eSaral - JEE | NEET | Class 8-10 Exam Prep App

Let f, g : R → R be defined by f(x)

Question: Let $f, g: \mathbf{R} \rightarrow \mathbf{R}$ be defined by $f(x)=2 x+\mid$ and $g(x)=x^{2}-2$ for all $x \in \mathbf{R}$, respectively. Then, find gof. [NCERT EXEMPLAR] Solution: We have, $f, g: \mathbf{R} \rightarrow \mathbf{R}$ are defined by $f(x)=2 x+\mid$ and $g(x)=x^{2}-2$ for all $x \in \mathbf{R}$, respectively Now, $g o f(x)=g(f(x))$ $=g(2 x+1)$ $=(2 x+1)^{2}-2$ $=4 x^{2}+4 x+1-2$ $=4 x^{2}+4 x-1$...

Find the 4th term from the beginning and 4th term from the end in the expansion of

Question: Find the 4 th term from the beginning and 4 th term from the end in the expansion of $\left(x+\frac{2}{x}\right)^{9}$. Solution: LetTr+1be the 4th term from the end. Then,Tr+1is (10 4 + 1)th, i.e., 7th, term from the beginning. $\therefore T_{7}=T_{6+1}$ $={ }^{9} C_{6}\left(x^{9-6}\right)\left(\frac{2}{x}\right)^{6}$ $=\frac{9 \times 8 \times 7}{3 \times 2}\left(x^{3}\right)\left(\frac{64}{x^{6}}\right)$ $=\frac{5376}{x^{3}}$ 4th term from the beginning $=T_{4}=T_{3+1}$ $\therefore T_...

In the given figure, ∠OAB = 110° and ∠BCD = 130° then ∠ABC is equal to

Question: In the given figure, OAB = 110 and BCD = 130 then ABC is equal to (a) 40(b) 50(c) 60(d) 70 Solution: In the given figure, OA || CD.Construction: Extend OA such that it intersects BC at E. Now, OE || CD and BC is a transversal.AEC =BCD = 130 (Pair of corresponding angles)Also,OAB +BAE =180 (Linear pair)110 +BAE =180⇒BAE =180 110 =70In∆ABE,AEC =BAE +ABE (In a triangle, exterior angle is equal to the sum of two opposite interior angles)130 =70 +x⇒x=13070=60Thus, the measure of angleABC is...

Find the 7th term in the expansion

Question: Find the 7 th term in the expansion of $\left(\frac{4 x}{5}+\frac{5}{2 x}\right)^{8}$. Solution: We need to find the 7th term in the given expression. $\because T_{7}=T_{6+1}$ $\therefore T_{7}=T_{6+1}$ $={ }^{8} C_{6}\left(\frac{4 x}{5}\right)^{8-6}\left(\frac{5}{2 x}\right)^{6}$ $=\frac{8 \times 7 \times 4 \times 4 \times 125 \times 125}{2 \times 1 \times 25 \times 64} x^{2}\left(\frac{1}{x^{6}}\right)$ $=\frac{4375}{x^{4}}$...

The angles of a triangle are in the ratio 2 : 3 : 4. The largest angle of the triangle is

Question: The angles of a triangle are in the ratio 2 : 3 : 4. The largest angle of the triangle is(a) 12(b) 100(c) 80(d) 60 Solution: Suppose∆ABC be such thatA :B :C =2 : 3 : 4.LetA = 2k,B= 3kandC= 4k, wherekis some constant.In ∆ABC,A +B +C = 180 (Angle sum property)⇒ 2k+ 3k+ 4k=180⇒9k=180⇒k=20 Measure of the largest angle = 4k= 4 20 = 80Hence, the correct answer is option (c)....

Find the 7th term in the expansion

Question: Find the 7 th term in the expansion of $\left(\frac{4 x}{5}+\frac{5}{2 x}\right)^{8}$. Solution: We need to find the 7th term in the given expression. $\because T_{7}=T_{6+1}$ $\therefore T_{7}=T_{6+1}$ $={ }^{8} C_{6}\left(\frac{4 x}{5}\right)^{8-6}\left(\frac{5}{2 x}\right)^{6}$ $=\frac{8 \times 7 \times 4 \times 4 \times 125 \times 125}{2 \times 1 \times 25 \times 64} x^{2}\left(\frac{1}{x^{6}}\right)$ $=\frac{4375}{x^{4}}$...

