Which of the following statements are true (T) and which are false (F)?

Question: Which of the following statements are true (T) and which are false (F)? (i) Sum of the three sides of a triangle is less than the sum of its three altitudes. (ii) Sum of any two sides of a triangle is greater than twice the median drown to the third side (iii) Sum of any two sides of a triangle is greater than the third side. (iv) Difference of any two sides of a triangle is equal to the third side. (v) If two angles of a triangle are unequal, then the greater angle has the larger side...

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In Fig., prove that:

Question: In Fig., prove that: (i) CD + DA + AB + BC 2AC (ii) CD + DA + AB BC Solution: To prove (i) CD + DA + AB + BC 2AC (ii) CD + DA+ AB BC From the given figure, We know that, in a triangle sum of any two sides is greater than the third side (i) So, InΔABC, we have AB + BC AC .... (i) InΔADC, we have CD + DA AC .... (ii) Adding (i) and (ii), we get AB + BC + CD + DA AC + AC AB + BC + CD + DA 2AC (ii) Now, InΔABC, we have, AB + AC BC ... (iii) And inΔADC, we have CD + DA AC Add AB on both sid...

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If f(x)=

Question: If $f(x)=\left\{\begin{array}{cc}x^{2}, \text { when } x0 \\ x, \text { when } 0 \leq x1 \\ \frac{1}{x}, \text { when } x \geq 1\end{array}\right.$ find: (a) $f(1 / 2)$, (b) $f(-2)$, (c) $f(1)$, (d) $f(\sqrt{3})$ and $($ e) $f(\sqrt{-3})$. Solution: Given: $f(x)= \begin{cases}x^{2}, \text { when } x0 \\ x, \text { when } 0 \leq x1 \\ \frac{1}{x}, \text { when } x \geq 1\end{cases}$ Now, (a) $f\left(\frac{1}{2}\right)=\frac{1}{2}$[ Usingf(x) =x, 0 x 1] (b) $f(-2)=(-2)^{2}=4$ (c) $f(1)=\...

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If f(x)=

Question: If $f(x)=\left\{\begin{array}{cc}x^{2}, \text { when } x0 \\ x, \text { when } 0 \leq x1 \\ \frac{1}{x}, \text { when } x \geq 1\end{array}\right.$ find: (a) $f(1 / 2)$, (b) $f(-2)$, (c) $f(1)$, (d) $f(\sqrt{3})$ and $($ e) $f(\sqrt{-3})$. Solution: Given: $f(x)= \begin{cases}x^{2}, \text { when } x0 \\ x, \text { when } 0 \leq x1 \\ \frac{1}{x}, \text { when } x \geq 1\end{cases}$ Now, (a) $f\left(\frac{1}{2}\right)=\frac{1}{2}$[ Usingf(x) =x, 0 x 1] (b) $f(-2)=(-2)^{2}=4$ (c) $f(1)=\...

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Prove that the perimeter of a triangle is greater than the sum of its altitudes.

Question: Prove that the perimeter of a triangle is greater than the sum of its altitudes. Solution: Given,ΔABC in which ADBC, BEAC and CFAB. To prove, AD + BE + CF AB + BC + AC Figure: Proof: We know that of all the segments that can be drawn to a given line, from a point not lying on it, the perpendicular distance i.e, the perpendicular line segment is the shortest. Therefore ADBC AB AD and AC AD AB + AC 2AD .... (i) BEAC BA BE and BC BE BA + BC 2BE ... (ii) CFAB CA CF and CB CF CA + CB 2CF .....

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Show that

Question: $x(\log x)^{2}$ Solution: $I=\int x(\log x)^{2} d x$ Taking $(\log x)^{2}$ as first function and $x$ as second function and integrating by parts, we obtain $I=(\log x)^{2} \int x d x-\int\left[\left\{\frac{d}{d x}(\log x)^{2}\right\} \int x d x\right] d x$ $=\frac{x^{2}}{2}(\log x)^{2}-\left[\int 2 \log x \cdot \frac{1}{x} \cdot \frac{x^{2}}{2} d x\right]$ $=\frac{x^{2}}{2}(\log x)^{2}-\int x \log x d x$ Again integrating by parts, we obtain $I=\frac{x^{2}}{2}(\log x)^{2}-\left[\log x ...

