1 + 2 + 22 + … 2n = 2n+1 – 1
Question: 1 + 2 + 22+ 2n= 2n+1 1 for all natural numbers n. Solution: According to the question, P(n) is 1 + 2 + 22+ 2n= 2n+1 1. So, substituting different values for n, we get, P(0) = 1 = 20+1 1 Which is true. P(1) = 1 + 2 = 3 = 21+1 1 Which is true. P(2) = 1 + 2 + 22= 7 = 22+1 1 Which is true. P(3) = 1 + 2 + 22+ 23= 15 = 23+1 1 Which is true. Let P(k) = 1 + 2 + 22+ 2k= 2k+1 1 be true; So, we get, ⇒P(k+1) is 1 + 2 + 22+ 2k+ 2k+1= 2k+1 1 + 2k+1 = 22k+1 1 = 2(k+1)+1 1 ⇒P(k+1) is true when P(k) is...
Read More →2 + 4 + 6 + …+ 2n = n2 + n for all natural
Question: 2 + 4 + 6 + + 2n = n2 + n for all natural numbers n. Solution: According to the question, P(n) is 2 + 4 + 6 + + 2n = n2+ n. So, substituting different values for n, we get, P(0) = 0 = 02+ 0 Which is true. P(1) = 2 = 12+ 1 Which is true. P(2) = 2 + 4 = 22+ 2 Which is true. P(3) = 2 + 4 + 6 = 32+ 2 Which is true. Let P(k) = 2 + 4 + 6 + + 2k = k2+ k be true; So, we get, ⇒P(k+1) is 2 + 4 + 6 + + 2k + 2(k+1) = k2+ k + 2k +2 = (k2+ 2k +1) + (k+1) = (k + 1)2+ (k + 1) ⇒P(k+1) is true when P(k)...
Read More →√n < 1/√1 + 1/√2 + … 1/√n,
Question: n 1/1 + 1/2 + 1/n, for all natural numbers n 2. Solution: According to the question, $P(n)$ is $\sqrt{n}\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\ldots \ldots \ldots . .+\frac{1}{\sqrt{n}} ; n \geq 2 P(n)$ is $\sqrt{n}\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\ldots \ldots \ldots . .+$ $\frac{1}{\sqrt{n}} ; n \geq 2$ $\mathrm{P}(2)$ is $\sqrt{2}\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}} \Rightarrow 1.4141.707$ It's true $\mathrm{P}(3)$ is $\sqrt{3}\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sq...
Read More →2n < (n + 2)! for all natural
Question: 2n (n + 2)! for all natural number n. Solution: According to the question, P(n) is 2n (n + 2)! So, substituting different values for n, we get, P(0)⇒0 2! P(1)⇒2 3! P(2)⇒4 4! P(3)⇒6 5! Let P(k) = 2k (k + 2)! is true; Now, we get that, ⇒P(k+1) = 2(k+1) ((k+1)+2))! We know that, [(k+1)+2)! = (k+3)! = (k+3)(k+2)(k+1)321] But, we also know that, = 2(k+1) (k+3)(k+2)31 2(k+1) Therefore,2(k+1) ((k+1) + 2)! ⇒P(k+1) is true when P(k) is true. Therefore, by Mathematical Induction, P(n) = 2n (n + ...
Read More →n2 < 2n for all natural numbers
Question: n2 2n for all natural numbers n 5. Solution: According to the question, P(n) is n2 2n for n5 Let P(k) = k2 2kbe true; ⇒P(k+1) = (k+1)2 = k2+ 2k + 1 2k+1= 2(2k) 2k2 Since,n2 2n + 1 for n 3 We get that, k2+ 2k + 1 2k2 ⇒(k+1)2 2(k+1) ⇒P(k+1) is true when P(k) is true. Therefore, by Mathematical Induction, P(n) = n2 2n is true for all natural numbers n 5....
Read More →n(n2 + 5) is divisible by 6,
Question: n(n2 + 5) is divisible by 6, for each natural number n. Solution: According to the question, P(n) = n(n2+ 5) is divisible by 6. So, substituting different values for n, we get, P(0) = 0(02+ 5) = 0 Which is divisible by 6. P(1) = 1(12+ 5) = 6 Which is divisible by 6. P(2) = 2(22+ 5) = 18 Which is divisible by 6. P(3) = 3(32+ 5) = 42 Which is divisible by 6. Let P(k) = k(k2+ 5) be divisible by 6. So, we get, ⇒k(k2+ 5) = 6x. Now, we also get that, ⇒ P(k+1) = (k+1)((k+1)2+ 5) = (k+1)(k2+2k...
