Solve the following equations for x:

Question: Solve the following equations for x: (i) $2^{2 x}-2^{x+3}+2^{4}=0$ (ii) $3^{2 x+4}+1=2.3^{x+2}$ Solution: (i) $2^{2 x}-2^{x+3}+2^{4}=0$ $\Rightarrow\left(2^{x}\right)^{2}-\left(2^{x} \times 2^{3}\right)+\left(2^{2}\right)^{2}=0$ $\Rightarrow\left(2^{x}\right)^{2}-2 \times 2^{x} \times 2^{2}+\left(2^{2}\right)^{2}=0$ $\Rightarrow\left(2^{x}-2^{2}\right)^{2}=0$ $\Rightarrow 2^{x}-2^{2}=0$ $\Rightarrow 2^{x}=2^{2}$ $\Rightarrow x=2$ (ii) $3^{2 x+4}+1=2.3^{x+2}$ $\Rightarrow\left(3^{x+2}\r...

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Find the derivative of the following functions

Question: Find the derivative of the following functions (it is to be understood that $a, b, c, d, p, q, r$ and $s$ are fixed non-zero constants and $m$ and $n$ are integers): $\frac{a}{x^{4}}-\frac{b}{x^{2}}+\cos x$ Solution: Let $f(x)=\frac{a}{x^{4}}-\frac{b}{x^{2}}+\cos x$ $f^{\prime}(x)=\frac{d}{d x}\left(\frac{a}{x^{4}}\right)-\frac{d}{d x}\left(\frac{b}{x^{2}}\right)+\frac{d}{d x}(\cos x)$ $=a \frac{d}{d x}\left(x^{-4}\right)-b \frac{d}{d x}\left(x^{-2}\right)+\frac{d}{d x}(\cos x)$ $=a\le...

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Find the least value of a such that the function

Question: Find the least value of a such that the function $f$ given $f(x)=x^{2}+a x+1$ is strictly increasing on $[1,2]$. Solution: We have, $f(x)=x^{2}+a x+1$ $\therefore f^{\prime}(x)=2 x+a$ Now, functionfis increasing on [1,2]. $\therefore f^{\prime}(x) \geq 0$ on $[1,2]$ Now, we have $1 \leqslant x \leqslant 2$ $\Rightarrow 2 \leqslant 2 x \leqslant 4$ $\Rightarrow 2+a \leqslant 2 x+a \leqslant 4+a$ $\Rightarrow 2+a \leqslant f^{\prime}(x) \leqslant 4+a$ Since $f^{\prime}(x) \geq 0$ $\Right...

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Find the derivative of the following functions

Question: Find the derivative of the following functions (it is to be understood thata,b,c,d,p, q,randsare fixed non-zero constants andmandnare integers):$\frac{p x^{2}+q x+r}{a x+b}$ Solution: Let $f(x)=\frac{p x^{2}+q x+r}{a x+b}$ By quotient rule, $f^{\prime}(x)=\frac{(a x+b) \frac{d}{d x}\left(p x^{2}+q x+r\right)-\left(p x^{2}+q x+r\right) \frac{d}{d x}(a x+b)}{(a x+b)^{2}}$ $=\frac{(a x+b)(2 p x+q)-\left(p x^{2}+q x+r\right)(a)}{(a x+b)^{2}}$ $=\frac{2 a p x^{2}+a q x+2 b p x+b q-a p x^{2}...

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Find the derivative of the following functions

Question: Find the derivative of the following functions (it is to be understood thata,b,c,d,p, q,randsare fixed non-zero constants andmandnare integers):$\frac{a x+b}{p x^{2}+q x+r}$ Solution: Let $f(x)=\frac{a x+b}{p x^{2}+q x+r}$ By quotient rule, $f^{\prime}(x)=\frac{\left(p x^{2}+q x+r\right) \frac{d}{d x}(a x+b)-(a x+b) \frac{d}{d x}\left(p x^{2}+q x+r\right)}{\left(p x^{2}+q x+r\right)^{2}}$ $=\frac{\left(p x^{2}+q x+r\right)(a)-(a x+b)(2 p x+q)}{\left(p x^{2}+q x+r\right)^{2}}$ $=\frac{a...

