How Golpo AI Turned IIT JEE Advanced 2026 Mathematics Paper 1 Into Video Solutions

Question 1: Calculus · derivatives and extrema

Format: Single-correct

Consider f:(0,∞)→ℝ given by f(x)=√x ln x−x+1. Which statement is true?

Answer choices

1. A. The derivative is decreasing on (0,1).
2. B. f has a local maximum at some positive point.
3. C. f has a local minimum at some positive point.
4. D. f has neither a local maximum nor a local minimum on (0,∞).

The opening shows the question and choices. The illustrated reasoning follows, then the verified answer.

Step-by-step solution

1. Differentiate twice

   $f^{\prime }(x)=\frac{\ln x+2}{2\sqrt{x}}-1,f^{″}(x)=-\frac{\ln x}{4x^{3/2}}$

   The second derivative is positive for 0<x<1 and negative for x>1.

2. Locate the maximum of the derivative

   $f^{\prime }(1)=0,f^{\prime }(x)<0(x\ne 1)$

   The derivative increases up to x=1 and decreases after it, so its unique maximum is zero. Option A is false.

3. Classify the stationary point

   f(1)=0

   The derivative has the same negative sign on both sides of 1. Thus f continues decreasing through its stationary point and has no local extremum. B and C are false; D is true.

Answer: Choice D — No local maximum or local minimum

Back to the answer key

Question 2: Coordinate geometry · tangents and circumcircles

Format: Single-correct

Let P on y=x² have tangent slope 4. Let Q in the first quadrant on x²+y²=2 have tangent slope −1. Let R in the first quadrant on x²+4y²=8 have tangent slope −1/2. Find the radius of the circle through P,Q,R.

Answer choices

1. A. √10
2. B. √5
3. C. √(5/2)
4. D. 2√5

The opening shows the question and choices. The illustrated reasoning follows, then the verified answer.

Step-by-step solution

1. Find P

   $\frac{dy}{dx}=2x=4⟹P=(2,4)$

   Substitute x=2 into y=x².

2. Find Q

   $-\frac{x}{y}=-1,x^2+y^2=2⟹Q=(1,1)$

   The first-quadrant condition selects the positive coordinates.

3. Find R

   $-\frac{x}{4y}=-\frac{1}{2}⟹x=2y,R=(2,1)$

   Substitute x=2y into the ellipse to get y=1.

4. Use the right angle

   $RP⟂RQ,PQ=\sqrt{1^2+3^2}=\sqrt{10}$

   RP is vertical and RQ is horizontal, so PQ is the diameter of the circumcircle.

5. Read the radius

   $r=\frac{\sqrt{10}}{2}=\sqrt{\frac{5}{2}}$

   The midpoint (3/2,5/2) is equidistant from all three points, providing an independent check. Choice C.

Answer: Choice C — √(5/2)

Back to the answer key

Question 3: Linear algebra · elementary row operations

Format: Single-correct

Which matrix can be obtained from the 3×3 identity matrix by elementary row transformations?

Answer choices

1. A. [[1,1,1],[1,1,1],[1,1,1]]
2. B. [[1,1,1],[2,3,4],[1,2,1]]
3. C. [[1,1,1],[2,3,4],[2,5,8]]
4. D. [[1,1,1],[-1,1,2],[0,2,3]]

The opening shows the question and choices. The illustrated reasoning follows, then the verified answer.

Step-by-step solution

1. Use the invariant

   $rank(I_3)=3$

   Elementary row operations preserve rank and are reversible. Exactly the invertible 3×3 matrices can be obtained from I₃.

2. Reject the singular choices

   $detA=0,detC=0,detD=0$

   A has repeated rows. In C, row 3 equals 3 row 2 minus 4 row 1. In D, row 3 equals row 1 plus row 2.

3. Check B

   $B~\left[\begin{array}{ccc}1& 1& 1\\ 0& 1& 2\\ 0& 0& -2\end{array}\right],detB=-2\ne 0$

   Subtract 2R₁ from R₂, subtract R₁ from R₃, then subtract the new R₂ from R₃. The triangular matrix has three nonzero pivots.

