Q. For the function $f(x)=x \cos \frac{1}{x}, x \geq 1$
(A) for at least one $x$ in the interval $[1, \infty), f(x+2)-f(x)<2$
(B) $\lim _{x \rightarrow \infty} f^{\prime}(x)=1$
(C) for all $x$ in the interval $[1, \infty), f(x+2)-f(x)>2$
(D) $f^{\prime}(x)$ is strictly decreasing in the interval $[1, \infty)$
[JEE 2009, 4]
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Sol. (B,C,D) $\mathrm{f}(\mathrm{x})=\mathrm{x} \cos \frac{1}{\mathrm{x}}, \mathrm{x} \geq 1$ $\mathrm{f}^{\prime}(\mathrm{x})=\cos \frac{1}{\mathrm{x}}+\frac{1}{\mathrm{x}} \sin \frac{1}{\mathrm{x}}$ $\lim _{x \rightarrow \infty} \mathrm{f}^{\prime}(\mathrm{x})=1$ $\mathrm{Now} \quad \mathrm{f}^{\prime \prime}(\mathrm{x})=-\sin \frac{1}{\mathrm{x}} \times-\frac{1}{\mathrm{x}^{2}}-\frac{1}{\mathrm{x}^{2}} \sin \frac{1}{\mathrm{x}}+\frac{1}{\mathrm{x}} \cos \frac{1}{\mathrm{x}} \times-\frac{1}{\mathrm{x}^{2}}$ $\therefore \mathrm{f}^{\prime}(\mathrm{x}+2)<\mathrm{f}^{\prime}(\mathrm{x})$ $\therefore \mathrm{f}^{\prime}(\mathrm{x}+2)<\mathrm{f}^{\prime}(\mathrm{x})$ Also $\lim _{\mathrm{x} \rightarrow \infty} \mathrm{f}(\mathrm{x}+2)-\mathrm{f}(\mathrm{x})$ $=\lim _{x \rightarrow \infty}\left[(x+2) \cos \frac{1}{(x+2)}-x \cos \frac{1}{x}\right]=2$ $\therefore f(x+2)-f(x)>2 \quad \forall x \geq 1$
Q. Let $f$ be a real-valued function defined on the interval $(0, \infty)$ by $\mathrm{f}(\mathrm{x})=\ell \mathrm{nx}+\int_{0}^{\mathrm{x}} \sqrt{1+\sin \mathrm{t} \mathrm{d} \mathrm{t}} \cdot$ Then which of the following statement (s) is/(are) true?
(A) $\mathrm{f}^{\prime \prime}(\mathrm{x})$ exists for all $\mathrm{x} \in(0, \infty)$
(B) $\mathrm{f}^{\prime}(\mathrm{x})$ exists for all $\mathrm{x} \in(0, \infty)$ and $\mathrm{f}^{\prime}$ is continuous on $(0, \infty),$ but not differentiable on $(0, \infty)$
(C) there exists $\alpha>1$ such that $\left|\mathrm{f}^{\prime}(\mathrm{x})\right|<|\mathrm{f}(\mathrm{x})|$ for all $\mathrm{x} \in(\alpha, \infty)$
(D) there exists $\beta>0$ such that $|\mathrm{f}(\mathrm{x})|+\left|\mathrm{f}^{\prime}(\mathrm{x})\right| \leq \beta$ for all $\mathrm{x} \in(0, \infty)$
[JEE 10, 3M]
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Sol. (B,C) $\mathrm{f}(\mathrm{x})=\ln \mathrm{x}+\int_{0}^{\mathrm{x}} \sqrt{1+\sin \mathrm{t}} \mathrm{d} \mathrm{t}$ $\mathrm{f}^{\prime}(\mathrm{x})=\frac{1}{\mathrm{x}}+\sqrt{1+\sin \mathrm{x}}$ $\mathrm{f}^{\prime}(\mathrm{x})=\frac{1}{\mathrm{x}}+\sqrt{2}\left|\cos \left(\frac{\mathrm{x}}{2}-\frac{\pi}{4}\right)\right|$ $\because\left|\cos \left(\frac{\mathrm{x}}{2}-\frac{\pi}{4}\right)\right|$ is non-derivable $\therefore \mathrm{f}^{\prime}(\mathrm{x})$ is non-derivable but continuous. hence option (A) is incorrect \& option (B) is correct. For option $\mathrm{C}$ $\mathrm{f}(\mathrm{x})=(\ell \mathrm{nx})+\int_{0}^{\mathrm{x}}(\sqrt{1+\sin \mathrm{x}}) \mathrm{d} \mathrm{x}$ since $\mathrm{f}(\mathrm{x})$ is positive increasing function for all $\mathrm{x}>1$ $\Rightarrow|\mathrm{f}(\mathrm{x})|=\mathrm{f}(\mathrm{x}) \&\left|\mathrm{f}^{\prime}(\mathrm{x})\right|=\mathrm{f}^{\prime}(\mathrm{x})$ Let $\mathrm{f}(\mathrm{x})=\mathrm{y}$ $\mathrm{f}^{\prime}(\mathrm{x})-\mathrm{f}(\mathrm{x})=\frac{1}{\mathrm{x}}-\ell \mathrm{nx}+\sqrt{1+\sin \mathrm{x}}-\int_{0}^{\mathrm{x}} \sqrt{1+\sin \mathrm{t}} \mathrm{d} \mathrm{t}$ $\mathrm{f}^{(\mathrm{x})}-\mathrm{f}(\mathrm{x})=\frac{1}{\mathrm{x}}-\ln \mathrm{x}+\sqrt{1+\sin \mathrm{x}}-\sqrt{2} \int_{0}^{\mathrm{x}}\left(\frac{\mathrm{t}}{2}-\frac{\pi}{4}\right) \| \mathrm{dt}$ $\frac{1}{\mathrm{x}}-\ln \mathrm{x}<0 ;$ when $\alpha>\mathrm{e}$ $0 \leq \sqrt{1+\sin \mathrm{x}} \leq \sqrt{2}$ $\int_{0}^{\mathrm{x}}\left|\cos \left(\frac{\mathrm{t}}{2}-\frac{\pi}{4}\right)\right| \mathrm{dt}>\sqrt{2} \forall \alpha>\frac{3 \pi}{2}$ $\Rightarrow \mathrm{f}^{\prime}(\mathrm{x})-\mathrm{f}(\mathrm{x})<0 \quad \forall \quad \alpha>\frac{3 \pi}{2}>1$ Hence option (C) is correct. For option (D) $|\mathrm{f}(\mathrm{x})|+\left|\mathrm{f}^{\prime}(\mathrm{x})\right| \rightarrow \infty$ when $\mathrm{x} \rightarrow \infty$ Therefore option (D) is incorrect. Alternate : $\mathrm{f}(\mathrm{x})=\ln \mathrm{x}+\int_{0}^{\mathrm{x}} \sqrt{1+\sin \mathrm{t}} \mathrm{dt}$ $\mathrm{f}^{\prime}(\mathrm{x})=\frac{1}{\mathrm{x}}+\sqrt{1+\sin \mathrm{x}}$ ……..(i) for $\mathrm{x}>1$ $\frac{1}{\mathrm{x}}+\sqrt{1+\sin \mathrm{x}}<1+\sqrt{2}$ but $\ell \mathrm{nx}+\int_{0}^{\mathrm{x}} \sqrt{1+\sin \mathrm{t}} \mathrm{dt}$ will always be more than $1+\sqrt{2}$ for some $\alpha>1$ $\because \int_{0}^{x} \sqrt{1+\sin t}>0 \quad \& \quad \ln x$ is increasing in $(1, \infty)$ $\Rightarrow \quad \mathrm{f}(\mathrm{x})>\mathrm{f}^{\prime}(\mathrm{x}) \forall \alpha>1$ $\therefore(\mathrm{C})$ is correct $\mathrm{f}^{\prime \prime}(\mathrm{x})=-\frac{1}{\mathrm{x}^{2}}+\frac{\cos \mathrm{x}}{2 \sqrt{1+\sin \mathrm{x}}}$ $\Rightarrow \mathrm{f}$ is not derivable on $(0, \infty)$ at $\frac{3 \pi}{2}, \frac{7 \pi}{2}$ $\therefore$ (B) is also correct $\mathrm{f}(\mathrm{x})$ is unbounded near $\mathrm{x}=0$ in $(0,1)$ hence $|\mathrm{f}(\mathrm{x})|$ can never be made less than a finite number hence $|\mathrm{f}(\mathrm{x})|+\left|\mathrm{f}^{\prime}(\mathrm{x})\right|$ can never be less than $\beta$
