JEE MATH APEX

Methods of differentiation-JEE-Mains-PYQ’s

Mathematics Questions

If \log _{e} y=3 \sin ^{-1} x, then \left(1-x^{2}\right) y^{\prime \prime}-x y^{\prime} at x=\frac{1}{2} is equal to
(A) 9 e^{\pi / 2}
(B) 9 e^{\pi / 6}
(C) 3 e^{\pi / 2}
(D) 3 e^{\pi / 6}
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Let f(x)=a x^{3}+b x^{2}+c x+41 be such that f(1)=40, f^{\prime}(1)=2 and f^{\prime \prime}(1)=4. Then a^{2}+b^{2}+c^{2} is equal to:
(A) 54
(B) 51
(C) 73
(D) 62
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Suppose for a differentiable function h, h(0)=0, h(1)=1 and h^{\prime}(0)=h^{\prime}(1)=2. If g(x)=h\left(e^{x}\right) e^{h(x)}, then g^{\prime}(0) is equal to:
(A) 4
(B) 5
(C) 3
(D) 8
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If f(x)=\begin{cases} x^{3} \sin \left(\frac{1}{x}\right), & x \neq 0 \\ 0, & x=0 \end{cases}, then
(A) f^{\prime \prime}(0)=0
(B) f^{\prime \prime}(0)=1
(C) f^{\prime \prime}\left(\frac{2}{\pi}\right)=\frac{24-\pi^{2}}{2 \pi}
(D) f^{\prime \prime}\left(\frac{2}{\pi}\right)=\frac{12-\pi^{2}}{2 \pi}
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Let f:(-\infty, \infty)-\{0\} \rightarrow \mathbb{R} be a differentiable function such that f^{\prime}(1)=\lim _{a \rightarrow \infty} a^{2} f\left(\frac{1}{a}\right). Then \lim _{a \rightarrow \infty} \frac{a(a+1)}{2} \tan ^{-1}\left(\frac{1}{a}\right)+a^{2}-2 \log _{e} a is equal to:
(A) \frac{5}{2}+\frac{\pi}{8}
(B) \frac{3}{8}+\frac{\pi}{4}
(C) \frac{3}{4}+\frac{\pi}{8}
(D) \frac{3}{2}+\frac{\pi}{4}
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If y(\theta)=\frac{2 \cos \theta+\cos 2 \theta}{\cos 3 \theta+4 \cos 2 \theta+5 \cos \theta+2}, then at \theta=\frac{\pi}{2}, y^{\prime \prime}+y^{\prime}+y is equal to:
(A) \frac{1}{2}
(B) 1
(C) \frac{3}{2}
(D) 2
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Let f: \mathbb{R}-\{0\} \rightarrow \mathbb{R} be a function satisfying f\left(\frac{x}{y}\right)=\frac{f(x)}{f(y)} for all x, y, f(y) \neq 0. If f^{\prime}(1)=2024, then:
(A) x f^{\prime}(x)+2024 f(x)=0
(B) x f^{\prime}(x)-2023 f(x)=0
(C) x f^{\prime}(x)-2024 f(x)=0
(D) x f^{\prime}(x)+f(x)=2024
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Let g: \mathbb{R} \rightarrow \mathbb{R} be a non-constant twice differentiable function such that g^{\prime}\left(\frac{1}{2}\right)=g^{\prime}\left(\frac{3}{2}\right). If a real-valued function f is defined as f(x)=\frac{1}{2}[g(x)+g(2-x)], then:
(A) f^{\prime \prime}(x)=0 for at least two x in (0,2)
(B) f^{\prime}\left(\frac{3}{2}\right)+f^{\prime}\left(\frac{1}{2}\right)=1
(C) f^{\prime \prime}(x)=0 for no x in (0,1)
(D) f^{\prime \prime}(x)=0 for exactly one x in (0,1)
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If f(x)=\begin{vmatrix} 2 \cos ^{4} x & 2 \sin ^{4} x & 3+\sin ^{2} 2 x \\ 3+2 \cos ^{4} x & 2 \sin ^{4} x & \sin ^{2} 2 x \\ 2 \cos ^{4} x & 3+2 \sin ^{4} x & \sin ^{2} 2 x \end{vmatrix}, then \frac{1}{5} f^{\prime}(0) is equal to:
(A) 2
(B) 1
(C) 0
(D) 6
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Let y=\log _{e}\left(\frac{1-x^{2}}{1+x^{2}}\right),-1<x<1. Then at x=\frac{1}{2}, the value of 225\left(y^{\prime}-y^{\prime \prime}\right) is equal to:
(A) 732
(B) 732
(C) 742
(D) 746
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