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On which of the following intervals is the function

Question: On which of the following intervals is the function $f$ given by $f(x)=x^{100}+\sin x-1$ strictly decreasing? (A) $(0,1)$ (B) $\left(\frac{\pi}{2}, \pi\right)$ (C) $\left(0, \frac{\pi}{2}\right)$ (D) None of these Solution: We have, $f(x)=x^{100}+\sin x-1$ $\therefore f^{\prime}(x)=100 x^{99}+\cos x$ In interval $(0,1), \cos x0$ and $100 x^{99}0$ $\therefore f^{\prime}(x)0$ Thus, functionfis strictly increasing in interval (0, 1). In interval $\left(\frac{\pi}{2}, \pi\right), \cos x0$ ...

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Solve the following equations for x:

Question: Solve the following equations forx: (i) $7^{2 x+3}=1$ (ii) $2^{x+1}=4^{x-3}$ (iii) $2^{5 x+3}=8^{x+3}$ (iv) $4^{2 x}=\frac{1}{32}$ (v) $4^{x-1} \times(0.5)^{3-2 x}=\left(\frac{1}{8}\right)^{x}$ (vi) $2^{3 x-7}=256$ Solution: (i) $7^{2 x+3}=1$ $\Rightarrow 7^{2 x+3}=7^{0}$ $\Rightarrow 2 x+3=0$ $\Rightarrow 2 x=-3$ $\Rightarrow x=-\frac{3}{2}$ (ii) $2^{x+1}=4^{x-3}$ $\Rightarrow 2^{x+1}=\left(2^{2}\right)^{x-3}$ $\Rightarrow 2^{x+1}=\left(2^{2 x-6}\right)$ $\Rightarrow x+1=2 x-6$ $\Righ...

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Find the derivative of the following functions

Question: Find the derivative of the following functions (it is to be understood thata,b,c,d,p, q,randsare fixed non-zero constants andmandnare integers):$\frac{1}{a x^{2}+b x+c}$ Solution: Let $f(x)=\frac{1}{a x^{2}+b x+c}$ By quotient rule, $f^{\prime}(x)=\frac{\left(a x^{2}+b x+c\right) \frac{d}{d x}(1)-\frac{d}{d x}\left(a x^{2}+b x+c\right)}{\left(a x^{2}+b x+c\right)^{2}}$ $=\frac{\left(a x^{2}+b x+c\right)(0)-(2 a x+b)}{\left(a x^{2}+b x+c\right)^{2}}$ $=\frac{-(2 a x+b)}{\left(a x^{2}+b ...

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Find the derivative of the following functions

Question: Find the derivative of the following functions (it is to be understood thata,b,c,d,p, q,randsare fixed non-zero constants andmandnare integers): $\frac{1+\frac{1}{x}}{1-\frac{1}{x}}$ Solution: Let $f(x)=\frac{1+\frac{1}{x}}{1-\frac{1}{x}}=\frac{\frac{x+1}{x-1}}{\frac{x-1}{x}}=\frac{x+1}{x-1}$, where $x \neq 0$ By quotient rule, $f^{\prime}(x)=\frac{(x-1) \frac{d}{d x}(x+1)-(x+1) \frac{d}{d x}(x-1)}{(x-1)^{2}}, x \neq 0,1$ $=\frac{(x-1)(1)-(x+1)(1)}{(x-1)^{2}}, x \neq 0,1$ $=\frac{x-1-x...

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Find the derivative of the following functions

Question: Find the derivative of the following functions (it is to be understood thata,b,c,d,p, q,randsare fixed non-zero constants andmandnare integers):$\frac{a x+b}{c x+d}$ Solution: Let $f(x)=\frac{a x+b}{c x+d}$ By quotient rule, $f^{\prime}(x)=\frac{(c x+d) \frac{d}{d x}(a x+b)-(a x+b) \frac{d}{d x}(c x+d)}{(c x+d)^{2}}$ $=\frac{(c x+d)(a)-(a x+b)(c)}{(c x+d)^{2}}$ $=\frac{a c x+a d-a c x-b c}{(c x+d)^{2}}$ $=\frac{a d-b c}{(c x+d)^{2}}$...

