27/4 is ___.**

Solution:

Answer: 2

For x ≥ 5,

(x + 1)

^{2}+ (x - 5) = 27/4⇒ x

^{2}+ 3x - 4 = 27/4⇒ x

^{2 }+ 3x - 43/4 = 0⇒ 4x

^{2}+ 12x - 43 = 0x = (-12 ± √(144 + 688))/8

x = (-12 ± √832)/8

= (-12 ± 28.8)/8

= (-3 ± 7.2)/2

= (-3 + 7.2)/2, (-3 - 7.2)/2 (therefore, no solution)

For x < 5,

(x + 1)

^{2}- (x - 5) = 27/4⇒ x

^{2 }+ x + 6 - 27/4=0⇒ 4x

^{2 }+ 4x - 3=0x = (-4 ± √(16 + 48))/8

x = (-4 ± 8)/8

⇒ x = -12/8, 4/8

∴ 2 real roots.

**Question 2: The students S _{1}, S_{2},... S_{10} are to be divided into 3 groups A, B, and C such that each group has at least one student and the group C has at most 3 students. Then the total number of possibilities of forming such groups is ___.**

Solution:

Answer: 31650

Number of ways

=

^{10}C_{1}[2^{9}- 2] +^{10}C_{2}[2^{8}- 2] +^{10}C_{3}[2^{7}- 2]= 2

^{7 }[^{10}C_{1}×4 +^{10}C_{2}×2 +^{10}C_{3}] - 20 - 90 - 240= 128 [40 + 90 + 120 ] - 350

= (128 × 250) – 350

= 10 [3165]

= 31650

**Question 3: If a + α = 1, b + β = 2 and af(x) + αf(1/x) = bx + β/2, x ≠ 0 then the value of the expression [f(x) + f(1/x) ]/(x + 1/x)**

Solution:

Answer: 2

af(x) + αf(1/x) = bx + β/x …(i)

Replace x by 1/x

af(1/x) + αf(x) = b/x + βx …(ii)

(i) + (ii)

(a + α)[f(x) + f(1/x) ]

= (x + 1/x)(b + β)

⇒ ( f(x) + f(1/x))/(x + 1/x)

= 2/1

= 2

**Question 4: If the variance of 10 natural numbers 1,1,1,…,1,k is less than 10, then the maximum possible value of k is ________.**

Solution:

Answer: 11

\(\sigma ^{2} = \frac{\Sigma x^{2}}{n} - \left (\frac{\Sigma x}{n} \right )^{2}\)⇒

\(\sigma ^{2} = \frac{9+k^{2}}{10} - \left (\frac{9+k}{10} \right )^{2}< 10\)⇒ 10(9 + k

^{2}) - (81 + k^{2 }+ 18k) < 1000⇒ 90 + 10k

^{2}- k^{2}- 18k - 81 < 1000⇒ 9k

^{2}- 18k + 9 < 1000⇒ (k - 1)

^{2 }< 1000/9⇒ k - 1< (10√10)/3

⇒ k < (10√10)/3 + 1

Maximum possible integral value of k is 11.

**Question 5: Let λ be an integer. If the shortest distance between the lines x - λ = 2y - 1 = -2z and x = y + 2λ = z - λ is √7/2√2, then the value of |λ| is __**

Solution:

Answer: 1

(x - λ)/1 = (y - 1/2)/(1/2) = z/(-1/2)

(x - λ)/2 = (y-1/2)/1 = z/(-1) ...(1) Point on line = (λ, 1/2, 0)

x/1 = (y + 2λ)/1 = (z - λ)/1 ...(2) Point on line = (0, -2λ, λ)

= |-5λ - 3/2|/√14

= √7/(2√2) (Given)

⇒ |10λ + 3| = 7

⇒ λ = -1 as λ is an integer.