4. Reverse the reduction

   $B~I_3$

   Scale the last row and eliminate above its pivot, then eliminate the entry above the middle pivot. Reversing these operations constructs B from I₃. Choice B.

Answer: Choice B — Only matrix B is invertible

Back to the answer key

Question 4: Trigonometry · principal inverse values

Format: Single-correct

Considering principal inverse trigonometric values, evaluate cot⁻¹(cot(−11))+10 sin(2 cos⁻¹(1/√2))+10 sin(2 tan⁻¹2).

Answer choices

1. A. 3π+7
2. B. 7
3. C. 4π+7
4. D. 3π−5

The opening shows the question and choices. The illustrated reasoning follows, then the verified answer.

Step-by-step solution

1. Choose the principal range

   ${\cot }^{-1}x\in (0,\pi )$

   All angles are in radians. An inverse function returns a principal angle; it is not a reciprocal.

2. Reduce the cotangent angle

   $3\pi <11<4\pi ⟹0<4\pi -11<\pi$

   Cotangent has period π, so cot⁻¹(cot(−11))=4π−11.

3. Evaluate the cosine term

   $10\sin \left(2{\cos }^{-1}\frac{1}{\sqrt{2}}\right)=10\sin \frac{\pi }{2}=10$

   The principal arccosine is π/4.

4. Evaluate the tangent term

   $\theta ={\tan }^{-1}2,\sin 2\theta =\frac{2\tan \theta }{1+{\tan }^2\theta }=\frac{4}{5}$

   The third term is 10×4/5=8. An acute right triangle with adjacent side 1, opposite side 2, and hypotenuse √5 verifies the identity.

5. Combine

   $(4\pi -11)+10+8=4\pi +7$

   The exact result matches choice C in the source key.

Answer: Choice C — 4π+7

Back to the answer key

Question 5: Probability: independence and conditional probability

Format: Multiple-correct

Suppose that Box I contains 6 red balls and 9 green balls, and Box II contains 8 red balls and 12 green balls. All the balls of Box I and Box II are mixed together and a ball is chosen at random from them. Let E1 be the event that the ball chosen belonged to Box I and let E2 be the event that the ball chosen belonged to Box II. Let F1 be the event that the ball chosen is red and let F2 be the event that the ball chosen is green. Then which of the following statements is (are) TRUE?

Answer choices

1. A. The events E1 and F1 are independent
2. B. The events E2 and F2 are dependent
3. C. The conditional probability P(F1 | E1) is equal to the conditional probability P(F1 | E2)
4. D. The conditional probability P(F1 | E1) is greater than the conditional probability P(F2 | E2)

The opening shows the question and choices. The illustrated reasoning follows, then the verified answer.

Step-by-step solution

1. Count each category

   N=6+9+8+12=35; P(E1)=15/35=3/7; P(E2)=20/35=4/7; P(F1)=14/35=2/5; P(F2)=21/35=3/5

   Uniform selection from the mixed balls weights each box by its number of balls.

2. Option A

   P(E1 ∩ F1)=6/35=(3/7)(2/5)=P(E1)P(F1)

   E1 and F1 are independent, so A is true.

3. Option B

   P(E2 ∩ F2)=12/35=(4/7)(3/5)=P(E2)P(F2)

   E2 and F2 are independent, so the assertion that they are dependent is false.

4. Options C and D

   P(F1|E1)=6/15=2/5; P(F1|E2)=8/20=2/5; P(F2|E2)=12/20=3/5

   The two red conditional probabilities are equal, making C true. Two fifths is less than three fifths, making D false.

Answer: Choices A, C — P(F1|E1)=P(F1|E2)=2/5; correct choices A and C

Back to the answer key

Question 6: Three-dimensional geometry: planes, distances, and angles

Format: Multiple-correct

Let P be the plane such that it contains the straight line (x-1)/2=(y-3)/3=(z+2)/1 and is perpendicular to the plane x+2y+3z=4. Let P1 be the plane which passes through the point (4,2,2) and is parallel to P. Then which of the following statements is (are) TRUE?