Read More →n3 – n is divisible by 6,
Question: n3 n is divisible by 6, for each natural number n 2. Solution: According to the question, P(n) = n3 n is divisible by 6. So, substituting different values for n, we get, P(0) = 03 0 = 0 Which is divisible by 6. P(1) = 13 1 = 0 Which is divisible by 6. P(2) = 23 2 = 6 Which is divisible by 6. P(3) = 33 3 = 24 Which is divisible by 6. Let P(k) = k3 k be divisible by 6. So, we get, ⇒k3 k = 6x. Now, we also get that, ⇒ P(k+1) = (k+1)3 (k+1) = (k+1)(k2+2k+11) = k3+ 3k2+ 2k = 6x+3k(k+1) [n(n...
Read More →For any natural number n,
Question: For any natural number n, xn yn is divisible by x y, where x integers with x y. Solution: According to the question, P(n) = xn ynis divisible by x y, x integers with x y. So, substituting different values for n, we get, P(0) = x0 y0= 0 Which is divisible by x y. P(1) = x y Which is divisible by x y. P(2) = x2 y2 = (x +y)(xy) Which is divisible by xy. P(3) = x3 y3 = (xy)(x2+xy+y2) Which is divisible by xy. Let P(k) = xk ykbe divisible by x y; So, we get, ⇒xk yk= a(xy). Now, we also get ...
Read More →Find the equation of the perpendicular drawn from the point P
Question: Find the equation of the perpendicular drawn from the point P(-2, 3) to the line x 4y + 7 = 0. Also, find the coordinates of the foot of the perpendicular. Solution: Let the equation of line $A B$ be $x-4 y+7=0$ and point $C$ be $(-2,3)$ CD is perpendicular to the line AB, and we need to find: 1) Equation of Perpendicular drawn from point $\mathrm{C}$ 2) Coordinates of D Let the coordinates of point D be (a, b) Also, point $D(a, b)$ lies on the line $A B$, i.e. point $(a, b)$ satisfy t...
Read More →For any natural number n,
Question: For any natural number n, 7n 2n is divisible by 5. Solution: According to the question, P(n) = 7n 2nis divisible by 5. So, substituting different values for n, we get, P(0) = 70 20= 0 Which is divisible by 5. P(1) = 71 21= 5 Which is divisible by 5. P(2) = 72 22= 45 Which is divisible by 5. P(3) = 73 23= 335 Which is divisible by 5. Let P(k) = 7k 2kbe divisible by 5 So, we get, ⇒7k 2k= 5x. Now, we also get that, ⇒ P(k+1)= 7k+1 2k+1 = (5 + 2)7k 2(2k) = 5(7k) + 2 (7k 2k) = 5(7k) + 2 (5x)...
Read More →32n – 1 is divisible by 8,
Question: 32n 1 is divisible by 8, for all natural numbers n. Solution: According to the question, P(n) = 32n 1 is divisible by 8. So, substituting different values for n, we get, P(0) = 30 1 = 0 which is divisible by 8. P(1) = 32 1 = 8 which is divisible by 8. P(2) = 34 1 = 80 which is divisible by 8. P(3) = 36 1 = 728 which is divisible by 8. Let P(k) = 32k 1 be divisible by 8 So, we get, ⇒32k 1 = 8x. Now, we also get that, ⇒ P(k+1) = 32(k+1) 1 = 32(8x + 1) 1 = 72x + 8 is divisible by 8. ⇒P(k+...
Read More →n3 – 7n + 3 is divisible by 3,
Question: n3 7n + 3 is divisible by 3, for all natural numbers n. Solution: According to the question, P(n) = n3 7n + 3 is divisible by 3. So, substituting different values for n, we get, P(0) = 03 70 + 3 = 3 which is divisible by 3. P(1) = 13 71 + 3 = 3 which is divisible by 3. P(2) = 23 72 + 3 = 3 which is divisible by 3. P(3) = 33 73 + 3 = 9 which is divisible by 3. Let P(k) = k3 7k + 3 be divisible by 3 So, we get, ⇒k3 7k + 3 = 3x. Now, we also get that, ⇒ P(k+1) = (k+1)3 7(k+1) + 3 = k3+ 3k...