Q. Let $f:(0,1) \rightarrow \mathrm{R}$ be defined by $f(\mathrm{x})=\frac{\mathrm{b}-\mathrm{x}}{1-\mathrm{bx}},$ where b is a constant such that $0<\mathrm{b}<1 .$ Then
(A) $f$ is not invertible on $(0,1)$
(B) $f \neq f^{-1}$ on $(0,1)$ and $f^{\prime}(b)=\frac{1}{f^{\prime}(0)}$
(C) $f=f^{-1}$ on $(0,1)$ and $f^{\prime}(b)=\frac{1}{f^{\prime}(0)}$
(D) $f^{-1}$ is differentiable on $(0,1)$
[JEE 2011, 4M]
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Sol. (A) $f:(0,1) \rightarrow \mathrm{R}$ $f(\mathrm{x})=\frac{\mathrm{b}-\mathrm{x}}{1-\mathrm{bx}}$ $\mathrm{b} \in(0,1)$ $\Rightarrow \quad f^{\prime}(x)=\frac{b^{2}-1}{(b x-1)^{2}}$ $\Rightarrow f^{\prime}(x)<0 \forall x \in(0,1)$ hence $f(x)$ is decreasing function hence its range $(-1, b)$ $\Rightarrow$ co-domain $\neq$ range $\Rightarrow f(x)$ is non-invertible function
Q. The number of distinct real roots of $x^{4}-4 x^{3}+12 x^{2}+x-1=0$ is
[JEE 2011, 4M]
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Sol. 2 Let $f(x)=x^{4}-4 x^{3}+12 x^{2}+x-1$ $f^{\prime}(x)=4 x^{3}-12 x^{2}+24 x$ $f^{\prime \prime}(x)=12 x^{2}-24 x+24$ $=12\left(x^{2}-2 x+2\right)>0$ $\Rightarrow f^{\prime}(x)$ is strictly increasing function $\because f^{\prime}(x)$ is cubic polynomial hence number of roots of $f^{\prime}(x)=0$ is 1 $\Rightarrow$ Number of maximum roots of $f(x)=0$ are 2 Now $f(0)=-1, f(1)=9, f(-1)=15$ $\Rightarrow f(x)$ has exactly 2 distinct real roots.
Paragraph for Question 5 and 6 Let $f(\mathrm{x})=(1-\mathrm{x})^{2} \sin ^{2} \mathrm{x}+\mathrm{x}^{2}$ for all $\mathrm{x} \in \mathbb{R},$ and let $\mathrm{g}(\mathrm{x})=\int_{1}^{\mathrm{x}}\left(\frac{2(\mathrm{t}-1)}{\mathrm{t}+1}-\ell \mathrm{nt}\right) f(\mathrm{t}) \mathrm{d} \mathrm{t}$ for all $\mathrm{x} \in$ $(1, \infty)$
Q. Consider the statements :
$\mathbf{P}:$ There exists some $\mathrm{x} \in \mathbb{R}$ such that $f(\mathrm{x})+2 \mathrm{x}=2\left(1+\mathrm{x}^{2}\right)$
$\mathrm{Q}:$ There exists some $\mathrm{x} \in \mathbb{R}$ such that $2 f(\mathrm{x})+1=2 \mathrm{x}(1+\mathrm{x})$ $\quad$ Then
(A) both P and Q are true
(B) P is true and Q is false
(C) P is false and Q is true
(D) both P and Q are false
[JEE 2012, 3M, –1M]
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Sol. (C) $f(x)=(1-x)^{2} \sin ^{2} x+x^{2}$ $P: f(x)+2 x=2\left(1+x^{2}\right)$ $\Rightarrow(1-x)^{2} \sin ^{2} x+x^{2}+2 x=2+2 x^{2}$ $\Rightarrow(1-x)^{2} \sin ^{2} x-x^{2}+2 x-2=0$ $(1-x)^{2} \cos ^{2} x+1=0$ which is not possible. $\therefore \mathrm{P}$ is false. $\mathrm{Q}: 2 f(\mathrm{x})+1=2 \mathrm{x}(1+\mathrm{x})$ $2 \mathrm{x}^{2}+2(1-\mathrm{x})^{2} \sin ^{2} \mathrm{x}+1=2 \mathrm{x}^{2}+2 \mathrm{x}$ $2(1-\mathrm{x})^{2} \sin ^{2} \mathrm{x}-2 \mathrm{x}+1=0$ Let $\mathrm{h}(\mathrm{x})=2(1-\mathrm{x})^{2} \sin ^{2} \mathrm{x}-2 \mathrm{x}+1$ clearly $\mathrm{h}(1)=-1$ and $\mathrm{h}(\mathrm{x})=2\left(\mathrm{x}^{2}-2 \mathrm{x}+1\right) \sin ^{2} \mathrm{x}-2 \mathrm{x}+1$ $=\mathrm{x}^{2}\left[2\left(1-\frac{2}{\mathrm{x}}+\frac{1}{\mathrm{x}^{2}}\right) \cdot \sin ^{2} \mathrm{x}-\frac{2}{\mathrm{x}}+\frac{1}{\mathrm{x}^{2}}\right]$ $\therefore \mathrm{h}(\mathrm{x}) \rightarrow \infty$ as $\mathrm{x} \rightarrow \infty$ $\therefore$ By intermediate value theorem $\mathrm{h}(\mathrm{x})=0$ has a root which is greater than 1 Hence $\mathrm{Q}$ is true.
Q. Which of the following is true ?
(A) $\mathrm{g}$ is increasing on $(1, \infty)$
(B) $\mathrm{g}$ is decreasing on $(1, \infty)$
(C) $\mathrm{g}$ is increasing on $(1,2)$ and decreasing on $(2, \infty)$
(D) $\mathrm{g}$ is decreasing on $(1,2)$ and increasing on $(2, \infty)$
[JEE 2012, 3M, –1M]
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Sol. (B) $\mathrm{g}(\mathrm{x})=\int_{1}^{\mathrm{x}}\left(\frac{2(\mathrm{t}-1)}{(\mathrm{t}+1)}-\ell \mathrm{nt}\right) f(\mathrm{t}) \mathrm{dt}$ $g^{\prime}(x)=\left(\frac{2(x-1)}{x+1}-\ln x\right) f(x)$ $f(\mathrm{x})>0 \quad \forall \mathrm{x} \in \mathrm{R}$ Suppose. $h(x)=\frac{2(x-1)}{x+1}-\ln x$ $h(x)=2-\left(\frac{4}{x+1}+\ln x\right)$ $h^{\prime}(x)=\frac{4}{(x+1)^{2}}-\frac{1}{x}$ $h^{\prime}(x)=-\frac{(x-1)^{2}}{x(x+1)^{2}}$ $h^{\prime}(x)<0$ So h(x) is decreasing so h(x) is decreasing so h(x) $<0 \quad$ (1). $\quad \forall x>1$ h(x) $<0 \quad \forall x>1$ $\{h(1)=0\}$ So $g^{\prime}(x)=h(x) f(x)$ $g^{\prime}(x)<0 \quad \forall x>1$ $\mathrm{g}(\mathrm{x})$ is decreasing in $(1, \infty)$
Q. If $f(\mathrm{x})=\int_{0}^{\mathrm{x}} \mathrm{e}^{\mathrm{t}^{2}}(\mathrm{t}-2)(\mathrm{t}-3) \mathrm{dt}$ for all $\mathrm{x} \in(0, \infty),$ then $-$
(A) $f$ has a local maximum at $x=2$
(B) $f$ is decreasing on $(2,3)$
(C) there exists some $c \in(0, \infty)$ such that $f^{\prime \prime}(c)=0$
(D) $f$ has a local minimum at $x=3$
[JEE 2012, 4M]
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Sol. (A,B,C,D) $f(\mathrm{x})=\int_{0}^{\mathrm{x}} e^{\mathrm{t}^{2}}(\mathrm{t}-2)(\mathrm{t}-3) \mathrm{dt}$
$\Rightarrow f^{\prime}(x)=e^{x^{2}}(x-2)(x-3)$
$\therefore f^{\prime}(2)=f^{\prime}(3)=0$
$\Rightarrow f^{\prime \prime}(\mathrm{c})=0$ for same $\mathrm{c} \in(2,3)$ (by Rolle’s theorem)
Q. The number of points in $(-\infty, \infty),$ for which $\mathrm{x}^{2}-\mathrm{x} \sin \mathrm{x}-\cos \mathrm{x}=0,$ is
(A) 6 (B) 4 (C) 2 (D) 0
[JEE 2013, 2M]