Answer choices

1. A. The equation of the plane P is 7x-5y+z=-10
2. B. The distance between the planes P and P1 is 30
3. C. The distance of the plane P from the origin is 2√3
4. D. The acute angle between the plane P and the plane 2x+2y+z=3 is cos^-1(1/(3√3))

The opening shows the question and choices. The illustrated reasoning follows, then the verified answer.

Step-by-step solution

1. Find a normal

   d=(2,3,1); m=(1,2,3); n=d×m=(7,-5,1); |n|=√75=5√3

   Containing the line requires n perpendicular to d; the perpendicular plane condition requires n perpendicular to m.

2. Plane equation and option A

   Q=(1,3,-2); n·Q=7-15-2=-10; P:7x-5y+z=-10

   A is true. Substituting Q+t d yields -10 for every real t.

3. Parallel plane and option B

   P1:7x-5y+z=20; distance=|20-(-10)|/√75=30/(5√3)=2√3

   The separation is 2√3, not 30, so B is false.

4. Origin distance and option C

   distance(O,P)=10/√75=2/√3=2√3/3

   This is not 2√3; C is false.

5. Acute angle and option D

   k=(2,2,1); |k|=3; |n·k|=|14-10+1|=5; cos θ=5/((5√3)3)=1/(3√3)

   The acute plane angle equals the acute angle between the normals, making D true.

Answer: Choices A, D — P: 7x-5y+z=-10; correct choices A and D

Back to the answer key

Question 7: Limits, continuity, differentiability, and counterexamples

Format: Multiple-correct

Let R denote the set of all real numbers. Let f:R→R be an arbitrary function and let g:R→R be the function defined by g(x)=x f(x), for all x∈R. Then which of the following statements is (are) TRUE?

Answer choices

1. A. The function g is always continuous at x=0
2. B. If f is continuous at x=0, then g is differentiable at x=0
3. C. If g is differentiable at x=0, then f is continuous at x=0
4. D. If g is differentiable at x=0, then lim(x→0) f(x) exists

The opening shows the question and choices. The illustrated reasoning follows, then the verified answer.

Step-by-step solution

1. Core derivative identity

   g(0)=0 · f(0)=0; [g(h)-g(0)]/h=f(h) for h≠0

   Consequently g is differentiable at zero if and only if the finite two-sided limit of f at zero exists. Its value need not equal f(0).

2. A is false

   f(0)=0; f(x)=1/x for x≠0. Then g(0)=0 and g(x)=1 for x≠0.

   The punctured limit of g is 1 but g(0)=0, so g need not be continuous at zero.

3. B is true

   f continuous at 0 ⇒ lim(h→0) f(h)=f(0) ⇒ g′(0)=f(0)

   Continuity of f supplies exactly the finite difference-quotient limit needed for g.

4. C is false

   f(0)=0; f(x)=1 for x≠0. Then g(x)=x for every real x, so g′(0)=1 while f is discontinuous at 0.

   g cannot detect the chosen value f(0), because multiplication by x erases it at zero.

5. D is true

   g differentiable at 0 ⇒ lim(h→0)[g(h)-g(0)]/h exists finitely ⇒ lim(h→0)f(h)=g′(0)

   The derivative implies existence of the punctured limit, but does not force equality with f(0).

Answer: Choices B, D — g′(0)=lim(h→0) f(h), whenever the limit exists; correct choices B and D

Back to the answer key

Question 8: Matrices: nilpotent decomposition, integer inverses, linear systems

Format: Multiple-correct

Consider M = [[2, -1], [1, 0]]. Let p,q,r,s,a,b,c,d be integers such that M^26 = [[p,q],[r,s]] and sum from k=1 to 26 of M^k = [[a,b],[c,d]]. Which statements are true?

Answer choices

1. A. There exists a 2 by 2 invertible matrix N with real entries such that MN = N[[1,1],[0,1]].
2. B. The value of a is 378.
3. C. For any two given integers m and n, there exist unique integers x and y such that px+qy=m and rx+sy=n.
4. D. For each positive real number t, the system (a+t)x+by=1, cx+(d+t)y=-1 has a unique solution.

Step-by-step solution

1. Step 1

   $K^2=0,M^k=I+kK$

   Set K=M-I=[[1,-1],[1,-1]]. Direct multiplication gives K^2=0, hence M^k=I+kK.

