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Exponential numerator that factorises as (a^x - 1)(b^x - 1) or a square

This type of question, asked 10 times: 21 Apr 2025 S1, 21 Apr 2025 S2, 23 Apr 2025 S1, 25 Apr 2025 S2, 16 Apr 2026 S2, 18 Apr 2026 S2, 12 May 2026 S1, 14 May 2026 S1, 15 May 2026 S1, 15 May 2026 S2. All 10 questions below, oldest first.

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  1. Question 1

    21 Apr 2025 S1

    143
    If f(x)={8x−4x−2x+1x2, if x>0exsin⁡x+x+λlog⁡4, if x⩽0f(x) =\left\{ \begin{matrix} \frac{8^{x}-4^{x}-2^{x}+ 1}{x^{2}}, & \text{~if~}x> 0 \\ e^{x}\sin x+x+\lambda \log4, & \text{~if~}x\leqslant 0 \end{matrix} \right. is continuous at x=0x= 0 then the value of 1000eλ=1000e^{\lambda}=

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  2. Question 2

    21 Apr 2025 S2

    109
    If f(x)={8x−4x−2x+1xx2, if x>0exsin⁡x+ix+λlog⁡4, if x⩽0,i∈Rf(x) =\left\{ \begin{matrix} \frac{8^{x}-4^{x}-2^{x}+1^{x}}{x^{2}}, & \text{~if~}x> 0 \\ e^{x}\sin x+ix+\lambda \log4, & \text{~if~}x\leqslant 0,i\in\mathbb{R} \end{matrix} \right. is continuous at x=0x= 0, then the value of 500eλ500e^{\lambda} is

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  3. Question 3

    23 Apr 2025 S1

    109
    If f(x)=10x+7x−14x−5x1−cos⁡x,x≠0f(x) =\frac{10^{x}+7^{x}-14^{x}-5^{x}}{1 - \cos x},x\neq 0 is continuous at x=0x= 0, then the value of f(0)f(0) is

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  4. Question 4

    25 Apr 2025 S2

    104
    lim⁡x→063x−9x−7x+12−1+cos⁡x=\lim_{x\rightarrow 0} \frac{63^{x}-9^{x}-7^{x}+ 1}{\sqrt{2}-\sqrt{1 + \cos x}}=...... .

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  5. Question 5

    16 Apr 2026 S2

    59
    If lim⁡x→045x−9x−5x+1(kx−1)(3x−1)=2\displaystyle\lim_{x \to 0}\dfrac{45^x - 9^x - 5^x + 1}{(k^x - 1)(3^x - 1)} = 2, then the value of kk is ...

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  6. Question 6

    18 Apr 2026 S2

    70
    If f(x)=3x+3−x−2tan⁡x⋅log⁡(1+x)f(x) = \dfrac{3^x + 3^{-x} - 2}{\tan x \cdot \log(1 + x)} for x≠0x \neq 0, is continuous at x=0x = 0, then the value of f(0)f(0) is equal to …\ldots

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  7. Question 7

    12 May 2026 S1

    63
    lim⁡x→1[22x−2−2x+1sin⁡2(x−1)]=\lim_{x \to 1}\left[\dfrac{2^{2x-2} - 2^x + 1}{\sin^2(x - 1)}\right] =

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  8. Question 8

    14 May 2026 S1

    61
    If the function f(x)f(x) is defined as below f(x)={(3x+3−x−2)log⁡(1+4x)(1−cos⁡4x)tan⁡(2x),x≠0(log⁡p)rq,x=0f(x) = \begin{cases} \dfrac{(3^x + 3^{-x} - 2)\log(1 + 4x)}{(1 - \cos 4x)\tan(2x)}, & x \neq 0 \\ \dfrac{(\log p)^r}{q}, & x = 0 \end{cases} is continuous at x=0x = 0 then p+q−r=…p + q - r = \ldots

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  9. Question 9

    15 May 2026 S1

    71
    Let f(x)=12x−4x−3x+11−cos⁡2xf(x) = \dfrac{12^x - 4^x - 3^x + 1}{1 - \cos 2x}, for x≠0x \ne 0. If f(x)f(x) is continuous at x=0x = 0 and f(0)=(log⁡a)⋅(log⁡3)4kf(0) = \dfrac{(\log a) \cdot (\log 3)}{4k}, then the value of aa in terms of kk is:

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  10. Question 10

    15 May 2026 S2

    65
    If k>5k > 5 and lim⁡x→0k2x+25x−2⋅(5k)xtan⁡3x⋅ln⁡(1+3x)=lim⁡n→∞(n+1)!+n!(n+1)!−n!\lim_{x \to 0}\dfrac{k^{2x} + 25^x - 2\cdot(5k)^x}{\tan 3x \cdot \ln(1 + 3x)} = \lim_{n \to \infty}\dfrac{(n + 1)! + n!}{(n + 1)! - n!}, then value of kk is...

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