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Find the value of k for which each of the following system of equations have infinitely many solutions :

Question: Find the value ofkfor which each of the following system of equations have infinitely many solutions : $x+(k+1) y=4$ $(k+1) x+9 y=5 k+2$ Solution: GIVEN: $x+(k+1) y=4$ $(k+1) x+9 y=5 k+2$ To find: To determine for what value ofkthe system of equation has infinitely many solutions We know that the system of equations $a_{1} x+b_{1} y=c_{1}$ $a_{2} x+b_{2} y=c_{2}$ For infinitely many solution $\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$ Here, $\frac{1}{k+1}=\frac{(k+1)}...

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O is any point in the interior of ΔABC. Prove that

Question: O is any point in the interior ofΔABC. Prove that (i) AB + AC OB + OC (ii) AB + BC + CA OA + QB + OC (iii) OA + OB + OC (1/2)(AB + BC +CA) Solution: Given that O is any point in the interior ofΔABC To prove (i) AB + AC OB + OC (ii) AB + BC + CA OA + QB + OC (iii) OA + OB + OC (1/2)(AB + BC +CA) We know that in a triangle the sum of any two sides is greater than the third side. So, we have InΔABC AB + BC AC BC + AC AB AC + AB BC InΔOBC OB + OC BC ... (i) InΔOAC OA + OC AC ... (ii) InΔOA...

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Show that

Question: $\tan ^{-1} x$ Solution: Let $I=\int 1 \cdot \tan ^{-1} x d x$ Taking $\tan ^{-1} x$ as first function and 1 as second function and integrating by parts, we obtain $I=\tan ^{-1} x \int 1 d x-\int\left\{\left(\frac{d}{d x} \tan ^{-1} x\right) \int 1 \cdot d x\right\} d x$ $=\tan ^{-1} x \cdot x-\int \frac{1}{1+x^{2}} \cdot x d x$ $=x \tan ^{-1} x-\frac{1}{2} \int \frac{2 x}{1+x^{2}} d x$ $=x \tan ^{-1} x-\frac{1}{2} \log \left|1+x^{2}\right|+\mathrm{C}$ $=x \tan ^{-1} x-\frac{1}{2} \log...

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If f(x)

Question: If $f(x)=\frac{x+1}{x-1}$, show that $f[f[(x)]]=x$ Solution: Given: $f(x)=\frac{x+1}{x-1}$ Therefore, $f[f\{(x)\}]=f\left(\frac{x+1}{x-1}\right)$ $=\frac{\left(\frac{x+1}{x-1}\right)+1}{\left(\frac{x+1}{x-1}\right)-1}$ $=\frac{\frac{x+1+x-1}{x-1}}{\frac{x+1-x+1}{x-1}}=\frac{\frac{2 x}{x-1}}{\frac{2}{x-1}}=\frac{2 x}{2}=x$ Thus, f[f{(x)}] =x Hence proved....

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Is it possible to draw a triangle with sides of length 2 cm, 3 cm and 7 cm?

Question: Is it possible to draw a triangle with sides of length 2 cm, 3 cm and 7 cm? Solution: Given lengths of sides are 2cm, 3cm and 7cm. To check whether it is possible to draw a triangle with the given lengths of sides We know that, A triangle can be drawn only when the sum of any two sides is greater than the third side. So, let's check the rule. 2 + 3≯7 or 2 + 3 7 2 + 7 3 and 3 + 7 2 Here 2 + 3 7 So, the triangle does not exit....

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Show that

Question: $x \sec ^{2} x$ Solution: Let $I=\int x \sec ^{2} x d x$ Taking $x$ as first function and $\sec ^{2} x$ as second function and integrating by parts, we obtain $\begin{aligned} I =x \int \sec ^{2} x d x-\int\left\{\left\{\frac{d}{d x} x\right\} \int \sec ^{2} x d x\right\} d x \\ =x \tan x-\int 1 \cdot \tan x d x \\ =x \tan x+\log |\cos x|+\mathrm{C} \end{aligned}$...