2. Step 2

   $detN=-1\ne 0,MN=NJ$

   A: N=[[1,1],[1,0]] has determinant -1 and MN=NJ for J=[[1,1],[0,1]]. Both sides equal [[1,2],[1,1]].

3. Step 3

   $S=26I+351K,a=377\ne 378$

   Sum k=1..26 is 351, so S=26I+351K=[[377,-351],[351,-325]]. Thus a=377, and B is false.

4. Step 4

   $x=-25m+26n,y=-26m+27n$

   M^26=[[27,-26],[26,-25]], determinant 1, inverse [[-25,26],[-26,27]]. Hence x=-25m+26n and y=-26m+27n are unique integers for any integer m,n. C is true.

5. Step 5

   $det(S+tI)=(t+26{)}^2>0(t>0)$

   det(S+tI)=(t+377)(t-325)+351^2=t^2+52t+676=(t+26)^2>0 for t>0. D is true, for the specified right-hand side and indeed any real right-hand side.

6. Step 6

   $\boxed{A,C,D}$

   All true statements are A, C and D.

Answer: Choices A, C, D — A, C and D

Back to the answer key

Question 9: Equivalence relations and set partitions

Format: Numerical-value

Let S={1,2,3,...,10}. Let X={R: R is an equivalence relation on S such that R has exactly 42 elements}. Find the number of elements in X.

The opening shows the question and choices. The illustrated reasoning follows, then the verified answer.

Step-by-step solution

1. Step 1

   ${\sum }_i{n}_i=10,{\sum }_i{n}_i^2=42$

   Equivalence relations correspond one-to-one to partitions of S. For class sizes n_i, sum(n_i)=10 and |R|=sum(n_i^2)=42, because every ordered pair inside a class belongs to R.

2. Step 2

   $L\in \{5,6\}$

   If largest size L<=4, sum n_i^2<=L sum n_i<=40, impossible. If L>=7, L^2>=49>42. Thus L=5 or 6.

3. Step 3

   (6,2,1,1)

   L=6 leaves total size 4 and square sum 6. All remaining sizes are <=2. If u classes have size 2 and v have size 1, 2u+v=4 and 4u+v=6, giving u=1,v=2. Pattern (6,2,1,1).

4. Step 4

   (5,4,1)

   L=5 leaves total size 5 and square sum 17. A remaining largest size <=3 would give square sum <=3*5=15. Therefore there is a size 4 and the remaining size 1. Pattern (5,4,1).

5. Step 5

   C(10,6)C(4,2)=1260

   For (6,2,1,1), choose the 6-class and then the 2-class: C(10,6)C(4,2)=210*6=1260. Equivalently 10!/(6!2!2!); the last 2! removes the order of equal singleton classes.

6. Step 6

   $1260+1260=\boxed{2520}$

   For (5,4,1), choose the 5-class and then the 4-class: C(10,5)C(5,4)=252*5=1260. Total 1260+1260=2520.

Answer: 2520

Back to the answer key

Question 10: Continuity, differentiability, absolute values and greatest-integer function

Format: Numerical-value

On (-π/2,π/2), f(x)=(|x|+|x-1|)sin x+floor(x sin x). Let α be the number of discontinuities and β the number of nondifferentiability points in this open interval. Find α+β.

Step-by-step solution

1. Step 1

   $g(x)=\{\begin{array}{cc}(1-2x)\sin x\hfill & x<0\hfill \\ \sin x\hfill & 0\le x\le 1\hfill \\ (2x-1)\sin x\hfill & x>1\hfill \end{array}$

   Write f=g+floor(h), with g=(|x|+|x-1|)sin x continuous, h=x sin x. The absolute-value coefficient is 1-2x for x<0, 1 for 0<=x<=1, and 2x-1 for x>1.

2. Step 2

   $h(\pm a)=1,1<a<\pi /2$

   h is even, h(0)=0, and h'(x)=sin x+x cos x>0 for 0<x<π/2. Its range is [0,π/2), with 1<π/2<2, so h=1 has exactly two roots ±a. Also h(1)=sin1<1, hence 1<a<π/2.