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Sol. (C) $f(x)=x^{2}-x \sin x-\cos x$ $\quad f^{\prime}(x)=2 x-x \cos x-\sin x+\sin x$ $\quad=x(2-\cos x)$
Q. Let $f(\mathrm{x})=\mathrm{x} \sin \pi \mathrm{x}, \mathrm{x}>0 .$ Then for all natural numbers $\mathrm{n}, f^{\prime}(\mathrm{x})$ vanishes at –
(A) a unique point in the interval $\left(\mathrm{n}, \mathrm{n}+\frac{1}{2}\right)$
(B) a unique point in the interval $\left(\mathrm{n}+\frac{1}{2}, \mathrm{n}+1\right)$
(C) a unique point in the interval $(\mathrm{n}, \mathrm{n}+1)$
(D) two points in the interval $(\mathrm{n}, \mathrm{n}+1)$
[JEE 2013, 4M, –1M]
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Sol. (B,C) $f^{\prime}(\mathrm{x})=\sin \pi \mathrm{x}+\pi \mathrm{x} \cos \pi \mathrm{x}=0$ $\tan \pi \mathrm{x}=-\pi \mathrm{x}$ $\mathrm{y}=\tan \mathrm{x} \pi \& \mathrm{y}=-\pi \mathrm{x}$
Q. Let $f, \mathrm{g}:[-1,2] \rightarrow \square$ be continuous function which are twice differentiable on the interval $(-1,2) .$ Let the values of $f$ and $\mathrm{g}$ at the points $-1,0$ and 2 be as given in the following table :
In each of the intervals (–1, 0) and (0, 2) the function (ƒ – 3g)” never vanishes. Then the correct statement(s) is(are)
(A) $f^{\prime}(x)-3 g^{\prime}(x)=0$ has exactly three solutions in $(-1,0) \cup(0,2)$
(B) $f^{\prime}(x)-3 g^{\prime}(x)=0$ has exactly one solution in $(-1,0)$
(C) $f^{\prime}(x)-3 g^{\prime}(x)=0$ has exactly one solutions in $(0,2)$
(D) $f^{\prime}(x)-3 g^{\prime}(x)=0$ has exactly two solutions in $(-1,0)$ and exactly two
solutions in $(0,2)$
[JEE 2015, 4M, –2M]
In each of the intervals (–1, 0) and (0, 2) the function (ƒ – 3g)” never vanishes. Then the correct statement(s) is(are)
(A) $f^{\prime}(x)-3 g^{\prime}(x)=0$ has exactly three solutions in $(-1,0) \cup(0,2)$
(B) $f^{\prime}(x)-3 g^{\prime}(x)=0$ has exactly one solution in $(-1,0)$
(C) $f^{\prime}(x)-3 g^{\prime}(x)=0$ has exactly one solutions in $(0,2)$
(D) $f^{\prime}(x)-3 g^{\prime}(x)=0$ has exactly two solutions in $(-1,0)$ and exactly two
solutions in $(0,2)$
[JEE 2015, 4M, –2M]
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Sol. (B,C) Let $F(x)=f(x)-3 g(x)$ $\therefore F(-1)=3 ; F(0)=3 \& F(2)=3$ $\therefore F^{\prime}(x)$ will vanish at least twice in $(-1,0) \cup(0,2)$ $\because F^{\prime \prime}(x)>0$ or $<0 \forall x \in(-1,0) \cup(0,2)$ so there will be exactly one solution in $(-1,0)$ and one in $(0,2)$ Alter Let $\mathrm{H}(\mathrm{x})=\mathrm{f}(\mathrm{x})-3 \mathrm{g}(\mathrm{x})$ $\mathrm{H}(-1)=\mathrm{H}(0)=\mathrm{H}(2)=3$ applying Rolle’s theorem in interval $[-1,0]$ $\mathrm{H}(\mathrm{x})=\mathrm{f}(\mathrm{x})-3 \mathrm{g}^{\prime}(\mathrm{x})=0$ for at least one c $\mathrm{c} \in(-1,0)$ As $\mathrm{H}^{\prime \prime}(\mathrm{x})$ never vanises in the interval $\Rightarrow$ exactly one $\mathrm{c} \in(-1,0)$ for which $\mathrm{H}^{\prime}(\mathrm{x})=0$ similarily apply rolle’s theorems in the interval $[0,2]$ $\mathrm{H}^{\prime}(\mathrm{x})=0$ has exactly one solution in $(0,2)$
Q. The total number of distinct $\mathrm{x} \in[0,1]$ for which $\int_{0}^{x} \frac{\mathrm{t}^{2}}{1+\mathrm{t}^{4}} \mathrm{dt}=2 \mathrm{x}-1$ is
[JEE(Advanced) 2016]
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Sol. 1 Let $f(\mathrm{x})=\int_{0}^{\mathrm{x}} \frac{\mathrm{t}^{2}}{1+\mathrm{t}^{4}} \mathrm{dt}-2 \mathrm{x}+1$ $\Rightarrow f^{\prime}(x)=\frac{x^{2}}{1+x^{4}}-2$ as $\quad \frac{1+x^{4}}{x^{2}} \geq 2$ $\Rightarrow \frac{\mathrm{x}^{2}}{1+\mathrm{x}^{4}} \leq \frac{1}{2}$ $\Rightarrow f^{\prime}(x) \leq-\frac{3}{2}$ $\Rightarrow f(\mathrm{x})$ is continuous and decreasing $f(0)=1$ and $f(1)=\int_{0}^{1} \frac{t^{2}}{1+t^{4}} d t-2 \leq-\frac{3}{2}$ by IVT $f(x)=0$ possesses exactly one solution in $[0,1]$
Q. Let $f(\mathrm{x})=\lim _{\mathrm{n} \rightarrow \infty}\left(\frac{\mathrm{n}^{\mathrm{n}}(\mathrm{x}+\mathrm{n})\left(\mathrm{x}+\frac{\mathrm{n}}{2}\right) \ldots .\left(\mathrm{x}+\frac{\mathrm{n}}{\mathrm{n}}\right)}{\mathrm{n} !\left(\mathrm{x}^{2}+\mathrm{n}^{2}\right)\left(\mathrm{x}^{2}+\frac{\mathrm{n}^{2}}{4}\right) \dots .\left(\mathrm{x}^{2}+\frac{\mathrm{n}^{2}}{\mathrm{n}^{2}}\right)}\right)^{\mathrm{x} / \mathrm{n}},$ for all $\mathrm{x}>0 .$ Then
(A) $f\left(\frac{1}{2}\right) \geq f(1)$
(B) $f\left(\frac{1}{3}\right) \leq f\left(\frac{2}{3}\right)$
(C) $f^{\prime}(2) \leq 0$
(D) $\frac{f^{\prime}(3)}{f(3)} \geq \frac{f^{\prime}(2)}{f(2)}$
[JEE(Advanced) 2016]
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Sol. (B,C)
Q. Let $f: \square \rightarrow(0,1)$ be a continuous function. Then, which of the following function(s) has (have) the value zero at some point in the interval $(0,1) ?$
(A) $\mathrm{e}^{\mathrm{x}}-\int_{0}^{\mathrm{x}} f(\mathrm{t}) \sin \mathrm{td} \mathrm{t}$
(B) $\mathrm{x}^{9}-f(\mathrm{x})$
(C) $f(\mathrm{x})+\int_{0}^{\frac{\pi}{2}} f(\mathrm{t}) \sin \mathrm{td} \mathrm{t}$
(D) $\quad x-\int_{0}^{\frac{\pi}{2}-x} f(t) \cos t d t$
[JEE(Advanced) 2017]