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In ΔABC, side AB is produced to D so that BD = BC. If ∠B = 60°

Question: In ΔABC, side AB is produced to D so that BD = BC. If B = 60 and A = 70. Prove that: (i) AD CD (ii) AD AC Solution: Given that, inΔABC, side AB is produced to D so that BD = BC. B = 60, andA = 70 To prove, (i) AD CD (ii) AD AC First join C and D We know that, Sum of angles in a triangle =180 A +B +C = 180 70 + 60 +C = 180 C = 180 - (130) = 50 C = 50 ACB = 50 ... (i) And also inΔBDC DBC =180-ABC [ABD is a straight angle] 180 - 60 = 120 and also BD = BC[given] BCD =BDC [Angles opposite t...

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Find the value of k for which each of the following system of equations have infinitely many solutions :

Question: Find the value ofkfor which each of the following system of equations have infinitely many solutions : $2 x+3 y=2$ $(k+2) x+(2 k+1) y=2(k-1)$ Solution: GIVEN: $2 x+3 y=2$ $(k+2) x+(2 x+1) y=2(k-1)$ To find: To determine for what value ofkthe system of equation has infinitely many solutions We know that the system of equations $a_{1} x+b_{1} y=c_{1}$ $a_{2} x+b_{2} y=c_{2}$ For infinitely many solution $\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$ Here, $\frac{2}{k+2}=\f...

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If f(x)

Question: If $f(x)=\frac{1}{1-x}$, show that $\left.f[f f(x)]\right]=x$ Solution: Given: $f(x)=\frac{1}{1-x}$ Thus, $f\{f(x)\}=f\left\{\frac{1}{1-x}\right\}$ $=\frac{1}{1-\frac{1}{1-x}}$ $=\frac{1}{\frac{1-x-1}{1-x}}$ $=\frac{1-x}{-x}$ $=\frac{x-1}{x}$ Again, $f[f\{f(x)\}]=f\left[\frac{x-1}{x}\right]$ $=\frac{1}{1-\left(\frac{x-1}{x}\right)}$ $=\frac{1}{\frac{x-x+1}{x}}$ $=\frac{x}{1}$ = x Therefore,f [ f {f (x)}] =x. Hence proved....

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Show that

Question: $\frac{x \cos ^{-1} x}{\sqrt{1-x^{2}}}$ Solution: Let $I=\int \frac{x \cos ^{-1} x}{\sqrt{1-x^{2}}} d x$ $I=\frac{-1}{2} \int \frac{-2 x}{\sqrt{1-x^{2}}} \cdot \cos ^{-1} x d x$ Taking $\cos ^{-1} x$ as first function and $\left(\frac{-2 x}{\sqrt{1-x^{2}}}\right)$ as second function and integrating by parts, we obtain $\begin{aligned} I =\frac{-1}{2}\left[\cos ^{-1} x \int \frac{-2 x}{\sqrt{1-x^{2}}} d x-\int\left\{\left(\frac{d}{d x} \cos ^{-1} x\right) \int \frac{-2 x}{\sqrt{1-x^{2}}...

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In a ΔABC, if ∠B = ∠C = 45°, which is the longest side?

Question: In a ΔABC, if B = C = 45, which is the longest side? Solution: Given that inΔABC, B = C = 45 We have to find longest side We know that. Sum of angles in a triangle =180 A +B +C = 180 A + 45 + 45 = 180 A = 180 - (45 + 45) = 180 - 90 = 90 A = 90...

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In ΔABC, if ∠A = 40° and ∠B = 60°.

Question: In ΔABC, if A = 40 and B = 60. Determine the longest and shortest sides of the triangle. Solution: Given that inΔABC,A = 40 andB = 60 We have to find longest and shortest side We know that, Sum of angles in a triangle 180 A +B +C = 180 40 + 60 +C = 180 C = 180 - ((10)0) = 80 C = 80 Now, ⟹ 40 60 80 =A B C ⟹C is greater angle andA is smaller angle. Now,A B C ⟹ BC AAC AB [Side opposite to greater angle is larger and side opposite to smaller angle is smaller] AB is longest and BC is smalle...