3. Step 3

   $\alpha =2$

   floor(h)=0 for |x|<a and 1 for a<=|x|<π/2. Thus it has jumps at -a and a only. At zero it remains zero on a full neighborhood, despite h(0)=0. Since g is continuous, f has exactly these two discontinuities. α=2.

4. Step 4

   f'_-(0)=f'_+(0)=1

   At x=0, g'_-(0)=g'_+(0)=1; floor(h) is locally zero. Hence f is differentiable at zero.

5. Step 5

   $\beta =3$

   At x=1, g'_-(1)=cos1 and g'_+(1)=cos1+2sin1, which differ because sin1>0. The floor is locally zero since a>1, so f is not differentiable at 1. Away from -a,a,0,1 both summands are smooth or locally constant. β=3, at -a,1,a.

6. Step 6

   $\boxed{\alpha +\beta =5}$

   α+β=2+3=5. The domain endpoints are excluded.

Answer: α + β = 2 + 3 = 5

Back to the answer key

Question 11: Combinatorics: bounded compositions, stars and bars, inclusion-exclusion

Format: Numerical-value

The number of ways to distribute 10 identical red pens and 14 identical blue pens among four persons such that each person gets 6 pens, is _____________.

The opening shows the question and choices. The illustrated reasoning follows, then the verified answer.

Step-by-step solution

1. Reduce to red counts

   r1+r2+r3+r4=10, 0<=ri<=6; bi=6-ri; sum bi=14

   A valid four-tuple of red counts determines exactly one complete distribution; the four persons are distinct.

2. Unrestricted nonnegative solutions

   C(10+4-1,4-1)=C(13,3)=286

   Stars and bars temporarily ignores the upper bound.

3. Remove invalid allocations

   ri>=7 => yi=ri-7; y1+y2+y3+y4=3; C(6,3)=20

   For each specified person, exactly 20 unrestricted allocations give that person at least 7 red pens.

4. No intersections and final count

   7+7>10; N=C(13,3)-4 C(6,3)=286-80=206

   Two simultaneous violations are impossible, so no overlap correction is needed.

Answer: 206

Back to the answer key

Question 12: Trigonometry: triple-angle identity and telescoping products

Format: Numerical-value

Let α=(1−2cos(π/11))(1−2cos(3π/11))(1−2cos(9π/11))(1−2cos(27π/11))(1−2cos(81π/11)). Then the value of 5−α² is _____________.

The opening shows the question and choices. The illustrated reasoning follows, then the verified answer.

Step-by-step solution

1. Ratio identity

   cos3t/cos t=4cos²t−3=2cos2t−1; 1−2cos x=−cos(3x/2)/cos(x/2)

   Apply the cosine triple-angle formula with t=x/2.

2. Denominators do not vanish

   ck=cos(3^k π/22), k=0,...,5; ck=0 =>3^k=11(2m+1), impossible

   A power of three is not divisible by eleven, so all relevant cosine ratios are defined.

3. Telescope five factors

   α=∏[k=0..4](−c(k+1)/ck)=−c5/c0=−cos(243π/22)/cos(π/22)

   Four intermediate factors cancel; five negative signs give one negative sign.

4. Evaluate the final ratio

   243π/22=11π+π/22; cos(243π/22)=−cos(π/22); α=1

   Cosine changes sign after adding an odd multiple of pi.

5. Compute the requested expression

   5−α²=5−1²=4 ∈[3.9,4.1]

   The exact value 4 agrees with the provisional key's accepted interval.

Answer: 4

Back to the answer key

Question 13: Complex roots, cyclic powers and Vieta formulas

Format: Matching-list

Match List-I with List-II. P: for distinct roots α,β of x²+x+1=0, find the quadratic with roots (α+1)^(-2026),(β+1)^(-2026). Q: the same with exponent -2027. R: for distinct roots γ,δ of x²−x+1=0, evaluate (γ−1)^(-2026)+(δ−1)^(-2026). S: for distinct roots p,r of x²+x−1=0, evaluate (p+1)^(-3)+(r+1)^(-3). List-II: (1) x²+x+1=0; (2) x²−x+1=0; (3) x²+x−1=0; (4) −1; (5) −4.