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Sol. (B,D) For option (A), Let $g(x)=e^{x}-\int_{0}^{x} f(t) \sin t d t$ $\therefore g^{\prime}(x)=e^{x}-(f(x) \cdot \sin x)>0 \forall x \in(0,1)$ $\Rightarrow g(x)=e^{x}-(f(x) \cdot \sin x)>0 \forall x \in(0,1)$ $\Rightarrow g(x)$ is strictly incrasing function. Also, $g(0)=1$ $\Rightarrow \mathrm{g}(\mathrm{x})>1 \forall \mathrm{x} \in(0,1)$ $\therefore$ option $(\mathrm{A})$ is not possible. For option $(\mathrm{B}),$ let $\mathrm{k}(\mathrm{x})=\mathrm{x}^{9}-f(\mathrm{x})$ Now, $\mathrm{k}(0)=-f(0)<0(\mathrm{As} f \in(0,1))$ Also, $\mathrm{k}(1)=1-f(1)>0(\mathrm{As} f \in(0,1))$ $\Rightarrow \mathrm{k}(0) . \mathrm{k}(1)<0$ So, option(B) is correct. For option (C), let $\mathrm{T}(\mathrm{x})=f(\mathrm{x})+\int_{0}^{\frac{\pi}{2}} f(\mathrm{t}) \cdot \sin \mathrm{t} \mathrm{dt}$ $\Rightarrow \mathrm{T}(\mathrm{x})>0 \forall \mathrm{x} \in(0,1) \text { (As } f \in(0,1))$ so, option(C) is not possible. For option (D) Let $\mathrm{M}(\mathrm{x})=\mathrm{x}-\int_{0}^{\frac{\pi}{2}-\mathrm{x}} f(\mathrm{t}) \cos \mathrm{t} \mathrm{dt}$ $\therefore \mathrm{M}(0)=0-\int_{0}^{\pi / 2} f(\mathrm{t}) \cdot \cos \mathrm{t} \mathrm{dt}<0$ Also, $\mathrm{M}(1)=1-\int_{0}^{\frac{\pi}{2}-1} f(\mathrm{t}) \cdot \cos \mathrm{td} \mathrm{t}>0$ $\Rightarrow \mathrm{M}(0) \cdot \mathrm{M}(1)<0$ $\therefore$ option (D) is correct.
Let $\mathrm{f}(\mathrm{x})=\mathrm{x}+\log _{\mathrm{e}} \mathrm{x}-\mathrm{x} \log _{\mathrm{e}} \mathrm{x}, \mathrm{x} \in(0,0)$ $*$ Column 1 contains information about zeros of $f(\mathrm{x}), f^{\prime}(\mathrm{x})$ and $f^{\prime \prime}(\mathrm{x})$ $*$ Column 2 contains information about the limiting behavior of $f(\mathrm{x}), f^{\prime}(\mathrm{x})$ and $$ f^{\prime \prime}(\mathrm{x}) \text { at infinity. } $$ $*$ Column 3 contains information about increasing/decreasing nature of $f(\mathrm{x})$ an $$ f^{\prime}(\mathrm{x}) $$
Q. Which of the following options is the only CORRECT combination ?
(A) (IV) (i) (S)
(B) (I) (ii) (R)
(C) (III) (iv) (P)
(D) (II) (iii) (S)
[JEE(Advanced) 2017]
Q. Which of the following options is the only CORRECT combination ?
(A) (III) (iii) (R)
(B) (I) (i) (P)
(C) (IV) (iv) (S)
(D) (II) (ii) (Q)
[JEE(Advanced) 2017]
Q. Which of the following options is the only INCORRECT combination ?
(A) (II) (iii) (P)
(B) (II) (iv) (Q)
(C) (I) (iii) (P)
(D) (III) (i) (R)
[JEE(Advanced) 2017]
Q. If $f: \square \rightarrow \square$ is a twice differentiable function such that $f^{\prime \prime}(x)>0$ for all $x \in \square,$ and $f\left(\frac{1}{2}\right)=\frac{1}{2}, f(1)=1,$ then
(A) $0<f^{\prime}(1) \leq \frac{1}{2}$
(B) $f^{\prime}(1) \leq 0$
(C) $f^{\prime}(1)>1$
(D) $\frac{1}{2}<f^{\prime}(1) \leq 1$
[JEE(Advanced) 2017]
When you tap hyperbola you will get continuity and differentiablity ….what is this man
above question are not of hyperbola chapter