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Day 41: JEE Mains Maths

5 questions. Try each one first, then open its answer.

  1. Question 1

    Vector Algebra. This type asked 4 times: 2021 Paper 19, 4 Apr 2024, 8 Apr 2024, 22 Jan 2026 S1. This one: 22 Jan 2026 S1 See all 4 of this type

    51
    Let AB→=2i^+4j^−5k^\overrightarrow{AB}= 2\widehat{i}+ 4\widehat{j}- 5\widehat{k} and AD→=i^+2j^+λk^,λ∈R\overrightarrow{AD}=\widehat{i}+ 2\widehat{j}+ \lambda\widehat{k},\lambda \in R. Let the projection of the vector v→=i^+j^+k^\overrightarrow{v}=\widehat{i}+\widehat{j}+\widehat{k} on the diagonal AC→\overrightarrow{AC} of the parallelogram ABCD be of length one unit. If α,β\alpha,\beta, where α>β\alpha > \beta, be the roots of the equation λ2x2−6λx+5=0\lambda^{2}x^{2}- 6\lambda x + 5 = 0, then 2α−β2\alpha - \beta is equal to

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  2. Question 2

    Conic Sections. This type asked 4 times: 2021 Paper 17, 2021 Paper 18, 4 Apr 2025, 21 Jan 2026 S2. This one: 21 Jan 2026 S2 See all 4 of this type

    54
    If the line αx+4y=7\alpha x + 4y =\sqrt{7}, where α∈R\alpha \in R, touches the ellipse 3x2+4y2=13x^{2}+ 4y^{2}= 1 at the point PP in the first quadrant, then one of the focal distances of P is :

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  3. Question 3

    Differentiation. This type asked 4 times: 24 Jan 2023, 1 Feb 2023, 27 Jan 2024, 21 Jan 2026 S2. This one: 21 Jan 2026 S2 See all 4 of this type

    53
    Let f(x)=x3+x2f′(1)+2xf′′(2)+f′′′(3)f(x) =x^{3}+x^{2}f^{'}(1) + 2xf^{''}(2) +f^{'''}(3), x∈Rx \in R. Then the value of f′(5)f^{'}(5) is :

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  4. Question 4

    Complex Numbers. This type asked 4 times: 2021 Paper 18, 26 Jun 2022, 4 Apr 2025, 21 Jan 2026 S1. This one: 21 Jan 2026 S1 See all 4 of this type

    67
    If x2+x+1=0x^{2}+ x + 1 = 0, then the value of (x+1x)4+(x2+1x2)4+(x3+1x3)4+…+(x25+1x25)4\left( x +\frac{1}{x} \right)^{4}+\left( x^{2}+\frac{1}{x^{2}} \right)^{4}+\left( x^{3}+\frac{1}{x^{3}} \right)^{4}+ \ldots +\left( x^{25}+\frac{1}{x^{25}} \right)^{4} is :

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  5. Question 5

    Conic Sections. This type asked 4 times: 2021 Paper 4, 27 Jul 2022, 27 Jan 2024, 7 Apr 2025. This one: 7 Apr 2025 See all 4 of this type

    143
    Let e1e_{1} and e2e_{2} be the eccentricities of the ellipse x2 b2+y225=1\frac{x^{2}}{{\text{ }b}^{2}}+\frac{y^{2}}{25}= 1 and the hyperbola x216−\frac{x^{2}}{16}- y2 b2=1\frac{y^{2}}{{\text{ }b}^{2}}= 1, respectively. If b<5b < 5 and e1e2=1e_{1}e_{2}= 1, then the eccentricity of the ellipse having its axes along the coordinate axes and passing through all four foci (two of the ellipse and two of the hyperbola) is:

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