In the given figure, AOB is a straight line. The value of x is

Question: In the given figure, AOB is a straight line. The value ofxis (a) 12(b) 15(c) 20(d) 25 Solution: It is given that,AOB is a straight line. 60 + (5x + 3x) = 180 (Linear pair)⇒8x = 18060 =120⇒x =15Thus, the value ofxis 15.Hence, the correct answer is option (b)....

Let A = {a, b, c, d} and f : A → A be given by

Question: Let $A=\{a, b, c, d\}$ and $f: A \rightarrow A$ be given by $f=\{(a, b),(b, d),(c, a),(d, c)\}$. Write $f^{-1}$. [NCERT EXEMPLAR] Solution: We have, $A=\{a, b, c, d\}$ and $f: A \rightarrow A$ be given by $f=\{(a, b),(b, d),(c, a),(d, c)\}$ Since, the elements of a function when interchanged gives inverse function. So, $f^{-1}=\{(b, a),(d, b),(a, c),(c, d)\}$...

Find the sum of all integers between 50 and 500,

Question: Find the sum of all integers between 50 and 500, which are divisible by 7. Solution: In this problem, we need to find the sum of all the multiples of 7 lying between 50 and 500. So, we know that the first multiple of 7 after 50 is 56 and the last multiple of 7 before 500 is 497. Also, all these terms will form an A.P. with the common difference of 7. So here, First term (a) = 56 Last term (l) = 497 Common difference (d) = 7 So, here the first step is to find the total number of terms. ...

Find the 8th term in the expansion

Question: Find the 8th term in the expansion of $\left(x^{3 / 2} y^{1 / 2}-x^{1 / 2} y^{3 / 2}\right)^{10} .$ Solution: We need to find the 8th term in the given expression. $\because T_{8}=T_{7+1}$ $\therefore T_{8}={ }^{10} C_{7}\left(x^{3 / 2} y^{1 / 2}\right)^{10-7}\left(-x^{1 / 2} y^{3 / 2}\right)^{7}$ $=-\frac{10 \times 9 \times 8}{3 \times 2} x^{9 / 2} y^{3 / 2}\left(x^{7 / 2} y^{21 / 2}\right.$) $=-120 x^{8} y^{12}$...

If one of the angles of a triangle is 130° then the angle between the bisectors of the other two angles can be

Question: If one of the angles of a triangle is 130 then the angle between the bisectors of the other two angles can be(a) 50(b) 65(c) 90(d) 155 Solution: Let ∆ABC be such that A = 130. Here, BP is the bisector of B and CP is the bisector of C. $\therefore \angle \mathrm{ABP}=\angle \mathrm{PBC}=\frac{1}{2} \angle \mathrm{B}$ .......(1) Also, $\angle \mathrm{ACP}=\angle \mathrm{PCB}=\frac{1}{2} \angle \mathrm{C}$ ............(2) In∆ABC,A + B +C=180 (Angle sum property)⇒130+B +C=180⇒ B + C = 180 ...

Write the domain of the real function f defined by

Question: Write the domain of the real function $f$ defined by $f(x)=\sqrt{25-x^{2}}$. Solution: We have, $f(x)=\sqrt{25-x^{2}}$ The function is defined only when $25-x^{2} \geq 0$ $\Rightarrow x^{2}-25 \leq 0$ $\Rightarrow(x+5)(x-5) \leq 0$ $\Rightarrow x \in[-5,5]$ So, the domain of the given function is $[-5,5]$....

Find the 5th term from the end in the expansion

Question: Find the 5 th term from the end in the expansion of $\left(3 x-\frac{1}{x^{2}}\right)^{10}$ Solution: Given: $\left(3 x-\frac{1}{x^{2}}\right)^{10}$ Clearly, the expression has 6 terms. The 5th term from the end is the (11 5 + 1)th, i.e., 7th, term from the beginning. Thus, we have: $T_{7}=T_{6+1}$ $={ }^{10} C_{6}(3 x)^{10-6}\left(\frac{-1}{x^{2}}\right)^{6}$ $={ }^{10} C_{6}\left(3^{4}\right)\left(x^{4}\right)\left(\frac{1}{x^{12}}\right)$ $=\frac{10 \times 9 \times 8 \times 7 \times...

Which one the following relations on A = {1, 2, 3} is a function?