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Simplify the following:

Question: Simplify the following: (i) $\frac{3^{n} \times 9^{n+1}}{3^{n-1} \times 9^{n-1}}$ (ii) $\frac{5 \times 25^{n+1}-25 \times 5^{2 n}}{5 \times 5^{2 n+3}-25^{n+1}}$ (iii) $\frac{5^{n+3}-6 \times 5^{n+1}}{9 \times 5^{x}-2^{2} \times 5^{n}}$ (iv) $\frac{6(8)^{n+1}+16(2)^{3 n-2}}{10(2)^{3 n+1}-7(8)^{n}}$ Solution: (i) $\frac{3^{n} \times 9^{n+1}}{3^{n-1} \times 9^{n-1}}$ $=\frac{3^{n} \times\left(3^{2}\right)^{(n+1)}}{3^{n-1} \times\left(3^{2}\right)^{n-1}}$ $=\frac{3^{n} \times 3^{2 n+2}}{3^...

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Which of the following functions are strictly decreasing on

Question: Which of the following functions are strictly decreasing on $\left(0, \frac{\pi}{2}\right) ?$ (A) cosx(B) cos 2x(C) cos 3x(D) tanx Solution: (A) Let $f_{1}(x)=\cos x$. $\therefore f_{1}^{\prime}(x)=-\sin x$ In interval $\left(0, \frac{\pi}{2}\right), f_{1}^{\prime}(x)=-\sin x0$ $\therefore f_{1}(x)=\cos x$ is strictly decreasing in interval $\left(0, \frac{\pi}{2}\right)$. (B) Let $f_{2}(x)=\cos 2 x$. $\therefore f_{2}^{\prime}(x)=-2 \sin 2 x$ Now, $0x\frac{\pi}{2} \Rightarrow 02 x\pi ...

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Find the derivative of the following functions

Question: Find the derivative of the following functions (it is to be understood thata,b,c,d,p, q,randsare fixed non-zero constants andmandnare integers):$(a x+b)(c x+d)^{2}$ Solution: Let $f(x)=(a x+b)(c x+d)^{2}$ By Leibnitz product rule, $f^{\prime}(x)=(a x+b) \frac{d}{d x}(c x+d)^{2}+(c x+d)^{2} \frac{d}{d x}(a x+b)$ $=(a x+b) \frac{d}{d x}\left(c^{2} x^{2}+2 c d x+d^{2}\right)+(c x+d)^{2} \frac{d}{d x}(a x+b)$ $=(a x+b)\left[\frac{d}{d x}\left(c^{2} x^{2}\right)+\frac{d}{d x}(2 c d x)+\frac...

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Find the derivative of the following functions

Question: Find the derivative of the following functions (it is to be understood thata,b,c,d,p, q,randsare fixed non-zero constants andmandnare integers):$(p x+q)\left(\frac{r}{x}+s\right)$ Solution: Let $f(x)=(p x+q)\left(\frac{r}{x}+s\right)$ By Leibnitz product rule, $f^{\prime}(x)=(p x+q)\left(\frac{r}{x}+s\right)^{\prime}+\left(\frac{r}{x}+s\right)(p x+q)^{\prime}$ $=(p x+q)\left(r x^{-1}+s\right)^{\prime}+\left(\frac{r}{x}+s\right)(p)$ $=(p x+q)\left(-r x^{-2}\right)+\left(\frac{r}{x}+s\ri...

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Find the derivative of the following functions

Question: Find the derivative of the following functions (it is to be understood thata,b,c,d,p, q,randsare fixed non-zero constants andmandnare integers): (x+a) Solution: Let $f(x)=x+a$. Accordingly, $f(x+h)=x+h+a$ By first principle, $f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$ $=\lim _{h \rightarrow 0} \frac{x+h+a-x-a}{h}$ $=\lim _{h \rightarrow 0}\left(\frac{h}{h}\right)$ $=\lim _{1 \rightarrow 0}(1)$ $=1$...