⇒ |λ| = 1

**Question 6: If i = √-1. If [(-1 + i√3) ^{21}/(1 - i)^{24} + (1 + i√3)^{21}/(1 + i)^{24} ] = k, and n = [|k| ] be the greatest integral part of |k|. Then **

Solution:

Answer: 310

= 2

^{9}e^{i(20π)}+2^{9 eiπ}= 2

^{9}+ 2^{9}(-1)= 0 = k

∴ n = 0

∑

_{(j=0)}^{5}(j + 5)^{2}- ∑_{(j=0)}^{5}(j + 5)= [5

^{2 }+ 6^{2 }+ 7^{2 }+ 8^{2 }+ 9^{2 }+ 10^{2}] - [5 + 6 + 7 + 8 + 9 + 10]= [(1

^{2}+ 2^{2 }+... + 10^{2}) - (1^{2 }+ 2^{2}+ 3^{2}+ 4^{2}) ] - [(1 + 2 + 3 + … + 10) - (1 + 2 + 3 + 4) ]= (385 - 30) -[55 - 10]

= 355 - 45

= 310

**Question 7: Let a point P be such that its distance from the point (5, 0) is thrice the distance of P from the point (-5, 0). If the locus of the point P is a circle of radius r, then 4r ^{2} is equal to ___.**

Solution:

Answer: 56.25

Let P be (h, k), A(5, 0) and B(-5, 0)

Given PA = 3PB

⇒ PA

^{2}= 9PB^{2}⇒ (h-5)

^{2 }+ k^{2 }= 9[(h + 5)^{2 }+ k^{2}]⇒ 8h

^{2}+ 8k^{2}+ 100h + 200 = 0∴ Locus of P is x

^{2 }+ y^{2 }+ (25/2)x + 25 = 0Centre = (-25/4, 0)

∴ r

^{2 }= (-25/4)^{2 }- 25= 625/16 - 25

= 225/16

∴4r

^{2 }= 4×225/16= 225/4

= 56.25

**Question 8: For integers n and r, let**

**The maximum value of k for which the sum**

**exists, is equal to ____.**

Solution:

Answer: Bonus

(1+x)

^{10}=^{10}C_{0}+^{10}C_{1}x +^{10}C_{2}x^{2}+ …… +^{10}C_{10}x^{10}(1+x)

^{15}=^{15}C_{0}+^{15}C_{1}x +^{15}C_{2}x^{2}+ …… +^{15}C_{k-1}x^{k-1}+^{15}C_{k+1}x^{k+1}+ ...^{15}C_{15}x^{15}Coefficient of x

^{k+1 }in (1+x)^{25}=

^{25}C_{k+1}^{25}C_{k}+^{25}C_{k+1}=^{26}C_{k+1}For maximum value

as per given, k can be as large as possible.

**Question 9: The sum of first four terms of a geometric progression (G.P.) is 65/12 and the sum of their respective reciprocals is 65/18. If the product of first three terms of the G.P. is 1, and the third term is α then 2α is____**

Solution:

Answer: 3

Let terms of GP are a, ar, ar

^{2}, ar^{3}a + ar + ar

^{2}+ ar^{3}= 65/12 …….(1)1/a + 1/ar + 1/(ar

^{2}) + 1/ar^{3}) = 65/18⇒1/a (r

^{3}+ r^{2 }+ r + 1)/r^{3}) = 65/18 …….(2)(1)/(2), we get

a

^{2}r^{3 }= 18/12= 3/2

Also, a

^{3}r^{3 }= 1⇒ a(3/2) = 1

⇒a = 2/3

(4/9) r

^{3 }= 3/2⇒ r

^{3 }= 3^{3}/2^{3}⇒ r = 3/2

α = ar

^{2}= 2/3.(3/2)

^{2 }= 3/2∴2α = 3

**Question 10: If the area of the triangle formed by the positive x-axis, the normal and the tangent to the circle (x - 2) ^{2}+ (y - 3)^{2 }= 25 at the point (5, 7) is A, then 24A is equal to ___.**

Solution:

Answer: Question is wrong

Equation of normal at P is

(y - 7) = ((7 - 3)/(5 - 2))(x - 5)

⇒ 3y - 21 = 4x – 20

⇒ 4x - 3y + 1 = 0

⇒ M is (-1/4, 0)

Equation of tangent at P is

(y - 7) = (-¾) (x - 5)

⇒ 4y - 28 = -3x + 15

⇒ 3x + 4y = 43

⇒ N is (43/3, 0)

The question is wrong. The normal cuts at a point on the negative axis.

JEE Main 2021 Maths Paper February 24 Shift 2