Question: Which one the following relations onA= {1, 2, 3} is a function? f= {(1, 3), (2, 3), (3, 2)},g= {(1, 2), (1, 3), (3, 1)} Solution: As, each element of the domain set has unique image in the relationf= {(1, 3), (2, 3), (3, 2)}So,fis a function.Also, the element 1 of the domain set has two images 2 and 3 of the range set in the relationg= {(1, 2), (1, 3), (3, 1)}So,gis not a function....

Let $f(x)=\frac{\alpha x}{x+1}, x \neq-1$. Then, for what value of $\alpha$ is $f(f(x))=x ?$

[question] Question. Let $f(x)=\frac{\alpha x}{x+1}, x \neq-1$. Then, for what value of $\alpha$ is $f(f(x))=x ?$ (a) $\sqrt{2}$ (b) $-\sqrt{2}$ (c) 1 (d) $-1$ [/question] [solution] Solution: (d) $-1$ $f(f(x))=x$ $\Rightarrow f\left(\frac{\alpha x}{x+1}\right)=x$ $\Rightarrow \frac{\alpha\left(\frac{a x}{x+1}\right)}{\left(\frac{a x}{x+1}\right)+1}=x$ $\Rightarrow \frac{\alpha^{2} x}{\alpha x+x+1}=x$ $\Rightarrow \alpha^{2} x=\alpha x^{2}+x^{2}+x$ $\Rightarrow \alpha^{2} x-\alpha x^{2}-x^{2}-x=...

Find the 7th term in the expansion of

Question: Find the 7 th term in the expansion of $\left(3 x^{2}-\frac{1}{x^{3}}\right)^{10}$. Solution: We need to find the 7th term of the given expression. Let it beT7 Now, we have $T_{7}=T_{6+1}$ $={ }^{10} C_{6}\left(3 x^{2}\right)^{10-6}\left(\frac{-1}{x^{3}}\right)^{6}$ $={ }^{10} C_{6}\left(3^{4}\right)\left(x^{8}\right)\left(\frac{1}{x^{18}}\right)$ $=\frac{10 \times 9 \times 8 \times 7 \times 81}{4 \times 3 \times 2 \times x^{10}}=\frac{17010}{x^{10}}$ Thus, the 7 th term of the given e...

The angles of a triangle are in the ration 3 : 5 : 7. The triangle is

Question: The angles of a triangle are in the ration 3 : 5 : 7. The triangle is(a) acute-angled(b) obtuse-angled(c) right-angled(d) an isosceles triangle Solution: (a) acute-angled Let the angles measure $(3 x)^{\circ},(5 x)^{\circ}$ and $(7 x)^{\circ}$. Then, $3 x+5 x+7 x=180^{\circ}$ $\Rightarrow 15 x=180^{\circ}$ $\Rightarrow x=12^{\circ}$ Therefore, the angles are $3(12)^{\circ}=\mathbf{3 6}^{\circ}, 5(12)^{\circ}=\mathbf{6 0}^{\circ}$ and $7(12)^{\circ}=\mathbf{8 4}^{\circ}$. Hence, the tri...

Find the 11th term from the beginning and the 11th term from the end in the expansion of

Question: Find the 11 th term from the beginning and the 11 th term from the end in the expansion of $\left(2 x-\frac{1}{x^{2}}\right)^{25}$. Solution: Given: $\left(2 x-\frac{1}{x^{2}}\right)^{25}$ Clearly, the given expression contains 26 terms. So, the 11thterm from the end is the (26 11 + 1)thterm from the beginning. In other words, the 11thterm from the end is the 16thterm from the beginning. Thus, we have: $T_{16}=T_{15+1}={ }^{25} C_{15}(2 x)^{25-15}\left(\frac{-1}{x^{2}}\right)^{15}$ $={...

Find the 11th term from the beginning and the 11th term from the end in the expansion of

Question: Find the 11 th term from the beginning and the 11 th term from the end in the expansion of $\left(2 x-\frac{1}{x^{2}}\right)^{25}$. Solution: Given: $\left(2 x-\frac{1}{x^{2}}\right)^{25}$ Clearly, the given expression contains 26 terms. So, the 11thterm from the end is the (26 11 + 1)thterm from the beginning. In other words, the 11thterm from the end is the 16thterm from the beginning. Thus, we have: $T_{16}=T_{15+1}={ }^{25} C_{15}(2 x)^{25-15}\left(\frac{-1}{x^{2}}\right)^{15}$ $={...