Read More →23n – 1 is divisible by7,
Question: 23n 1 is divisible by7, for all natural numbers n. Solution: According to the question, P(n) = 23n 1 is divisible by 7. So, substituting different values for n, we get, P(0) = 20 1 = 0 which is divisible by 7. P(1) = 23 1 = 7 which is divisible by 7. P(2) = 26 1 = 63 which is divisible by 7. P(3) = 29 1 = 512 which is divisible by 7. Let P(k) = 23k 1 be divisible by 7 So, we get, ⇒23k 1 = 7x. Now, we also get that, ⇒ P(k+1) = 23(k+1) 1 = 23(7x + 1) 1 = 56x + 7 = 7(8x + 1) is divisible ...
Read More →4n – 1 is divisible by 3, for each natural number n.
Question: 4n 1 is divisible by 3, for each natural number n. Solution: According to the question, P(n) = 4n 1 is divisible by 3. So, substituting different values for n, we get, P(0) = 40 1 = 0 which is divisible by 3. P(1) = 41 1 = 3 which is divisible by 3. P(2) = 42 1 = 15 which is divisible by 3. P(3) = 43 1 = 63 which is divisible by 3. Let P(k) = 4k 1 be divisible by 3, So, we get, ⇒4k 1 = 3x. Now, we also get that, ⇒ P(k+1) = 4k+1 1 = 4(3x + 1) 1 = 12x + 3 is divisible by 3. ⇒P(k+1) is tr...
Read More →33 Prove that the function f(x)= cos x is :
Question: 33 Prove that the function $f(x)=\cos x$ is : i. strictly decreasing on $(0, \pi)$ ii. strictly increasing in $(\pi, 2 \pi)$ iii. neither increasing nor decreasing in $(0,2 \pi)$ Solution: Given $f(x)=\cos x$ $\therefore f^{\prime}(x)=-\sin x$ (i) Since for each $x \in(0, \pi), \sin x0$ $\Rightarrow \therefore f^{\prime}(x)0$ So $f$ is strictly decreasing in $(0, \pi)$ (ii) Since for each $x \in(\pi, 2 \pi), \sin x0$ $\Rightarrow \therefore f^{\prime}(x)0$ So $f$ is strictly increasing...
Read More →Give an example of a statement P(n)
Question: Give an example of a statement P(n) which is true for all n. Justify your answer. Solution: According to the question, P(n) which is true for all n. Let P(n) be, $1+2+3+\cdots \ldots \ldots+n=\frac{n(n+1)}{2}$ $\mathrm{P}(0)$ is $0=\frac{0(0+1)}{2}=0 ;$ it's true $\mathrm{P}(1)$ is $1=\frac{1(1+1)}{2}=1 ;$ it's true $P(2)$ is $1+2=\frac{2(2+1)}{2} ;$ it's true $P(k)$ is $1+2+3+\cdots \ldots \ldots+k=\frac{k(k+1)}{2}$ $\mathrm{P}(\mathrm{k})$ is $1+2+3+\cdots \ldots \ldots+\mathrm{k}+1=...
Read More →Give an example of a statement P(n)
Question: Give an example of a statement P(n) which is true for all n 4 but P(1), P(2) and P(3) are not true. Justify your answer. Solution: According to the question, P(n) which is true for all n 4 but P(1), P(2) and P(3) are not true Let P(n) be 2n n! So, the examples of the given statements are, P(0)⇒20 0! i.e 1 1 ⇒ not true P(1)⇒21 1! i.e 2 1 ⇒ not true P(2)⇒22 2! i.e 4 2 ⇒ not true P(3)⇒23 3! i.e 8 6 ⇒ not true P(4)⇒24 4! i.e 16 24 ⇒ true P(5)⇒25 5! i.e 32 60 ⇒ true, etc....
Read More →Prove that the function f given by
Question: Prove that the function f given by $f(x)=x^{3}-3 x^{2}+4 x$ is strictly increasing on $R$ ? Solution: given $f(x)=x^{3}-3 x^{2}+4 x$ $\therefore f(x)=3 x^{2}-6 x+4$ $=3\left(x^{2}-2 x+1\right)+1$ $=3(x-1)^{2}+10$ for all $x \in R$ Hence $f(x)$ is strickly increasing on $R$...