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Prove that the function

Question: Prove that the function $f$ given by $f(x)=x^{2}-x+1$ is neither strictly increasing nor strictly decreasing on $(-1,1)$. Solution: The given function is $f(x)=x^{2}-x+1$. $\therefore f^{\prime}(x)=2 x-1$ Now, $f^{\prime}(x)=0 \Rightarrow x=\frac{1}{2}$. The point $\frac{1}{2}$ divides the interval $(-1,1)$ into two disjoint intervals i.e., $\left(-1, \frac{1}{2}\right)$ and $\left(\frac{1}{2}, 1\right)$. Now, in interval $\left(-1, \frac{1}{2}\right), f^{\prime}(x)=2 x-10$. Therefore,...

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Find the derivative of the following functions from first principle:

Question: Find the derivative of the following functions from first principle: (i) $-x$ (ii) $(-x)^{-1}$ (iii) $\sin (x+1)$ (iv) $\cos \left(x-\frac{\pi}{8}\right)$ Solution: (i) Let $f(x)=-x .$ Accordingly, $f(x+h)=-(x+h)$ By first principle, $f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$ $=\lim _{h \rightarrow 0} \frac{-(x+h)-(-x)}{h}$ $=\lim _{h \rightarrow 0} \frac{-x-h+x}{h}$ $=\lim _{h \rightarrow 0} \frac{-h}{h}$ $=\lim _{h \rightarrow 0}(-1)=-1$ (ii) Let $f(x)=(-x)^{-1}=\f...

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If abc = 1, show that

Question: If $a b c=1$, show that $\frac{1}{1+a+b^{-1}}+\frac{1}{1+b+c^{-1}}+\frac{1}{1+c+a^{-1}}=1$ Solution: Consider the left hand side: $\frac{1}{1+a+b^{-1}}+\frac{1}{1+b+c^{-1}}+\frac{1}{1+c+a^{-1}}$ $=\frac{1}{1+a+\frac{1}{b}}+\frac{1}{1+b+\frac{1}{c}}+\frac{1}{1+c+\frac{1}{a}}$ $=\frac{1}{\frac{b+a b+1}{b}}+\frac{1}{1+b+a b}+\frac{1}{1+\frac{1}{a b}+\frac{1}{a}} \quad(a b c=1)$ $=\frac{b}{b+a b+1}+\frac{1}{1+b+a b}+\frac{a b}{a b+1+b}$ $=\frac{b+1+a b}{b+a b+1}$ $=1$ Hence proved....

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Prove that the logarithmic function is strictly increasing on (0, ∞).

Question: Prove that the logarithmic function is strictly increasing on $(0, \infty)$. Solution: The given function is $f(x)=\log x$. $\therefore f^{\prime}(x)=\frac{1}{x}$ It is clear that for $x0, f^{\prime}(x)=\frac{1}{x}0$. Hence, $f(x)=\log x$ is strictly increasing in interval $(0, \infty)$....

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Prove that is an increasing function of θ in.

Question: Prove that $y=\frac{4 \sin \theta}{(2+\cos \theta)}-\theta$ is an increasing function of $\theta$ in $\left[0, \frac{\pi}{2}\right]$. Solution: We have, $y=\frac{4 \sin \theta}{(2+\cos \theta)}-\theta$ $\therefore \frac{d y}{d x}=\frac{(2+\cos \theta)(4 \cos \theta)-4 \sin \theta(-\sin \theta)}{(2+\cos \theta)^{2}}-1$ $=\frac{8 \cos \theta+4 \cos ^{2} \theta+4 \sin ^{2} \theta}{(2+\cos \theta)^{2}}-1$ $=\frac{8 \cos \theta+4}{(2+\cos \theta)^{2}}-1$ Now, $\frac{d y}{d x}=0$ $\Rightarro...

