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Day 165: JEE Mains Maths

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

  1. Question 1

    Relations and Functions. Asked once: 4 Apr 2025 (same type asked 4 times) See all 4 of this type

    133
    Let the domains of the functions f(x)=log⁡4log⁡3log⁡7(8−log⁡2(x2+4x+5))f(x) =\log_{4}\log_{3}\log_{7}\left( 8 -\log_{2}\left( x^{2}+ 4x+ 5 \right) \right) and g(x)=sin⁡−1(7x+10x−2)g(x) =\sin^{- 1}\left( \frac{7x+ 10}{x- 2} \right) be (α,β)(\alpha,\beta) and [γ,δ]\lbrack\gamma,\delta\rbrack, respectively. Then α2+β2+γ2+δ2\alpha^{2}+\beta^{2}+\gamma^{2}+\delta^{2} is equal to :-

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  2. Question 2

    Vector Algebra. Asked once: 27 Jan 2024 (same type asked 4 times) See all 4 of this type

    81
    The least positive integral value of α\alpha, for which the angle between the vectors αi^−2j^+2k^\alpha\widehat{i}- 2\widehat{j}+ 2\widehat{k} and αi^+2αj^−2k^\alpha\widehat{i}+ 2\alpha\widehat{j}- 2\widehat{k} is acute, is.

  3. Question 3

    Vector Algebra. Asked once: 8 Apr 2024 (same type asked 4 times) See all 4 of this type

    152
    Let  a→=i^+2j^+3k^, b→=2i^+3j^−5k^\ \overrightarrow{a}=\widehat{i}+ 2\widehat{j}+ 3\widehat{k},\ \overrightarrow{b}= 2\widehat{i}+ 3\widehat{j}- 5\widehat{k} and c→=3i^−j^+λk^\overrightarrow{c}= 3\widehat{i}-\widehat{j}+ \lambda\widehat{k} be three vectors. Let r→\overrightarrow{r} be a unit vector along b→+c→\overrightarrow{b}+\overrightarrow{c}. If r→⋅a→=3\overrightarrow{r}\cdot\overrightarrow{a}= 3, then 3λ3\lambda is equal to:

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  4. Question 4

    Conic Sections. Asked once: 4 Apr 2025 (same type asked 4 times) See all 4 of this type

    139
    Let for two distinct values of pp the lines y=x+py=x+p touch the ellipse E:x242+y232=1E:\frac{x^{2}}{4^{2}}+\frac{y^{2}}{3^{2}}= 1 at the points A and B . Let the line y=xy = x intersect E at the points C and D . Then the area of the quadrilateral ABCD is equal to.

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  5. Question 5

    Differentiation. Asked once: 1 Feb 2023 (same type asked 4 times) See all 4 of this type

    86
    If f(x)=x2+g′(1)x+g′′(2)f(x) =x^{2}+g^{'}(1)x +g^{''}(2) and g(x)=f(1)x2+xf′(x)+f′′(x)g(x) = f(1)x^{2}+ xf^{'}(x) +f^{''}(x), then the value of f(4)−f(4) - g(4)g(4) is equal to