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If y=f(x)

Question: If $y=f(x)=\frac{a x-b}{b x-a}$, show that $x=f(y)$. Solution: Given: $f(x)=\frac{a x-b}{b x-a}$ Lety=f(x) . ⇒y(bx-a) =axb ⇒xybay=axb ⇒xybax=ayb ⇒x(bya) =ayb $\Rightarrow x=\frac{a y-b}{b y-a}$ ⇒x=f(y) Hence proved....

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Show that

Question: $\left(\sin ^{-1} x\right)^{2}$ Solution: Let $I=\int\left(\sin ^{-1} x\right)^{2} \cdot 1 d x$ Taking $\left(\sin ^{-1} x\right)^{2}$ as first function and 1 as second function and integrating by parts, we obtain $I=\left(\sin ^{-1} x\right) \int 1 d x-\int\left\{\frac{d}{d x}\left(\sin ^{-1} x\right)^{2} \cdot \int 1 \cdot d x\right\} d x$ $=\left(\sin ^{-1} x\right)^{2} \cdot x-\int \frac{2 \sin ^{-1} x}{\sqrt{1-x^{2}}} \cdot x d x$ $=x\left(\sin ^{-1} x\right)^{2}+\int \sin ^{-1} x...

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If f(x)

Question: Iff(x) = (xa)2(xb)2, findf(a+b). Solution: Given: f(x) = (xa)2(xb)2 Thus, $f(a+b)=(a+b-a)^{2}(a+b-b)^{2}$ $=b^{2} a^{2}$ Hence, $f(a+b)=a^{2} b^{2}$....

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Fill the blanks In the following so that each of the following statements is true.

Question: Fill the blanks In the following so that each of the following statements is true. (i) Sides opposite to equal angles of a triangle are ___ (ii) Angle opposite to equal sides of a triangle are ___ (iii) In an equilateral triangle all angles are ___ (iv) InΔABC, ifA =C, then AB = (v) If altitudes CE and BF of a triangle ABC are equal, then AB ___ (vi) In an isosceles triangle ABC with AB = AC, if BD and CE are its altitudes, then BD is ___ CE. (vii) In right triangles ABC and DEF, if hy...

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Which of the following statements are true (T) and which are false (F):

Question: Which of the following statements are true (T) and which are false (F): (i) Sides opposite to equal angles of a triangle may be unequal. (ii) Angles opposite to equal sides of a triangle are equal (iii) The measure of each angle of an equilateral triangle is 60 (iv) If the altitude from one vertex of a triangle bisects the opposite side, then the triangle may be isosceles. (v) The bisectors of two equal angles of a triangle are equal. (vi) If the bisector of the vertical angle of a tri...

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If f(x)

Question: Iff(x) =x2 3x+ 4, then find the values ofxsatisfying the equationf(x) =f(2x+ 1). Solution: Given: f(x) =x2 3x+ 4 Therefore, f(2x+ 1) = (2x+ 1)2 3(2x+ 1) + 4 = 4x2+ 1 + 4x 6x 3 + 4 = 4x2 2x + 2 Now,f(x) =f(2x+ 1) ⇒x2 3x+ 4 = 4x2 2x+ 2 ⇒ 4x2x2 2x+ 3x+ 2 4 = 0 ⇒ 3x2+x 2 = 0 ⇒ 3x2+ 3x 2x 2 = 0 ⇒ 3x(x+ 1) 2(x+1) = 0 ⇒ (3x 2)(x+1) = 0 ⇒ (x+ 1) = 0 or ( 3x 2) = 0 $\Rightarrow x=-1$ or $x=\frac{2}{3}$ Hence, $x=-1, \frac{2}{3}$....

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Find the value of k for which each of the following system of equations have infinitely many solutions :

Question: Find the value ofkfor which each of the following system of equations have infinitely many solutions : $2 x-3 y=7$ $(k+2) x-(2 k+1) y=3(2 k-1)$ Solution: GIVEN: $2 x-3 y=7$ $(k+2) x-(2 k+1) y=3(2 k-1)$ To find: To determine for what value ofkthe system of equation has infinitely many solutions We know that the system of equations $a_{1} x+b_{1} y=c_{1}$ $a_{2} x+b_{2} y=c_{2}$ For infinitely many solution $\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$ Here $\frac{2}{k+2}...

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