Answer choices

1. A. P→1, Q→2, R→5, S→4
2. B. P→3, Q→1, R→4, S→5
3. C. P→1, Q→2, R→4, S→5
4. D. P→2, Q→3, R→5, S→4

Step-by-step solution

1. P: transform to sixth roots

   u=α+1,v=β+1; u+v=uv=1; u²−u+1=v²−v+1=0; u³=v³=−1; u⁶=v⁶=1. 2026≡4 (mod 6), so u^(−2026)=u², v^(−2026)=v². Sum=−1, product=1.

   P gives x²+x+1=0, List-II entry 1.

2. Q: reciprocal power is the original shifted root

   2027≡5 (mod 6); u^(−2027)=u, v^(−2027)=v; sum=1, product=1.

   Q gives x²−x+1=0, entry 2.

3. R: shifted roots are cube roots of unity

   a=γ−1,b=δ−1; a+b=−1,ab=1; a³=b³=1. 2026≡1 (mod 3), so a^(−2026)+b^(−2026)=1/a+1/b=(a+b)/(ab)=−1.

   R matches entry 4.

4. S: reciprocal substitution removes denominators

   p(p+1)=r(r+1)=1; (p+1)^(-3)+(r+1)^(-3)=p³+r³=(p+r)³−3pr(p+r)=−1−3=−4.

   S matches entry 5. The entire mapping uniquely gives C.

Answer: Choice C — P→1, Q→2, R→4, S→5

Back to the answer key

Question 14: Trigonometric equations and counting on closed intervals

Format: Matching-list

Match the cardinalities in List-I to List-II. P: {x∈[−π,π]: sin⁶x+cos⁴x=1}; Q: {x∈[−π/2,π/2]: sin²x+cos⁶x=1}; R: {x∈[−π,π]: cos²(x/2)−sin²x=1/2}; S: {x∈[−2π,2π]: 6sin²(x/2)−cos(3x)=3}. List-II entries (1),(2),(3),(4),(5) mean counts 1,2,3,4,5 respectively.

Answer choices

1. A. P→2, Q→5, R→3, S→4
2. B. P→5, Q→3, R→2, S→4
3. C. P→5, Q→4, R→1, S→3
4. D. P→4, Q→3, R→2, S→5

Step-by-step solution

1. P: polynomial in sin²x

   t=sin²x∈[0,1]; t³+(1−t)²=1 ⇒ t(t+2)(t−1)=0 ⇒ t=0 or 1. x∈{−π,−π/2,0,π/2,π}.

   Five distinct real x values, including both closed endpoints: P→5.

2. Q: polynomial in cos²x

   u=cos²x∈[0,1]; 1−u+u³=1 ⇒ u(u−1)(u+1)=0 ⇒ u=0 or 1. x∈{−π/2,0,π/2}.

   Three values, including both closed endpoints: Q→3.

3. R: reject the cosine root outside [−1,1]

   c=cos x; (1+c)/2−(1−c²)=1/2 ⇒ 2c²+c−2=0 ⇒ c=(−1±√17)/4. c₋<−1, while 0<c₊<1. x=±arccos((√17−1)/4).

   Exactly two solutions within [−π,π]: R→2.

4. S: triple-angle cancellation

   3(1−cos x)−cos 3x=3 ⇒ 3cos x+cos 3x=0 ⇒ 4cos³x=0. x∈{−3π/2,−π/2,π/2,3π/2}.

   Exactly four solutions within [−2π,2π]: S→4. The complete map uniquely selects B.

Answer: Choice B — P→5, Q→3, R→2, S→4

Back to the answer key

Question 15: Orthogonal matrices and vector products

Format: Matching-list

For real numbers α, β, γ, δ and μ, let M=[[α,1/√2,−1/√2],[1/√3,β,1/√3],[γ,δ,μ]] and suppose MMᵀ=I₃. Let u=(α,1/√3,γ), v=(1/√2,β,δ), w=(−1/√2,1/√3,μ). Match List I to List II and choose the correct option.