Read More →Prove that the function f given by
Question: Prove that the function $f$ given by $f(x)=\log \cos x$ is strictly increasing on $(-\pi / 2,0)$ and strictly decreasing on $(0, \pi / 2)$ ? Solution: we have, $\mathrm{f}(\mathrm{x})=\log \cos \mathrm{x}$ $\therefore f^{\prime}(x)=\frac{1}{\cos x}(-\sin x)=-\tan x$ In Interval $\left(0, \frac{\pi}{2}\right), \tan x0 \Rightarrow-\tan x0$ $\therefore \mathrm{f}^{\prime}(\mathrm{x})0$ on $\left(0, \frac{\pi}{2}\right)$ $\therefore \mathrm{f}$ is strickly decreasing on $\left(0, \frac{\pi...
Read More →Find the equation of the perpendicular drawn from the origin to the line
Question: Find the equation of the perpendicular drawn from the origin to the line 4x 3y + 5 = 0. Also, find the coordinates of the foot of the perpendicular. Solution: Let the equation of line $A B$ be $4 x-3 y+5=0$ and point C be (0, 0) CD is perpendicular to the line AB, and we need to find: 1) Equation of Perpendicular drawn from point $C$ 2) Coordinates of $D$ Let the coordinates of point D be (a, b) Also, point D(a, b) lies on the line AB, i.e. point (a, b) satisfy the equation of line AB ...
Read More →The value of tan 75°– cot 75°
Question: The value of tan 75 cot 75 is equal to A. 23 B.2 + 3 C. 2 3 D. 1 Solution: A. 23 Explanation: According to the question, We have, tan 75 cot 75 $=\frac{\sin 75^{\circ}}{\cos 75^{\circ}}-\frac{\cos 75^{\circ}}{\sin 75^{\circ}}$ $=\frac{\sin ^{2} 75^{\circ}-\cos ^{2} 75^{\circ}}{\cos 75^{\circ} \sin 75^{\circ}}$ $=\frac{2\left(\sin ^{2} 75^{\circ}-\cos ^{2} 75^{\circ}\right)}{2 \cos 75^{\circ} \sin 75^{\circ}}$ $=\frac{-2 \cos 150^{\circ}}{\sin 150^{\circ}}$ = -2cot150 = -2 cot (180-30) ...
Read More →Prove that the following function is increasing on r ?
Question: Prove that the following function is increasing on $r$ ? i. $f(x)=3 x^{5}+40 x^{3}+240 x$ ii. $f(x)=4 x^{3}-18 x^{2}+27 x-27$ Solution: (i) we have $f(x)=3 x^{5}+40 x^{3}+240 x$ $\therefore \mathrm{f}^{\prime}(\mathrm{x})=15 \mathrm{x}^{4}+120 \mathrm{x}^{2}+240$ $=15\left(x^{4}+8 x^{2}+16\right)$ $=15\left(x^{2}+4\right)^{2}$ Now, $x \in R$ $\Rightarrow\left(\mathrm{x}^{2}+4\right)^{2}0$ $\Rightarrow 15\left(\mathrm{x}^{2}+4\right)^{2}0$ $\Rightarrow \mathrm{f}^{\prime}(\mathrm{x})0$ ...
Read More →Prove that the function
Question: Prove that the function $f$ given by $f(x)=x-[x]$ is increasing in $(0,1) ?$ Solution: we have, $f(x)=x-[x]$ $\therefore f^{\prime}(x)=10$ $\therefore f(x)$ is an increasing function on $(0,1)$...
Read More →If tan θ = 3 and θ lies in third quadrant,
Question: If tan = 3 and lies in third quadrant, then the value of sin is A. 1/10 B. 1/10 C. 3/10 D. 3/10 Solution: C. 3/10 Explanation: According to the question, Given that, tan = 3 and lies in third quadrant ⇒ cot = 1/3 We know that, Cosec2 = 1+cot2 $=1+\left(\frac{1}{3}\right)^{2}=1+\frac{1}{9}=\frac{10}{9}$ $\Rightarrow \sin ^{2} \theta=\frac{9}{10}$ $\Rightarrow \sin \theta=\pm \frac{3}{\sqrt{10}}$ $\Rightarrow \sin \theta=-\frac{3}{\sqrt{10}}$, since $\theta$ lies in third quadrant. Thus,...
Read More →Show that the function
Question: Show that the function $f$ given by $f(x)=10^{x}$ is increasing for all $x$ ? Solution: we have, $f(x)=10^{x}$ $\therefore f^{\prime}(x)=10^{x} \log 10$ Now, $X \in R$ $\Rightarrow 10^{x}0$ $\Rightarrow 10^{x} \log 100$ $\Rightarrow f^{\prime}(x)0$ Hence, $f(x)$ in an increasing function for all $x$...
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