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Prove that:

Question: Prove that: (i) $\frac{a+b+c}{a^{-1} b^{-1}+b^{-1} c^{-1}+c^{-1} a^{-1}}=a b c$ (ii) $\left(a^{-1}+b^{-1}\right)^{-1}=\frac{a b}{a+b}$ Solution: (i) Consider the left hand side: $\frac{a+b+c}{a^{-1} b^{-1}+b^{-1} c^{-1}+c^{-1} a^{-1}}$ $=\frac{a+b+c}{\frac{1}{a b}+\frac{1}{b c}+\frac{1}{c a}}$ $=\frac{a+b+c}{\frac{c+a+b}{a b c}}$ $=(a+b+c) \times\left(\frac{a b c}{a+b+c}\right)$ $=a b c$ Therefore left hand side is equal to the right hand side. Hence proved. (ii)Consider the left hand ...

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Prove that:

Question: Prove that: (i) $\frac{1}{1+x^{a-b}}+\frac{1}{1+x^{b-a}}=1$ (ii) $\frac{1}{1+x^{b-a}+x^{c-a}}+\frac{1}{1+x^{a-b}+x^{c-b}}+\frac{1}{1+x^{b-c}+x^{a-c}}=1$ Solution: (i) Consider the left hand side: $\frac{1}{1+x^{a-b}}+\frac{1}{1+x^{b-a}}$ $=\frac{1}{1+\frac{x^{a}}{x^{b}}}+\frac{1}{1+\frac{x^{b}}{x^{a}}}$ $=\frac{1}{\frac{x^{b}+x^{a}}{x^{b}}}+\frac{1}{\frac{x^{a}+x^{b}}{x^{a}}}$ $=\frac{x^{b}}{x^{b}+x^{a}}+\frac{x^{a}}{x^{a}+x^{b}}$ $=\frac{x^{b}+x^{a}}{x^{b}+x^{a}}$ $=1$ Therefore left ...

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Find the values of

Question: Find the values of $x$ for which $y=[x(x-2)]^{2}$ is an increasing function. Solution: We have, $y=[x(x-2)]^{2}=\left[x^{2}-2 x\right]^{2}$ $\therefore \frac{d y}{d x}=y^{\prime}=2\left(x^{2}-2 x\right)(2 x-2)=4 x(x-2)(x-1)$ $\therefore \frac{d y}{d x}=0 \Rightarrow x=0, x=2, x=1$ The points $x=0, x=1$, and $x=2$ divide the real line into four disjoint intervals i.e., $(-\infty, 0),(0,1)(1,2)$, and $(2, \infty)$. In intervals $(-\infty, 0)$ and $(1,2), \frac{d y}{d x}0$. $\therefore y$...

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Show that

Question: Show that $y=\log (1+x)-\frac{2 x}{2+x}, x-1$, is an increasing function of $x$ throughout its domain. Solution: We have, $y=\log (1+x)-\frac{2 x}{2+x}$ $\therefore \frac{d y}{d x}=\frac{1}{1+x}-\frac{(2+x)(2)-2 x(1)}{(2+x)^{2}}=\frac{1}{1+x}-\frac{4}{(2+x)^{2}}=\frac{x^{2}}{(1+x)(2+x)^{2}}$ Now, $\frac{d y}{d x}=0$ $\Rightarrow \frac{x^{2}}{(1+x)(2+x)^{2}}=0$ $\Rightarrow x^{2}=0 \quad[(2+x) \neq 0$ as $x-1]$ $\Rightarrow x=0$ Since $x-1$, point $x=0$ divides the domain $(-1, \infty)$...

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Prove that:

Question: Prove that: Solution: Left hand side is equal to right hand side.Hence proved. (ii) $\left(\frac{x^{a}}{x^{b}}\right)^{c} \times\left(\frac{x^{b}}{x^{c}}\right)^{a} \times\left(\frac{x^{c}}{x^{a}}\right)^{b}=1$ Consider the left hand side: $=\frac{x^{a c}}{x^{b c}} \times \frac{x^{b a}}{x^{c a}} \times \frac{x^{c b}}{x^{a b}}$ $=\frac{x^{a c}}{x^{b c}} \times \frac{x^{b a}}{x^{c a}} \times \frac{x^{c b}}{x^{a b}}$ $=\frac{x^{a c} \times x^{b a} \times x^{c b}}{x^{b c} \times x^{c a} \t...

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