List I

1. P. γ²+δ²
2. Q. x, where xu+yv+zw=ĵ for real x,y,z
3. R. |u·(v×w)|
4. S. |u×(v×w)|

List II

1. 1. 0
2. 2. 1
3. 3. 1/√2
4. 4. 1/√3
5. 5. 5/6

Answer choices

1. A. P→5, Q→4, R→2, S→1
2. B. P→4, Q→5, R→1, S→2
3. C. P→5, Q→3, R→2, S→1
4. D. P→5, Q→4, R→1, S→2

Step-by-step solution

1. Rows and columns are orthonormal

   MMᵀ=I ⇒ M⁻¹=Mᵀ ⇒ MᵀM=I

   Because M is square, the hypothesis implies invertibility and makes both its rows and its columns orthonormal.

2. P: use the third column and third row

   1/2+1/3+μ²=1 ⇒ μ²=1/6; γ²+δ²+μ²=1 ⇒ γ²+δ²=5/6

   P matches List II entry 5.

3. Q: project onto u

   u·(xu+yv+zw)=u·ĵ ⇒ x=1/√3

   u has unit length and is perpendicular to v and w. Q matches entry 4.

4. R: determinant magnitude

   |u·(v×w)|=|det M|; (det M)²=det(MMᵀ)=1 ⇒ |det M|=1

   The determinant may have either sign, but its absolute value is always 1. R matches entry 2.

5. S: vector triple product

   u×(v×w)=v(u·w)−w(u·v)=0 ⇒ |u×(v×w)|=0

   Both scalar products vanish by orthogonality. S matches entry 1.

Answer: Choice A — P→5, Q→4, R→2, S→1; values (5/6,1/√3,1,0)

Back to the answer key

Question 16: Coordinate geometry: circle, parabola, ellipse and hyperbola

Format: Matching-list

Match each entry in List I to the correct point in List II and choose the correct option.

List I

1. P. The circle with centre (1,2), touching 3x+4y=1, passes through which point?
2. Q. The common tangent to x²+y²=2 and y²=8x with positive slope passes through which point?
3. R. M is the first-quadrant endpoint of the latus rectum of 3x²+4y²=48. The normal at M passes through which point?
4. S. H is the hyperbola centred at the origin, with one focus (5,0) and one directrix 5x+16=0. H passes through which point?

List II

1. 1. (1,1)
2. 2. (7,9)
3. 3. (3,2)
4. 4. (2,5)
5. 5. (8,3√3)

Answer choices

1. A. P→3, Q→4, R→1, S→2
2. B. P→3, Q→2, R→1, S→5
3. C. P→3, Q→2, R→4, S→5
4. D. P→4, Q→1, R→2, S→3

Step-by-step solution

1. P: circle radius from a tangent line

   r=|3(1)+4(2)−1|/√(3²+4²)=2; (x−1)²+(y−2)²=4

   The point (3,2) satisfies the circle equation, so P→3.

2. Q: common tangent with positive slope

   y=mx+2/m; (2/m)²/(1+m²)=2 ⇒ m⁴+m²−2=0 ⇒ (m²−1)(m²+2)=0 ⇒ m=1; y=x+2

   The only positive real slope is m=1. The line passes through (7,9), so Q→2. Its distance from the origin is √2, and substitution into y²=8x gives (x−2)²=0, confirming tangency to both curves.

3. R: normal at the first-quadrant latus rectum endpoint

   x²/16+y²/12=1; a=4,b²=12,c=2; M=(c,b²/a)=(2,3); dy/dx=−3x/(4y)=−1/2; normal: y−3=2(x−2)

   The normal is y=2x−1 and passes through (1,1), so R→1.

4. S: hyperbola from a focus and a directrix

   c=5; a²/c=16/5 ⇒ a²=16; b²=c²−a²=9; x²/16−y²/9=1

   At (8,3√3), the left side is 4−3=1, so S→5.

5. Match to the choices and source key

   P→3, Q→2, R→1, S→5 ⇒ B

   This complete mapping is exactly option B, matching the supplied provisional answer key B.

Answer: Choice B — P→3, Q→2, R→1, S→5

Back to the answer key