Find the sum of all integers between 84 and 719, which are multiples of 5.

Question: Find the sum of all integers between 84 and 719, which are multiples of 5. Solution: In this problem, we need to find the sum of all the multiples of 5 lying between 84 and 719. So, we know that the first multiple of 5 after 84 is 85 and the last multiple of 5 before 719 is 715. Also, all these terms will form an A.P. with the common difference of 5. So here, First term (a) = 85 Last term (l) = 715 Common difference (d) = 5 So, here the first step is to find the total number of terms. ...

Let f : R → R be the function defined by

Question: Let $f: \mathbf{R} \rightarrow \mathbf{R}$ be the function defined by $f(x)=4 x-3$ for all $x \in \mathbf{R}$. Then write $f^{-1}$. Solution: We have, $f: \mathbf{R} \rightarrow \mathbf{R}$ is the function defined by $f(x)=4 x-3$ for all $x \in \mathbf{R}$ Let $f(x)=y$. Then, $y=4 x-3$ $\Rightarrow 4 x=y+3$ $\Rightarrow x=\frac{y+3}{4}$ So, $f^{-1}(y)=\frac{y+3}{4}$ or, $f^{-1}(x)=\frac{x+3}{4}$...

Show that

Question: Show that $2^{4 n+4}-15 n-16$, where $n \in \mathbb{N}$ is divisible by 225 . Solution: We have: $2^{4 n+4}-15 n-16=2^{4(n+1)}-15 n-16$ $=16^{n+1}-15 n-16$ $=(1+15)^{n+1}-15 n-16$ $={ }^{n+1} \mathrm{C}_{0} 15^{0}+{ }^{\mathrm{n}+1} \mathrm{C}_{1} 15^{1}+{ }^{\mathrm{n}+1} \mathrm{C}_{2} 15^{2}+\ldots+{ }^{\mathrm{n}+1} \mathrm{C}_{\mathrm{n}+1} 15^{\mathrm{n}+1}-15 \mathrm{n}-16$ $=1+(\mathrm{n}+1) 15+{ }^{\mathrm{n}+1} \mathrm{C}_{2} 15^{2}+\ldots+{ }^{\mathrm{n}+1} \mathrm{C}_{\math...

An exterior angle of a triangle is 110° and its two interior opposite angles are equal. Each of these equal angles is

Question: An exterior angle of a triangle is 110 and its two interior opposite angles are equal. Each of these equal angles is (a) $70^{\circ}$ (b) $55^{\circ}$ (c) $35^{\circ}$ (d) $27 \frac{1^{\circ}}{2}$ Solution: Let the measure of each of the two equal interior opposite angles of the triangle bex.In a triangle, the exterior angle is equal to the sum of the two interior opposite angles.x+x= 110⇒ 2x=110⇒x=55Thus, the measure of each of these equal angles is55.Hence, the correct answer is opti...

If one angle of a triangle is equal to the sum of the other two angles, then the triangle is

Question: If one angle of a triangle is equal to the sum of the other two angles, then the triangle is(a) an isosceles triangle(b) an obtuse triangle(c) an equilateral triangle(d) a right triangle Solution: Let∆ABC be such thatA =B +C.In∆ABC,A +B +C = 180 (Angle sum property)⇒A +A= 180 (A =B +C)⇒ 2A = 180⇒ A = 90Therefore,∆ABC is a right triangle.Thus,if one angle of a triangle is equal to the sum of the other two angles, then the triangle is a right triangle.Hence, the correct answer is option ...

Find the value of (1.01)10 + (1 − 0.01)

Question: Find the value of (1.01)10 + (1 0.01)10correct to 7 places of decimal. Solution: $(1.01)^{10}+(1-0.01)^{10}$ $=(1+0.01)^{10}+(1-0.01)^{10}$ $=2\left[{ }^{10} C_{0} \times(0.01)^{0}+{ }^{10} C_{2} \times(0.01)^{2}+{ }^{10} C_{4} \times(0.01)^{4}+{ }^{10} C_{6} \times(0.01)^{6}+{ }^{10} C_{8} \times(0.01)^{8}+{ }^{10} C_{10} \times(0.01)^{10}\right]$ $=2(1+45 \times 0.0001+210 \times 0.00000001+\ldots)$ $=2(1+0.0045+0.00000210+\ldots)$ $=2.0090042+\ldots$ Hence, the value of (1.01)10 + (...

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